What does the following algorithm approximate?
x = m; y = 1; while (x - y > e)
{ x = (x + y)/2;
y = m/x;
} print(x); |
(Assume m > 1, e > 0).
(A) log m
(B) m2
(C) m1/2
(D) m1/3
Answer: (C)
Explanation: The given code is implementation of Babylonian method for square root
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