CMI2015-A-08

Question

How many times is the comparison $i \geq n$ performed in the following program?

int i=85, n=5;
main() {
    while (i >= n) {
        i=i-1;
        n=n+1;
    }
}

• A. $40$
• B. $41$
• C. $42$
• D. $43$

Comments

We can re-write the code of while loop as follows
while( i>=n)
{

i=i+2;
}
i=87, n=5;
after 40 iterations value of i will be $85 - 40*2 = 5$.
For 41st iteration, i>=n evaluates true, and i will become 3.
for 42nd iteration i>=5 fails as 3 < 5.

So, In total, 'while' condition is checked 42 times, which results 41 times in success and 1 time in failure.

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the formula for this would be = $\left \lfloor \frac{x-y}{2} \right \rfloor +2$

addition of 2(1 for checking loop condition for false and 1 for including last integer)

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i thought it slightly different way :- since i'm average student so i always do it in first go like this->

it is similar to without changing variable n like n = n+1 why we not do like step size of i = i - 2

then while(i >= n)
{

      i = i - 2; // same step size decremented by 2

}

then we got  -> 85, 83 , 81 , 79........, 9, 7, 5, 3(terminate cond) n=5(no change)

85 = 3+(n-1)* 2// simple AP simply got n = 42.

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I mentally solved this is a weird way. Can someone verify if this approach is generally all right?

 

i is moving towards n with a speed of 1. And n is moving towards i with a speed of 1, as well. Relative speed of i is, therefore, 2. So, to reach 5, i takes $\frac{85-5}{2}=40$ steps. When they're equal, it's a non-issue, so do +1.

Now, they're unequal and the condition will be tested once more and it will fail. So, $40+1+1=42$

Answer 1