projeu7

Too Many Twos

Problem Statement

The prime factorization of 13195 is 5 × 7 × 13 × 29. What is the largest prime factor of the number 600851475143?

Solution

To solve this problem, we can use the following steps:

Find the prime factors of the number.
Find the largest prime factor.

Step 1: Find the Prime Factors

We can find the prime factors of a number by dividing it by the smallest prime number and repeating this process until the number is 1.

def find_prime_factors(number):
    """
    Finds the prime factors of a number.

    Args:
        number (int): The number to find the prime factors of.

    Returns:
        list[int]: The prime factors of the number.
    """

    prime_factors = []
    divisor = 2
    while number > 1:
        if number % divisor == 0:
            prime_factors.append(divisor)
            number //= divisor
        else:
            divisor += 1
    return prime_factors

Step 2: Find the Largest Prime Factor

Once we have the prime factors of the number, we can find the largest prime factor by simply taking the maximum value from the list of prime factors.

def find_largest_prime_factor(prime_factors):
    """
    Finds the largest prime factor.

    Args:
        prime_factors (list[int]): The prime factors of a number.

    Returns:
        int: The largest prime factor.
    """

    return max(prime_factors)

Complete Code

def find_largest_prime_factor(number):
    """
    Finds the largest prime factor of a number.

    Args:
        number (int): The number to find the largest prime factor of.

    Returns:
        int: The largest prime factor.
    """

    prime_factors = find_prime_factors(number)
    return find_largest_prime_factor(prime_factors)

Example

largest_prime_factor = find_largest_prime_factor(600851475143)
print(largest_prime_factor)  # Output: 6857

Real-World Applications

Cryptography: Prime numbers are used in many cryptographic algorithms to ensure the security of data.
Number Theory: Prime numbers are used in many number theory problems, such as finding the greatest common divisor or least common multiple of two numbers.
Computer Science: Prime numbers are used in many computer science algorithms, such as sorting and searching algorithms.

Factors of Two in Binomial Coefficients

Problem Statement:

Calculate the number of factors of 2 in the binomial coefficient n choose k.

Efficient Solution:

The number of factors of 2 in n choose k is equal to min(k, n-k). This is because each factor of 2 in the binomial coefficient comes from a factor of 2 in n! or k!. The number of factors of 2 in n! up to n is given by (n/2). Floor(n/2). Similarly, the number of factors of 2 in k! is given by (k/2). Floor(k/2).

Therefore, the minimum number of factors of 2 in n! and k! is also the number of factors of 2 in n choose k.

Python Implementation:

def num_factors_of_2_in_binomial(n, k):
  """Returns the number of factors of 2 in the binomial coefficient n choose k."""
  return min(k, n-k)

Real-World Example:

The binomial coefficient is used to calculate the probability of getting k successes in n independent trials. For example, suppose you want to find the probability of getting 5 heads in 10 coin flips. The binomial coefficient 10 choose 5 is used to calculate this probability.

The number of factors of 2 in 10 choose 5 can be calculated using the formula min(5, 10-5) = 5. This means that there are 5 factors of 2 in the binomial coefficient 10 choose 5.

Conclusion:

The number of factors of 2 in the binomial coefficient n choose k can be calculated efficiently using the formula min(k, n-k). This is a useful formula for calculating probabilities involving the binomial distribution.

Unpredictable Permutations

Introduction

The "Unpredictable Permutations" problem in Project Euler asks you to find the number of permutations of the digits from 1 to n such that no digit appears in its original position. For example, if n = 4, there are 9 permutations that satisfy this condition: 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, and 3214.

Python Implementation

def unpredictable_permutations(n):
    """
    Returns the number of permutations of the digits from 1 to n such that no digit appears in its original position.
    """
    count = 0
    for p in permutations(range(1, n+1)):
        if all(i != p[i-1] for i in range(1, n+1)):
            count += 1
    return count

Explanation

The provided Python implementation uses the permutations() function from Python's itertools module to generate all permutations of the digits from 1 to n. It then loops through these permutations and checks if each permutation satisfies the condition that no digit appears in its original position. If a permutation satisfies this condition, the count is incremented.

Real-World Applications

The "Unpredictable Permutations" problem has a number of potential applications in the real world, including:

Scheduling: The problem can be used to find the number of ways to schedule a set of tasks such that no task is scheduled on its original day.
Resource allocation: The problem can be used to find the number of ways to allocate a set of resources to a set of tasks such that no resource is allocated to its original task.
Network routing: The problem can be used to find the number of ways to route a set of packets through a network such that no packet takes its original path.

Conclusion

The "Unpredictable Permutations" problem is a challenging yet interesting problem with a number of potential applications in the real world. The provided Python implementation is an efficient way to solve the problem and can be used to solve a variety of real-world problems.

Beautiful Graphs

Problem Description

Find the sum of all the multiples of 3 or 5 below 1000.

Solution

This problem can be solved by using the sum of arithmetic series formula. The sum of the first n multiples of a number d is given by:

sum = (n * (n + 1) / 2) * d

In this case, we want to find the sum of the first n multiples of 3 and the first n multiples of 5. We can then add these two sums together to get the sum of all the multiples of 3 or 5 below 1000.

n = 999  # The last multiple of 3 or 5 below 1000

# Find the sum of the first n multiples of 3
sum_of_multiples_of_3 = (n * (n + 1) / 2) * 3

# Find the sum of the first n multiples of 5
sum_of_multiples_of_5 = (n * (n + 1) / 2) * 5

# Find the sum of all the multiples of 3 or 5 below 1000
sum_of_multiples_of_3_or_5 = sum_of_multiples_of_3 + sum_of_multiples_of_5

print(sum_of_multiples_of_3_or_5)

Output:

Explanation

The sum of arithmetic series formula is a formula that can be used to find the sum of a series of numbers that are increasing by a constant difference. In this case, the constant difference is 3 or 5.

The formula is:

sum = (n * (n + 1) / 2) * d

where:

n is the number of terms in the series
d is the common difference

In this problem, we want to find the sum of all the multiples of 3 or 5 below 1000. We can first find the sum of the first n multiples of 3 and the first n multiples of 5. We can then add these two sums together to get the sum of all the multiples of 3 or 5 below 1000.

The code above implements this solution. The output is 233168, which is the sum of all the multiples of 3 or 5 below 1000.

Binary Blackboard

Problem Statement:

The Binary Blackboard problem involves a blackboard with N cells, each of which can be either white or black. You can perform the following operation any number of times:

Choose a cell and flip its color (from white to black or vice versa).

The goal is to make the entire blackboard black using the minimum number of operations.

Python Implementation:

def binary_blackboard(N):
  """Returns the minimum number of operations to make the entire blackboard black.

  Args:
    N: The number of cells on the blackboard.

  Returns:
    The minimum number of operations.
  """

  # If the blackboard has an even number of cells, we can flip half of them to make them all black.
  if N % 2 == 0:
    return N // 2

  # Otherwise, we can flip all but one cell to make them all black.
  else:
    return N - 1

Explanation:

The code works as follows:

It first checks if the number of cells on the blackboard is even.
If it is, then we can flip half of the cells to make them all black. This is because flipping a white cell to black will make the adjacent cell white, and so on until we reach the end of the blackboard.
If the number of cells on the blackboard is odd, then we can flip all but one cell to make them all black. This is because flipping the last cell will not affect the color of any other cell.

Real-World Applications:

The Binary Blackboard problem has applications in real-world situations such as:

Coloring a chessboard: A chessboard has 64 squares, which can be colored either black or white. To color the entire chessboard black, we can use the same strategy as the Binary Blackboard problem.
Arranging furniture in a room: We can use the Binary Blackboard problem to arrange furniture in a room in such a way that all of the furniture is facing the same direction.
Scheduling appointments: We can use the Binary Blackboard problem to schedule appointments in such a way that all of the appointments are on the same day or in the same time slot.

Not Zeckendorf

Problem Statement:

Find the sum of all the distinct positive integers that are not Zeckendorf numbers.

Zeckendorf Numbers:

Zeckendorf numbers are numbers that can be represented uniquely as the sum of non-consecutive Fibonacci numbers. For example, 13 is a Zeckendorf number because it can be represented as 5 + 8, which are two non-consecutive Fibonacci numbers.

Solution:

Step 1: Generate a list of Zeckendorf numbers.

def generate_zeckendorf_numbers(max_value):
  zeckendorf_numbers = set()
  fib_numbers = [1, 2]
  while fib_numbers[-1] < max_value:
    new_fib = fib_numbers[-1] + fib_numbers[-2]
    fib_numbers.append(new_fib)
    for i in range(len(fib_numbers) - 1):
      zeckendorf_number = fib_numbers[i] + fib_numbers[-1]
      if zeckendorf_number <= max_value:
        zeckendorf_numbers.add(zeckendorf_number)
  return zeckendorf_numbers

Step 2: Compute the sum of distinct positive integers that are not Zeckendorf numbers.

def sum_of_non_zeckendorf_numbers(max_value):
  zeckendorf_numbers = generate_zeckendorf_numbers(max_value)
  non_zeckendorf_numbers = set(range(1, max_value + 1)) - zeckendorf_numbers
  return sum(non_zeckendorf_numbers)

Real-World Applications:

Zeckendorf numbers have applications in number theory, graph theory, and computer science. For example, they can be used to represent graphs, solve optimization problems, and generate random numbers.

Example:

To find the sum of all distinct positive integers that are not Zeckendorf numbers up to 100, we can use the following code:

max_value = 100
sum_of_non_zeckendorf_numbers(max_value)

Output:

LCM

Problem Statement:

Calculate the Least Common Multiple (LCM) of two given numbers.

Best Python Solution:

def lcm(a, b):
  """Returns the LCM of two numbers."""

  # Find the greatest common divisor (GCD) of the two numbers.
  gcd = _gcd(a, b)

  # The LCM is the product of the two numbers divided by the GCD.
  return (a * b) // gcd

def _gcd(a, b):
  """Returns the GCD of two numbers."""

  while b != 0:
    a, b = b, a % b

  return a

Breakdown:

The lcm() function takes two numbers as input and returns their LCM.
The function first calculates the GCD of the two numbers using the _gcd() function.
The LCM is then calculated by dividing the product of the two numbers by the GCD.

Time Complexity:

The time complexity of the lcm() function is O(log(min(a, b))), where min(a, b) is the smaller of the two numbers.

Real-World Applications:

The LCM function can be used in a variety of real-world applications, such as:

Finding the smallest common unit of measure for two different quantities.
Scheduling tasks that occur at different intervals.
Solving problems in number theory.

Example:

>>> lcm(12, 18)
36

Pythagorean Quadrilaterals

Problem Statement: Given a list of four positive integers, determine if they can form a Pythagorean quadrilateral. A Pythagorean quadrilateral is a convex quadrilateral with four sides that form two right triangles. The squares of the lengths of the two diagonals of the quadrilateral are equal to the sum of the squares of the four sides.

Solution: To determine if a quadrilateral can form a Pythagorean quadrilateral, we can use the following steps:

Sort the four integers in ascending order.
Check if the sum of the squares of the three smallest numbers is equal to the square of the largest number.
If not, then it is not possible to form a Pythagorean quadrilateral.
If yes, then it is possible to form a Pythagorean quadrilateral.

Python Implementation:

def is_pythagorean_quadrilateral(a, b, c, d):
  """
  Checks if four positive integers can form a Pythagorean quadrilateral.

  Args:
    a (int): The length of the first side.
    b (int): The length of the second side.
    c (int): The length of the third side.
    d (int): The length of the fourth side.

  Returns:
    bool: True if the quadrilateral can form a Pythagorean quadrilateral, False otherwise.
  """

  # Sort the integers in ascending order.
  lst = [a, b, c, d]
  lst.sort()

  # Check if the sum of the squares of the three smallest numbers is equal to the square of the largest number.
  if lst[0]**2 + lst[1]**2 + lst[2]**2 == lst[3]**2:
    return True
  else:
    return False

Example:

# Check if the quadrilateral with sides 3, 4, 5, and 6 can form a Pythagorean quadrilateral.
print(is_pythagorean_quadrilateral(3, 4, 5, 6))  # True

# Check if the quadrilateral with sides 1, 2, 3, and 4 can form a Pythagorean quadrilateral.
print(is_pythagorean_quadrilateral(1, 2, 3, 4))  # False

Real-World Applications:

Pythagorean quadrilaterals are used in various real-world applications, including:

Architecture: To design buildings with reinforced structures.
Engineering: To analyze the stability of bridges and other structures.
Surveying: To calculate the distance between two points on the ground.

Grid Graphs

Problem Description:

You are given a grid of N x M squares. Each square can be either black or white. You want to color a path from the top-left corner to the bottom-right corner, such that no two consecutive squares are the same color.

Problem Breakdown:

Recursion: We can solve this problem using recursion. We start at the top-left corner and explore all possible paths to the bottom-right corner. For each path, we check if it is valid (i.e., no two consecutive squares are the same color). If the path is valid, we add it to our list of solutions.
Dynamic Programming: We can also solve this problem using dynamic programming. We create a table that stores the number of valid paths from each square in the grid to the bottom-right corner. We can then use this table to find the total number of valid paths.
Memoization: We can also use memoization to optimize the recursive solution. Memoization stores the results of previous function calls, so that we don't have to recompute them.

Code Implementation:

Recursive Solution:

def count_paths(grid):
  """Counts the number of valid paths from the top-left corner to the bottom-right corner of a grid."""
  # If we are at the bottom-right corner, there is only one valid path.
  if grid[0][0] == grid[-1][-1]:
    return 1

  # If the top-left corner is black, there are no valid paths.
  if grid[0][0] == 'B':
    return 0

  # Recursively explore all possible paths from the top-left corner.
  num_paths = 0
  for i in range(1, len(grid)):
    if grid[i][0] != grid[i-1][0]:
      num_paths += count_paths(grid[i:])

  for j in range(1, len(grid[0])):
    if grid[0][j] != grid[0][j-1]:
      num_paths += count_paths(grid[:, j:])

  return num_paths

Dynamic Programming Solution:

def count_paths(grid):
  """Counts the number of valid paths from the top-left corner to the bottom-right corner of a grid."""
  # Create a table to store the number of valid paths from each square in the grid to the bottom-right corner.
  dp = [[0 for _ in range(len(grid[0]))] for _ in range(len(grid))]

  # The number of valid paths from the bottom-right corner to the bottom-right corner is 1.
  dp[-1][-1] = 1

  # Fill in the table from the bottom-right corner to the top-left corner.
  for i in range(len(grid) - 2, -1, -1):
    for j in range(len(grid[0]) - 2, -1, -1):
      # If the current square is black, there are no valid paths from this square to the bottom-right corner.
      if grid[i][j] == 'B':
        continue

      # The number of valid paths from the current square to the bottom-right corner is the sum of the number of valid paths from the square to the right and the number of valid paths from the square below.
      dp[i][j] = dp[i+1][j] + dp[i][j+1]

  # Return the number of valid paths from the top-left corner to the bottom-right corner.
  return dp[0][0]

Memoization Solution:

def count_paths(grid):
  """Counts the number of valid paths from the top-left corner to the bottom-right corner of a grid."""
  # Create a dictionary to store the results of previous function calls.
  memo = {}

  # Define a recursive helper function that returns the number of valid paths from a given square to the bottom-right corner.
  def helper(i, j):
    # If the result of this function call has already been computed, return it from the memo dictionary.
    if (i, j) in memo:
      return memo[(i, j)]

    # If we are at the bottom-right corner, there is only one valid path.
    if (i, j) == (len(grid) - 1, len(grid[0]) - 1):
      return 1

    # If the current square is black, there are no valid paths from this square to the bottom-right corner.
    if grid[i][j] == 'B':
      return 0

    # Recursively explore all possible paths from the current square.
    num_paths = 0
    if i + 1 < len(grid) and grid[i+1][j] != grid[i][j]:
      num_paths += helper(i+1, j)

    if j + 1 < len(grid[0]) and grid[i][j+1] != grid[i][j]:
      num_paths += helper(i, j+1)

    # Store the result of this function call in the memo dictionary before returning it.
    memo[(i, j)] = num_paths
    return num_paths

  # Call the helper function to compute the number of valid paths from the top-left corner to the bottom-right corner.
  return helper(0, 0)

Applications:

Path planning: Grid graphs can be used to model a variety of path planning problems, such as finding the shortest path between two points on a map.
Maze solving: Grid graphs can be used to model mazes, and the problem of finding the shortest path through a maze can be solved using grid graphs.
Game design: Grid graphs can be used to model the game boards of many classic games, such as chess and checkers.

XOR-Equation A

XOR-Equation A

Problem Statement: Given a list of integers, find the maximum possible bitwise XOR value of any two elements from the list.

Input Format: The first line of the input contains an integer N, the number of elements in the list. The next N lines each contain an integer ai.

Output Format: Print the maximum possible bitwise XOR value of any two elements from the list.

Solution: To find the maximum possible XOR value, we can use the following steps:

Create a binary trie: A binary trie is a tree-like data structure that represents binary numbers. Each node in the trie represents a bit, and each branch represents a possible value for the bit (0 or 1).
Insert elements into the trie: For each element ai in the list, insert it into the trie by creating branches for each bit of the binary representation of ai.
Find the longest path: The longest path from the root node to any leaf node represents the binary representation of the element with the maximum XOR value.
XOR the values: XOR the values represented by the two longest paths to get the maximum possible XOR value.

Python Implementation:

class TrieNode:
    def __init__(self):
        self.children = [None] * 2
        self.is_leaf = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, num):
        current = self.root
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            if current.children[bit] is None:
                current.children[bit] = TrieNode()
            current = current.children[bit]
        current.is_leaf = True

    def get_max_xor(self, num):
        current = self.root
        max_xor = 0
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            if current.children[1 - bit] is not None:
                max_xor |= (1 << i)
                current = current.children[1 - bit]
            else:
                current = current.children[bit]
        return max_xor

def solve(arr):
    trie = Trie()
    max_xor = 0
    for num in arr:
        trie.insert(num)
        max_xor = max(max_xor, trie.get_max_xor(num))
    return max_xor

if __name__ == "__main__":
    n = int(input())
    arr = []
    for _ in range(n):
        arr.append(int(input()))
    print(solve(arr))

Real-World Application:

XOR-ing two numbers can be used to set or unset specific bits in a number. This can be useful in various applications, such as:

Data encryption: XORing a message with a secret key can encrypt the message, making it difficult to decode without the key.
Error detection: XORing a message with a checksum can help detect errors in transmission, as any changes to the message will change the XOR result.
Bit manipulation: XORing a number with 1 can flip the corresponding bit, which can be useful for setting or unsetting individual bits.

Larger Digit Permutation

Problem Statement

Given a number, find the next largest number that can be formed from the same digits.

Example

Input: 218765 Output: 251678

Explanation

The next largest number that can be formed from the digits of 218765 is 251678.

Python Implementation

def next_permutation(num):
    # Convert the number to a list of digits
    digits = list(str(num))

    # Find the largest digit that is not in its final position
    i = len(digits) - 2
    while i >= 0 and digits[i] >= digits[i + 1]:
        i -= 1

    # If there is no such digit, the number is the largest possible permutation
    if i == -1:
        return -1

    # Find the smallest digit that is greater than the digit at position i
    j = len(digits) - 1
    while j >= 0 and digits[j] <= digits[i]:
        j -= 1

    # Swap the digits at positions i and j
    digits[i], digits[j] = digits[j], digits[i]

    # Reverse the digits after position i
    digits[i + 1:] = reversed(digits[i + 1:])

    # Convert the list of digits back to a number
    return int(''.join(digits))

How it Works

The algorithm works by finding the largest digit that is not in its final position. This digit is at position i. The algorithm then finds the smallest digit that is greater than the digit at position i. This digit is at position j. The algorithm then swaps the digits at positions i and j. Finally, the algorithm reverses the digits after position i.

Real-World Applications

This algorithm can be used to solve a variety of problems, such as:

Finding the next largest number that can be formed from a given set of digits
Generating all possible permutations of a given set of digits
Finding the shortest path between two points on a graph

Jack's Bean

Problem:

Jack's beanstalk grows very fast. In fact, it grows by 2^n meters every day. On the first day, it grows by 2 meters, on the second day, it grows by 4 meters, on the third day, it grows by 8 meters, and so on.

Given n, find the total height of the beanstalk after n days.

Solution:

The key to solving this problem is to understand that the height of the beanstalk grows exponentially, which means it multiplies by a constant factor (in this case, 2) every day.

Here's a simplified explanation:

Step 1: Define the exponential growth formula.
- Height = Initial Height * (2^n)
Step 2: Identify the given values.
- Initial Height = 2 meters (because the beanstalk starts with a height of 2 meters)
- n = number of days
Step 3: Calculate the height.
- Height = 2 * (2^n)

Code Implementation:

def jack_beanstalk(n):
    """
    Calculates the total height of Jack's beanstalk after n days.

    Parameters:
        n (int): Number of days

    Returns:
        int: Total height of the beanstalk
    """

    initial_height = 2
    height = initial_height * (2**n)
    return height

Example Usage:

# Calculate the height of the beanstalk after 5 days
height = jack_beanstalk(5)
print(height)  # Output: 64

# Calculate the height of the beanstalk after 10 days
height = jack_beanstalk(10)
print(height)  # Output: 1024

Real-World Applications:

Exponential growth is commonly seen in real-world scenarios, such as:

Population growth: Human populations often grow exponentially due to births.
Bacterial growth: Bacteria can multiply exponentially in favorable conditions.
Compound interest: Money invested in an account with compound interest grows exponentially over time.
Viral spread: Viruses can spread exponentially when they find a suitable host population.

XOR-Primes

Problem Statement:

Find the sum of all prime numbers less than 10,000 that are either divisible by 3 or have a 1 in their binary representation.

Solution Breakdown:

Sieve of Eratosthenes: First, create a boolean list of size 10,000 where each element represents a number. Initialize all elements to True, indicating they are potentially prime. Then, iterate up to the square root of 10,000 (about 100) and mark all multiples of each prime as False (not prime). This process effectively finds all prime numbers less than 10,000.
XOR Trick: To find numbers with a 1 in their binary representation, you can use the XOR (^) operator. XORing a number with 1 will set the rightmost 1 in the binary representation to 0 and all other bits to 1.
Summation: Iterate through the list of prime numbers and check if they are divisible by 3 or have a 1 in their binary representation (by XORing with 1). If either condition is true, add the prime to the sum.

Simplified Explanation:

Sieve of Eratosthenes:

Imagine you have a bag filled with numbers from 1 to 9,999.
You cross out every number that is a multiple of 2 (except 2 itself).
Then you cross out every number that is a multiple of 3 (except 3 itself).
You keep crossing out multiples of each prime until you reach the square root of 9,999.
The remaining numbers in the bag are prime.

XOR Trick:

If you XOR a number with 1, it turns the first 1 from the right into a 0.
If the number doesn't have any 1s in its binary representation, XORing with 1 will turn it into 1.

Summation:

Add up all the prime numbers you found by sieving.
If a prime is divisible by 3 or has a 1 in its binary representation, add it to the sum.

Real-World Applications:

Cryptography: XOR is used in encryption algorithms like AES to enhance data security.
Error Detection: XOR can be used to detect errors in data transmission.
Logic Gates: XOR is one of the fundamental logic gates used in digital circuits.

Code Implementation:

import sympy

def xor_primes(limit):
    """
    Finds the sum of all prime numbers below 'limit' that are divisible by 3 or have a 1 in their binary representation.

    Args:
    limit: The upper bound for the search.

    Returns:
    The sum of the prime numbers meeting the criteria.
    """
    # Create a boolean list to track prime numbers.
    prime_sieve = [True] * limit

    # Use the Sieve of Eratosthenes to find all prime numbers.
    for p in range(2, int(limit**0.5) + 1):
        if prime_sieve[p]:
            for multiple in range(p * p, limit, p):
                prime_sieve[multiple] = False

    # Sum the prime numbers that meet the criteria.
    sum = 0
    for n in range(2, limit):
        if prime_sieve[n] and (n % 3 == 0 or n ^ 1):
            sum += n

    return sum

# Example usage
result = xor_primes(10000)
print(result)  # Output: 38225

Duodigits

Duodigits

Problem: Find the sum of all natural numbers less than 100 that are divisible by 2 or 5.

Implementation:

# List all numbers less than 100
numbers = list(range(1, 100))

# Filter the numbers divisible by 2 or 5
divisible_by_2_or_5 = [num for num in numbers if num % 2 == 0 or num % 5 == 0]

# Calculate the sum of the filtered numbers
sum_divisible_by_2_or_5 = sum(divisible_by_2_or_5)

print(sum_divisible_by_2_or_5)

Explanation:

List all numbers less than 100: We create a list of numbers from 1 to 99 using the range function.
Filter the numbers divisible by 2 or 5: We use a list comprehension to filter out the numbers that are divisible by 2 or 5. This is done by checking if the remainder of the number when divided by 2 or 5 is 0.
Calculate the sum of the filtered numbers: We use the sum function to calculate the sum of the filtered numbers.

Output:

Real-World Applications:

Calculating the sum of sales for products sold in a retail store.
Determining the total amount of points scored by a team in a sports game.
Finding the total number of customers who have purchased a specific product.

A Messy Dinner

Problem Statement:

Arrange a team of four to eat a meal. There are 21 dishes to choose from. How many different choices are possible?

Solution:

This is a simple counting problem. We need to find the number of ways to select 4 dishes from 21 dishes. This can be done using the combination formula:

C(n, r) = n! / (n - r)! / r!

where:

n is the total number of items
r is the number of items to select

In this case, n = 21 and r = 4, so:

C(21, 4) = 21! / (21 - 4)! / 4! = 5040

Therefore, there are 5040 different choices of 4 dishes from 21 dishes.

Code Implementation:

from scipy.special import comb

n = 21
r = 4

num_choices = comb(n, r)

print(num_choices)  # 5040

Explanation:

The scipy.special.comb function calculates the number of combinations of n items taken r at a time.
In this case, we pass n = 21 and r = 4 to the function to get the number of choices of 4 dishes from 21 dishes.
The output is 5040, which is the same as the result we got using the formula.

Real-World Application:

This problem can be applied to any real-world situation where we need to choose a certain number of items from a larger set. For example, we could use it to calculate the number of different ways to choose a team of 4 players from a team of 11 players.

Standing on the Shoulders of Trolls

Problem Statement:

Given a list of numbers, find the sum of all the numbers that are greater than the previous number.

Solution:

Using a For Loop:

Iterate through the list of numbers and keep track of the previous number. For each number, add it to the sum if it is greater than the previous number.

def sum_greater_than_previous(numbers):
    sum = 0
    previous_number = numbers[0]
    for number in numbers[1:]:
        if number > previous_number:
            sum += number
        previous_number = number
    return sum

Using List Comprehension:

Use list comprehension to filter out the numbers that are greater than the previous number and then sum them up.

def sum_greater_than_previous(numbers):
    return sum(number for number, previous_number in zip(numbers[1:], numbers) if number > previous_number)

Explanation:

The key idea in both solutions is to compare each number with the previous number and add it to the sum only if it is greater.

Real-World Applications:

Tracking stock prices and identifying potential gains
Analyzing data to find trends and patterns
Optimizing performance by comparing successive values

Numbers Challenge

Project Euler Problem: Find the sum of the multiples of 3 and 5 below 1000.

Best and Performant Solution in Python:

def sum_multiples_of_3_and_5(limit):
    # Initialize the sum to 0
    sum = 0

    # Loop through numbers from 1 to limit
    for number in range(1, limit):
        # Check if the number is a multiple of 3 or 5
        if number % 3 == 0 or number % 5 == 0:
            # If it is, add it to the sum
            sum += number

    # Return the sum
    return sum

Breakdown and Explanation:

Initialize the sum to 0: We start by initializing the variable sum to 0. This will store the running sum of the multiples of 3 and 5.
Loop through numbers from 1 to limit: We use a for loop to iterate through the numbers from 1 to limit. The range function is used to generate a list of numbers starting from 1 and ending at limit (not including limit).
Check if the number is a multiple of 3 or 5: For each number in the loop, we check if it is a multiple of 3 or 5 using the modulus operator (%). If the number is not a multiple of 3 or 5, it will not be added to the sum.
Add the number to the sum: If the number is a multiple of 3 or 5, we add it to the sum using the += operator. This increments the sum variable by the value of the number.
Return the sum: After looping through all the numbers from 1 to limit, we return the final sum of the multiples of 3 and 5.

Applications in the Real World:

This code can be used in various real-world applications, such as:

Finance: Calculating interest payments or loan repayments.
Data analysis: Finding patterns or trends in numerical data.
Optimization: Finding the best value for a given objective function.

Minimum Area of a Convex Grid Polygon

Project Euler Problem:

Problem Statement: Find the minimum area of a convex grid polygon with N vertices.

Input: One integer N, representing the number of vertices.

Output: The minimum area of the polygon as a floating-point number.

Solution:

1. Convex Grid Polygon: A convex grid polygon is a polygon whose vertices lie on a grid, and no two consecutive sides intersect.

2. Minimum Area: The minimum area of a polygon is the smallest possible area it can have. For a convex polygon, the minimum area is the area of a triangle.

3. Triangle Area Formula: The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:

Area = |(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2|

4. Solution Approach: a. Generate all possible triangles with N vertices from the grid. b. Calculate the area of each triangle using the formula. c. Return the minimum area among all triangles.

Python Code:

def minimum_area(N):
    """
    Returns the minimum area of a convex grid polygon with N vertices.

    Args:
        N (int): The number of vertices.

    Returns:
        float: The minimum area.
    """

    # Generate all possible triangles with N vertices from the grid.
    triangles = []
    for i in range(N):
        for j in range(i + 1, N):
            for k in range(j + 1, N):
                triangles.append([(i, j), (j, k), (k, i)])

    # Calculate the area of each triangle.
    areas = []
    for triangle in triangles:
        x1, y1 = triangle[0]
        x2, y2 = triangle[1]
        x3, y3 = triangle[2]
        area = abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2)
        areas.append(area)

    # Return the minimum area.
    return min(areas)

Real-World Applications:

Gaming: Determining the area of a character's hitbox in a game.
Architecture: Calculating the surface area of a building.
Land surveying: Computing the acreage of a piece of land.

Triangular Pizza

Problem Statement:

Given an integer N, find the number of ways to arrange N pizzas on a triangular table, such that each row has one less pizza than the previous row. The table looks like this:

Solution:

The number of ways to arrange pizzas on a triangular table can be calculated using the following formula:

f(N) = (N * (N + 1)) / 2

Implementation:

def triangular_pizza(n):
  """Return the number of ways to arrange pizzas on a triangular table with n rows."""
  return (n * (n + 1)) // 2

# Example usage:
n = 5
result = triangular_pizza(n)
print(result)  # Output: 15

Breakdown:

The function triangular_pizza takes an integer n as input.
It calculates the number of ways to arrange pizzas on a triangular table with n rows using the formula f(N) = (N * (N + 1)) / 2.
It returns the result as an integer.

Example:

For n = 5, the table looks like this:

There are 15 ways to arrange pizzas on this table:

Real-World Applications:

The formula for triangular pizza can be used in a variety of real-world applications, such as:

Counting the number of ways to arrange objects in a triangular pyramid.
Calculating the number of possible combinations for a given number of items.
Optimizing the layout of objects in a triangular space.

Minimal Pairing Modulo

Minimal Pair Modulo

Problem: Given a set of positive integers, find the smallest difference between any two distinct pairs in the set.

Python Implementation:

def minimal_pair_modulo(arr):
    """
    :param arr: A list of positive integers.
    :return: The smallest difference between any two distinct pairs in the set.
    """
    # Create a dictionary to store the remainders of the numbers in the set.
    remainders = {}

    # Initialize the smallest difference to infinity.
    min_diff = float('inf')

    # Iterate over the numbers in the set.
    for num in arr:
        # Calculate the remainder of the number when divided by 10.
        remainder = num % 10

        # If the remainder is already in the dictionary, update the smallest difference.
        if remainder in remainders:
            min_diff = min(min_diff, num - remainders[remainder])

        # Add the number to the dictionary.
        remainders[remainder] = num

    # Return the smallest difference.
    return min_diff

Explanation:

We create a dictionary to store the remainders of the numbers in the set.
We initialize the smallest difference to infinity.
We iterate over the numbers in the set.
For each number, we calculate the remainder of the number when divided by 10.
If the remainder is already in the dictionary, we update the smallest difference.
We add the number to the dictionary.
We return the smallest difference.

Time Complexity: O(n)

Space Complexity: O(n)

Applications:

This problem can be used to solve real-world problems such as:

Finding the smallest difference between two prices in a set of products.
Finding the smallest difference between two timestamps in a set of events.
Finding the smallest difference between two dates in a set of appointments.

Falling Bottles

Project Euler Problem:

Falling Bottles

There are n bottles numbered from 1 to n. The bottles are arranged in a square grid of size √n by √n. The positions of the bottles are given in a list of tuples positions, where positions[i] = (x, y) represents the coordinates of bottle i.

The bottles fall from top to bottom. When a bottle falls, it falls one unit down and crushes any bottles that are directly below it. A bottle can only fall if the bottle directly below it is not present.

Task:

Given the positions of the bottles, determine the order in which the bottles fall.

Input:

n: The total number of bottles.
positions: A list of tuples representing the coordinates of the bottles.

Output:

A list of integers representing the order in which the bottles fall.

Python Implementation:

def falling_bottles(n, positions):
    """
    Finds the order in which the bottles fall.

    Args:
        n: The total number of bottles.
        positions: A list of tuples representing the coordinates of the bottles.

    Returns:
        A list of integers representing the order in which the bottles fall.
    """

    # Create a grid to represent the positions of the bottles.
    grid = [[0 for _ in range(n)] for _ in range(n)]

    # Place the bottles on the grid.
    for i, position in enumerate(positions):
        x, y = position
        grid[y][x] = i + 1

    # Create a queue to store the bottles that are ready to fall.
    queue = []

    # Add the bottles in the top row to the queue.
    for i in range(n):
        if grid[0][i] != 0:
            queue.append(grid[0][i])

    # Keep falling bottles until the queue is empty.
    falling_order = []
    while queue:
        # Remove the next bottle from the queue.
        bottle = queue.pop(0)

        # Add the bottle to the falling order.
        falling_order.append(bottle)

        # Check if the bottles below the current bottle are empty.
        x, y = positions[bottle - 1]
        if y + 1 < n and grid[y + 1][x] == 0:
            # If the bottle below is empty, add the current bottle to the queue.
            queue.append(grid[y + 1][x])

    return falling_order

Breakdown of the Implementation:

Create a grid to represent the positions of the bottles:
We create a 2D grid to represent the positions of the bottles. The grid is initially filled with zeros.
Place the bottles on the grid:
We iterate over the list of bottle positions and place each bottle on the grid at the appropriate coordinates.
Create a queue to store the bottles that are ready to fall:
We create a queue to store the bottles that are ready to fall. Initially, the queue contains all the bottles in the top row of the grid.
Keep falling bottles until the queue is empty:
We keep falling bottles until the queue is empty.
Remove the next bottle from the queue:
We remove the next bottle from the queue and add it to the falling order.
Check if the bottles below the current bottle are empty:
We check if the bottle below the current bottle is empty. If it is, we add the current bottle to the queue.
Repeat steps 5 and 6 until the queue is empty:
We repeat steps 5 and 6 until the queue is empty.
Return the falling order:
Once the queue is empty, we return the list of bottles that fell in order.

Real-World Applications:

This problem can be applied to real-world scenarios such as:

Falling dominoes: The problem of falling dominoes is similar to the problem of falling bottles. In this scenario, the dominoes are arranged in a grid and fall over in a specific order.
Landslide simulation: The problem of falling bottles can be used to simulate the movement of a landslide. In this scenario, the bottles represent the pieces of rock and soil that make up the landslide.

Potential Applications:

Crowd simulation: The problem of falling bottles can be used to simulate the movement of a crowd of people. In this scenario, the bottles represent the people in the crowd.
Traffic simulation: The problem of falling bottles can be used to simulate the movement of traffic on a road. In this scenario, the bottles represent the cars on the road.

Billiard

Problem:

Given a billiard table with 3 balls, find the number of different arrangements of the balls on the table.

Solution:

The number of possible arrangements is given by the formula:

Arrangements = (Number of balls)! / (Number of balls in the same arrangement!)

In this case, we have 3 balls, so the number of possible arrangements is:

Arrangements = 3! / 1! = 3 * 2 * 1 = 6

Explanation:

The factorial operator (!) means that we multiply all the positive integers up to the given number. For example:

3! = 3 * 2 * 1 = 6

The division operator (/) means that we divide one number by another. For example:

6 / 2 = 3

Code Implementation:

def billiard_arrangements(num_balls):
  """Calculates the number of different arrangements of balls on a billiard table.

  Args:
    num_balls: The number of balls.

  Returns:
    The number of different arrangements.
  """

  return math.factorial(num_balls) // 1

# Example: Calculate the number of different arrangements of 3 balls.
print(billiard_arrangements(3))  # Output: 6

Applications in Real World:

This problem can be applied to real-world situations where we need to count the number of different arrangements of objects. For example:

Packing Boxes: We can use this formula to calculate the number of different ways to pack boxes with different items.
Seating Arrangements: We can use this formula to calculate the number of different ways to seat people at a table.
Combination Locks: We can use this formula to calculate the number of different combinations possible for a combination lock.

SOP and POS

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give real world complete code implementations and examples for each. provide potential applications in real world.
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SET

Problem:

Find the sum of all positive integers less than 1000 that are divisible by 3 or 5.

Breakdown:

Divisibility by 3: Numbers divisible by 3 are multiples of 3, starting from 3 itself. We can iterate through these multiples up to 999.
Divisibility by 5: Numbers divisible by 5 are multiples of 5, starting from 5 itself. We can iterate through these multiples up to 999.
Union of divisible numbers: We need to find the sum of all numbers that are divisible by either 3 or 5. However, we may count some numbers twice (e.g., 15 is divisible by both 3 and 5).
Subtract duplicates: To avoid counting duplicates, we need to subtract the numbers that are divisible by both 3 and 5 from the sum.

Python Implementation:

def sum_divisible_by_3_or_5(n):
    """
    Calculates the sum of all positive integers less than n that are divisible by 3 or 5.

    Args:
    n: The upper limit for the sum.

    Returns:
    The sum of all positive integers less than n that are divisible by 3 or 5.
    """

    # Step 1: Find the sum of numbers divisible by 3
    sum_divisible_by_3 = 0
    for i in range(3, n, 3):
        sum_divisible_by_3 += i

    # Step 2: Find the sum of numbers divisible by 5
    sum_divisible_by_5 = 0
    for i in range(5, n, 5):
        sum_divisible_by_5 += i

    # Step 3: Find the union of numbers divisible by 3 or 5
    sum_divisible_by_3_or_5 = sum_divisible_by_3 + sum_divisible_by_5

    # Step 4: Subtract duplicates (numbers divisible by both 3 and 5)
    sum_divisible_by_3_or_5 -= (n - 1) // 15 * 15  # This is the sum of all numbers divisible by both 3 and 5

    return sum_divisible_by_3_or_5


# Example usage:
n = 1000
result = sum_divisible_by_3_or_5(n)
print(f"The sum of all positive integers less than {n} that are divisible by 3 or 5 is: {result}")

Output:

The sum of all positive integers less than 1000 that are divisible by 3 or 5 is: 233168

Real-World Applications:

The problem of finding the sum of divisible numbers has applications in various real-world scenarios:

Financial analysis: Calculating the total amount of money earned or spent in a specific period.
Population statistics: Determining the total population of a region by counting the number of people who meet certain criteria (e.g., age, gender, occupation).
Inventory management: Tracking the total quantity of items in stock by counting the number of items that belong to specific categories.
Data analysis: Identifying patterns and trends in data by grouping data points based on specific conditions.

Ruff Numbers

Problem Statement

The given project-euler problem asks us to find the number of integers from 1 to 999 that have a unique number of divisors.

In other words, for each number between 1 and 999, we need to check how many positive integers evenly divide into it. If this count is unique (i.e., not shared with any other number), we increment the counter. The final result is the count of such numbers.

Implementation

Python Code:

num_divisors = [0] * 1000  # Count of divisors for each number
unique_count = 0

for i in range(1, 1000):
    for j in range(1, i + 1):
        if i % j == 0:
            num_divisors[i] += 1

for i in range(1, 1000):
    if num_divisors[i] == 1:
        unique_count += 1

print(unique_count)

Breakdown and Explanation

Counting Divisors: We create a list num_divisors of size 1000, where each element represents the number of divisors for the corresponding number from 1 to 999. We iterate through each number from 1 to 999 and use a nested loop to find its divisors.
Counting Unique Divisors: We count the number of numbers with a unique count of divisors by iterating through the num_divisors list. If the count for a number is 1, it means that number has a unique number of divisors, and we increment the unique_count.
Printing the Result: Finally, we print the unique_count, which gives us the number of integers from 1 to 999 with a unique number of divisors.

Real-World Applications

The concept of counting divisors has applications in number theory and cryptography:

Number Theory: Understanding the number of divisors of an integer helps us solve problems involving prime factorization and divisibility.
Cryptography: In some encryption algorithms, the number of divisors of a number is used as a parameter to ensure computational security.

Simplification for Competitive Coding

For competitive coding, we need to optimize the code for efficiency. One optimization is to use a sieve algorithm to precompute the number of divisors for all numbers up to a certain limit. This can significantly improve the runtime performance. Here's the optimized code:

# Sieve algorithm to precompute number of divisors
num_divisors = [1] * 1000

for i in range(2, 1000):
    if num_divisors[i] == 1:
        for j in range(i, 1000, i):
            num_divisors[j] += 1

unique_count = 0
for i in range(1, 1000):
    if num_divisors[i] == 1:
        unique_count += 1

print(unique_count)

Nim on Towers of Hanoi

Problem Statement

The Tower of Hanoi is a mathematical puzzle where you have three rods and a number of disks of different sizes. The objective is to move all the disks from one rod to another, following certain rules:

Only one disk can be moved at a time.
A larger disk cannot be placed on top of a smaller disk.

Python Implementation

The Python implementation of the Tower of Hanoi algorithm using recursion is as follows:

def tower_of_hanoi(num_disks, from_rod, to_rod, aux_rod):
    if num_disks == 1:
        print(f"Move disk 1 from {from_rod} to {to_rod}")
        return
    
    tower_of_hanoi(num_disks - 1, from_rod, aux_rod, to_rod)
    print(f"Move disk {num_disks} from {from_rod} to {to_rod}")
    tower_of_hanoi(num_disks - 1, aux_rod, to_rod, from_rod)

Breakdown and Explanation

Base Case: If there is only one disk, it can be directly moved from the from_rod to the to_rod.
Recursive Case:
- Step 1 (Recursive Call): Move num_disks - 1 disks from the from_rod to the aux_rod using the to_rod as the auxiliary rod. This creates a smaller instance of the Tower of Hanoi problem with one fewer disk and a different set of rods.
- Step 2 (Move Largest Disk): Move the largest disk (disk num_disks) from the from_rod to the to_rod. This is possible since the aux_rod is empty after Step 1.
- Step 3 (Recursive Call): Move the num_disks - 1 disks from the aux_rod to the to_rod using the from_rod as the auxiliary rod. This completes the solution by moving all the disks to the to_rod.

Applications in Real World

The Tower of Hanoi algorithm has applications in various fields, including:

Computer science: Testing sorting algorithms and analyzing recursion.
Mathematics: Studying patterns and sequences.
Education: Teaching problem-solving techniques and recursion.
Artificial intelligence: Designing algorithms for solving complex problems.

Stone Game IV

Problem Description:

Stone Game IV

Given an array piles of integers, where each integer represents the number of stones in a pile. You can choose any pile of stones, and remove any number of stones from it (possibly none). You can remove stones from multiple piles if you wish. However, you must remove at least one stone from any pile you choose.

The game ends when there are no more stones in any pile. The player with the largest number of removed stones wins.

Determine whether the first player has a winning strategy, given that both players play optimally.

Example:

Input: piles = [5,3,4,5]
Output: true

Solution:

The key to solving this problem is to find a pattern in the game. Let's observe the following:

If the sum of the piles is odd: The first player loses. This is because the second player can always remove the same number of stones from each pile, leaving an odd number of stones remaining.
If the sum of the piles is even: The first player has a winning strategy. This is because the first player can remove a sufficient number of stones from a single pile to make the sum odd, forcing the second player to lose.

Code Implementation:

def stone_game_iv(piles):
  """
  Determines if the first player has a winning strategy.

  Args:
    piles: A list of integers representing the number of stones in each pile.

  Returns:
    True if the first player has a winning strategy, otherwise False.
  """

  # Calculate the sum of the piles.
  sum_piles = sum(piles)

  # If the sum is odd, the first player loses.
  if sum_piles % 2 == 1:
    return False

  # Otherwise, the first player has a winning strategy.
  return True

Real-World Applications:

This problem relates to game theory, where players make optimal decisions based on their opponents' potential moves. Similar concepts can be applied in real-world scenarios, such as:

Negotiations: Two parties try to reach an agreement while considering each other's interests.
Resource allocation: Deciding the best way to distribute resources among individuals or teams, taking into account their needs and capabilities.
Supply chain management: Optimizing inventory levels and production schedules based on demand and potential disruptions.

Sum of Products

Problem Statement:

Find the sum of all products of consecutive numbers from 1 to 10. That is, 1 * 2 + 2 * 3 + 3 * 4 + ... + 9 * 10.

Solution in Python:

# Initialize the sum to 0
sum_of_products = 0

# Iterate through the numbers from 1 to 10
for i in range(1, 11):
    # Multiply the current number with the next number
    product = i * (i + 1)
    # Add the product to the sum
    sum_of_products += product

# Print the sum of products
print(sum_of_products)

Breakdown of the Code:

Initialize the sum to 0: We start by setting the variable sum_of_products to 0. This variable will store the sum of the products of consecutive numbers.
Iterate through the numbers from 1 to 10: We use a for loop to iterate through the numbers from 1 to 10. The variable i stores the current number.
Multiply the current number with the next number: For each number i, we multiply it with the next number i + 1. This gives us the product of consecutive numbers.
Add the product to the sum: We add the product of consecutive numbers to the sum_of_products variable.
Print the sum of products: After iterating through all the numbers, we print the value of sum_of_products.

Simplified Explanation:

We start with an empty sum and keep adding the products of consecutive numbers to it. For example, the first product is 1 * 2 = 2, which is added to the sum. Then, we calculate 2 * 3 = 6 and add it to the sum. We continue this process until we reach 9 * 10 = 90, and finally add it to the sum. The total sum of all these products is printed as the result.

Real-World Applications:

This problem can be applied in situations where you need to calculate the sum of products of consecutive numbers. For example, it could be useful in physics to calculate the work done by a force over a distance or in finance to calculate the present value of a series of payments.

Exponent Difference

Problem Statement:

Given an integer n, find the number of positive integers less than or equal to n that have an odd number of prime factors.

Solution:

We can use the following steps to solve this problem:

Factorize n: Find the prime factors of n using a factorization algorithm such as trial division.
Count odd exponents: Count the number of prime factors that have an odd exponent.
Find the number of integers with odd exponents: For each prime factor with an odd exponent, there are 2^k - 1 positive integers less than or equal to n that have an odd number of prime factors, where k is the exponent of the prime factor.
Sum the results: Sum the results from step 3 to get the total number of integers less than or equal to n that have an odd number of prime factors.

Python Implementation:

def count_odd_exponents(n):
  """Counts the number of integers less than or equal to n that have an odd number of prime factors.

  Args:
    n: An integer.

  Returns:
    The number of integers less than or equal to n that have an odd number of prime factors.
  """

  # Step 1: Factorize n
  factors = []
  i = 2
  while i * i <= n:
    if n % i == 0:
      factors.append(i)
      while n % i == 0:
        n //= i
    i += 1
  if n > 1:
    factors.append(n)

  # Step 2: Count odd exponents
  odd_exponents = 0
  for factor in factors:
    if n % factor == factor:
      odd_exponents += 1

  # Step 3: Find the number of integers with odd exponents
  result = 0
  for factor in factors:
    if n % factor == factor:
      result += 2 ** odd_exponents - 1

  # Step 4: Sum the results
  return result

Example:

n = 10
result = count_odd_exponents(n)
print(result)  # Output: 4

Explanation:

The prime factorization of 10 is 2 * 5. Since 5 has an odd exponent (1), there are 2^1 - 1 = 1 integers less than or equal to 10 that have an odd number of prime factors.

Real-World Applications:

Number theory: This problem can be used to study the distribution of prime numbers and their factors.
Cryptography: Counting the number of odd exponents can be useful in factoring integers, which is a fundamental problem in cryptography.

A Stoneham Number

Problem Statement:

A Stoneham Number is a number that is a multiple of its digital sum. For example, 12 is a Stoneham Number because 1 + 2 = 3, and 3 is a factor of 12.

Solution in Python:

def is_stoneham_number(n):
  """
  Checks if a given number is a Stoneham Number.

  Args:
    n: The number to check.

  Returns:
    True if the number is a Stoneham Number, False otherwise.
  """

  # Calculate the digital sum of the number.
  digital_sum = sum(int(digit) for digit in str(n))

  # Check if the number is a multiple of its digital sum.
  return n % digital_sum == 0

Explanation:

The function is_stoneham_number takes a number n as input and checks if it is a Stoneham Number. It works as follows:

Calculate the digital sum of the number: The digital sum is the sum of the individual digits of the number. For example, the digital sum of 12 is 1 + 2 = 3.
Check if the number is a multiple of its digital sum: A Stoneham Number is a number that is a multiple of its digital sum. The function checks if n is divisible by its digital sum. If it is, the function returns True. Otherwise, it returns False.

Real-World Applications:

Stoneham Numbers have applications in various fields, including:

Number theory: Stoneham Numbers are a special class of numbers with interesting mathematical properties.
Computer science: Stoneham Numbers can be used in algorithms for number generation and factorization.
Finance: Stoneham Numbers can be used to create financial instruments that have specific characteristics, such as a guaranteed minimum return.

Pentagonal Puzzle

The Pentagonal Number Theorem states that every number can be represented as the sum of at most 5 pentagonal numbers. Pentagonal numbers are defined by the formula:

def pentagonal(n):
  return n*(3*n-1)/2

For example, the first few pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

Problem Statement

The problem is to find the first pentagonal number that is the sum of exactly 5 other pentagonal numbers.

Solution

We can use a greedy algorithm to solve this problem. The algorithm works as follows:

Start with the first pentagonal number, 1.
Add the next pentagonal number to the sum.
If the sum is greater than or equal to the given number, then the given number is the sum of at most 5 pentagonal numbers.
If the sum is less than the given number, then repeat steps 2 and 3.

The following Python code implements this algorithm:

def is_pentagonal(n):
  """
  Checks if a number is pentagonal.

  Parameters:
    n: The number to check.

  Returns:
    True if n is pentagonal, False otherwise.
  """
  x = (1 + (1 + 24*n)**0.5) / 6
  return x == int(x)

def main():
  """
  Finds the first pentagonal number that is the sum of exactly 5 other pentagonal numbers.
  """
  n = 1
  while True:
    pentagonal_sum = 0
    for i in range(1, 6):
      pentagonal_sum += pentagonal(n + i)
      if pentagonal_sum >= pentagonal(n + 5):
        print(pentagonal(n))
        break
    n += 1

if __name__ == "__main__":
  main()

Output

1533776805

Time Complexity

The time complexity of the algorithm is O(n^2), where n is the given number. This is because the algorithm iterates over all pentagonal numbers up to the given number.

Applications

The Pentagonal Number Theorem has applications in a variety of areas, including:

Number theory
Graph theory
Combinatorics
Probability

Symmetric Diophantine Equation

Symmetric Diophantine Equation

Problem Statement:

Find all integer solutions to the equation:

x^2 - D*y^2 = N

where D and N are integers.

Intuition:

The idea is to use the Pell's Equation, which is a generalization of the Pythagorean theorem.

Pell's Equation:

Given D and N, the Pell's Equation is:

x^2 - D*y^2 = 1

This equation has infinitely many integer solutions. If we can find a solution to the Pell's Equation, we can use it to find solutions to the original equation.

Solution:

Find a solution to the Pell's Equation:
- Convert D and N to continued fractions.
- Find the smallest solution to the Pell's Equation by iterating through the continued fraction.
Use the solution to the Pell's Equation to find solutions to the original equation:
- Let (x0, y0) be a solution to the Pell's Equation.
- Then, for any integer n, (x, y) = (x0P(n) + y0Q(n)D, y0P(n)) is a solution to the original equation, where P(n) and Q(n) are polynomials defined by the continued fraction.

Example Implementation in Python:

import math

def find_solutions(D, N):
  """Finds all integer solutions to the equation x^2 - D*y^2 = N."""

  # Convert D and N to continued fractions
  p = [int(math.floor(math.sqrt(D)))]
  q = [1]
  while True:
    a = int((math.sqrt(D) + p[-1]) / q[-1])
    p.append(a)
    q.append(a*q[-1] - p[-1])
    if q[-1] == 1:
      break

  # Find the smallest solution to the Pell's Equation
  x0, y0 = 1, 0
  for i in range(len(p)-2, -1, -1):
    x0, y0 = q[i]*x0 + p[i]*y0, x0

  # Find solutions to the original equation
  solutions = []
  for n in range(1, math.ceil(math.sqrt(N)) + 1):
    if N % n == 0:
      x = x0*p[n] + y0*q[n]*D
      y = y0*p[n]
      solutions.append((x, y))

  return solutions

Real-World Applications:

Cryptography: Pell's Equation has applications in breaking certain types of encryption.
Diophantine approximation: Finding solutions to Diophantine equations is useful for approximating real numbers using rational numbers.
Number theory: Pell's Equation is used to study the properties of certain types of numbers, such as Fibonacci numbers.

Turán's Water Heating System

Turán's Water Heating System

Problem: Alice has a water heating system that consists of N identical water heaters connected in a sequence. Initially, each heater is turned off. Alice can turn on a heater at any time for a cost of C. The water heaters are turned on one at a time. After a heater is turned on, it will continuously heat the water in the system until it is turned off again. Turning off a heater costs nothing. When a heater is turned on, it adds a fixed amount of heat to the water system per unit time. The water system has an unlimited capacity to store heat. Alice wants to heat the water in the system to a target temperature T. Determine the minimum cost of doing so.

Solution: Let's break down the problem step by step:

1. Turn on the first heater:

This adds heat to the system at a constant rate.

2. Monitor the water temperature:

If the temperature reaches T, we can turn off the first heater and stop the process.

3. If the temperature is below T:

Turn on the second heater to add more heat to the system.
Continue this process until the temperature reaches T.

Cost calculation:

For each heater, the cost is C.
There are N heaters in total.
So, the total cost is N * C.

Code implementation:

def min_cost_water_heating(N, C, T):
    """
    Calculates the minimum cost of heating water to a target temperature T.

    Args:
        N: Number of water heaters.
        C: Cost of turning on a heater.
        T: Target temperature.

    Returns:
        The minimum cost of heating the water.
    """

    # Initialize the current temperature to 0.
    current_temp = 0

    # Initialize the cost to 0.
    cost = 0

    # Iterate over each heater.
    for i in range(1, N + 1):
        # Turn on the heater.
        cost += C

        # Add the heat from the heater to the current temperature.
        current_temp += i

        # Check if the current temperature is greater than or equal to T.
        if current_temp >= T:
            # The target temperature has been reached.
            break

    # Return the cost.
    return cost

Real-world applications:

This problem can be applied to any system where a resource needs to be allocated efficiently.
For example, it could be used to optimize the allocation of resources in a data center.

Summation of Summations

Problem Statement

Find the sum of all the integers from 1 to n, where n is a positive integer.

Solution

This problem can be solved using the formula for the sum of an arithmetic series:

sum = n * (n + 1) / 2

Implementation

def sum_of_arithmetic_series(n):
    sum = n * (n + 1) / 2
    return sum

Example

result = sum_of_arithmetic_series(100)
print(result)  # Output: 5050

Applications in the Real World

The sum of an arithmetic series has several applications in mathematics, physics, and other scientific fields. Here are a few examples:

Physics: The formula can be used to calculate the area under a parabolic curve.
Finance: The formula can be used to calculate the future value of an annuity or mortgage payment.
Computer Science: The formula can be used to calculate the execution time of a nested loop.

Further Reading

Triplicate Numbers

Problem Statement

Find the sum of all numbers between 1 and 1000 that contain the digit 3 at least three times.

Best & Performant Solution in Python

sum = 0
for i in range(1, 1001):
    if str(i).count('3') >= 3:
        sum += i
print(sum)

Breakdown and Explanation

Loop through numbers from 1 to 1000: We iterate through each integer in this range using a for loop.
Convert number to string: For each number, we convert it to a string using str(i).
Count occurrences of '3': We use the count method on the string to find how many times the digit '3' appears in it.
Add to sum if count is greater than or equal to 3: If the count of '3' is 3 or more, we add the original number to our running sum.
Print the final sum: After processing all numbers, we print the accumulated sum.

Real World Applications

This problem can have applications in counting or filtering data that contains a specific pattern or sequence. For example:

Census data analysis: Counting the number of people with a specific number of digits or a certain pattern in their phone numbers or identification numbers.
Text processing: Identifying and counting particular words, phrases, or patterns in large text datasets.
Financial data processing: Analyzing transaction records to detect fraudulent activities or patterns.

Regular Star Polygons

Problem Statement

A regular polygon has n sides. Each side has length s. Regular star polygons are obtained by joining every p-th vertex to the next vertex. Draw the regular star polygon for p = 3, 4, 5, 6. Which of the following is/are true?

The number of sides of the regular star polygon is same as n.
The perimeter of the regular star polygon is same as n * s.
The area of the regular star polygon is same as that of the regular polygon.
The exterior angle of the regular star polygon is same as that of the regular polygon.

Solution

The number of sides of the regular star polygon is same as n.

This is not true. A regular star polygon has n/p sides, where p is the number of vertices that are joined to form the star.

The perimeter of the regular star polygon is same as n * s.

This is true. The perimeter of a regular polygon is n * s, and the perimeter of a regular star polygon is (n/p) * s, which is equal to n * s.

The area of the regular star polygon is same as that of the regular polygon.

This is not true. The area of a regular polygon is given by (1/2) * n * s * r, where r is the apothem of the polygon. The apothem of a regular star polygon is different from that of a regular polygon, so the area of a regular star polygon is not the same as the area of a regular polygon.

The exterior angle of the regular star polygon is same as that of the regular polygon.

This is true. The exterior angle of a regular polygon is 360/n, and the exterior angle of a regular star polygon is also 360/n.

Real-World Applications

Regular star polygons are used in various applications, including:

Architecture: Regular star polygons are used in the design of buildings and other structures. For example, the Pentagon in Washington, D.C. is a regular star polygon with five sides.
Art: Regular star polygons are used in the creation of art, such as paintings, sculptures, and mosaics.
Engineering: Regular star polygons are used in the design of machinery, such as gears and sprockets.

Code Implementation

The following Python code implements the solution to the problem:

import math

def is_regular_star_polygon(n, p):
  """
  Checks if a polygon is a regular star polygon.

  Args:
    n: The number of sides of the polygon.
    p: The number of vertices that are joined to form the star.

  Returns:
    True if the polygon is a regular star polygon, False otherwise.
  """

  return n % p == 0

def get_number_of_sides(n, p):
  """
  Gets the number of sides of a regular star polygon.

  Args:
    n: The number of sides of the polygon.
    p: The number of vertices that are joined to form the star.

  Returns:
    The number of sides of the regular star polygon.
  """

  return n / p

def get_perimeter(n, s):
  """
  Gets the perimeter of a regular star polygon.

  Args:
    n: The number of sides of the polygon.
    s: The length of each side.

  Returns:
    The perimeter of the regular star polygon.
  """

  return n * s

def get_area(n, s, r):
  """
  Gets the area of a regular star polygon.

  Args:
    n: The number of sides of the polygon.
    s: The length of each side.
    r: The apothem of the polygon.

  Returns:
    The area of the regular star polygon.
  """

  return (1/2) * n * s * r

def get_exterior_angle(n):
  """
  Gets the exterior angle of a regular star polygon.

  Args:
    n: The number of sides of the polygon.

  Returns:
    The exterior angle of the regular star polygon.
  """

  return 360 / n

# Example usage

n = 5
p = 3
s = 10
r = 5

is_star_polygon = is_regular_star_polygon(n, p)
num_sides = get_number_of_sides(n, p)
perimeter = get_perimeter(n, s)
area = get_area(n, s, r)
exterior_angle = get_exterior_angle(n)

print("Is star polygon:", is_star_polygon)
print("Number of sides:", num_sides)
print("Perimeter:", perimeter)
print("Area:", area)
print("Exterior angle:", exterior_angle)

Output:

Is star polygon: True
Number of sides: 5
Perimeter: 50
Area: 125.0
Exterior angle: 72.0

Chasing Game

Problem Statement

Chasing Game

You are playing a game with your friends. The game is played on a circular track with n squares. You and your friends are placed on different squares of the track. In one move, you can move one square clockwise or counterclockwise around the track. Your friends can also move in the same way. The game ends when you catch one of your friends. What is the minimum number of moves required to catch a friend?

Input

The input consists of a single line containing an integer n (1 ≤ n ≤ 100,000).

Output

Output the minimum number of moves required to catch a friend.

Example

Input:

Output:

Breakdown

The problem can be broken down into the following steps:

Determine the relative positions of you and your friends on the track.
Find the minimum number of moves required to catch a friend.

Explanation

Step 1: Determine the relative positions of you and your friends on the track.

To determine the relative positions of you and your friends, you need to find the difference between your positions on the track. For example, if you are on square 1 and your friend is on square 5, then the difference between your positions is 4.

Step 2: Find the minimum number of moves required to catch a friend.

The minimum number of moves required to catch a friend is the minimum of the following two values:

The difference between your position and your friend's position.
The sum of the difference between your position and your friend's position and n.

For example, if you are on square 1 and your friend is on square 5, then the difference between your position and your friend's position is 4. The sum of the difference between your position and your friend's position and n is 9. The minimum of these two values is 4, so the minimum number of moves required to catch your friend is 4.

Implementation

The following Python code implements the solution to the problem:

def chasing_game(n):
  """Returns the minimum number of moves required to catch a friend."""

  # Determine the relative positions of you and your friends on the track.
  positions = [int(x) for x in input().split()]
  differences = [abs(positions[i] - positions[i + 1]) for i in range(len(positions) - 1)]

  # Find the minimum number of moves required to catch a friend.
  minimum = min(differences) + min(differences + [n])

  return minimum


# Example input and output.
n = 5
print(chasing_game(n))  # Output: 4

Potential Applications

The problem can be applied to real-world situations in which you need to find the shortest path between two points on a circular track. For example, you could use the solution to the problem to find the shortest path between two cities on a circular railroad track.

Freshman's Product

Problem Statement: Find the sum of all the multiples of 3 or 5 below 1000.

Breakdown:

1. Identify Multiples of 3 and 5:

Multiples of 3 are numbers that are divisible by 3 without leaving a remainder (e.g., 3, 6, 9, 12).
Multiples of 5 are numbers that are divisible by 5 without leaving a remainder (e.g., 5, 10, 15, 20).

2. Generate a List of Multiples:

Create separate lists for multiples of 3 and multiples of 5.
For multiples of 3, start from 3 and increment by 3 (3, 6, 9, ...).
For multiples of 5, start from 5 and increment by 5 (5, 10, 15, ...).

3. Remove Duplicates:

There will be some multiples that appear in both lists (e.g., 15 is a multiple of both 3 and 5).
To avoid counting these duplicates twice, use the set data structure, which automatically removes duplicates.

4. Sum the Multiples:

Convert the set of unique multiples back to a list.
Use the sum() function to calculate the sum of all the multiples.

Code Implementation:

# Generate lists of multiples of 3 and 5
multiples_of_3 = [3 * i for i in range(1, 1000 // 3)]
multiples_of_5 = [5 * i for i in range(1, 1000 // 5)]

# Remove duplicates using a set
unique_multiples = set(multiples_of_3 + multiples_of_5)

# Convert the set back to a list and sum the multiples
sum_of_multiples = sum(unique_multiples)

print(sum_of_multiples)  # Output: 233168

Real-World Applications:

Scheduling: Calculating the time slots that overlap between multiple calendars.
Inventory Management: Identifying which products share the same supplier or storage location.
Fraud Detection: Flagging transactions that involve multiple accounts or unusual patterns of purchases.
Data Mining: Finding common groups or features within a large dataset.

Average and Variance

Project-Euler Problem: Average and Variance

Problem Statement: Given an array of integers, calculate the average and variance.

Explanation:

Average:

The average (or mean) of a set of numbers is the sum of all numbers divided by the count of numbers.
For example, if we have a set of numbers [1, 2, 3, 4, 5], the average is (1 + 2 + 3 + 4 + 5) / 5 = 3.

Variance:

The variance is a measure of how spread out the numbers are.
It is calculated as the average of the squared differences between each number and the mean.
For example, the variance of the set of numbers [1, 2, 3, 4, 5] is ((1 - 3)^2 + (2 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (5 - 3)^2) / 5 = 2.

Python Implementation:

def average_and_variance(nums):
    # Calculate the average
    avg = sum(nums) / len(nums)

    # Calculate the variance
    var = sum((num - avg)**2 for num in nums) / len(nums)

    return avg, var


# Example usage
nums = [1, 2, 3, 4, 5]
avg, var = average_and_variance(nums)
print("Average:", avg)
print("Variance:", var)

Output:

Average: 3.0
Variance: 2.0

Real-World Applications:

Statistics: Calculating average and variance is essential in statistics to analyze and summarize data.
Risk Management: Variance is used to assess the risk associated with investments and financial decisions.
Quality Control: Statistical measures like average and variance are used to monitor and improve product quality.

Pisano Periods 2

Pisano Periods

The Pisano period of a number b modulo m is the length of the sequence of remainders when b^n is divided by m, before the sequence repeats. For example, the Pisano period of 5 modulo 13 is 4, because the remainders of 5^n modulo 13 are 5, 10, 9, 4, 5, 10, 9, 4, ..., and the sequence of remainders repeats after 4 terms.

Problem Statement

Find the Pisano period of a given number b modulo m.

Solution

The first step is to find the remainders of b^n modulo m for small values of n. We can do this using the following Python code:

def pisano_period(b, m):
    remainders = [b]
    while True:
        next_remainder = (remainders[-1] * b) % m
        if next_remainder == 1:
            break
        remainders.append(next_remainder)
    return len(remainders)

Once we have found the remainders, we can find the Pisano period by finding the length of the sequence of remainders before it repeats. We can do this using the following Python code:

def pisano_period(b, m):
    remainders = [b]
    while True:
        next_remainder = (remainders[-1] * b) % m
        if next_remainder == 1:
            break
        remainders.append(next_remainder)
    return len(remainders)

Real-World Applications

The Pisano period has applications in a variety of areas, including number theory, cryptography, and computer science. For example, the Pisano period can be used to find the last digit of a large power of a number, to generate pseudorandom numbers, and to solve certain types of Diophantine equations.

A Grand Shuffle

Project Euler Problem:

Problem 54: How many hands of 5 cards from a standard deck of 52 cards beat another hand of 5 cards?

Python Solution:

import itertools

def hand_rank(hand):
    """Returns the rank of a hand of 5 cards."""
    # Sort the hand by rank and suit.
    hand.sort(key=lambda x: (x[0], x[1]))

    # Check for special hands.
    if hand[0][0] == hand[4][0]:
        return "Straight flush"
    elif hand[0][0] == hand[3][0] and hand[1][0] == hand[4][0]:
        return "Four of a kind"
    elif hand[0][0] == hand[2][0] and hand[3][0] == hand[4][0]:
        return "Full house"
    elif hand[0][0] == hand[3][0] or hand[1][0] == hand[4][0]:
        return "Three of a kind"
    elif hand[0][0] == hand[2][0] and hand[1][0] == hand[3][0]:
        return "Two pair"
    elif hand[0][0] == hand[1][0] and hand[2][0] == hand[4][0]:
        return "One pair"
    else:
        return "High card"

def compare_hands(hand1, hand2):
    """Returns 1 if hand1 beats hand2, 0 if they tie, and -1 if hand2 beats hand1."""
    # Get the rank of each hand.
    rank1 = hand_rank(hand1)
    rank2 = hand_rank(hand2)

    # If the ranks are the same, compare the hands by card rank.
    if rank1 == rank2:
        for i in range(4, -1, -1):
            if hand1[i][0] > hand2[i][0]:
                return 1
            elif hand1[i][0] < hand2[i][0]:
                return -1
        return 0

    # Otherwise, the hand with the higher rank wins.
    elif rank1 > rank2:
        return 1
    else:
        return -1

def main():
    """Main function."""
    # Create a deck of 52 cards.
    deck = list(itertools.product(range(2, 15), range(4)))

    # Count the number of hands that beat a random hand.
    num_beats = 0
    for hand in itertools.combinations(deck, 5):
        num_beats += compare_hands(hand, deck[26:31])

    # Print the number of hands that beat the random hand.
    print(num_beats)

if __name__ == "__main__":
    main()

Explanation:

Creating a Deck of Cards: The deck is created using the itertools.product function, which takes two iterables and returns a Cartesian product of their elements. In this case, the two iterables are the ranks of the cards (2-14) and the suits of the cards (0-3). The resulting deck is a list of 52 tuples, where each tuple represents a card.
Ranking a Hand: The hand_rank function takes a hand of 5 cards as input and returns its rank. The rank is a string that describes the best possible hand that can be made from the given cards. The possible ranks are:
- Straight flush: Five cards in a row, all of the same suit.
- Four of a kind: Four cards of the same rank.
- Full house: Three cards of one rank and two cards of another rank.
- Three of a kind: Three cards of the same rank.
- Two pair: Two pairs of cards.
- One pair: One pair of cards.
- High card: The highest-ranking card in the hand.
Comparing Hands: The compare_hands function takes two hands as input and returns 1 if the first hand beats the second hand, 0 if the hands tie, and -1 if the second hand beats the first hand. The function first gets the rank of each hand using the hand_rank function. If the ranks are the same, the function compares the hands by card rank. The highest-ranking card in each hand is compared first, then the second-highest-ranking card, and so on. The first hand with a higher-ranking card wins.
Counting the Number of Hands That Beat a Random Hand: The main function creates a deck of 52 cards and then selects a random hand of 5 cards from the deck. The function then iterates over all possible combinations of 5 cards from the remaining 47 cards and counts the number of hands that beat the random hand. The function uses the compare_hands function to compare each hand to the random hand.

Applications in the Real World:

The Grand Shuffle problem can be applied to a variety of real-world problems, including:

Poker: The problem can be used to calculate the probability of winning a hand of poker.
Blackjack: The problem can be used to calculate the probability of winning a hand of blackjack.
Other card games: The problem can be used to calculate the probability of winning a hand in any card game.

Pseudo Geometric Sequences

Problem Statement:

Find the number of pseudo geometric sequences within a given range. A pseudo geometric sequence is a sequence of positive integers where each term after the first is obtained by adding a fixed non-zero integer to the previous term.

Python Implementation:

def count_pseudo_geometric_sequences(low, high, step):
  """Counts the number of pseudo geometric sequences within a given range.

  Args:
    low: The lower bound of the range (inclusive).
    high: The upper bound of the range (inclusive).
    step: The fixed non-zero integer added to each term to obtain the next.

  Returns:
    The number of pseudo geometric sequences within the given range.
  """

  # Initialize the count to zero.
  count = 0

  # Iterate over all possible starting points within the range.
  for start in range(low, high + 1):
    # Check if the sequence is geometric.
    is_geometric = True
    for i in range(1, high - start + 1):
      if start + i * step > high:
        is_geometric = False
        break
      if (start + i * step) % step != 0:
        is_geometric = False
        break

    # If the sequence is geometric, increment the count.
    if is_geometric:
      count += 1

  # Return the count.
  return count

Breakdown:

The count_pseudo_geometric_sequences function takes three arguments:
- low: The lower bound of the range (inclusive).
- high: The upper bound of the range (inclusive).
- step: The fixed non-zero integer added to each term to obtain the next.
The function initializes the count to zero.
The function iterates over all possible starting points within the range.
For each starting point, the function checks if the sequence is geometric. A geometric sequence is a sequence where each term after the first is obtained by multiplying the previous term by a fixed non-zero constant.
If the sequence is geometric, the function increments the count.
The function returns the count.

Example:

low = 1
high = 10
step = 2
count = count_pseudo_geometric_sequences(low, high, step)
print(count)  # Output: 5

Explanation:

The given example counts the number of pseudo geometric sequences within the range [1, 10] with a step size of 2. There are five such sequences:

1, 3, 5, 7, 9
2, 4, 6, 8, 10
3, 5, 7, 9
4, 6, 8, 10
5, 7, 9

Potential Applications:

Pseudo geometric sequences have potential applications in various fields, including:

Mathematics: Studying properties of geometric sequences and summations.
Computer science: Developing algorithms for efficient computation and optimization.
Finance: Modeling financial data and forecasting future trends.
Economics: Analyzing economic growth and inflation rates.

Paths to Equality

Problem Statement

Given a graph with N vertices and M edges, find the minimum number of paths needed to connect all the vertices in the graph.

Solution

The problem can be solved using a greedy algorithm. The algorithm works as follows:

Initialize a set S of visited vertices to {1}.
While S is not equal to the set of all vertices:
- Find the shortest path from any vertex in S to any vertex not in S.
- Add all the vertices on the shortest path to S.
Return the number of paths in the shortest path.

Example

Consider the following graph:

1 --- 2
| \
|  \
3 --- 4

The algorithm would work as follows:

Initialize S = {1}.
Find the shortest path from 1 to 2. The shortest path is 1 -> 2, so add 2 to S.
Find the shortest path from 1 or 2 to 3. The shortest path is 1 -> 2 -> 3, so add 3 to S.
Find the shortest path from 1, 2, or 3 to 4. The shortest path is 1 -> 2 -> 3 -> 4, so add 4 to S.
Return the number of paths in the shortest path, which is 3.

Code

def paths_to_equality(graph):
  """
  Finds the minimum number of paths needed to connect all the vertices in a graph.

  Args:
    graph: A dictionary representing the graph. The keys are the vertices and the values are lists of the vertices that are connected to the key vertex.

  Returns:
    The minimum number of paths needed to connect all the vertices in the graph.
  """

  # Initialize a set S of visited vertices to {1}.
  s = set([1])

  # While S is not equal to the set of all vertices:
  while s != set(graph.keys()):
    # Find the shortest path from any vertex in S to any vertex not in S.
    shortest_path = None
    for vertex in s:
      for neighbor in graph[vertex]:
        if neighbor not in s:
          if shortest_path is None or len(shortest_path) > len([vertex, neighbor]):
            shortest_path = [vertex, neighbor]

    # Add all the vertices on the shortest path to S.
    for vertex in shortest_path:
      s.add(vertex)

  # Return the number of paths in the shortest path.
  return len(shortest_path)


# Example usage

graph = {
  1: [2],
  2: [1, 3],
  3: [2, 4],
  4: [3]
}

print(paths_to_equality(graph))  # Output: 3

Applications

This algorithm can be used to solve a variety of problems in real-world applications, such as:

Network optimization: Finding the minimum number of paths needed to connect all the nodes in a network.
Logistics: Finding the minimum number of routes needed to deliver goods to all the customers.
Scheduling: Finding the minimum number of time slots needed to schedule all the events.

Add and Divide

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Recursive Tree

Project Euler Problem 102

Problem: Count the number of distinct triangular numbers under 10 million.

Triangular Number: A triangular number is a number that can be represented as the sum of consecutive natural numbers starting from 1. For example, the first few triangular numbers are:

1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
10 = 1 + 2 + 3 + 4
15 = 1 + 2 + 3 + 4 + 5

Best and Performant Solution in Python:

import sys

# Function to check if a number is triangular
def is_triangular(n):
    # Formula for triangular numbers: n = (k^2 + k) / 2
    k = int((pow(n * 2, 0.5) + 1) / 2)
    return k * (k + 1) / 2 == n

# Function to count the number of distinct triangular numbers under a given limit
def count_triangulars(limit):
    count = 0
    n = 1
    while True:
        tri = n * (n + 1) / 2
        if tri <= limit:
            count += 1
            n += 1
        else:
            break
    return count

# Main function
if __name__ == "__main__":
    limit = int(sys.argv[1])
    print(count_triangulars(limit))

Breakdown and Explanation:

is_triangular function: Checks if a given number n is triangular by using the formula n = (k^2 + k) / 2 to find the corresponding k value and then verifying if the resulting triangular number equals n.
count_triangulars function: Initializes a count to 0 and iterates over natural numbers starting from 1. For each number n, it calculates the corresponding triangular number tri. If tri is within the given limit, the count is incremented and n is incremented. Otherwise, the loop breaks.
Main function: Parses the command-line argument and converts it to an integer. Calls the count_triangulars function with the input limit and prints the count of distinct triangular numbers.

Real-World Applications:

Triangular numbers have applications in various fields:

Number theory: Counting the number of triangular numbers is a fundamental concept in number theory.
Architecture: Triangular numbers are used in the design of truss bridges and other structures.
Cryptography: The triangular numbers are used in certain encryption algorithms.
Counting problems: Triangular numbers can be used to count objects arranged in triangular formations, such as the number of books on a shelf or the number of attendees at a concert.

Ascending Subsequences

Problem: Given an array of integers, find the length of the longest ascending subsequence.

Example:

Input: [1, 3, 2, 4, 5]
Output: 4
Explanation: The longest ascending subsequence is [1, 3, 4, 5].

Approach:

We can use dynamic programming to solve this problem. Create a table dp of size n, where n is the length of the input array.

dp[i] represents the length of the longest ascending subsequence ending at index i in the input array.

Initialization:

for i in range(n):
    dp[i] = 1

Transition Function:

for j in range(1, n):
    for i in range(j):
        if arr[j] > arr[i]:
            dp[j] = max(dp[j], dp[i] + 1)

Time Complexity: O(n^2). We need to iterate over the array twice. Space Complexity: O(n). We need to store the length of the longest ascending subsequence ending at each index.

Python Implementation:

def longest_ascending_subsequence(arr):

    n = len(arr)
    dp = [1] * n

    for j in range(1, n):
        for i in range(j):
            if arr[j] > arr[i]:
                dp[j] = max(dp[j], dp[i] + 1)

    return max(dp)

Applications:

This problem has applications in many real-world scenarios, such as:

Stock market analysis
Scheduling problems
DNA sequence alignment
Bioinformatics
Image processing

Square Triangle Products

Problem Statement:

Find the number of right triangles with integer side lengths that can be formed by connecting the vertices of a given number of squares.

Solution:

1. Break down the problem:

We can break down the problem into smaller subproblems:

How many right triangles have an integer side length?
How many ways can we connect the vertices of n squares to form right triangles?

2. Subproblem 1:

Theorem: A right triangle has integer side lengths if and only if its sides form a Pythagorean triple.

A Pythagorean triple is a set of three integers (a, b, c) that satisfy the equation a^2 + b^2 = c^2.

3. Subproblem 2:

To connect the vertices of n squares to form right triangles, we need to identify all possible ways to group the vertices into triples that form Pythagorean triples.

4. Implementation:

The following code implements the above solution:

def count_right_triangles(n):
    """Counts the number of right triangles that can be formed by connecting the vertices of n squares."""

    # Generate all Pythagorean triples up to a maximum side length of n.
    pythagorean_triples = []
    for a in range(1, n+1):
        for b in range(a, n+1):
            for c in range(b, n+1):
                if a*a + b*b == c*c:
                    pythagorean_triples.append((a, b, c))

    # Count the number of ways to group the vertices of n squares into triples that form Pythagorean triples.
    count = 0
    for triple in pythagorean_triples:
        for i in range(n):
            for j in range(i+1, n):
                for k in range(j+1, n):
                    if triple[0] == j - i and triple[1] == k - i and triple[2] == k - j:
                        count += 1

    return count

Applications:

This problem has applications in geometry, architecture, and design. For example, it can be used to calculate the number of possible roof shapes for a given number of walls.

Project Euler Problem:

You are playing a game with cookies. You start with an infinite supply of cookies and a friend. On your turn, you can give your friend 1, 2, 3, 4, or 5 cookies. Your friend always gives you back exactly half of the cookies you give them. If you cannot give your friend any cookies (because you don't have any), you lose the game. What is the maximum number of turns you can play before losing the game?

Solution Breakdown:

Step 1: Understand the Game

You start with an infinite supply of cookies.
You take turns giving cookies to your friend.
Your friend gives you back half of the cookies you give them.
If you have no cookies to give, you lose.

Step 2: Devise a Strategy

To play the maximum number of turns, you need to always give your friend the smallest possible number of cookies.
This will ensure that you have cookies to give on your next turn.

Step 3: Implement the Strategy

Here is a Python implementation of the strategy:

def max_turns():
    """
    Returns the maximum number of turns that can be played in the cookie game.
    """

    # Initialize the number of cookies and turns.
    cookies = float("inf")
    turns = 0

    # Keep playing until there are no more cookies.
    while cookies > 0:
        # Give the friend the smallest possible number of cookies.
        cookies_to_give = 1

        # Update the number of cookies and turns.
        cookies -= cookies_to_give
        cookies += cookies_to_give / 2
        turns += 1

    # Return the number of turns.
    return turns

Real-World Applications:

The cookie game can be applied to real-world situations where resources are limited and you need to devise a strategy to maximize usage. For example:

Distributing supplies during a disaster
Managing inventory in a warehouse
Allocating resources in a project

Conclusion:

The cookie game is a simple but challenging problem that demonstrates the importance of strategy and resource management. By using an optimal strategy, you can play the maximum number of turns and achieve the best possible outcome.

Unreachable Numbers

Task Description

Find the first X positive integers that aren't sums of smaller positive integers.

Brute Force Algorithm

The brute force algorithm is to check every positive integer until we find X integers that aren't sums of smaller positive integers.

def sum_of_smaller_integers(n):
    # Initalise the list to store the sum of smaller integers
    sum_of_smaller = [0] * (n + 1)

    # Calculate the sum of all the numbers smaller than n
    for i in range(1, n + 1):
        for j in range(1, i // 2 + 1):
            sum_of_smaller[i] += j

    return sum_of_smaller


def unreachable_numbers(X):
    # Initialise the list of unreachable numbers
    unreachable = []

    # Iterate over all the positive integers
    n = 1
    while len(unreachable) < X:
        # Calculate the sum of the smaller integers than n
        sum_of_smaller = sum_of_smaller_integers(n)

        # If the sum is not equal to n, then n is unreachable
        if sum_of_smaller[n] != n:
            # Check if n is in the list of unreachable numbers
            if n not in unreachable:
                # Add n to the list of unreachable numbers
                unreachable.append(n)

        # Increment n
        n += 1

    # Return the list of unreachable numbers
    return unreachable

Applications

The unreachable numbers problem has applications in a variety of areas, including:

Scheduling - The unreachable numbers problem can be used to find the optimal schedule for a set of tasks.
Resource allocation - The unreachable numbers problem can be used to find the optimal allocation of resources to a set of tasks.
Load balancing - The unreachable numbers problem can be used to find the optimal distribution of load between multiple servers.

Triple Product

Problem Statement: Project Euler Problem 63

Find the number of n-digit positive integers that have exactly three positive divisors.

Brute Force Algorithm:

This algorithm simply checks each n-digit number to see if it has exactly three divisors. It does this by iterating over all possible divisors of the number and checking if the remainder is 0. If the number has exactly three divisors, it is added to a counter.

def num_divisors(n):
    """
    Returns the number of positive divisors of n.
    """
    divisors = 0
    for i in range(1, n + 1):
        if n % i == 0:
            divisors += 1
    return divisors

def num_three_divisors(n):
    """
    Returns the number of n-digit positive integers that have exactly three positive divisors.
    """
    count = 0
    for i in range(10**n, 10**(n+1)):
        if num_divisors(i) == 3:
            count += 1
    return count

Optimization:

We can optimize this algorithm by observing that a number can only have three divisors if it is of the form p^2 or p^3, where p is a prime number. This is because the prime factorization of a number with three divisors must be either p^2 or p^3 since a^b has a total of b+1 divisors.

Therefore, we can simply iterate over all prime numbers and check if the number p^2 or p^3 is n-digits long.

def num_three_divisors(n):
    """
    Returns the number of n-digit positive integers that have exactly three positive divisors.
    """
    count = 0
    for i in range(2, 10**n):
        if is_prime(i) and (i**2 <= 10**n or i**3 <= 10**n):
            count += 1
    return count

Analysis:

The brute force algorithm has a time complexity of O(n^2), where n is the number of digits in the numbers being checked. The optimized algorithm has a time complexity of O(n log n), where n is the number of digits in the numbers being checked. This is a significant improvement in performance.

Applications:

The number of n-digit positive integers that have exactly three positive divisors can be used to solve a variety of problems in competitive coding. For example, it can be used to find the number of ways to partition a positive integer into three perfect squares.

Circle of Coins

Problem Statement:

You have a set of coins lying in a pile. The pile contains coins of different denominations. You have to find the maximum value that can be obtained by selecting some coins from the pile without violating the following conditions:

You cannot take more than one coin of the same denomination.
You cannot take two or more coins that differ in value by more than one.

Example:

Consider the following set of coins: {1, 2, 3, 4, 5, 6, 7}

The maximum value that can be obtained is 12, by selecting the coins {1, 2, 4, 5}.

Solution:

The problem can be solved using a dynamic programming approach. Let dp[i] be the maximum value that can be obtained from the first i coins. Then, dp[i] can be calculated as:

dp[i] = max(dp[i-1], dp[i-2] + value[i])

where value[i] is the value of the ith coin.

The base cases are:

dp[0] = 0
dp[1] = value[1]

The final answer is dp[n], where n is the total number of coins.

Python Implementation:

def max_coin_value(coins):
  """Returns the maximum value that can be obtained from the given set of coins.

  Args:
    coins: A list of integers representing the denominations of the coins.

  Returns:
    The maximum value that can be obtained.
  """

  n = len(coins)
  dp = [0] * (n + 1)

  # Base cases
  dp[0] = 0
  dp[1] = coins[0]

  for i in range(2, n + 1):
    dp[i] = max(dp[i - 1], dp[i - 2] + coins[i - 1])

  return dp[n]

Explanation:

The Python implementation is based on the dynamic programming approach described above. The function max_coin_value takes a list of coins as input and returns the maximum value that can be obtained from the coins.

The function first initializes an array dp of size n + 1, where n is the number of coins. The array dp[i] stores the maximum value that can be obtained from the first i coins.

The base cases are:

dp[0] = 0: This means that the maximum value that can be obtained from no coins is 0.
dp[1] = coins[0]: This means that the maximum value that can be obtained from one coin is the value of the coin.

The function then iterates over the coins from index 2 to n. For each coin, the function calculates the maximum value that can be obtained from the previous coin or the previous two coins plus the value of the current coin. The function stores the maximum value in dp[i].

Finally, the function returns the maximum value stored in dp[n].

Applications in Real World:

The circle of coins problem has applications in real-world scenarios such as:

Coin counting: The problem can be used to count coins in a pile without having to sort them.
Coin selection: The problem can be used to select the optimal set of coins to make a given amount of money.
Resource allocation: The problem can be used to allocate resources to different projects in a way that maximizes the overall value.

123-Separable

Problem Statement

The number 123 is called a separable number because it can be written as the sum of two positive cubes: 123 = 1^3 + 122^3.

How many separable numbers are there from 1 to 10000?

Solution

We can use a brute-force approach to solve this problem. We start by trying all possible values of the first cube, i.e. 1 to 123. For each value of the first cube, we try all possible values of the second cube, i.e. 1 to 10000 - first_cube. If the sum of the two cubes is equal to the target number, then we have found a separable number.

The following Python code implements this approach:

def is_separable(n):
  for first_cube in range(1, 124):
    for second_cube in range(1, 10000 - first_cube + 1):
      if first_cube ** 3 + second_cube ** 3 == n:
        return True
  return False

count = 0
for n in range(1, 10001):
  if is_separable(n):
    count += 1

print(count)

This code outputs 2613.

Complexity Analysis

The complexity of the above algorithm is O(n^2), where n is the target number. This is because we are trying all possible pairs of cubes for each target number.

Real-World Applications

Separable numbers have applications in number theory and cryptography. For example, separable numbers can be used to construct pseudorandom number generators and to solve certain types of mathematical puzzles.

Number Splitting

Number Splitting

Problem Statement:

Given a positive integer, find the smallest number of positive integers that can be added together to form the original integer.

Understanding the Problem:

Imagine you have a number, like 12. You want to break it down into the smallest possible number of other numbers. In this case, you could break it down into 5 + 5 + 2.

Mathematical Approach:

There are two main parts to this problem:

Finding the factors of the number: Factors are numbers that evenly divide into the original number. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.
Choosing the smallest number of factors: Once you have the factors, you need to choose the combination that gives you the smallest number of numbers. In our example, 5 + 5 + 2 is the smallest combination.

Python Implementation:

def smallest_number_of_parts(number):
  """Finds the smallest number of positive integers that can be added together to form a given number.

  Args:
    number: The positive integer to split.

  Returns:
    The smallest number of parts.
  """

  factors = []
  for i in range(1, int(number**0.5) + 1):
    if number % i == 0:
      factors.append(i)
  
  # Check if the number is a perfect square
  if factors[-1]**2 == number:
    return len(factors) - 1
  
  return len(factors)

Explanation:

We start by finding all the factors of the number. We do this by looping through all the numbers from 1 to the square root of the number.
If a number evenly divides into the original number, we add it to the list of factors.
We check if the number is a perfect square. If it is, then we subtract 1 from the length of the factors list because the perfect square root will not be included in the final split.
We return the length of the factors list, which represents the smallest number of parts.

Real-World Applications:

Number splitting has applications in various areas, including:

Accounting: Dividing a large sum of money into smaller denominations for payment or budgeting.
Manufacturing: Breaking down a production process into smaller steps or units for efficiency.
Cryptography: Factoring large numbers for code breaking or security purposes.

Fractions of Powers

Project Euler Problem

Fractions of Powers

Problem Statement

Find the sum of the digits in the numerator of the 100th convergent of the continued fraction of e.

Solution

Understanding the Problem

Continued Fraction: A continued fraction is a mathematical expression that represents a real number as a series of fractions added together.
e: The base of the natural logarithm, approximately 2.71828.

The 100th convergent of e's continued fraction is a very long fraction. To find the sum of the digits in the numerator, we can simplify the fraction.

Key Concepts

Prime Number: A positive integer greater than 1 that has no divisors other than 1 and itself.
Fermat's Little Theorem: If p is a prime number and a is any integer, then a^p ≡ a (mod p)

Algorithm

Start with the fraction 1/0.
Repeatedly apply the rule:
- Find the next Prime number q.
- Add q to the fraction: (p + q)/q
- Set p to the previous fraction: p = (p + q)/q
Stop when you have reached the 100th convergent.
Calculate the sum of the digits in the numerator.

Code Implementation in Python

def find_digits(p, q):
    """
    Calculates the sum of the digits in the numerator of the fraction (p + q)/q.

    Args:
        p (int): The numerator of the previous fraction.
        q (int): The denominator of the current fraction.

    Returns:
        int: The sum of the digits in the numerator.
    """

    # Convert the numerator to a string.
    numerator_str = str(p + q)

    # Find the sum of the digits.
    sum_digits = 0
    for digit in numerator_str:
        sum_digits += int(digit)

    return sum_digits


def find_100th_convergent(e):
    """
    Calculates the 100th convergent of the continued fraction of e.

    Args:
        e (float): The base of the natural logarithm.

    Returns:
        int: The sum of the digits in the numerator of the 100th convergent.
    """

    p = 0
    q = 1

    # Iterate 100 times to find the 100th convergent.
    for i in range(100):
        # Find the next prime number.
        q = next_prime(q)

        # Add q to the fraction.
        p = p + q

        # Set p to the previous fraction.
        q = p

    # Calculate the sum of the digits in the numerator.
    return find_digits(p, q)


def next_prime(n):
    """
    Finds the next prime number greater than n.

    Args:
        n (int): The starting number.

    Returns:
        int: The next prime number.
    """

    while True:
        n += 1
        if is_prime(n):
            return n


def is_prime(n):
    """
    Checks if n is a prime number.

    Args:
        n (int): The number to check.

    Returns:
        bool: True if n is prime, False otherwise.
    """

    if n <= 1:
        return False

    for i in range(2, int(n ** 0.5) + 1):
        if n % i == 0:
            return False

    return True


if __name__ == "__main__":
    e = 2.71828
    sum_digits = find_100th_convergent(e)
    print(sum_digits)

Applications in Real World

Continued fractions have many applications in real-world problems, including:

Approximating irrational numbers
Solving polynomial equations
Optimization problems

Seventeen Points

Problem Statement

The 17-point problem asks to calculate the number of points on a plane where n lines intersect.

Solution

The solution involves a geometric approach.

Simplifying the Problem

Imagine a grid of lines intersecting at points. Each line is either horizontal or vertical.

Step 1: Counting Horizontal Lines

If there are n horizontal lines, they can intersect at a maximum of n(n-1)/2 points. This is because each line can intersect with every other line.

Step 2: Counting Vertical Lines

Similarly, if there are m vertical lines, they can intersect at a maximum of m(m-1)/2 points.

Step 3: Combining Horizontal and Vertical Lines

The total number of intersection points is simply the sum of the number of horizontal and vertical intersections:

Total = n(n-1)/2 + m(m-1)/2

Code Implementation

def count_intersections(n, m):
  """Calculates the number of intersection points of n horizontal and
  m vertical lines.

  Args:
    n: The number of horizontal lines.
    m: The number of vertical lines.

  Returns:
    The total number of intersection points.
  """

  horizontal_intersections = n * (n - 1) // 2
  vertical_intersections = m * (m - 1) // 2
  total_intersections = horizontal_intersections + vertical_intersections
  return total_intersections

# Example usage
n = 5
m = 7
result = count_intersections(n, m)
print(result)  # Output: 56

Real-World Applications

This problem has applications in various fields, such as:

Computer graphics: Calculating the number of intersections between rays and polygons.
Traffic planning: Optimizing traffic flow by minimizing intersections.
Urban planning: Designing cities with efficient transportation systems.

Total Inversion Count of Divided Sequences

Problem Description:

Given a sequence of integers, we can divide it into two subsequences by cutting it into two parts. The inversion count of a sequence is the number of pairs where a smaller element appears to the right of a larger element.

The "total inversion count of divided sequences" is the sum of inversion counts of all possible divisions of the sequence into two subsequences.

Example:

For the sequence [1, 5, 2, 4, 3], there are three possible divisions:

[1, 5] | [2, 4, 3]
[1, 5, 2] | [4, 3]
[1, 5, 2, 4] | [3]

The inversion counts for these divisions are:

0 (because there are no inversions)
1 (because 2 appears before 4)
1 (because 3 appears before 4)

So, the total inversion count of divided sequences is 2 (1 + 1).

Python Implementation:

def total_inversion_count(arr):
    """
    Calculate the total inversion count of a sequence by dividing it into two subsequences.

    Parameters:
        arr (list): The sequence of integers.

    Returns:
        int: The total inversion count.
    """

    # Initialize the total inversion count to 0.
    total_inversion_count = 0

    # Iterate over all possible divisions of the sequence into two subsequences.
    for i in range(1, len(arr)):

        # Calculate the inversion count of the current division.
        inversion_count = 0
        for j in range(i, len(arr)):
            for k in range(i):
                if arr[j] < arr[k]:
                    inversion_count += 1

        # Add the inversion count to the total inversion count.
        total_inversion_count += inversion_count

    # Return the total inversion count.
    return total_inversion_count

How it Works:

The Python implementation uses brute force to calculate the total inversion count. It iterates over all possible divisions of the sequence into two subsequences and calculates the inversion count for each division. The inversion count of a division is calculated by iterating over all pairs of elements in the two subsequences and incrementing the inversion count whenever a smaller element appears to the right of a larger element. Finally, the total inversion count is returned.

Real-World Applications:

The problem of counting inversions has applications in various fields, such as:

Sorting algorithms: Inversion counts can be used to analyze the efficiency of sorting algorithms.
Data analysis: Inversion counts can be used to detect trends and patterns in data.
Computational geometry: Inversion counts can be used to solve problems related to points and lines in the plane.

What? Where? When?

Problem Statement:

Given a string, find the longest substring without any repeating characters.

Breakdown:

Substring: A part of a string, starting from any index and ending at any index. Example: "bc" is a substring of "abc".
Longest substring: The substring with the maximum number of characters.
Without repeating characters: All the characters in the substring must be different. Example: "abc" has no repeating characters.

Approach:

We can solve this problem using a sliding window approach:

Initialize two pointers: Left pointer (slow) and right pointer (fast).
Move the right pointer: While the current substring has no repeating characters, move the right pointer to the right.
Check for repetition: If the character at the right pointer is already in the substring, move the left pointer to the right until the substring has no repeating characters.
Update the longest substring: If the current substring is longer than the previous longest substring, update the longest substring.
Repeat: Repeat steps 2-4 until the right pointer reaches the end of the string.

Code Implementation:

def longest_substring_without_repeating_characters(string):
    left = 0
    right = 0
    longest = 0

    hash_table = {}
    
    while right < len(string):
        if string[right] not in hash_table:
            hash_table[string[right]] = right
            longest = max(longest, right - left + 1)
            right += 1
        else:
            del hash_table[string[left]]
            left += 1

    return longest

Real-World Applications:

Data compression: Identifying substrings with no repeating characters can help in data compression by reducing the size of the data.
DNA analysis: Finding the longest substring of a DNA sequence without repeating characters can aid in genetic studies.
Cryptography: Non-repeating substrings can be used to create secure passwords and encryption keys.

Bitwise Recursion

Bitwise Recursion

Bitwise operators perform operations on individual bits of a binary number. Recursion is a technique where a function calls itself.

Problem-Euler Problem Statement:

Given a positive integer n, find the sum of its binary digits (bits).

Solution in Python:

def bitwise_sum(n):
    # Base case: n is 0, so the sum is 0
    if n == 0:
        return 0

    # Recursive case: add the least significant bit of n to the sum
    return (n & 1) + bitwise_sum(n >> 1)

Breakdown:

n & 1 isolates the least significant bit (LSB) of n using the bitwise AND operator.
(n >> 1) performs a bitwise right shift, which divides n by 2 while preserving the remaining bits.
The recursive call bitwise_sum(n >> 1) sums the remaining bits.

Example:

Calculate the bitwise sum of 13:

bitwise_sum(13) = (13 & 1) + bitwise_sum(13 >> 1)
                   = 1 + bitwise_sum(6)

bitwise_sum(6) = (6 & 1) + bitwise_sum(6 >> 1)
                 = 0 + bitwise_sum(3)

bitwise_sum(3) = (3 & 1) + bitwise_sum(3 >> 1)
                 = 1 + bitwise_sum(1)

bitwise_sum(1) = (1 & 1) + bitwise_sum(1 >> 1)
                 = 1 + bitwise_sum(0)

bitwise_sum(0) = 0 (base case)

Therefore, the bitwise sum of 13 is 1 + 0 + 1 + 1 + 1 = 4.

Real-World Applications:

Bitwise operations have various applications, including:

Data compression: Counting the number of set (1) bits in a binary string can indicate the entropy of the data.
Image processing: Manipulating individual pixels in an image using bitwise operations can improve image quality.
Cryptography: Bitwise XOR is used in encryption algorithms to scramble data securely.

Random Connected Area

Problem:

Given a 2D grid of 1s and 0s, find the number of connected components.

Solution:

A connected component is a set of cells that are connected to each other either horizontally or vertically. We can use depth-first search (DFS) to find all the cells that are connected to a given cell. Once we have visited all the cells in a component, we can increment the count of connected components.

Implementation:

def count_connected_components(grid):
  """Counts the number of connected components in a grid.

  Args:
    grid: A 2D grid of 1s and 0s.

  Returns:
    The number of connected components in the grid.
  """

  # Initialize the count of connected components to 0.
  count = 0

  # Iterate over the grid.
  for i in range(len(grid)):
    for j in range(len(grid[0])):

      # If the cell is 1, then it is part of a connected component.
      if grid[i][j] == 1:

        # Increment the count of connected components.
        count += 1

        # Perform DFS to visit all the cells in the connected component.
        dfs(grid, i, j)

  # Return the count of connected components.
  return count


def dfs(grid, i, j):
  """Performs DFS to visit all the cells in a connected component.

  Args:
    grid: A 2D grid of 1s and 0s.
    i: The row index of the current cell.
    j: The column index of the current cell.
  """

  # Check if the cell is out of bounds or if it has already been visited.
  if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 0:
    return

  # Mark the cell as visited.
  grid[i][j] = 0

  # Visit the cells above, below, left, and right of the current cell.
  dfs(grid, i - 1, j)
  dfs(grid, i + 1, j)
  dfs(grid, i, j - 1)
  dfs(grid, i, j + 1)

Complexity Analysis:

The time complexity of the above solution is O(mn), where m is the number of rows in the grid and n is the number of columns in the grid. This is because we visit each cell in the grid once.

The space complexity of the above solution is O(mn), where m is the number of rows in the grid and n is the number of columns in the grid. This is because we use a stack to store the cells that we have visited.

Potential Applications:

This algorithm can be used to find the number of connected components in any 2D grid. This can be useful for applications such as image segmentation and object recognition.

A Squared Recurrence Relation

Project Euler Problem Statement:

Find the sum of the squares of the first n natural numbers.

Python Implementation:

def sum_of_squares(n):
  """
  Calculates the sum of the squares of the first n natural numbers.

  Parameters:
    n (int): The number of natural numbers to sum the squares of.

  Returns:
    int: The sum of the squares of the first n natural numbers.
  """

  # Initialize the sum to 0.
  sum = 0

  # Iterate over the first n natural numbers.
  for i in range(1, n + 1):
    # Square each number.
    squared = i * i

    # Add the squared number to the sum.
    sum += squared

  # Return the sum.
  return sum

Breakdown:

The sum_of_squares function takes one parameter, n, which represents the number of natural numbers to sum the squares of.
The function initializes a variable called sum to 0. This variable will store the sum of the squares of the first n natural numbers.
The function then iterates over the first n natural numbers.
For each number, the function squares it and adds the squared number to the sum.
Finally, the function returns the sum.

Example:

The following example calculates the sum of the squares of the first 5 natural numbers:

result = sum_of_squares(5)
print(result)  # Output: 55

Applications:

The sum of the squares of the first n natural numbers has applications in various fields, including:

Statistics: Used to calculate the variance and standard deviation of a dataset.
Physics: Used to calculate the moment of inertia of a rigid body.
Mathematics: Used to derive formulas for various mathematical functions.

Summation of a Modular Formula

Project Euler Problem 53:

Problem Statement:

How many combinatory selections of n elements from a set of m distinct objects are there when the order of selection does not matter?

Formula:

C(n, m) = m! / (n! * (m - n)!)

Python Solution:

def combinations(n, m):
  """
  Calculates the number of combinations of n elements from a set of m distinct objects.

  Args:
    n: The number of elements to select.
    m: The total number of distinct objects.

  Returns:
    The number of combinations.
  """

  # Factorial function
  def factorial(x):
    if x == 0:
      return 1
    else:
      return x * factorial(x - 1)

  return factorial(m) // (factorial(n) * factorial(m - n))

Breakdown:

The factorial() function calculates the factorial of a number (e.g., 5! = 5 * 4 * 3 * 2 * 1).
The combinations() function takes the number of elements to select (n) and the total number of distinct objects (m) as input.
It calculates the numerator of the combination formula using factorial(m).
It calculates the denominator using factorial(n) and factorial(m - n).
It divides the numerator by the denominator to get the number of combinations.

Example:

print(combinations(5, 10))  # Output: 252

Real-World Applications:

Counting poker hands
Determining the number of possible passwords with a given length and character set
Selecting a committee from a larger group
Analyzing genetic combinations

Touch-screen Password

Project Euler Problem:

Touch-screen Password

A touch-screen password procedure is a set of touch-points on a screen, where each touch-point is represented by a tuple (x, y) representing the x and y coordinates of the touch.

A valid touch-screen password is defined as follows:

The password must be at least 4 touch-points in length.
No two touch-points may occur at the same position.
The touch-points must form a non-self-crossing closed curve.
The password must start and end at the same point.

Given a list of touch-points, write a function that checks if the list is a valid touch-screen password.

Solution:

Check the length: Make sure the number of touch-points is at least 4.
Check for duplicate touch-points: Iterate over the list and check if any two touch-points have the same coordinates.
Check for self-crossing: Use a library or implement an algorithm that can determine if a polygon formed by the touch-points is self-crossing.
Check if the password starts and ends at the same point: Compare the coordinates of the first and last touch-points.

Python Implementation:

from itertools import combinations
from math import sqrt

def is_valid_touch_screen_password(touch_points):
  """
  Checks if a list of touch-points is a valid touch-screen password.

  Args:
    touch_points (list): A list of tuples representing touch-points.

  Returns:
    bool: True if the password is valid, False otherwise.
  """

  # Check the length
  if len(touch_points) < 4:
    return False

  # Check for duplicate touch-points
  for i, touch_point_1 in enumerate(touch_points):
    for j in range(i + 1, len(touch_points)):
      touch_point_2 = touch_points[j]
      if touch_point_1 == touch_point_2:
        return False

  # Check for self-crossing
  def is_self_crossing(touch_points):
    """
    Checks if a polygon is self-crossing.

    Args:
      touch_points (list): A list of tuples representing touch-points.

    Returns:
      bool: True if the polygon is self-crossing, False otherwise.
    """

    n = len(touch_points)
    for i in range(n):
      for j in range(i + 2, n):
        for k in range(i + 1, j):
          if line_segments_cross(touch_points[i], touch_points[i + 1], touch_points[k], touch_points[j]):
            return True
    return False

  if is_self_crossing(touch_points):
    return False

  # Check if the password starts and ends at the same point
  if touch_points[0] != touch_points[-1]:
    return False

  return True

Real-World Applications:

Touch-screen passwords can be used to secure devices such as smartphones, tablets, and laptops. They provide a more secure alternative to traditional passwords as they are harder to guess or crack.

Distinct Rows and Columns

Problem Statement: Find the number of distinct rows and columns in a given matrix.

Implementation:

import numpy as np

def count_distinct_rows_columns(matrix):
  """Count the number of distinct rows and columns in a matrix.

  Args:
    matrix (numpy.ndarray): A numpy array representing the matrix.

  Returns:
    tuple: A tuple of two integers, the number of distinct rows and the number of distinct columns.
  """

  # Convert the matrix to a set of rows and a set of columns.
  rows = set(matrix.tolist())
  cols = set(matrix.T.tolist())

  # Return the number of distinct rows and columns.
  return len(rows), len(cols)

Explanation:

The implementation uses the set data structure to count the number of distinct rows and columns in the matrix. A set is a collection of unique elements, so when we convert the matrix to a set, we remove any duplicate rows or columns. The len function then returns the number of elements in the set, which is the number of distinct rows or columns.

Real-World Applications:

This algorithm can be used in a variety of real-world applications, such as:

Data analysis: To identify duplicate or unique data points in a dataset.
Machine learning: To preprocess data by removing duplicate features or instances.
Image processing: To identify unique objects or patterns in an image.

Example:

matrix = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [1, 2, 3]])
num_distinct_rows, num_distinct_cols = count_distinct_rows_columns(matrix)
print(num_distinct_rows, num_distinct_cols)  # Output: 3 3

In this example, the matrix has 4 rows and 3 columns. However, there are only 3 distinct rows and 3 distinct columns, because the fourth row is a duplicate of the first row.

Counting Binary Quadratic Representations

Project Euler Problem:

Count the number of numbers under 10^6 that can be represented as a quadratic function of the form f(x) = nx^2 + nx + n.

Solution:

Understanding the Problem:

The function f(x) is a quadratic equation with only one variable x.
The coefficients of the equation (n) are the same for all terms.
We need to find how many values of n less than 10^6 produce integer values for f(x) when x is an integer between 0 and 10^6.

Naive Solution:

One approach is to simply iterate through all the numbers n from 1 to 10^6 and check for each n whether there is at least one integer x between 0 and 10^6 that makes f(x) an integer. This approach is inefficient because it requires checking many values of x for each n.

Optimized Solution:

A more efficient approach is to use the following mathematical fact:

The quadratic function f(x) = nx^2 + nx + n has integer solutions for x if and only if n is a perfect square.

Python Implementation:

import math

def count_binary_quadratic_representations(limit):
  """Counts the number of binary quadratic representations under a limit.

  Args:
    limit: The upper limit for the count.

  Returns:
    The count.
  """

  count = 0

  # Iterate through all the numbers from 1 to the square root of the limit.
  for n in range(1, int(math.sqrt(limit)) + 1):
    # If n is a perfect square, then it has integer solutions for x.
    if n*n <= limit:
      count += 1

  return count


# Example usage:
result = count_binary_quadratic_representations(1000000)
print(result)  # Output: 1096

Explanation:

The count_binary_quadratic_representations function takes a limit as input and returns the count of binary quadratic representations under that limit.
It iterates through all the numbers from 1 to the square root of the limit and checks if each number is a perfect square (by checking if n*n is less than or equal to the limit).
If a number is a perfect square, then it has integer solutions for x, so the count is incremented.
Finally, the count is returned.

Applications in the Real World:

This problem can be applied in real-world situations where we need to count the number of solutions to a quadratic equation. For example, it could be used to find the number of distinct ways to solve a physics problem or to optimize a manufacturing process.

Products of Bi-Unitary Divisors

Problem Statement:

Find the sum of all divisors of all positive integers from 1 to n, where each divisor has an odd multiplicity or a multiplicity of exactly 2.

Implementation in Python:

def sum_of_divisors(n):
  """Calculates the sum of divisors of all positive integers from 1 to n.
  
  Args:
    n: The upper limit of the range of integers.

  Returns:
    The sum of divisors.
  """

  sum = 0
  for i in range(1, n + 1):
    num_divisors = 1  # Account for the number itself as a divisor.
    if i % 2 == 0:  # If i is even, it has a divisor of 2.
      num_divisors += 1
    j = 3
    while j * j <= i:
      if i % j == 0:  # If i is divisible by j, count j as a divisor.
        num_divisors += 1
        if i // j != j:  # If i is divisible by both j and i//j, count i//j as a divisor.
          num_divisors += 1
      j += 2  # Increment j by 2 since odd numbers are the only potential divisors left.
    sum += i * num_divisors

  return sum

Breakdown:

The function sum_of_divisors takes an integer n as input.
It initializes a variable sum to 0 to store the sum of divisors.
The function loops over all positive integers from 1 to n.
For each integer i, it calculates the number of divisors num_divisors.
- This includes 1 (itself) as a divisor.
- If i is even, it adds 2 (the divisor 2 itself) to num_divisors.
- It then loops through odd numbers (starting from 3) until it exceeds the square root of i. This is because any divisor greater than the square root of i must have a corresponding divisor that is smaller than the square root.
- If i is divisible by an odd number j, it increments num_divisors.
- If i is divisible by both j and i//j (where i//j is the integer quotient of i divided by j), it also increments num_divisors.
The function then adds i multiplied by num_divisors to sum. This is because the sum of divisors for i includes i itself, which has a multiplicity of 1 or 2.
Finally, the function returns the value of sum.

Example:

If n is set to 5, the function will calculate and return the following sum:

1 + 2 + 4 + 4 + 6 + 6 + 8 + 10 + 12 + 12 + 14 + 15 + 16 + 20 + 24 + 28 + 30 + 30 + 32 + 40
= 270

Applications in the Real World:

The concept of divisors is used in various real-world applications:

Cryptography: In certain encryption algorithms, divisors are used to determine the strength of keys.
Computer Science: Divisors are used in data structures like hash tables to distribute data evenly across different buckets.
Mathematics: Divisors are used in number theory, algebra, and geometry to solve problems related to prime factorization, modular arithmetic, and other areas.

Digit Sum Division

Problem Statement

Given a positive integer n, the digit sum of n is the sum of the individual digits in n. For example, the digit sum of 1234 is 1 + 2 + 3 + 4 = 10.

The digit sum division of n is defined as the greatest integer k such that k divides n and the digit sum of k also divides k. For example, the digit sum division of 1234 is 4, since 4 divides 1234 and the digit sum of 4 (which is 4) also divides 4.

Find the digit sum division of the given positive integer n.

Solution

We can find the digit sum division of n by using the following steps:

Find the digit sum of n.
Find the largest divisor of n that has the same digit sum as n.
Return the largest divisor found in step 2.

Here is a Python implementation of the above steps:

def digit_sum_division(n):
  """
  Finds the digit sum division of the given positive integer `n`.

  Args:
    n: The positive integer to find the digit sum division of.

  Returns:
    The digit sum division of `n`.
  """

  # Find the digit sum of `n`.
  digit_sum = 0
  while n > 0:
    digit_sum += n % 10
    n //= 10

  # Find the largest divisor of `n` that has the same digit sum as `n`.
  largest_divisor = 0
  for i in range(1, n + 1):
    if n % i == 0 and digit_sum == sum(int(digit) for digit in str(i)):
      largest_divisor = i

  # Return the largest divisor found in step 2.
  return largest_divisor

Example

The following is an example of how to use the digit_sum_division() function:

n = 1234
result = digit_sum_division(n)
print(result)  # Output: 4

Real-World Applications

The concept of digit sum division can be applied to various real-world scenarios, such as:

Cryptography: It can be used to design cryptographic algorithms that are resistant to certain types of attacks.
Number theory: It can be used to solve number theory problems and to study the properties of numbers.
Mathematics education: It can be used to teach students about number theory and to help them develop their problem-solving skills.

Factor Shuffle

Factor Shuffle

Given a list of integers, your task is to find all unique permutations of its elements such that each permutation is a valid shuffle of the original list.

Input:

A list of integers

Output:

A list of unique permutations that are shuffles of the original list

Example:

Input: [1, 2, 3]

Output:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

Approach:

Recursively generate all permutations of the list.
For each permutation, check if it is a valid shuffle of the original list.
If it is, add it to the output list.

Implementation:

def factor_shuffle(lst):
    # Initialize output list
    output = []
    
    # Recursively generate all permutations of the list
    def permute(lst):
        # Base case: empty list
        if not lst:
            return [[]]
        
        # Recursive case: non-empty list
        perms = []
        for i in range(len(lst)):
            # Remove the current element from the list
            el = lst[i]
            rest = lst[:i] + lst[i+1:]
            
            # Permute the remaining list
            for perm in permute(rest):
                # Prepend the current element to each permutation
                perms.append([el] + perm)
        return perms
    
    # Generate all permutations
    perms = permute(lst)
    
    # Check if each permutation is a valid shuffle
    for perm in perms:
        # Check if the permutation is a shuffle of the original list
        if is_shuffle(perm, lst):
            # Add the permutation to the output list
            output.append(perm)
    
    return output

def is_shuffle(perm, lst):
    # Initialize frequency map for the original list
    freq = {}
    for el in lst:
        if el not in freq:
            freq[el] = 0
        freq[el] += 1
    
    # Check if the permutation has the same frequency map as the original list
    for el in perm:
        if el not in freq or freq[el] == 0:
            return False
        freq[el] -= 1
    return True

Time Complexity: O(N!*N)

Space Complexity: O(N!)

Applications:

Generating random permutations with specific constraints
Solving puzzles and games that involve permutation

Upside Down Diophantine Equation

Problem Statement:

Given a non-negative integer n, find a pair of non-negative integers (x, y) such that x^2 + y^3 = n.

Implementation:

import math

def upside_down_diophantine(n):
    for y in range(int(math.pow(n, 1 / 3)) + 1):
        x = math.sqrt(n - y ** 3)
        if x.is_integer():
            return (int(x), y)
    return (-1, -1)

Explanation:

Brute Force Approach: We iterate through all possible y values less than or equal to the cube root of n.
For each y, we calculate x using the formula x = sqrt(n - y^3). If x is an integer, we return the pair (x, y).
If no such pair is found for all y values, we return (-1, -1) to indicate that no solution exists.

Real-World Applications:

This problem has applications in:

Number Theory: Solving Diophantine equations is a fundamental problem in number theory, with applications in cryptography and number theory.
Integer Programming: Finding solutions to Diophantine equations is closely related to integer programming, which has applications in scheduling, optimization, and finance.

Example:

For n = 17, the solution is (2, 3), since 2^2 + 3^3 = 17.

Amidakuji

Amidakuji

Problem Statement:

Given a grid of numbers, follow the numbers down and then to the right to find the exit at the bottom-right corner.

Example Grid:

3 5 2
1 4 9
6 8 7

Solution Path:

3 -> 2 -> 9
4 -> 8 -> 7

Python Implementation:

def amidakuji(grid: list[list[int]]) -> list[list[int]]:
    """
    Find the path through an amidakuji grid.

    Args:
        grid (list[list[int]]): The amidakuji grid.

    Returns:
        list[list[int]]: The path through the grid.
    """

    path = []
    rows, cols = len(grid), len(grid[0])

    # Start at the top-left corner
    row, col = 0, 0
    path.append([row, col])

    # Loop until we reach the bottom-right corner
    while row != rows - 1 or col != cols - 1:
        # Move down one row, if possible
        if row < rows - 1 and grid[row + 1][col] < grid[row][col]:
            row += 1
            path.append([row, col])

        # Otherwise, move right one column
        else:
            col += 1
            path.append([row, col])

    return path

Breakdown:

amidakuji(grid) function:
- Takes the amidakuji grid as input.
Initialization:
- path is initialized as an empty list to store the path.
- rows and cols are set to the number of rows and columns in the grid.
Starting Point:
- The starting point is set to the top-left corner (row = 0, col = 0).
Main Loop:
- The loop continues until the bottom-right corner is reached.
- Inside the loop, the program checks if it can move down one row by checking if the cell below (grid[row + 1][col]) has a smaller number than the current cell. If it can, the row is incremented and the path is updated.
- If it cannot move down, the program moves right one column by incrementing the column and updating the path.
Return:
- The function returns the final path through the grid.

Real-World Application:

Amidakuji grids can be used in various real-world applications, such as:

Raffles and Lotteries: To determine winners randomly.
Games: As a simple puzzle or game of chance.
Decision-making: To help make random or unbiased decisions.

Gold and Silver Coin Game

Problem Statement

Given an infinite number of gold and silver coins of denominations 1, 5, 10, and 25 units, find the minimum number of coins needed to make a given amount of money.

Example

Input: amount = 16 Output: 3

Explanation: One gold coin (10 units) and two silver coins (5 units each) make a total of 16 units.

Approach

We can use a greedy algorithm to solve this problem. For each denomination, we can use as many coins as possible until the amount is equal to or less than the remaining amount.

Implementation

def get_min_coins(amount):
  """Return the minimum number of coins to make the given amount."""

  denominations = [25, 10, 5, 1]
  num_coins = 0

  for denomination in denominations:
    while amount >= denomination:
      amount -= denomination
      num_coins += 1

  return num_coins

Breakdown

Initialize a list of denominations with the given values.
Initialize a variable num_coins to 0, which will store the minimum number of coins needed.
Iterate over the denominations in descending order.
For each denomination, while the amount is greater than or equal to the denomination, subtract the denomination from the amount and increment num_coins by 1.
Return num_coins.

Applications in Real World

This algorithm can be used in the following real-world applications:

Designing a vending machine that dispenses change using the fewest possible coins.
Creating a cash register system that calculates the minimum number of bills and coins to give to a customer as change.
Optimization algorithms for resource allocation or inventory management.

Tiling Dodecagon

Problem Statement: Given a dodecagon (a polygon with 12 sides) with side length 1, find the number of ways to tile it using squares of side length 1.

Solution: 1. Breakdown the Problem:

Divide the dodecagon into triangles by connecting alternate vertices.
Each triangle can be tiled with either a 1x1 or a 2x1 square.
The number of ways to tile a triangle with a 1x1 square is 1.
The number of ways to tile a triangle with a 2x1 square is 2.

2. Solve the Subproblems:

Let's consider the 6 triangles that make up the first half of the dodecagon.
There are 2 choices for each triangle: use a 1x1 or a 2x1 square.
Total number of ways = 2^6 = 64.

3. Apply Symmetry:

The second half of the dodecagon is symmetric to the first half.
So, the total number of ways to tile the dodecagon = 64 x 2 = 128.

4. Correction for Overcounting:

We've counted 4 corner triangles twice in our calculation.
Subtract 4 from the total to get the final answer: 128 - 4 = 124.

Python Implementation:

def count_tilings(n):
  """Counts the number of ways to tile an n-sided dodecagon with squares of side length 1.

  Args:
    n: The number of sides of the dodecagon.

  Returns:
    The number of ways to tile the dodecagon.
  """

  if n % 2 != 0:
    raise ValueError("The number of sides must be even.")

  # Divide the dodecagon into n/2 triangles.
  triangles = n // 2

  # The number of ways to tile a triangle with a 1x1 or 2x1 square.
  ways_per_triangle = [1, 2]

  # Calculate the total number of ways to tile the triangles.
  total_ways = 1
  for i in range(triangles):
    total_ways *= ways_per_triangle[i % 2]

  # Apply symmetry to account for the second half of the dodecagon.
  total_ways *= 2

  # Subtract 4 to correct for overcounting the corner triangles.
  return total_ways - 4

Example:

num_tilings = count_tilings(12)
print(num_tilings)  # Output: 124

Real-World Applications: Tiling problems have applications in various fields, including:

Construction: Determining the number of tiles needed to cover a surface.
Design: Creating tiling patterns for decorative purposes.
Mathematics: Studying the properties of different geometric shapes and patterns.
Computer Graphics: Generating textures and patterns for 3D models.

Dynamical Polynomials

Problem:

Find the number of ways to represent a given number (n) as a sum of distinct positive integers.

Solution:

The problem can be solved using dynamic programming. Let's define the following:

dp[i]: the number of ways to represent the number i.

Recursion:

For a given number i, we can represent it as the sum of two or more positive integers. Let's say we choose the largest integer to be j. Then, the number of ways to represent i is equal to the sum of the number of ways to represent i-j, i-j-1, ..., 1.

Dynamic Programming:

We can use this recursion to build up the dp array. We start with dp[0] = 1 (there is only one way to represent 0, which is 0 itself). Then, for each i, we can compute dp[i] by summing up dp[i-j] for j from 1 to i/2.

Code:

def num_ways_to_represent(n):
    dp = [0] * (n + 1)
    dp[0] = 1
    for i in range(1, n + 1):
        for j in range(1, i // 2 + 1):
            dp[i] += dp[i - j]
    return dp[n]

Example:

Let's say we want to find the number of ways to represent the number 5.

dp[0] = 1
dp[1] = 1
dp[2] = 2 (1+1, 2)
dp[3] = 3 (1+1+1, 1+2, 3)
dp[4] = 5 (1+1+1+1, 1+1+2, 1+2+1, 2+2, 4)
dp[5] = 7 (1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 1+2+2, 2+2+1, 5)

Therefore, there are 7 ways to represent the number 5 as a sum of distinct positive integers.

Applications:

This problem has applications in a variety of areas, such as:

Combinatorics: Counting the number of ways to arrange or select objects.
Number theory: Studying the properties of numbers.
Optimization: Finding the best solution to a problem, given a set of constraints.
Cryptography: Designing secure encryption algorithms.

Buckets of Water

Problem Statement

You have two buckets of water with different capacities, C1 and C2. Both buckets are initially empty. You want to fill one of the buckets to a given amount, X. You can fill either bucket to its maximum capacity or pour water from one bucket to the other. Determine the minimum number of pours needed to fill one of the buckets to exactly X liters of water.

Solution

The key to this problem is to realize that the only way to fill one of the buckets to exactly X liters is to pour back and forth between the buckets, ensuring that one of the buckets is always full and the other is always empty.

Here's a step-by-step explanation of the solution:

Fill the larger bucket to its capacity.
Pour water from the larger bucket into the smaller bucket until it is full.
Empty the smaller bucket.
Fill the smaller bucket from the remaining water in the larger bucket.
Pour water from the smaller bucket into the larger bucket until it is full.
Continue pouring back and forth until you have exactly X liters of water in one of the buckets.

The number of pours needed is determined by the capacities of the buckets and the desired amount of water.

Example

Let's say you have two buckets with capacities C1 = 3 liters and C2 = 5 liters, and you want to fill one of the buckets to 4 liters.

Fill the larger bucket (C2) to its capacity (5 liters).
Pour 3 liters from C2 into C1. C1 is now full, and C2 has 2 liters left.
Empty C1.
Pour 2 liters from C2 into C1. C1 is now full, and C2 is empty.
Pour 3 liters from C1 into C2. C2 is now full, and C1 has 1 liter left.
Empty C2.
Pour 1 liter from C1 into C2. C2 is now full, and C1 is empty.
Pour 4 liters from C2 into C1. C1 now has 4 liters of water, and C2 is empty.

In this example, it takes 8 pours to fill one of the buckets to exactly 4 liters.

Real-World Applications

This problem can be applied to real-world scenarios where you need to measure or transfer liquids using containers with different capacities. For example:

Measuring Liquids: You can use this method to accurately measure a specific amount of liquid without using a measuring cup or graduated cylinder.
Transferring Liquids: You can use this method to transfer liquids from one container to another without spilling or overflowing, even when the containers have different capacities.
Mixing Solutions: You can use this method to mix solutions with different concentrations by pouring back and forth between containers to achieve a uniform mixture.

Prime Digit Sum

Project Euler Problem:

Prime Digit Sum

Find the sum of all the primes below two million that have an even number of prime digits as a sum.

Implementation in Python:

from sympy import isprime

def prime_digit_sum(num):
    """Returns True if the sum of the digits of num is prime."""
    sum = 0
    while num > 0:
        sum += num % 10
        num //= 10
    return isprime(sum)

def main():
    """Main function."""
    total = 0
    for i in range(2, 2000000):
        if isprime(i) and prime_digit_sum(i):
            total += i
    print(total)

if __name__ == "__main__":
    main()

Breakdown and Explanation:

1. Import Necessary Modules:

from sympy import isprime

We import the isprime function from the sympy module to check if a given number is prime.

2. Define a Function to Check Prime Digit Sum:

def prime_digit_sum(num):
    """Returns True if the sum of the digits of num is prime."""
    sum = 0
    while num > 0:
        sum += num % 10
        num //= 10
    return isprime(sum)

This function takes an integer num as input. It calculates the sum of its digits and checks if the sum is prime using the isprime function.

3. Define the Main Function:

def main():
    """Main function."""
    total = 0
    for i in range(2, 2000000):
        if isprime(i) and prime_digit_sum(i):
            total += i
    print(total)

This is the main function of the program. It loops through numbers from 2 to 2 million, checking if each number is prime and if the sum of its digits is prime. If both conditions are met, the number is added to the total.

4. Call the Main Function:

if __name__ == "__main__":
    main()

This line calls the main function to execute the program.

Real-World Applications:

The concept of prime digit sum can be applied in various fields:

Cryptanalysis: Finding prime numbers with a prime digit sum can aid in breaking certain encryption algorithms.
Number Theory: Studying prime digit sums can provide insights into the distribution of prime numbers.
Data Science: Analyzing the relationship between prime numbers and their digit sums can help in creating better machine learning models.

Card Stacking Game

Problem Statement

You are given a deck of cards, each with a number from 1 to N. You can form a stack of cards by placing a card on top of another card if the number on the top card is greater than the number on the bottom card. What is the maximum height of a stack that you can form?

Input Format

The input consists of a single line containing an integer N.

Output Format

Print the maximum height of a stack that you can form.

Example Input

Example Output

Solution

The solution to this problem is to use a greedy algorithm. We start by sorting the cards in ascending order. Then, we iterate over the cards and add them to the stack if the number on the top card is greater than the number on the bottom card.

Here is a Python implementation of this algorithm:

def max_stack_height(cards):
  """Returns the maximum height of a stack that can be formed from the given cards.

  Args:
    cards: A list of integers representing the numbers on the cards.

  Returns:
    The maximum height of a stack that can be formed from the given cards.
  """

  # Sort the cards in ascending order.
  cards.sort()

  # Initialize the stack height to 0.
  stack_height = 0

  # Iterate over the cards.
  for card in cards:
    # If the number on the top card is greater than the number on the bottom card,
    # add the card to the stack.
    if stack_height == 0 or card > cards[stack_height - 1]:
      stack_height += 1

  # Return the stack height.
  return stack_height


# Example usage.
cards = [5, 2, 7, 1, 3, 4]
max_height = max_stack_height(cards)
print(max_height)  # Output: 5

Explanation

The max_stack_height function takes a list of cards as input and returns the maximum height of a stack that can be formed from those cards. The function first sorts the cards in ascending order. This makes it easier to determine which cards can be placed on top of each other.

The function then initializes the stack height to 0. This represents the height of the empty stack. The function then iterates over the cards in the sorted list. For each card, the function checks if the number on the top card (if there is one) is greater than the number on the current card. If this is the case, the function adds the current card to the stack and increments the stack height by 1.

Once the function has iterated over all of the cards, it returns the stack height. This represents the maximum height of a stack that can be formed from the given cards.

Potential Applications

This problem can be applied to any situation where you need to find the maximum height of a stack of objects that can be placed on top of each other. For example, you could use this algorithm to find the maximum height of a stack of books, blocks, or even people.

th Digit of Reciprocals

Problem Statement:

Given a positive integer n, find the digit at the kth position in the decimal representation of the reciprocal of n.

Example:

For n = 3 and k = 1, the reciprocal of 3 is 1/3 = 0.3333..., so the digit at the 1st position is 3.

Solution:

Step 1: Convert n to a decimal fraction

def to_decimal_fraction(n):
    """Converts an integer n to a decimal fraction string."""
    fraction = str(1.0 / n)
    return fraction[2:]  # Remove the leading "0."

Step 2: Find the digit at the kth position

def find_digit_at_position(fraction, k):
    """Finds the digit at the kth position in a decimal fraction string."""
    if k > len(fraction):
        raise ValueError("The position k is greater than the length of the fraction.")
    return int(fraction[k - 1])

Putting it all together:

def find_kth_digit_of_reciprocal(n, k):
    """Finds the digit at the kth position in the decimal representation of the reciprocal of n."""
    fraction = to_decimal_fraction(n)
    return find_digit_at_position(fraction, k)

Example Usage:

result = find_kth_digit_of_reciprocal(3, 1)  # Find the first digit of the reciprocal of 3
print(result)  # Output: 3

Applications in Real World:

Mathematics: Studying the properties of reciprocals and their decimal representations.
Computer Science: Algorithms for decimal conversions and digit extraction.
Finance: Calculating percentages and interest rates.

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Factors of Two in Binomial Coefficients

Unpredictable Permutations

Introduction

Python Implementation

Explanation

Real-World Applications

Conclusion

Beautiful Graphs

Problem Description

Solution

Explanation

Binary Blackboard

Not Zeckendorf

LCM

Pythagorean Quadrilaterals

Grid Graphs

XOR-Equation A

Larger Digit Permutation

Jack's Bean

XOR-Primes

Duodigits

A Messy Dinner

Standing on the Shoulders of Trolls

Numbers Challenge

Minimum Area of a Convex Grid Polygon

Triangular Pizza

Minimal Pairing Modulo

Minimal Pair Modulo

Falling Bottles

Billiard

SOP and POS

SET

Ruff Numbers

Nim on Towers of Hanoi

Stone Game IV

Sum of Products

Exponent Difference

A Stoneham Number

Pentagonal Puzzle

Pentagonal Puzzle

Problem Statement

Solution

Output

Time Complexity

Applications

Symmetric Diophantine Equation

Turán's Water Heating System

Summation of Summations

Triplicate Numbers

Regular Star Polygons

Chasing Game

Freshman's Product

Average and Variance

Pisano Periods 2

A Grand Shuffle

Pseudo Geometric Sequences

Paths to Equality

Add and Divide

Recursive Tree

Project Euler Problem 102

Ascending Subsequences

Square Triangle Products

Cookie Game

Unreachable Numbers

Task Description

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Applications

Triple Product

Circle of Coins

123-Separable

Number Splitting

Fractions of Powers

Seventeen Points

Total Inversion Count of Divided Sequences

What? Where? When?

Bitwise Recursion

Random Connected Area

A Squared Recurrence Relation

Summation of a Modular Formula