Solve quadratic constraints #2619
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It seems figuring out |
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Oh, maybe it could work using "only known constant is known"-mode |
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chriseth
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Superseded by #2659 |
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Apr 23, 2025
Adds a quick-and-dirty way to return bit constraints when running `QuadraticSymbolicExpression::solve()`, by looking for a specific expression (`X * (X - 1)`). As far as I can see, this is orthogonal to #2619, which would return a `Branch` effect instead of a range constraint. With this PR, we remove the need for "global range constraints". This will help progressing on the new solver (#2651), which tries to rely only on `QuadraticSymbolicExpression`. Then, we can have a bit decomposition example even before implementing machine calls. --------- Co-authored-by: chriseth <[email protected]>
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Strategy to solve
X * Y = 0:Go into both branches, figure out that we can solve both for
var_8, on the left we havevar_8 = Land the right we havevar_8 = R. We figure out thatL - Ris a constant that is at least as large as the range constraint ofvar_8. This means thatLandRcannot both fall into the allowed range forvar_8at the same time. and thus we generate a branch:if L in_range_for var_8 { var_8 = L } else { var_8 = R}