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Approaches to Determine if a Number is a Power of Two (includes bit manipulations and logarithm)

2020

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Approaches to Determine if a Number is a Power of Two

This document outlines two effective approaches to determine if a given integer ( n ) is a power of two: the Logarithmic Approach and the Bit Manipulation Approach. Each approach is explained with intuition, code examples, and dry runs.

1. Logarithmic Approach

Intuition

The logarithmic approach leverages the mathematical property of logarithms. A number ( n ) is a power of two if there exists an integer ( k ) such that ( n = 2^k ). This can be expressed using logarithms:

  • If ( n ) is a power of two, then ( \log_2(n) ) must yield an integer value.
  • We can check this by comparing ( \log_2(n) ) with its floored value.

Purpose of the Comparison

  1. Identifying Integer Values:

    • The logarithm ( \log_2(n) ) gives us the exponent to which ( 2 ) must be raised to obtain ( n ).
    • If ( n ) is a power of two, ( \log_2(n) ) will be an integer (e.g., 0, 1, 2, 3,…).
    • If ( n ) is not a power of two, ( \log_2(n) ) will yield a non-integer value (e.g., approximately 1.585 for 3).

  2. Using floor to Check for Integers:

    • The floor function rounds down any non-integer value to the nearest whole number.
    • By comparing ( \log_2(n) ) with its floored value, we can check if ( \log_2(n) ) is an integer:
      • If ( \log_2(n) ) equals floor(logValue), it indicates that ( \log_2(n) ) is indeed an integer.
      • If they are not equal, it means ( \log_2(n) ) is a non-integer, confirming that ( n ) is not a power of two.

Code

#include <cmath>

bool isPowerOfTwo(int n) {
    return n > 0 && (log2(n) == floor(log2(n)));
}

Example to Illustrate the Comparison

Example 1: ( n = 8 ) (which is ( 2^3 ))

  • Calculation:
    • ( \log_2(8) = 3 )
    • floor(3) = 3
  • Comparison:
    • 3 == 3 → True (indicating ( 8 ) is a power of two)

Example 2: ( n = 3 ) (not a power of two)

  • Calculation:
    • ( \log_2(3) \approx 1.585 )
    • floor(1.585) = 1
  • Comparison:
    • 1.585 == 1 → False (indicating ( 3 ) is not a power of two)

Summary of Dry Runs

Case 1: ( n = 8 )

  1. Input: isPowerOfTwo(8)
  2. Execution:
    • Check: 8 > 0 → True
    • Calculate: log2(8) = 3
    • Calculate: floor(3) = 3
    • Check: 3 == 3 → True
  3. Output: True

Case 2: ( n = 3 )

  1. Input: isPowerOfTwo(3)
  2. Execution:
    • Check: 3 > 0 → True
    • Calculate: log2(3) \approx 1.585
    • Calculate: floor(1.585) = 1
    • Check: 1.585 == 1 → False
  3. Output: False

2. Bit Manipulation Approach

Intuition

The bit manipulation approach utilizes the properties of binary representation. A number ( n ) is a power of two if it has exactly one set bit in its binary form. The expression n & (n - 1) clears the lowest set bit of ( n ):

  • If ( n ) is a power of two, then ( n > 0 ) and n & (n - 1) will equal 0.

Understanding Binary Representation

  1. Binary Representation:

    • Every integer can be represented in binary (base-2) format, where each bit represents a power of two.
    • For example, the number ( 6 ) in binary is 110, which means ( 4 + 2 = 6 ).

  2. Set Bits:

    • A "set bit" is a bit that is 1. In the binary representation of a number, the position of the set bits indicates which powers of two are included in the number.
    • For example, in the binary representation of ( 6 ) (110), the lowest set bit is the rightmost 1, which corresponds to ( 2^1 ).

The Expression n & (n - 1)

  1. Subtracting One:

    • When you subtract ( 1 ) from ( n ), it changes the rightmost set bit and all bits to the right of it.
    • For example:
      • If ( n = 6 ) (binary 110), then ( n - 1 = 5 ) (binary 101).
      • The binary representation changes from 110 to 101, effectively flipping the rightmost 1 to 0 and all bits to the right of it to 1.

  2. Bitwise AND Operation:

    • The bitwise AND operation (&) compares each bit of ( n ) and ( n - 1 ). A bit in the result is 1 only if both corresponding bits in ( n ) and ( n - 1 ) are 1.
    • Continuing with our example:
      • ( n = 6 ) (binary 110)
      • ( n - 1 = 5 ) (binary 101)
      • Performing the AND operation: 
        110
& 101
------
100  (which is 4 in decimal)

  3. Clearing the Lowest Set Bit:

    • The result of n & (n - 1) effectively clears the lowest set bit of ( n ).
    • In our example, 6 & 5 results in 4, which is 100 in binary. The lowest set bit (the rightmost 1 in 110) has been cleared.

Summary

  • The expression n & (n - 1) is a powerful technique in bit manipulation that clears the lowest set bit of ( n ).
  • This operation is useful for various applications, including:
    • Checking if a number is a power of two (since powers of two have exactly one set bit).
    • Counting the number of set bits in a number.
    • Efficiently performing operations that require manipulation of individual bits.

Example

To illustrate further, let's consider a few more examples:

Example 1: ( n = 12 ) (binary 1100)

  • Calculation:
    • ( n - 1 = 11 ) (binary 1011)
    • 12 & 11: 
      1100
& 1011
------
1000  (which is 8 in decimal)
  • Result: The lowest set bit (the second 1 from the right) is cleared.

Example 2: ( n = 5 ) (binary 0101)

  • Calculation:
    • ( n - 1 = 4 ) (binary 0100)
    • 5 & 4: 
      0101
& 0100
------
0100  (which is 4 in decimal)
  • Result: The lowest set bit (the rightmost 1) is cleared.

Code for Bit Manipulation Approach

bool isPowerOfTwo(int n) {
    return n > 0 && (n & (n - 1)) == 0;
}

Dry Run for Bit Manipulation Approach

Case 1: ( n = 8 )

  1. Input: isPowerOfTwo(8)
  2. Execution:
    • Check: 8 > 0 → True
    • Calculate: 8 & (8 - 1) → 8 & 7 → 0
    • Check: 0 == 0 → True
  3. Output: True

Case 2: ( n = 3 )

  1. Input: isPowerOfTwo(3)
  2. Execution:
    • Check: 3 > 0 → True
    • Calculate: 3 & (3 - 1) → 3 & 2 → 2
    • Check: 2 == 0 → False
  3. Output: False

Summary of Approaches

  • Logarithmic Approach:

    • Pros: Intuitive and straightforward for mathematical understanding.
    • Cons: Requires floating-point operations, which may introduce precision issues.

  • Bit Manipulation Approach:

    • Pros: Efficient and operates directly on binary representations, making it fast.
    • Cons: May be less intuitive for those unfamiliar with bitwise operations.

Both approaches are valid and can be used based on the context and requirements of the problem. The bit manipulation approach is generally preferred for its efficiency in competitive programming and low-level applications.