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35 votes
35 votes

Suppose $c = \langle c[0], \dots, c[k-1]\rangle$ is an array of length $k$, where all the entries are from the set $\{0, 1\}$. For any positive integers $a \text{ and } n$, consider the following pseudocode.

DOSOMETHING (c, a, n)

$z \leftarrow 1$
for $i \leftarrow 0 \text{ to } k-1$
     do $z \leftarrow z^2 \text{ mod } n$
     if $c[i]=1$
           then $z \leftarrow (z \times a) \text{ mod } n$
return z

If $k=4, c = \langle 1, 0, 1, 1 \rangle , a = 2, \text{ and } n=8$, then the output of DOSOMETHING(c, a, n) is _______.

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7 Answers

Best answer
39 votes
39 votes
Initially $k = 4$, $c = [1, 0, 1, 1]$, $a = 2$, $n = 8$.

Now let's iterate through the function step by step :

$z = 1$ (at the start of do-something)

$i = 0$ (start of external for loop)

In the do loop

$z = 1*1 % 8 = 1$ (non zero value so considered as true and continue)

$c[0] = 1$, so in the if clause, $z = 1*2 \% 8 = 2$

In the do loop

$z = 2*2 \% 8 = 4$ (since now $z = 2$) (non zero value so considered as true and continue)

$c[0] = 1$, so in the if clause, $z = 4*2 \% 8 = 0$

Now no need to check further :

Reason :  All the operations that update $Z$ are multiplicative operations and hence the value of $Z$ will never change from $0$.
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20 votes
20 votes

z=1  k = 4, c = [1, 0, 1, 1], a = 2, n = 8

now if we analyze the code we will get table like this Hence Ans is 0.

i z
0 2
1 4
2 0
3 0
6 votes
6 votes
Answer is 0. By manually iterating through the steps by pencil and paper we can get this answer
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4 votes
4 votes
we can do it mentally , at one stage value of z will be zero , beyond that , for any value of K it will be 0 only.

Ans :0
Answer:

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