How to find factor combinations

When faced with an integer, it’s natural to think of its factors—the numbers that divide it without leaving a remainder. From basic arithmetic to advanced number theory, factors are crucial in mathematics and appear in various contexts. In this Answer, we’ll explore the intriguing problem of finding all possible combinations of factors for a given integer.

Problem statement

Given an integer $n$ , we’re tasked with finding all possible combinations of its factors within the range of $[2,n−1]$ . The goal is to identify pairs of integers that multiply to produce $n$ , excluding $1$ and $n$ itself.

Example

For instance, if $𝑛=12$ , the possible factor combinations are $(2,6)$ and $(3,4)$ , as $2×6=12$ and $3×4=12$ . We need to explore and enumerate all such combinations for any given $n$ .

Solution

Approach 1: Backtracking

Backtracking systematically explores the potential factors of the given number. It starts with the smallest factor and checks if the number is divisible by it. If it is, the factor is added to a stack, and the number is updated by dividing it by the factor. If the number is not divisible by the current factor, the algorithm proceeds to the next potential factor. When all possible factors up to the square root of the number have been considered, the algorithm records the current set of factors. It then continues the process by backtracking the previous factor and exploring the next potential factor. This process repeats until all combinations of factors have been explored, ensuring no duplicates are included in the final result.

Let’s look at the Python code for this approach:

def factor_combinations(n):
  factors, stack, num = [], [], 2
  
  while True:
    if num > n / num:
      if not stack:
          return factors
      
      factors.append(stack + [int(n)])
      num = stack.pop()
      n *= num
      num += 1
    
    elif n % num == 0:
      stack.append(num)
      n /= num
    
    else:
      num += 1
def main():
    numbers = [12, 48, 100, 250, 486]
    for number in numbers:
        print("Number:", number)
        print("\nFactor combinations: ", factor_combinations(number))
        print("-" * 100)
if __name__ == "__main__":
    main()

Here are the algorithmic steps for the factor_combinations function:

Line 2: Initialize an empty list factors to store factor combinations, an empty stack stack, and set num to 2.
Lines 4–19: Start an infinite loop.
Lines 5–12: Check if num is greater than the square root of n.
- If true and the stack is empty, returns factors.
- Otherwise, append the current factor combination (stack + [n]) to factors.
- Pop the top element from the stack and update num to its value.
- Multiply n by the popped value from the stack.
- Increment num by 1.
Lines 14–16: If n is divisible by num with no remainder:
- Append num to the stack.
- Divide n by num.
Lines 18–19: If n is not divisible by num:
- Increment num by 1.

Approach 2: Recursive DFS

This approach utilizes a depth-first search (DFS) strategy to systematically explore and identify all possible combinations of factors for a given number. Starting with the number as the initial factor combination, the algorithm iteratively divides the number by increasing integers (starting from $2$ ) to find its factors. At each step, it checks if the current integer is a factor and recursively explores the remaining factors. The algorithm efficiently generates a list containing all valid factor combinations for the given number by traversing possible factors and updating the factor combinations accordingly.

Let’s look at the Python code for this approach:

def factor_combinations(n):
    ret = []
    def DFS(cur, i):
        num = cur.pop()
        
        while i * i <= num:
            div = int(num / i)
            if num % i == 0:
                ret.append(cur + [i, div])
                DFS(cur + [i, div], i)
            i += 1
    DFS([n], 2)
    
    return ret
def main():
    numbers = [12, 48, 100, 250, 486]
    for number in numbers:
        print("Number:", number)
        print("\nFactor combinations: ", factor_combinations(number))
        print("-" * 100)
if __name__ == "__main__":
    main()

Here are the algorithmic steps for the factor_combinations function:

Line 2: Initialize an empty list ret to store the factor combinations.
Line 4: Define a helper function DFS that takes two parameters: cur (a list representing the current factor combination) and i (the starting factor to consider).
Line 5: Pop the last element num from the cur list. This num represents the current number being factored in.
Lines 7–12: While the square of i is less than or equal to num, iterate through the loop:
- Calculate the integer division div of num by i.
- Check if i evenly divides num (i.e., num % i == 0).
  - If divisible:
    - Append the factor combination cur extended with [i, div] to the ret list. This represents a valid factor pair (i, div) for num.
    - Recursively call DFS with the updated factor combination cur extended with [i, div] and i unchanged.
- Increment i by 1 to explore the next potential factor.
Line 14: Call the DFS function with the initial factor combination [n] (where n is the input number to be factored) and the starting factor 2.
Line 15: Return the ret list containing all the factor combinations found during the DFS traversal.

Conclusion

Finding all possible factor combinations is an interesting exploration of number theory and recursion. By breaking down the problem into smaller subproblems and leveraging the properties of factors, we can efficiently identify and enumerate these combinations. This problem tests our algorithmic skills and deepens our understanding of basic number theory concepts.

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