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Number of WAR dependencies possible

R1 <- R3 + R2

R3 <- R3 + R5

R3 <- R2 + R3

Is it 2 or 3?
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it's 2
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Answer is 3.  They have taken I1-I3 also as WAR.
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3 is correct ...

you should consider all cases ... (combination)

As there are 3 WAR dependencies I1 I2,I2 I3 and I1 I3.But the WAR dependency I1 I3 is using register R3 whose value is updated in I2.So we can neglect I1 I3 dependency.

I think 2 must be the ans.
by (307 points)
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What if it would have been R4 <- R3 + R5 as second instruction instead of R3 <- R3 + R5
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then it would be two   instruction 1 to 3 and intrusion  2 to 3
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Why is 1-3 not a WAR in original question ? Reordering of instruction can take place and hence dependency is also possible.

I need confirmation on this. Maybe @Arjun sir can help.
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it is 2 . As 1 time write done then after that all dependency depend on that only.

So, R1 has no dependency on R3
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Are you sure? because in one answer @habib khan sir considered every possible dependency even the one in which variable was updated already.