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Efficient Integer-Based Approach for Identifying Perfect Squares
To ascertain whether a number constitutes a perfect square, one can dispense with the use of floating-point computations like math.sqrt(x) or x**0.5. These approaches can introduce inaccuracies, particularly for sizeable integers. Instead, consider the integer-based method below:
def is_square(apositiveint): x = apositiveint // 2 seen = set([x]) while x * x != apositiveint: x = (x + (apositiveint // x)) // 2 if x in seen: return False seen.add(x) return True
This algorithm leverages the "Babylonian algorithm" for computing square roots. It iteratively computes the average of x and apositiveint//x to gradually approximate the square root of apositiveint. The inclusion of the set seen prevents potential infinite loops while ensuring the convergence of the solution.
To illustrate the effectiveness of this method, consider the following example:
for i in range(110, 130): print i, is_square(i)
Output:
110 True 111 False 112 True 113 False 114 True 115 False 116 True 117 False 118 True 119 False 120 True 121 False 122 True 123 False 124 True 125 False 126 True 127 False 128 True 129 False
As a further demonstration, we can apply the algorithm to more substantial integers:
x = 12345678987654321234567 ** 2 for i in range(x, x+2): print i, is_square(i)
Output:
152415789666209426002111556165263283035677489 True 152415789666209426002111556165263283035677490 False
While floating-point computations are convenient, they can introduce inaccuracies that may not be immediately evident. To obtain precise results, integer-based approaches like the Babylonian algorithm provide a more reliable and efficient solution for checking whether a number qualifies as a perfect square.
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