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.

Given an integer

For instance, if

**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 = [], [], 2while True:if num > n / num:if not stack:return factorsfactors.append(stack + [int(n)])num = stack.pop()n *= numnum += 1elif n % num == 0:stack.append(num)n /= numelse:num += 1def 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.

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

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 += 1DFS([n], 2)return retdef 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.

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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