Circuit cutting: Remove circuit cuts that do not result in a disconnection #2260
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…pennylane into qcut_remove_unneeded_subgraphs
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Codecov Report
@@ Coverage Diff @@
## master #2260 +/- ##
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Coverage 99.31% 99.31%
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Files 237 237
Lines 18899 18902 +3
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+ Hits 18769 18772 +3
Misses 130 130
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pennylane/transforms/qcut.py
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| # to worry about adding back an edge between the predecessor to node1 and the successor | ||
| # to node2 because edge connectivity no longer matters in the subgraphs. |
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Sorry, maybe this comment doesn't need to change but I'm personally not sure what "edge connectivity no longer matters" means here.
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I clarified the comment here: b358d1d
The idea is that after fragmenting we are happy to convert back to a tape for further processing. Going from graph to tape only requires looking at the nodes and the total order (which is a node attribute), so we don't care about edges in this conversion.
| @@ -2095,7 +2112,7 @@ def circuit(x): | |||
| res = circuit(x) | |||
| assert np.allclose(res, np.cos(x)) | |||
| assert len(spy.call_args[0][0]) == 1 # there should be 2 tensors for wire 0 | |||
| assert spy.call_args[0][0][0].shape == (4, 4) | |||
| assert spy.call_args[0][0][0].shape == () | |||
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Why did this have to change?
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The circuit in this test actually has a fully contained cut. Beforehand, we were going over all measurement and preparation configurations even though the cut didn't result in a disconnection. So we'd have a (4, 4) tensor with contraction equation a,a
pennylane/transforms/qcut.py
Outdated
| # to worry about adding back an edge between the predecessor to node1 and the successor | ||
| # to node2 because edge connectivity no longer matters in the subgraphs. |
There was a problem hiding this comment.
I clarified the comment here: b358d1d
The idea is that after fragmenting we are happy to convert back to a tape for further processing. Going from graph to tape only requires looking at the nodes and the total order (which is a node attribute), so we don't care about edges in this conversion.
| @@ -2095,7 +2112,7 @@ def circuit(x): | |||
| res = circuit(x) | |||
| assert np.allclose(res, np.cos(x)) | |||
| assert len(spy.call_args[0][0]) == 1 # there should be 2 tensors for wire 0 | |||
| assert spy.call_args[0][0][0].shape == (4, 4) | |||
| assert spy.call_args[0][0][0].shape == () | |||
There was a problem hiding this comment.
The circuit in this test actually has a fully contained cut. Beforehand, we were going over all measurement and preparation configurations even though the cut didn't result in a disconnection. So we'd have a (4, 4) tensor with contraction equation a,a
Context:
Not all place
WireCutoperations lead to a disconnection between a pair of subgraphs. Sometimes aWireCutwill live within a subgraph. For example, in this circuit:This PR removes the
MeasureNodeandPrepareNodepairs resulting fromWireCuts that do not lead to a disconnection.