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The Fourier-Motzkin elimination (FME) is a procedure that reduces an $n$-variable problem to an equivalent $(n − 1)$-variable problem. It is analogous to Gaussian elimination but for a system of inequalities. In this post, it was shown why Gaussian elimination does not change the solution set. Does Fourier-Motzkin elimination also not change the solution set? Why or why not?

Adrian
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  • Without having actually look at what it is, I must assume that reducing a problem to an equivalent problem means precisely that the solution set does not change. – Marc van Leeuwen May 05 '21 at 19:18
  • @MarcvanLeeuwen Based on what I've read, the definition of "equivalent" is "Equivalent means that one system of inequalities has a feasible solution if and only if the other one does." Does this imply that the solution set does not change? I'm not sure if this "feasible" solution is the same across both systems of inequalities. – Adrian May 05 '21 at 20:12
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    I don't know what the adjective "feasible" adds to a solution. The main problem with saying that FME does not change the solution set is that it eliminates a variable, so that a solution is an assignment of values to a smaller set of variables, and a solution set is therefore a difference kind of value that the solution set for the original problem (it is an (infinite) set of $n-1$-tuples rather than of $n$-tuples). But if one removes (projects away) the eliminated variable from the original solution set, the result is most certainly equal to the solution set of the reduced system. – Marc van Leeuwen May 06 '21 at 08:05

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I think the answer is affirmative. By what I know of the Fourier-Motzkin elimination: FME first classifies the inequalities into three kinds and then change the coefficient of the first variable $x_1$ to $1$ or 0, this first step does not change the solution set apparently; then FME kicks $x_1$ out of the game by observing that all possible values of $x_1$ is already obtained for a fixed $x' = (x_2,...,x_n)$, this doesn't change the solution set too. Hence, in conclusion, Fourier-Motzkin elimination also not change the solution set.

penseur_32
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