Mixing
Difference flow dependence & anti dependence
I'm trying to understand a paper concerning the detection of dependence for parallelizing tasks and I'm struggling a bit with an example. Considering A, B, C and D are vectors object:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = A(I + 1) + 1
S3: C(I) = D(I)
Focusing on dependence between S1 and S2. To me, it is a flow dependence S1 -> S2, A is in OUT(S1) and IN(S2) and A is indeed used in S2, verifying the flow dependence definition given by Woolfe & Banerjee. However, the explication states that it is an antidependence S2 -> S1. Is it because we're inside a loop ? And in that case, does it mean that in a loop, any flow dependence is reverse antidependence and inversely ?
optimization compilation
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
add a comment |
I'm trying to understand a paper concerning the detection of dependence for parallelizing tasks and I'm struggling a bit with an example. Considering A, B, C and D are vectors object:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = A(I + 1) + 1
S3: C(I) = D(I)
Focusing on dependence between S1 and S2. To me, it is a flow dependence S1 -> S2, A is in OUT(S1) and IN(S2) and A is indeed used in S2, verifying the flow dependence definition given by Woolfe & Banerjee. However, the explication states that it is an antidependence S2 -> S1. Is it because we're inside a loop ? And in that case, does it mean that in a loop, any flow dependence is reverse antidependence and inversely ?
optimization compilation
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
add a comment |
I'm trying to understand a paper concerning the detection of dependence for parallelizing tasks and I'm struggling a bit with an example. Considering A, B, C and D are vectors object:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = A(I + 1) + 1
S3: C(I) = D(I)
Focusing on dependence between S1 and S2. To me, it is a flow dependence S1 -> S2, A is in OUT(S1) and IN(S2) and A is indeed used in S2, verifying the flow dependence definition given by Woolfe & Banerjee. However, the explication states that it is an antidependence S2 -> S1. Is it because we're inside a loop ? And in that case, does it mean that in a loop, any flow dependence is reverse antidependence and inversely ?
optimization compilation
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
I'm trying to understand a paper concerning the detection of dependence for parallelizing tasks and I'm struggling a bit with an example. Considering A, B, C and D are vectors object:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = A(I + 1) + 1
S3: C(I) = D(I)
Focusing on dependence between S1 and S2. To me, it is a flow dependence S1 -> S2, A is in OUT(S1) and IN(S2) and A is indeed used in S2, verifying the flow dependence definition given by Woolfe & Banerjee. However, the explication states that it is an antidependence S2 -> S1. Is it because we're inside a loop ? And in that case, does it mean that in a loop, any flow dependence is reverse antidependence and inversely ?
optimization compilation
optimization compilation
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
edited Nov 22 '18 at 10:30
Sam The Sid
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
asked Nov 21 '18 at 18:51
Sam The SidSam The Sid
216
216
add a comment |
add a comment |
1 Answer
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votes
Okay, finally it depends not of the object, but of the memory location being used. In this example, A(I) and A(I + 1) may not refer to the same memory (in case there is no aliasing). So there is no dependence between them in the current iteration. However, if we unroll the loop a bit, we get:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = -> A(I + 1) + 1
S3: C(I) = D(I)
S1: -> A(I + 1) = B(I + 1) + C(I + 1)
S2: D(I + 1) = A(I + 2) + 1
S3: C(I + 1) = D(I + 1)
Now, the anti dependence appears clearly.
Seems obvious now..
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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votes
Okay, finally it depends not of the object, but of the memory location being used. In this example, A(I) and A(I + 1) may not refer to the same memory (in case there is no aliasing). So there is no dependence between them in the current iteration. However, if we unroll the loop a bit, we get:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = -> A(I + 1) + 1
S3: C(I) = D(I)
S1: -> A(I + 1) = B(I + 1) + C(I + 1)
S2: D(I + 1) = A(I + 2) + 1
S3: C(I + 1) = D(I + 1)
Now, the anti dependence appears clearly.
Seems obvious now..
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
add a comment |
Okay, finally it depends not of the object, but of the memory location being used. In this example, A(I) and A(I + 1) may not refer to the same memory (in case there is no aliasing). So there is no dependence between them in the current iteration. However, if we unroll the loop a bit, we get:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = -> A(I + 1) + 1
S3: C(I) = D(I)
S1: -> A(I + 1) = B(I + 1) + C(I + 1)
S2: D(I + 1) = A(I + 2) + 1
S3: C(I + 1) = D(I + 1)
Now, the anti dependence appears clearly.
Seems obvious now..
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
add a comment |
Okay, finally it depends not of the object, but of the memory location being used. In this example, A(I) and A(I + 1) may not refer to the same memory (in case there is no aliasing). So there is no dependence between them in the current iteration. However, if we unroll the loop a bit, we get:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = -> A(I + 1) + 1
S3: C(I) = D(I)
S1: -> A(I + 1) = B(I + 1) + C(I + 1)
S2: D(I + 1) = A(I + 2) + 1
S3: C(I + 1) = D(I + 1)
Now, the anti dependence appears clearly.
Seems obvious now..
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
Okay, finally it depends not of the object, but of the memory location being used. In this example, A(I) and A(I + 1) may not refer to the same memory (in case there is no aliasing). So there is no dependence between them in the current iteration. However, if we unroll the loop a bit, we get:
do I=2,N
S1: A(I) = B(I) + C(I)
S2: D(I) = -> A(I + 1) + 1
S3: C(I) = D(I)
S1: -> A(I + 1) = B(I + 1) + C(I + 1)
S2: D(I + 1) = A(I + 2) + 1
S3: C(I + 1) = D(I + 1)
Now, the anti dependence appears clearly.
Seems obvious now..
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
answered Nov 22 '18 at 10:30
Sam The SidSam The Sid
216
216
add a comment |
add a comment |
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