5 Best Ways to Check Whether a Number Can Be a Perfect Square After Adding 1 in Python

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πŸ’‘ Problem Formulation: The task is to determine if, by simply adding 1 to a given integer, the result can become a perfect square in Python. This means that for an input number n, there needs to be an integer x such that (n+1) = x^2. For example, given the input 8, adding 1 gives us 9, which is 3^2, hence the output is True. Conversely, for the input 10, adding 1 gives us 11, which is not a perfect square, hence the output is False.

Method 1: Using the Square Root Function

An approach to determine if the incremented number is a perfect square is to use the square root function math.sqrt() and check if the square root of n+1 is an integer. This is a straightforward method using the Python standard library.

Here’s an example:

import math

def is_perfect_square_plus_one(n):
    return math.isqrt(n + 1)**2 == n + 1

print(is_perfect_square_plus_one(8))
print(is_perfect_square_plus_one(10))

Output:

True
False

This snippet defines a function that adds 1 to the input number n, takes the integer square root of the result with math.isqrt(), and then squares it to check if it equals n+1. This confirms if the number after adding 1 is a perfect square. Notably, math.isqrt() ensures the square root result is an integer.

Method 2: Iterative Guess and Check

One can iteratively check each number starting from 1, squaring it, until the square is equal to n+1 or has surpassed it. This method is simple but not efficient for large numbers.

Here’s an example:

def is_perfect_square_plus_one(n):
    x = 1
    while x*x <= n + 1:
        if x*x == n + 1:
            return True
        x += 1
    return False

print(is_perfect_square_plus_one(8))
print(is_perfect_square_plus_one(10))

Output:

True
False

In this example, the function starts with x=1 and incrementally checks if x^2 equals n+1. If it finds such an x, it returns True; otherwise, it continues until x^2 surpasses n+1, then it returns False.

Method 3: Binary Search

To improve efficiency, especially with larger numbers, one can use binary search to find if n+1 is a perfect square. Starting with a range that holds the potential square root, the method progressively halves the search space.

Here’s an example:

def is_perfect_square_plus_one(n):
    left, right = 0, n+1
    while left <= right:
        mid = (left + right) // 2
        mid_squared = mid * mid
        if mid_squared == n + 1:
            return True
        elif mid_squared < n + 1:
            left = mid + 1
        else:
            right = mid - 1
    return False

print(is_perfect_square_plus_one(8))
print(is_perfect_square_plus_one(10))

Output:

True
False

This function uses binary search logic to narrow down the possible square root of n+1. By comparing and halving the interval, it efficiently determines if n+1 is a perfect square, greatly reducing the time complexity compared to linear search methods.

Method 4: Using List Comprehension and Any

By utilizing list comprehension in combination with the any function, one can test if any number squared equals n+1 within a specified range. It’s a more Pythonic and compact way of doing a linear search method.

Here’s an example:

def is_perfect_square_plus_one(n):
    return any(x * x == n + 1 for x in range(1, n + 2))

print(is_perfect_square_plus_one(8))
print(is_perfect_square_plus_one(10))

Output:

True
False

This code utilizes the succinctness of list comprehensions and the any function to find if there is any integer x in the range from 1 to n+1 such that x^2 equals n+1. Although it’s more elegant, it has similar performance characteristics to the iterative guess and check method.

Bonus One-Liner Method 5: Using Power and Modulo Operator

A concise one-liner method involves incrementing the number by 1 and checking if (n+1)**0.5 % 1 == 0, which asserts that the square root is an integer.

Here’s an example:

is_perfect_square_plus_one = lambda n: (n + 1)**0.5 % 1 == 0

print(is_perfect_square_plus_one(8))
print(is_perfect_square_plus_one(10))

Output:

True
False

The lambda function checks whether the square root of n+1 is an integer by using the modulo operator on (n+1)**0.5. If there’s no remainder, then the number is a perfect square. This method is the most concise but could run into floating-point arithmetic issues for very large numbers.

Summary/Discussion

  • Method 1: Using the Square Root Function. Straightforward and efficient for all ranges of integers. Utilizes Python’s built-in math library functions.
  • Method 2: Iterative Guess and Check. Simple to understand. However, it is inefficient for large numbers as it involves a linear search.
  • Method 3: Binary Search. Efficient for large numbers. Reduces computation time by continually halving the search space. Preferred for high-performance needs.
  • Method 4: Using List Comprehension and Any. Elegant and Pythonic. Offers no performance benefits over the linear search but is more readable.
  • Method 5: Using Power and Modulo Operator. Most concise option. Good for quick checks and small numbers but potentially inaccurate for very large numbers due to floating-point errors.