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Integ Trans Syste 27634 pair #381738961
details
property
value
status
complete
benchmark
ex12.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.77090382576 seconds
cpu usage
0.810317135
max memory
1.2357632E7
stage attributes
key
value
output-size
2134
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f3#(I0, 0) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, 0) f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20] f1(I6, I7) -> f2(I6, I7) [20 <= I7] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f3#(I0, 0) 2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] 3) f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, 0) f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20] f1(I6, I7) -> f2(I6, I7) [20 <= I7] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2)] = 20 + -1 * (1 + z2) NU[f3#(z1,z2)] = 20 + -1 * (1 + (2 + z2)) This gives the following inequalities: rnd1 = 2 + 2 + I3 ==> 20 + -1 * (1 + (2 + I3)) >= 20 + -1 * (1 + (2 + I3)) 1 + I5 <= 20 ==> 20 + -1 * (1 + I5) > 20 + -1 * (1 + (2 + I5)) with 20 + -1 * (1 + I5) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, 0) f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20] f1(I6, I7) -> f2(I6, I7) [20 <= I7] The dependency graph for this problem is: 2 -> Where: 2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3] We have the following SCCs.
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