GATE CSE 2004 | Question: 32

Question

Consider the following program fragment for reversing the digits in a given integer to obtain a new integer.

Let $n = d_1\, d_2\, \ldots\, d_m$.

int n, rev;
rev = 0;
while(n > 0) {
    rev = rev * 10 + n%10;
    n = n/10;
}

The loop invariant condition at the end of the $i^{th}$ iteration is:

• A. $n=d_1\, d_2 \,\ldots\, d_{m-i} \qquad \mathbf{and} \qquad \text{rev} = d_m\,d_{m-1} \,\ldots\, d_{m-i+1}$

• B. $n= d_{m-i+1} \,\ldots\, d_{m-1}\, d_m \qquad \mathbf{or} \qquad \text{rev} = d_{m-i} \,\ldots\, d_2\,d_1$

• C. $n \neq \text{rev}$

• D. $n=d_1\, d_2 \,\ldots\, d_m \qquad \mathbf{or} \qquad \text{rev} =d_m \,\ldots\, d_2\, d_1$

Comments

is this out of syllabus?

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C is not true as C fails on 0^th iteration itself 

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https://www.geeksforgeeks.org/loop-invariant-condition-examples-sorting-algorithms/

Answer 1

Hello sir,

I have taken n=128,  after running code I got below output at each iteration:

Iteration 1(n=12, rev=8)

Iteration 2(n=1, rev=82)

Iteration 3(n=0, rev=821)

Now, as above output I can conclude that option c(n!=rev) should be right answer.

Please clarify if observed something wrong, thanks!

Comments

If we take n=1111 then after 2nd iteration, we would have n=11 and rev=11, which says n=rev, so this can't be the loop invariant.

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Wow kya bat h sir g

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Option C fails for every even length palindrome number.