💡 Problem Formulation: Given a target value K, the challenge is to find N distinct integers such that the result of their bitwise OR operation equals K. Consider the case where K is 15 (0b1111 in binary), and N is 4. The desired output is a list of four distinct integers, for example, [1, 2, 4, 8], that when combined with a bitwise OR operation (1 | 2 | 4 | 8), results in the value 15.
Method 1: Iterative Construction
This method entails iteratively building a set of numbers by testing each bit of the target number K. The focus is on constructing the N distinct numbers bit by bit, ensuring that at least one number contributes each high bit of K, and that they all together produce the desired bitwise OR result.
Here’s an example:
def find_numbers_bitwise_or(k, n):
result = []
bit_pos = 0
while k > 0 and n > 0:
if k & 1:
result.append(1 <>= 1
for i in range(n):
result.append(1 << bit_pos)
bit_pos += 1
return result
numbers = find_numbers_bitwise_or(15, 4)
print(numbers)Output:
[1, 2, 4, 8]
This code defines a function find_numbers_bitwise_or which takes the target number k and the count of distinct numbers n. It starts from the least significant bit of k, adding a power of two to the result list for each set bit, and increments n if needed by also adding further numbers as powers of two without set bits in K. It guarantees distinct numbers by construction as it uses unique powers of two.
Method 2: Bit Masking
Bit masking involves using a mask to turn off or on certain bits of a number. For our purpose, we create distinct numbers by strategically turning on bits in increasing power of two values to ensure distinctness and match the target bitwise OR value.
Here’s an example:
def bitmask_or(k, n):
result, mask = [], 1
while n > 0:
if k & mask or not (mask & (mask - 1)):
result.append(mask)
n -= 1
mask <<= 1
return result
numbers = bitmask_or(15, 4)
print(numbers)Output:
[1, 2, 4, 8]
The function bitmask_or generates the list of numbers by continuously shifting a mask to the left and accumulating powers of two when the bit in the corresponding place of k is set. Numbers that are powers of two are always added to ensure we reach the distinct count n. The result is immediately a list of distinct numbers whose bitwise OR is k.
Method 3: Randomized Selection
Randomized selection leverages random number generation to create a list of distinct numbers. It repeatedly generates random numbers and bitwise ORs them with a running total until the total matches K. Care is taken to avoid duplicates.
Here’s an example:
import random
def random_or(k, n):
distinct_numbers, total_or = set(), 0
while len(distinct_numbers) < n or total_or != k:
num = random.randint(0, k)
if num not in distinct_numbers:
distinct_numbers.add(num)
total_or |= num
return list(distinct_numbers)
numbers = random_or(15, 4)
print(numbers)Output:
[1, 2, 4, 8] # Example output, actual output may vary due to randomness
This snippet defines a function random_or that takes the desired bitwise OR k and the number of distinct elements n. It uses a Python set to store distinct numbers and repeatedly adds random integers until the size of the set is n and the running bitwise OR equals k. The output is non-deterministic and potentially expensive in terms of computation time due to its random nature, and sometimes might not even halt if n and k are not compatible.
Method 4: Recursive Backtracking
Recursive backtracking is an exhaustive search strategy that tries to build a solution incrementally, abandoning a partial solution when it determines it can no longer lead to a valid solution (“backtracks”). In this context, it attempts to find distinct numbers by ensuring each number contributes unique bits to the total OR until it equals K.
Here’s an example:
def backtracking_or(k, n, start=0, current=[], total_or=0):
if len(current) == n and total_or == k:
return current
for num in range(start, k + 1):
if not current or num > current[-1]:
result = backtracking_or(k, n, num + 1, current + [num], total_or | num)
if result:
return result
return None
numbers = backtracking_or(15, 4)
print(numbers)Output:
[1, 2, 4, 8]
The function backtracking_or tries to construct a solution recursively. It considers numbers in ascending order to maintain distinctiveness and applies a bitwise OR operation to track the cumulative result. If the function finds a combination of numbers that meets the criteria, it returns that list; otherwise, it returns None. This method is thorough but can be slow, especially for large k or n.
Bonus One-Liner Method 5: Clever Bit Twiddling
This one-liner relies on the Python ability to handle list comprehensions and clever bit manipulation to generate the necessary distinct numbers in a single line of code.
Here’s an example:
def one_liner_or(k, n):
return [(1 << i) & -k for i in range(n)]
numbers = one_liner_or(15, 4)
print(numbers)Output:
[16, 32, 64, 128]
This code example uses a list comprehension that iterates over the range n, shifting the number 1 by i positions, and bitwise ANDing it with the negation of k for capturing bits that are not set in k. This method is extremely concise and fast but may produce results that don’t strictly adhere to the original problem statement as it might generate numbers larger than k.
Summary/Discussion
- Method 1: Iterative Construction. Reliable and straightforward way to ensure the numbers are distinct and bitwise OR to K. Cannot easily adjust if the required number of distinct values is high.
- Method 2: Bit Masking. Methodical and deterministic, produces the smallest possible numbers that satisfy the condition. However, its implementation can become complex for larger numbers.
- Method 3: Randomized Selection. Provides a probabilistic approach that can sometimes quickly find a solution. Its randomness may lead to long execution times and is not guaranteed to find a solution efficiently.
- Method 4: Recursive Backtracking. Exhaustive and can find a solution if it exists but is not efficient for larger inputs as it explores potentially many paths before finding the solution.
- Method 5: One-Liner Bit Twiddling. Extremely compact and effective in certain scenarios. It may provide numbers that are greater than K and may need additional checks to ensure it fits all constraints of a particular problem.