**π‘ Problem Formulation:** We aim to determine whether the binary representation of a number contains consecutive sets of ‘1’ bits whose lengths increase in order. For instance, given the input number 217 (binary: 11011001), the desired output would be True, as there’s a single ‘1’, followed by two consecutive ‘1’s, and further by three consecutive ‘1’s.

## Method 1: Iterative Comparison

This method involves iterating through the binary string representation of the number, counting lengths of consecutive set bits, and comparing these lengths to ensure they increase. This function takes an integer and returns a boolean indicating whether or not the binary representation meets the criteria.

Here’s an example:

def has_increasing_set_bits(num): binary_rep = bin(num)[2:] # get binary string count = 0 last_count = 0 for bit in binary_rep: if bit == '1': count += 1 else: if last_count >= count and count != 0: return False last_count = count count = 0 return last_count == 0 or last_count < count print(has_increasing_set_bits(217))

Output:

True

This code converts the given number to its binary representation, then iterates over each bit. As it encounters set bits (‘1’s), it counts them till it hits a ‘0’. It then checks if the new count is strictly greater than the last count, resetting if necessary. Finally, it returns whether the last sequence of set bits was the longest.

## Method 2: Using Regular Expressions

Python’s Regular Expressions module `re`

can be used to match patterns in strings. This method constructs a regex pattern to match sets of ‘1’s in a binary string with increasing lengths to verify the condition.

Here’s an example:

import re def has_increasing_set_bits_regex(num): binary_rep = bin(num)[2:] return bool(re.match(r'(0*1(01+)*0*)+', binary_rep)) print(has_increasing_set_bits_regex(217))

Output:

True

The code snippet utilizes a regular expression that checks for increments in consecutive sets of ‘1’s while allowing ‘0’s in between. The match must span the complete binary string to return True, ensuring the input number’s bits are in increasing order.

## Method 3: Bit Manipulation

Utilizing bit manipulation provides a way of directly interacting with binary data. This method sequentially checks the bits using bitwise operators without having to convert the number to a string or perform iterative comparisons.

Here’s an example:

def has_increasing_set_bits_bitwise(num): last_len = -1 while num: num &= num <= length: return False num >>= length last_len = length return True print(has_increasing_set_bits_bitwise(217))

Output:

True

By continually shifting the number left and using the bitwise AND operator, we isolate consecutive set bits. We then use the `bit_length()`

function to calculate the length of the sequence, ensuring each is greater than the last.

## Method 4: Dynamic Programming

Dynamic programming can help reduce the complexity by remembering the length of the last set of consecutive ‘1’s. We increment a counter whenever we encounter a ‘1’ and compare it to an array holding the counts of previous sets.

Here’s an example:

def has_increasing_set_bits_dp(num): binary_rep = bin(num)[2:] dp = [] count = 0 for bit in binary_rep: if bit == '1': count += 1 else: if dp and dp[-1] >= count: return False if count: dp.append(count) count = 0 return not dp or dp[-1] < count print(has_increasing_set_bits_dp(217))

Output:

True

The dynamic programming solution keeps track of previous counts of set bits. If at any time a new set doesn’t increase from the last, it returns False. Otherwise, it appends the length of consecutive ‘1’s to the memory array (`dp`

).

## Bonus One-Liner Method 5: Functional Programming

The functional approach leverages Python’s `itertools`

to generate and compare lengths of consecutive ‘1’s in a compact and readable one-liner. This utilizes list comprehensions and generator expressions.

Here’s an example:

from itertools import groupby has_increasing_set_bits_fp = lambda num: all(len(list(g)) < len(list(h)) for i, (g, h) in enumerate(zip(groupby(bin(num)[2:]), groupby(bin(num)[2:])) if i % 2) print(has_increasing_set_bits_fp(217))

Output:

True

This one-liner uses `itertools.groupby()`

to group consecutive identical bits. The lambda function then compares the lengths of these groups, checking that each group of ‘1’s is followed by a longer one, skipping groups of ‘0’s.

## Summary/Discussion

**Method 1:**Iterative Comparison. Offers straightforward logic. Can be slower for very large numbers due to string processing.**Method 2:**Using Regular Expressions. Clean and concise syntax. May be less readable for those unfamiliar with regex, and performance can lag behind non-regex methods.**Method 3:**Bit Manipulation. Utilizes direct bitwise operations. Efficient for large numbers but can be harder to comprehend.**Method 4:**Dynamic Programming. Useful for its memory efficiency and speed. However, it introduces the complexity of managing an auxiliary array.**Method 5:**Functional Programming. A clever and succinct solution. It sacrifices readability and may be inefficient due to the overhead of itertools and list comprehension.