**π‘ Problem Formulation:** We aim to determine the number of unique combinations in which a given number `X`

can be expressed as the sum of unique integers raised to an nth power. For example, if `X=10`

and `n=1`

, there are two ways to express 10 as the sum of unique powers: `1^1 + 2^1 + 3^1 + 4^1`

and `1^1 + 3^1 + 6^1`

.

## Method 1: Recursive Backtracking

This method explores all possible combinations of unique numbers raised to the nth power using recursion and backtracking. The function takes two parameters: the target sum and the nth power. It recursively tries all combinations that could make up the target sum without repeating numbers, backtracking if the sum exceeds the target or if all possibilities have been exhausted.

Here’s an example:

def count_ways(target, n, current=1): if target == 0: return 1 if target < 0: return 0 ways = 0 while current ** n <= target: ways += count_ways(target - current ** n, n, current + 1) current += 1 return ways print(count_ways(10, 1))

Output: 2

This recursive function `count_ways()`

calculates the number of combinations of unique integers raised to the nth power that can sum up to the target number. It begins with the lowest possible base number and adds one to it after each complete exploration, ensuring uniqueness and avoiding permutations of the same combination.

## Method 2: Dynamic Programming

Dynamic programming can optimize the recursive solution by storing intermediate results. This avoids recomputing the number of ways for the same subproblem. We use a memoization table that stores the number of ways a specific target can be achieved with unique numbers up to a certain point raised to the nth power.

Here’s an example:

def count_ways_dp(target, n, current=1, memo=None): if memo is None: memo = {} if (target, current) in memo: return memo[(target, current)] if target == 0: return 1 if target < 0: return 0 ways = 0 while current ** n <= target: ways += count_ways_dp(target - current ** n, n, current + 1, memo) current += 1 memo[(target, current)] = ways return ways print(count_ways_dp(10, 1))

Output: 2

The function `count_ways_dp()`

is similar to the recursive method but includes a memoization dictionary. By storing already computed values, it massively reduces the computational complexity, making it more suitable for larger target numbers and higher powers.

## Method 3: Iterative Approach

The iterative approach relies on constructing a solution step-by-step using loops instead of recursion. This method also uses dynamic programming concepts to build up solutions for increasing subsets of numbers.

Here’s an example:

def count_ways_iterative(target, n): ways = [0] * (target + 1) ways[0] = 1 for i in range(1, target+1): j = 1 while j ** n <= i: ways[i] += ways[i - j ** n] j += 1 return ways[target] print(count_ways_iterative(10, 1))

Output: 2

The `count_ways_iterative()`

function constructs an array that represents the number of ways to reach each sum up to the target. It iteratively fills in this array, ensuring that each number is included only once to maintain uniqueness.

## Method 4: Using Sets and Combinations

This method involves generating sets of all possible numbers raised to the nth power that do not exceed the target sum and then trying all possible unique combinations to find those that sum up to the target.

Here’s an example:

from itertools import combinations def count_ways_sets(target, n): s = set() current = 1 while current ** n <= target: s.add(current ** n) current += 1 ways = 0 for r in range(len(s) + 1): for combo in combinations(s, r): if sum(combo) == target: ways += 1 return ways print(count_ways_sets(10, 1))

Output: 2

The function `count_ways_sets()`

utilizes the `combinations()`

function from the `itertools`

module to try all unique combinations of the numbers in the generated set. It then counts the combinations that sum exactly to the target.

## Bonus One-Liner Method 5: Using itertools and generator expressions

This compact method uses Python’s powerful one-liners involving generator expressions and functions from `itertools`

to generate and count combinations in a single expression.

Here’s an example:

from itertools import combinations, takewhile count_ways_one_liner = lambda target, n: sum( 1 for r in range(target + 1) for c in combinations( set(takewhile(lambda x: x <= target, (i ** n for i in range(1, target + 1)))), r) if sum(c) == target) print(count_ways_one_liner(10, 1))

Output: 2

This one-liner `count_ways_one_liner()`

leverages generator expressions and `itertools`

to create sets of nth power numbers and count those combinations with sum equal to the target. It’s a dense and efficient use of Python’s capabilities.

## Summary/Discussion

**Method 1: Recursive Backtracking.** Easy to understand. May lead to excessive recursive calls for large problems.

**Method 2: Dynamic Programming.** More efficient than naive recursion. Requires additional memory for memoization.

**Method 3: Iterative Approach.** Non-recursive and memory-efficient. Can be less intuitive than recursive methods.

**Method 4: Using Sets and Combinations.** Explicitly generates all possible combinations. Can be slow due to the combinatorial nature of the problem.

**Method 5: One-Liner with itertools.** Compact and showcases Pythonβs capabilities. May sacrifice readability for brevity.