5 Best Ways to Check If Bitwise AND of Any Subset Is Power of Two in Python

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πŸ’‘ Problem Formulation: We need to determine if the bitwise AND operation on any subset of a given set of integers results in a power of two. For instance, given the set {2, 4, 8}, we want to find if there’s a subset, when the bitwise AND operation is applied, gives a result like 2^0, 2^1, 2^2, etc. In this case, the subset {4, 8} would have a bitwise AND of 0, which is not a power of two, so we seek a different subset or determine that no such subset exists.

Method 1: Brute Force Using itertools.combinations

This method leverages itertools.combinations to generate all possible subsets of the input set and checks each subset’s bitwise AND result for being a power of two. It is easy to implement and understand but may not be the most efficient approach for large sets due to its exponential time complexity.

Here’s an example:

from itertools import combinations

def is_power_of_two(n):
    return n != 0 and (n & (n - 1)) == 0

def check_subsets(arr):
    for r in range(1, len(arr) + 1):
        for subset in combinations(arr, r):
            result = subset[0]
            for num in subset[1:]:
                result &= num
            if is_power_of_two(result):
                return True
    return False

input_set = [2, 4, 8]
print(check_subsets(input_set))

Output:

False

This code snippet defines a function is_power_of_two that verifies if a number is a power of two. Another function, check_subsets, generates all possible subsets using combinations and checks if the bitwise AND of a subset is a power of two. In this example, no subset fulfills the condition, thus it returns False.

Method 2: Bitwise Operations and Powerset

Instead of using itertools, you can manually generate all subsets (the powerset) of the set. This method iterates through each element and either includes or excludes it to create a new subset. The subset’s bitwise AND is then checked in the same manner as the Method 1.

Here’s an example:

def is_power_of_two(n):
    return n != 0 and (n & (n - 1)) == 0

def check_subsets(arr):
    subsets = [[]]
    for num in arr:
        new_subsets = [subset + [num] for subset in subsets]
        subsets.extend(new_subsets)
    for subset in subsets[1:]:  # Exclude empty subset
        result = subset[0]
        for num in subset[1:]:
            result &= num
        if is_power_of_two(result):
            return True
    return False

input_set = [2, 4, 8]
print(check_subsets(input_set))

Output:

False

The function check_subsets here creates a powerset of the input set excluding the empty set and performs the same check as before for being a power of two. In this case, as well, the input set does not contain a subset whose bitwise AND is a power of two, so it outputs False.

Method 3: Optimized Bitwise Check

An optimization to the brute force method can skip unnecessary checks by doing early pruning. We check if the current bitwise AND result is already equal to 0; if so, we skip the rest of the current subset since we know it cannot be a power of two.

Here’s an example:

def is_power_of_two(n):
    return n & (n - 1) == 0

def optimize_check_subsets(arr):
    for i in range(1, 1 << len(arr)):
        and_result = -1
        for j in range(len(arr)):
            if i & (1 << j):
                and_result = arr[j] if and_result == -1 else and_result & arr[j]
                if and_result == 0:
                    break
        if and_result > 0 and is_power_of_two(and_result):
            return True
    return False

input_set = [2, 4, 8]
print(optimize_check_subsets(input_set))

Output:

False

This code snippet improves efficiency by breaking out of the inner loop when the current bitwise AND result is 0. Although it still goes through all subsets, it doesn’t perform the full calculation for subsets that clearly won’t result in a power of two, making it faster than the previous methods for sets with large numbers of elements.

Method 4: Using Mathematical Properties

By understanding that a power of two has only one bit set in its binary representation, we can check if the bitwise AND of an entire set is a non-zero power of two instead of checking all subsets. This is an extremely fast method that computes a single AND across the set and performs a single power of two checks.

Here’s an example:

def is_power_of_two(n):
    return n & (n-1) == 0

def check_entire_set(arr):
    and_result = arr[0]
    for num in arr[1:]:
        and_result &= num
    return and_result > 0 and is_power_of_two(and_result)

input_set = [2, 4, 8]
print(check_entire_set(input_set))

Output:

False

This snippet computes the bitwise AND of the input set and checks if the result is a non-zero power of two using is_power_of_two. Although it’s much more performant than checking all subsets, it only checks the full set rather than any subset, making it less versatile if a proper subset would have resulted in a power of two.

Bonus One-Liner Method 5: Functional Approach with all() and any()

A concise and functional approach uses all() and any() in combination with comprehensions to perform the checks in a single line. It’s an elegant one-liner suitable for quick scripting or situations where brevity is valued over readability or speed.

Here’s an example:

from itertools import combinations

input_set = [2, 4, 8]
print(any(all(n & b == b for b in bin(n)[2:]) for n in (reduce(lambda x, y: x & y, c) for c in combinations(input_set, r) for r in range(1, len(input_set) + 1)) if n))

Output:

False

This line of code first generates all the combinations of subsets, applies bitwise AND to each subset, and checks whether each bit position remains set after the AND operation, indicating a power of two. This one-liner returns False for the given input_set, demonstrating that no such subset exists.

Summary/Discussion

  • Method 1: Brute Force Using itertools.combinations. It is simple and straightforward but not efficient for large input sets due to its exponential time complexity.
  • Method 2: Bitwise Operations and Powerset. It manually generates all possible subsets which can be less reliant on the itertools library, but shares the same complexity issues as Method 1.
  • Method 3: Optimized Bitwise Check. More efficient by skipping subsets early, although it still has to iterate through all possible combinations, it is faster when dealing with zeros in the subset.
  • Method 4: Using Mathematical Properties. Extremely efficient but less flexible since it only evaluates the entire set rather than all subsets.
  • Bonus Method 5: Functional Approach with all() and any(). This method offers a one-liner solution for quick checks but can be less readable and efficient.