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Consider the following three address code:

T1= a+b

T2= c+d

T3=a-b

T4= T1+T3

T5= T2+T1

T6 = T5+T4

T7 = c + T6

Number of nodes and edges in DAG of the above code?

If we'll do simplification , then T4 would be T4 = a+b+a-b => a+a , after this simplification T3 is not used anywhere in the program, so it can be eliminated or we'll include it in DAG??

T1= a+b

T2= c+d

T3=a-b

T4= T1+T3

T5= T2+T1

T6 = T5+T4

T7 = c + T6

Number of nodes and edges in DAG of the above code?

If we'll do simplification , then T4 would be T4 = a+b+a-b => a+a , after this simplification T3 is not used anywhere in the program, so it can be eliminated or we'll include it in DAG??

Ankita87077 Iām getting ans as 10 vertices and 12 edges. I started from T7 and then going backwards upto T1. final equation will be T7=a+a+a+b+d+c+c. Now draw dag for this , you will get 10 vertices and 12 edges. Sorry I tried but not able to add pictures.

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