A New Look at the Automatic Synthesis of Linear Ranking Functions$

2. If (13) is feasible, let 〈ˇy, ˇµ〉 ∈ Q m+n be any of its feasible solutions.Choosing ˇµ for the values of the parameters, (11) is feasible. There aretwo further possibilities:(a) either (11) is unbounded, so (5) trivially terminates;(b) or it is bounded by a rational q ≥ 1 and the same holds for itsdual (10).In both cases, ˇµ, possibly multiplied by a positive natural in order to geta tuple of naturals, defines, via (7), a ranking function for (5).The above case analysis boils down to the following algorithm:1. Use the simplex algorithm to determine the feasibility of (13), ignoringthe objective function. If it is feasible, then any feasible solution inducesa linear ranking function for (5); exit with success.2. If (13) is unfeasible, then try to determine the feasibility of (9) (e.g.,by using the simplex algorithm again to test whether the relaxation (10)is feasible). If (9) is unfeasible then (5) trivially terminates; exit withsuccess.3. Exit with failure (the analysis is inconclusive).An example should serve to better clarify the methodology we have employed.Example 4.1. In the CLP(N) programp(x 1 , x 2 ) :− x 1 ≤ 1 ∧ x 2 = 0,p(x 1 , x 2 ) :− x 1 ≥ 2 ∧ 2x ′ 1 + 1 ≥ x 1 ∧ 2x ′ 1 ≤ x 1 ∧ x ′ 2 + 1 = x 2 , p(x ′ 1, x ′ 2),p(x 1 , x 2 ) is equivalent tox 2 ={⌊log 2 (x 1 )⌋, if x 1 ≠ 0;0, otherwise.The relaxed optimization problem in LP notation (10) is 8minimize 〈µ 1 , µ 2 , −µ 1 , −µ 2 〉 T 〈x 1 , x 2 , x ′ 1, x ′ 2〉⎛⎞ ⎛ ⎞1 0 0 0 ⎛ ⎞ 2−1 0 2 0x 1subject to⎜ 1 0 −2 0⎜x 2⎟⎟ ⎝⎝ 0 1 0 −1⎠x ′ ⎠ ≥ −1⎜ 0⎟1 ⎝x0 −1 0 1′ 1 ⎠2−1〈x 1 , x 2 , x ′ 1, x ′ 2〉 ≥ 0,8 We will tacitly replace an equality in the form α = β by the equivalent pair of inequalitiesα ≥ β and −α ≥ −β whenever the substitution is necessary to fit our framework.14