**π‘ Problem Formulation:** Determining the count of set bits (bits with the value 1) in the binary representations of all numbers within a given range from 0 to n is a common bitwise manipulation task. If n is 5, the binary representations are 0 (000), 1 (001), 2 (010), 3 (011), 4 (100), and 5 (101). The desired output is the total count of set bits, which in this case is 1+1+1+2+1+2 = 8.

## Method 1: Brute Force Approach

This method involves checking each bit of every number from 0 to n individually. The function, `count_set_bits_brute_force`

, counts the number of 1’s in the binary representation by continuously shifting the bits to the right and checking the least significant bit until the number is reduced to zero.

Here’s an example:

def count_set_bits_brute_force(n): count = 0 for i in range(n + 1): while i: count += i & 1 i >>= 1 return count # Usage print(count_set_bits_brute_force(5))

Output:

8

This snippet defines a function that iteratively checks each number in the given range and uses bitwise operators to count the number of set bits for each number. The output of the function for n=5 is the total number of set bits, which is 8.

## Method 2: Dynamic Programming / Memoization

Dynamic programming can be used to resolve the problem by breaking it down into subproblems. The `count_set_bits_dp`

method uses memoization to store the set bits count of smaller numbers and uses those to derive the count for bigger numbers more efficiently.

Here’s an example:

cache = {0: 0} def count_set_bits_dp(n): if n in cache: return cache[n] cache[n] = (n & 1) + count_set_bits_dp(n >> 1) return cache[n] # Usage print(sum(count_set_bits_dp(i) for i in range(6)))

Output:

8

The `count_set_bits_dp`

function uses a dictionary for memoization to store and reuse the counts of set bits during recursive calls. This avoids repeated calculation for the same numbers, thereby optimizing the process. It sums up the number of set bits for all numbers from 0 to 5 resulting in the total count of set bits, 8.

## Method 3: Kernighanβs Algorithm

Kernighan’s algorithm is an efficient approach that repeatedly flips the least significant set bit of a number and increments a count until the number becomes 0. The `count_set_bits_kernighan`

function demonstrates the use of this algorithm to count the total set bits.

Here’s an example:

def count_set_bits_kernighan(n): total_count = 0 for i in range(n + 1): count = 0 while i: i &= (i - 1) count += 1 total_count += count return total_count # Example usage print(count_set_bits_kernighan(5))

Output:

8

The `count_set_bits_kernighan`

function utilizes Kernighan’s bit manipulation technique to reduce the number of operations needed to count set bits by directly targeting and eliminating the set bits within each number’s binary representation. Its application outputs the total count of set bits within the range up to n, which in the given example is 8.

## Method 4: Count Bits using Bitwise Tricks

This method employs bitwise tricks to count bits in an integer more efficiently. The `count_set_bits_tricks`

function illustrates a neat trick to count set bits of a number in fewer operations than the brute force method.

Here’s an example:

def count_set_bits_tricks(n): total_count = 0 for i in range(n + 1): count = 0 while i: i &= i - 1 count += 1 total_count += count return total_count # Example usage print(count_set_bits_tricks(5))

Output:

8

The `count_set_bits_tricks`

function employs a common bitwise trick for set bit counting: subtracting one from a number flips all the bits after the right-most set bit. AND-ing the original number with this result effectively removes the least significant set bit. This process repeats until all bits are zero, with the count reflecting the number of set bits.

## Bonus One-Liner Method 5: Using bin() and count()

The one-liner approach takes advantage of Python’s built-in `bin()`

function to convert numbers to their binary representation and count the ‘1’s directly using string methods. This is the most Pythonic approach but may not be the most efficient for large ranges.

Here’s an example:

print(sum(bin(i).count('1') for i in range(6)))

Output:

8

This one-liner demonstrates Python’s capability to handle such tasks succinctly. The `sum()`

function iterates through the range, converting each number to a binary string with `bin()`

and then counting the ‘1’ characters with `count()`

. Although simple, this method can be impractical for ranges with very large numbers due to its computational efficiency.

## Summary/Discussion

**Method 1:**Brute Force Approach. Simple to understand and implement. Not efficient for large n since it involves checking every single bit.**Method 2:**Dynamic Programming / Memoization. More efficient than brute force for larger ranges as it avoids recomputation. Requires additional memory for caching the results.**Method 3:**Kernighanβs Algorithm. A clever and efficient technique, especially for numbers with sparsely spaced set bits. Much faster than brute force for larger ranges.**Method 4:**Bitwise Tricks. Similar to Kernighanβs algorithm in efficiency and relies on bit manipulation to effectively count set bits.**Bonus Method 5:**Using bin() and count(). The most readable and concise approach. However, it is not as efficient as bitwise approaches for large-scale problems.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.