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Track whether or not let expressions failed to solve in solver#7982
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After mutating an expression, the solver needs to know two things: 1) Did the expression contain the variable we're solving for 2) Was the expression successfully "solved" for the variable. I.e. the variable only appears once in the leftmost position. We need to know this to know property 1 of any subexpressions (i.e. does the right child of the expression contain the variable). This drives what transformations we do in ways that are guaranteed to terminate and not take exponential time. We were tracking property 1 through lets but not property 2, and this meant we were doing unhelpful transformations in some cases. I found a case in the wild where this made a pipeline take > 1 hour to compile (I killed it after an hour). It may have been in an infinite transformation loop, or it might have just been exponential. Not sure.
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I sure which C++ had a terse way to express this pattern (that wasn't an RIAA class)...
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Does this need new test(s)? |
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It's a failure to scale gracefully with large inputs. For the correctness of it, the existing solver tests are fine. |
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Looks like there are some real failures to investigate |
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I assume this is still in-progress? |
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Yes |
…ess_through_solver_lets
…//github.com/halide/Halide into abadams/track_failedness_through_solver_lets
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The test failure was caused by use of an uninitialized value because I didn't check a return value. I added the must use result macro to reduce_expr_modulo to prevent this problem in future. |
…ess_through_solver_lets
…ess_through_solver_lets
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Failure unrelated. Just needs an approval |
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"must use result" should really be the default in the language, with the current behavior opt-in :-/
steven-johnson
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Jan 26, 2024
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* Track whether or not let expressions failed to solve in solver After mutating an expression, the solver needs to know two things: 1) Did the expression contain the variable we're solving for 2) Was the expression successfully "solved" for the variable. I.e. the variable only appears once in the leftmost position. We need to know this to know property 1 of any subexpressions (i.e. does the right child of the expression contain the variable). This drives what transformations we do in ways that are guaranteed to terminate and not take exponential time. We were tracking property 1 through lets but not property 2, and this meant we were doing unhelpful transformations in some cases. I found a case in the wild where this made a pipeline take > 1 hour to compile (I killed it after an hour). It may have been in an infinite transformation loop, or it might have just been exponential. Not sure. * Remove surplus comma * Fix use of uninitialized value that could cause bad transformation
ardier
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Mar 3, 2024
…e#7982) * Track whether or not let expressions failed to solve in solver After mutating an expression, the solver needs to know two things: 1) Did the expression contain the variable we're solving for 2) Was the expression successfully "solved" for the variable. I.e. the variable only appears once in the leftmost position. We need to know this to know property 1 of any subexpressions (i.e. does the right child of the expression contain the variable). This drives what transformations we do in ways that are guaranteed to terminate and not take exponential time. We were tracking property 1 through lets but not property 2, and this meant we were doing unhelpful transformations in some cases. I found a case in the wild where this made a pipeline take > 1 hour to compile (I killed it after an hour). It may have been in an infinite transformation loop, or it might have just been exponential. Not sure. * Remove surplus comma * Fix use of uninitialized value that could cause bad transformation
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After mutating an expression, the solver needs to know two things:
We were tracking property 1 through lets but not property 2, and this meant we were doing unhelpful transformations in some cases. I found a case in the wild where this made a pipeline take > 1 hour to compile (I killed it after an hour). It may have been in an infinite transformation loop, or it might have just been exponential. Not sure.
As a drive-by while trying to figure out why it was taking so long I also made remainder checking cheaper in the common case of division by a constant, by avoiding a call to can_prove.