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Const Assignment Map Fusion: If two maps assign the same value for every element in a subset of the underlying array (and the subset is not dependent on the array in any way), then we can often fuse the two maps (not always possible) #1685
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CPU Results (comment might be updated)
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GPU Results (comment might be updated)
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The PR contains all the working ideas I had about the constant assignment map fusion:
If it is too many feature at once, I can remove the non-essential pieces for future PRs too. Please let me know if you'd see more tests, different implementation or organization etc. |
depends on the map parameters.
grid-stride loop as a guard.
maps, create empty ones.
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Const assignment map fusion
Const Assignment Map Fusion: If two maps assign the same value for every element in a subset of the underlying array (and the subset is not dependent on the array in any way), then we can often fuse the two maps (not always possible)
Oct 29, 2024
hand-crafted graphs.
cover more cases in the state fusion.
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Since scheduling multiple map kernels with very little internal operation can have a large overhead, sometimes we would like to fuse two such maps if possible. Constant assignment maps are such a case. If two maps assign the same value for every element in a subset of the underlying array (and the subset is not dependent on the array in any way), then:
Motivating Example
Consider the following graphs, all representing a computation that assigns
1
to the boundary of a 2D domain. The first table represents the graphs scheduled for CPU, the second for GPU.Performance
We have profiled a 2D and a 3D boundary initialization, both on CPU and GPU (both on Davinci cluster).
Benchmark scripts and reports are to be found in https://github.com/pratyai/dace/tree/bench-const-assignment-fusion
I will be quoting the performance summaries in further comments.
Comment on GPU performance
The GPU transformation adds additional operation copying the entire array to and from GPU memory, resulting in
O(n^d)
main <=> GPU movement, whereas the assignment itself only touchesO(n^{d-1})
elements. However, this is because the benchmark itself does not do anything else but the assignment. In real computations, we would likely need to move the entire array anyway.Because of this, it is probably better to just focus on the combined performance of the map kernels here.