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A system with a product equation is solved branch by branch - #2129
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Rafael-SOWNet merged 3 commits intoOct 26, 2026
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The eliminator solved x * y = 0 for y as 0, which is the answer only where x is not zero, and in a system the triangularisation does not take nothing followed the branch where x is. Each row that came back satisfied the system, so the answer read as all of the solutions: x*y; sin(x) + y - 1 missing (0, 1) (x - 1)*(y - 2); e^x - y missing (1, e) x*e^y; x^2*y [[0, 0]], where every (0, t) is a solution A system with a product in more than one of its unknowns is now solved once for each way of choosing a factor of each such product, up to 64 systems, and its solutions are theirs together. A product whose factors all mention the same unknowns is left alone, since solving it reads every factor. A branch that answers null is taken to have no solutions, which is what Solve says of it. Every row of the union is put back into the system as written, because a factor can be zero where another is not defined: the branch x = 0 of x ln(y), x - y is (0, 0), where ln(0) is no number. A row whose residual has no symbol left and is not a finite number is now dropped, on the square path too, as a root of one equation is. A branch can leave an unknown that no equation mentions, and the eliminator had no equation to take it from and answered nothing. Such a system goes to the non-square solver, which makes the unknown a parameter: x - 1, x^2 - 1 in x, y was null and is [[1, t_1]]. SolveSystem.TestTrigSystem4 pinned null for a - 1, b * 0, where b * 0 is zero for every b; it now expects the one family (1, t_1). Measured with the generated check from #2126 on 1190 square systems in x and y built from a known integer solution: the known solution missing 22 times on master and 13 here, with no other change in its findings and no wrong row on either. The 13 left are roots found by Newton's method, one of several, and systems with abs. Fixes #2128 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…is-solved-branch-by-branch
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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October 26, 2026 23:41
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The eliminator solved
x * y = 0foryas0, which is the answer only wherexis not zero. In a system the triangularisation does not take, nothing followed the branch wherexis. Each row that came back satisfied the system, so the check from #2125 passed it, and the answer read as all of the solutions.x*y,sin(x) + y - 1[[pi / 2 + 2 * pi * n_1, 0], [pi - pi / 2 + 2 * pi * n_1, 0]][[0, 1], [pi / 2 + 2 * pi * n_1, 0], [pi - pi / 2 + 2 * pi * n_1, 0]](x - 1)*(y - 2),e^x - y[[ln(2), 2]][[1, e], [ln(2), 2]]x*e^y,x^2*y[[0, 0]][[0, t_1], [0, 0]]x - 1,x^2 - 1inx, ynull[[1, t_1]]x*ln(y),x - y[[1, 1]][[1, 1]]x (x - 1)is solved forxas0and1. A branch that answersnullis taken to have no solutions, which is whatSolvesays of it.x = 0ofx ln(y),x - ygives(0, 0), whereln(0)is no number. A row whose residual has no symbol left and is not a finite number is now dropped, on the square path too, as a root of one equation already is. That is why the last row of the table is unchanged.SolveSystem.TestTrigSystem4pinnednullfora - 1,b * 0, whereb * 0is zero for everyb. It now expects the one family(1, t_1).Measured with the generated check from #2126, on 1190 square systems in
xandybuilt from a known integer solution: the known solution was missing 22 times on master and 13 times here. Nothing else in its findings changed, and neither arm had a wrong row. The 13 left are roots found by Newton's method, where it finds one of several, and systems withabs.ProductSystemTestcovers each branch, the family, the undefined factor and the parameter, and checks every row at several values of what is left in it. Entry inBREAKING-CHANGES.md.Fixes #2128
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura