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solve with "excess" equations

asked 2013-04-15 09:54:19 +0100

andre gravatar image

updated 2020-08-07 13:13:50 +0100

slelievre gravatar image

Why does the following work

solve([a + b - 1, a - b], [a, b])

but this

solve([a + b - 1, a - b, c + d], [a, b])

gives an empty solution?

Can solve be convinced to ignore unnecessary equations?

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answered 2013-04-15 10:51:48 +0100

slelievre gravatar image

updated 2020-08-07 13:15:13 +0100

You could solve for a, b, c, d and then ignore c and d.

sage: a, b, c, d = var('a b c d')
sage: solns = solve([a + b - 1, a - b, c + d], [a, b, c, d]); solns
[[a == (1/2), b == (1/2), c == -r1, d == r1]]
sage: [s[:2] for s in solns]
[[a == (1/2), b == (1/2)]]

Or you can tweak solve to only use the equations which involve the variables that you want to solve for.

sage: def smart_solve(eqns, vars):
....:     return solve([eqn for eqn in eqns if
....:                   any(v in eqn.variables() for v in vars)], vars)
sage: smart_solve([a + b - 1, a - b, c + d], [a, b])
[[a == (1/2), b == (1/2)]]

I don't know what would be the pros and cons of having Sage's solve behave in this way (either by default or as an option).

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Thank you, I kind of went the "smart_solve" way, it just meant yet another iteration of ifs and fors in my application which I'd have liked to avoid. I had expected "solve" to be a bit smarter by itself but perhaps there would be too many implications if you'd expect a full test for solvability. Perhaps an error would be better instead of an empty solution though.

andre gravatar imageandre ( 2013-04-15 12:02:48 +0100 )edit

answered 2020-08-08 01:12:05 +0100

rburing gravatar image

Excess variables are interpreted as parameters, and only solutions that are valid for all values of the parameters are sought. This explains the empty list in the second case.

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Asked: 2013-04-15 09:54:19 +0100

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Last updated: Aug 07 '20