5 Best Ways to Check if Two Numbers Differ at One Bit Position Only in Python

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๐Ÿ’ก Problem Formulation: This article explores methods to determine whether two integer numbers differ at exactly one bit position. For instance, if we consider the binary representations of 14 (1110) and 10 (1010), these two numbers differ by just one bit at the third position from the right.

Method 1: XOR and Count Set Bits

This method involves taking the XOR of the two numbers, which highlights the differing bits, and then counting if there is only one bit set in the result. The built-in Python function bin() converts a number to its binary string representation, and the count() method is used to count the occurrence of ‘1’s.

Here’s an example:

def differ_by_one_bit(num1, num2):
    return bin(num1 ^ num2).count('1') == 1

# Test the function
print(differ_by_one_bit(14, 10)) # True
print(differ_by_one_bit(14, 9))  # False

Output:

True
False

This snippet defines a function differ_by_one_bit() that applies XOR to its arguments and checks if the binary result contains exactly one ‘1’. In the given examples, 14 and 10 differ at one bit, while 14 and 9 do not.

Method 2: Bit Shifting and Comparison

The second method uses bit shifting to iterate over each bit position and compares the bits at that position for both numbers. We shift both numbers right by a variable number of positions and then apply the bitwise AND operation with 1 to obtain the last bit of each. If only one bit differs, the method returns True.

Here’s an example:

def differ_one_bit_bitshift(num1, num2):
    diff_count = 0
    while num1 or num2:
        if (num1 & 1) != (num2 & 1):
            diff_count += 1
        num1 >>= 1
        num2 >>= 1
    return diff_count == 1

# Test the function
print(differ_one_bit_bitshift(14, 10)) # True
print(differ_one_bit_bitshift(14, 9))  # False

Output:

True
False

This code creates a function that loops through each bit of the given numbers, shifting them to the right and comparing the least significant bit until either number becomes zero. It keeps a count of differing bits and checks if this count equals one.

Method 3: Direct Bit Comparison

Direct bit comparison involves checking each bit at each position simultaneously for both numbers. This is similar to Method 2 but uses simultaneous bit manipulation to compare the bits without shifting.

Here’s an example:

def differ_one_bit_direct(num1, num2):
    diff_count = 0
    mask = 1
    while num1 or num2:
        if (num1 & mask) != (num2 & mask):
            diff_count += 1
        num1 -= num1 & mask
        num2 -= num2 & mask
        mask <<= 1
    return diff_count == 1

# Test the function
print(differ_one_bit_direct(14, 10)) # True
print(differ_one_bit_direct(14, 9))  # False

Output:

True
False

The function differ_one_bit_direct() uses a mask to check each bit position of both numbers at the same time. It compares each bit, counts the differences, and moves the mask left for the next bit position. The result indicates if there was only one differing position.

Method 4: Use of Built-in bit_count() in Python 3.10+

Python 3.10 introduced a new method bit_count() for integers which returns the number of bits set to 1 directly without converting to a binary string. This method is more efficient and straightforward than manually counting ‘1’s.

Here’s an example:

def differ_by_one_bit_bitcount(num1, num2):
    return (num1 ^ num2).bit_count() == 1

# Test the function
print(differ_by_one_bit_bitcount(14, 10)) # True
print(differ_by_one_bit_bitcount(14, 9))  # False

Output:

True
False

The function differ_by_one_bit_bitcount() uses XOR operation followed by calling bit_count() to quickly find out if there is only one differing bit between two given numbers. It offers a performance advantage for Python versions 3.10 and above.

Bonus One-Liner Method 5: Lambda Expression

A lambda expression can be used to create a succinct one-liner function. It combines the XOR approach and the counting of ‘1’s by using the expression inline, achieving the same result without explicitly defining a separate function.

Here’s an example:

differ_by_one_bit = lambda num1, num2: bin(num1 ^ num2).count('1') == 1

# Test the lambda function
print(differ_by_one_bit(14, 10)) # True
print(differ_by_one_bit(14, 9))  # False

Output:

True
False

This code introduces a lambda function that takes two integers, performs a bitwise XOR, converts the result into a binary string and checks if there’s only one ‘1’. It’s a compact version of Method 1.

Summary/Discussion

  • Method 1: XOR and Count Set Bits. Straightforward and easy to understand. Inefficient for large integers due to string conversion.
  • Method 2: Bit Shifting and Comparison. Provides bit-level control and understanding. More verbose and potentially slower due to loops.
  • Method 3: Direct Bit Comparison. Avoids bit shifting. Can be less intuitive and has a similar performance to Method 2.
  • Method 4: Use of Built-in bit_count(). Most efficient method, but only available in Python 3.10 or newer.
  • Method 5: Lambda Expression. Elegant and concise. Not always the most readable, especially for those new to lambda expressions.