What does the following algorithm approximate? (Assume m

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What does the following algorithm approximate?
(Assume m > 1, ∈ > 0).
x = m; y = 1; while (x– y > ∈ )
{ x = (x + y)/ 2;
y = m / x;
}
print (x);

1. log m
2. m²
3. m^{1 / 2}
4. m^{1 / 3}

Correct Option: C

Here we take let x = 16
Loop will stop when x – y = 0 or > 0

Iteration	X	Y
1	(16 + 1)/ 2 = 8	16/ 8 = 2
2	(8 + 2)/ 2 = 5	16/ 5 = 3
3	(5 + 3)/ 2 = 4	16/ 4 = 4

Here X = Y
Then take X which is 4.
(m)^{1 / 2} = 4 = (16)^{1 / 2}
Hence (c) is correct option.