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Suppose you are given arrays $p [1......N]$  and $q [1......N]$ both uninitialized, that is, each location may contain an arbitrary value), and a variable count, initialized to $0$. Consider the following procedures $set$ and $is\_set$:

set(i) {
count = count + 1;
q[count] = i;
p[i] = count;
}
is_set(i) {
if (p[i] ≤ 0 or p[i] > count)
return false;
if (q[p[i]] ≠ i)
return false;
return true;
}

1. Suppose we make the following sequence of calls:
$set(7)$; $set(3)$; $set(9)$;
After these sequence of calls, what is the value of count, and what do $q[1], q[2] ,q[3], p[7], p[3]$ and $p[9]$ contain?
2. Complete the following statement "The first count elements of __________contain values i such that set (_________________) has been called".
3. Show that if $set(i)$ has not been called for some $i$, then regardless of what $p[i]$ contains, $is\_set(i)$ will return false.
edited | 788 views
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Question C. Can be understand by taking specific example though it is told for arbitrary
Question A's Solution's call to
Set(3)

Cz @ time
i=3
Count=2
q[2]=3
p[3]=2
....
and if set(3) is not called then it will always return false.
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For the (c) part of the question which says us to show that $P\rightarrow Q$ .  Here P is set(i) has not been called for some i and Q is regardless of what p[i] contains, is_set(i) will return false.

So we have to prove just that if set(i) has not been called for some i then regardless of what p[i] contains, is_set(i) will return false. We do not concern ourselves to the case when set(i) has been called for some i .
So having assumed that set(i) has not been called, we can say that it can return true only when the first if condition and second condition are both false.
The first if condition will only be false when p[i]>0 && p[i] <= count. Let garbage value that p[i] contains is such that p[i]>0 && p[i]<= count. Now for this to return true second condition should be made false too which cannot be, because if p[i]<=count it means Array Q has been filled from index 1 to index count and q contain only the i values for which set(i) has been called , so there is no possibility for q[p[i]]=i because set(i) has not been called.

1. Initially $count= 0$;

When we call $set(7) - count=1, q[1] =7, p[7]= 1$;
when we call $set(3) - count=2, q[2]=3, p[3] =2$;
when we call  $set(9) - count=3, q[3]=9, p[9] = 3$;

2. Ans-  "The first count elements of  $array \ q$ contain values $i$ such that $set(i)$ has been called".

3. If $set(i)$ has not been called for some $i$, then  regardless of what $p[i]$ contains, When we call $is\_set(i)$ then
if (q[p[i]] ≠ i)
return false;


will always execute, because if set(i) is not called then p[i] ≠ count(any) and for then same count q[count] ≠ i. So if statement will be true and will return false.

edited by
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@vijaycs ..

variable count is not static...the why the value of count persist...when calling set(I) with different values
+6
count must be global variable as both functions are using this. It is not  local to any function.
+2
@bikram sir, Isn't there a slightest possibility that the garbage value that the arrays contain are in such a way that is_set becomes true?
+1

I am having the same doubt.

if (q[p[i]] ≠ i)
return false;
will always execute, because if set(i) is not called then p[i] ≠ count(any) and for then same count q[count] ≠ i. So if statement will be true and will return false.

may be p[5] contains 5 as garbage value and q[5] also contains 5.So the statement can become false.We cannot assume what garbage can contains and what it cannot contain as assumed in answer.

I think the first if will always be true.

if (p[i] ≤ 0 or p[i] > count)
return false; 

If p[i] is <=0 then first condition is true and if is true. If p[i]>0 ,then we know count is zero and hence second condition will be true.SO whatever p[i] may contain it will always enter in this if and return false.Please correct me if wrong

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same doubt
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We can get out of first IF only if p[i]>0 &&p[i]<= count

lets say i= 5 we r checking for 3rd question

so if p[5]< count let's suppose, for that to happen some other value has already been called with set(i) and incremented count more than p[5] because count increases only on calls to set(i)  so q[p[i]]!=i become it's been set with other value to increase count to have condition p[5]< count hence false returned

case 2 (this is a little tricky)

P[5]= count=7 suppose

than that means count has same value as p[5] so so in array q has been filled with 7 values and set(i) called 7 times to have value of p[5] as 7 we have to call set(5) at after count =6 at 7th call to set(I) becoz this can only set p[i] to 7 which is not possible acc to question hence false returned
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@Laconia can you please elaborate case 2, I didn't get it.
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@rahul sharma 5  :-
The case you say :- "may be p[5] contains 5 as garbage value and q[5] also contains 5.So the statement can become false.We cannot assume what garbage can contains and what it cannot contain as assumed in answer"

This could not be so because if q[5] has to be 5 then set(5) should have been called because array q values can only be from the list of i values for which set(i) has been called. Say for example, that q[5]=5 then that means set(5) must have been called that is why q[5]=5. But the question says - "if set(i) has not been called for some i, then regardless of what p[i] contains, is_set(i) will return false" which means we have to just prove this statement for the case when set(5) hasn't been called.

+1

@skywalker_19:-  This could not be so because if q[5] has to be 5 then set(5) should have been called because array q values can only be from the list of i values for which set(i) has been called.

Question says q is uninitialized.So it q[5] may contain 5 as garbage value.It is not necessary that it will contain 5 only when we set this.In c you print any variable without assigning value it print garbage,it means it can be anything