GATE2014-3-34

Question

Consider the basic block given below. 

a = b + c 
c = a + d 
d = b + c 
e = d - b 
a = e + b 

The minimum number of nodes and edges present in the DAG representation of the above basic block respectively are

• A. $6$ and $6$
• B. $8$ and $10$
• C. $9$ and $12$
• D. $4$ and $4$

Comments

Notice - No node no node for '=' operator

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THIS IS IN SYLLABUS?

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Yeah it is in syllabus. Question can be asked from anywhere if it is even a little bit in syllabus. So leave recent out of syllabus topics carefully.

Answer 1

A normal DAG construction will give $8$ nodes and $10$ edges as shown below.

Since, this question asks for minimum possible, we can assume algebraic simplification is allowed. So, $d = b + c, e = d - b$; can be simplified to $d = b + c$; $e = c$; Similarly, $e = d - b$; $a = e + b$; can be simplified to $a = d$. This gives the following DAG with $6$ nodes and $6$ edges.

https://cs.nyu.edu/~gottlieb/courses/2000s/2006-07-fall/compilers/lectures/lecture-14.html [working link]

Correct Answer: $A$

Comments

Why is definition d added to node a? In 3rd statement c is getting redefined and hence the new value of c would be used for defining d. Hence a new node for d is created. What am I missing, Arjun Sir?

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Sorry. It was a mistake. I have corrected it now.

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sir. the lnk is dead i need to study DAG . plz provide a gud link .

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thanx arjun sir

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@Arjun Sir, The link here is not working,have any alternative?

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Is DAG in syllabus?  but can questions of this type be expected anyone reply.... Isn't DAG used for optimisation?

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is this in syllabus???

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yes. it's in syllabus.

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@Arjun Sir

In 2nd Figure am unable to find edge of  e = d-b; except this am understand ...I think 7 edges will be there?plzz look if you have time?

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but code optimization is not, then how is it in syllabus? :(

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what sort of question can we expect with dag because it involves code optimization techniques which are nt in syllabus?

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@sushmita

I think it is intermediate code and DAG combination question and both of them are in syllabus..

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can we do it like this-

a=b+c

c=a+d

d=b+c

e=d-b

a=e+b

Now d is common subexpression so we need not compute it again-

Hence it will become-

a=b+c

c=a+d

e=a-b

a=e+b=a-b+b=a -----> redundant

Finally it becomes-

a=b+c

c=a+d

e=a-b

It will also require 6 nodes and 6 edges.

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isn't it right?

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It is also correct!!!

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Where is the assingment operator?

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Had minimum not been asked, would be then go with 8 and 10 option?

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@prayas

It is not required  because in DAG x:=y type of assignments need not to be performed (not required) untill and unless it is used for computation of further expression.

Answer 2

Simplifying the given equations : 
 
d = b + c (given) e = d – b (given) => d = b + c and e = c 
 
e = d - b (given) a = e + b (given) => a = d 
 
 
 
Thus, the given DAG has 6 nodes and 6 edges. 

Answer 3

Best approach of doing this question is "Go in reverse order". and it has asked for minimum so we can simplify wherever possible.

A = (E+B)

((D-B)+B) = (D) Because both + , - has same precedence

(B+C)

(B+(A+D))

(B+((B+C)+D))

 

Answer 4

option B

Answer 5

answer - B