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Integ Trans Syste 27634 pair #381740014
details
property
value
status
complete
benchmark
ex11.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n072.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
4.69018006325 seconds
cpu usage
4.524124469
max memory
2.3547904E7
stage attributes
key
value
output-size
2425
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f7#(x1, x2) -> f6#(x1, x2) f6#(I0, I1) -> f2#(0, I1) f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2] f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0] f4#(I6, I7) -> f3#(I6, I7) [1 <= I7] f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4] f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4] f3#(I14, I15) -> f1#(I14, I15) [5 <= I14] f1#(I16, I17) -> f2#(1 + I16, I17) R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f2(0, I1) f2(I2, I3) -> f4(I2, rnd2) [rnd2 = rnd2] f4(I4, I5) -> f3(I4, I5) [1 + I5 <= 0] f4(I6, I7) -> f3(I6, I7) [1 <= I7] f4(I8, I9) -> f5(I8, I9) [0 <= I9 /\ I9 <= 0] f3(I10, I11) -> f1(0, I11) [4 <= I10 /\ I10 <= 4] f3(I12, I13) -> f1(I12, I13) [1 + I12 <= 4] f3(I14, I15) -> f1(I14, I15) [5 <= I14] f1(I16, I17) -> f2(1 + I16, I17) The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3, 4 3 -> 5, 6, 7 4 -> 5, 6, 7 5 -> 8 6 -> 8 7 -> 8 8 -> 2 Where: 0) f7#(x1, x2) -> f6#(x1, x2) 1) f6#(I0, I1) -> f2#(0, I1) 2) f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2] 3) f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0] 4) f4#(I6, I7) -> f3#(I6, I7) [1 <= I7] 5) f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4] 6) f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4] 7) f3#(I14, I15) -> f1#(I14, I15) [5 <= I14] 8) f1#(I16, I17) -> f2#(1 + I16, I17) We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2] f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0] f4#(I6, I7) -> f3#(I6, I7) [1 <= I7] f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4] f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4] f3#(I14, I15) -> f1#(I14, I15) [5 <= I14] f1#(I16, I17) -> f2#(1 + I16, I17) R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f2(0, I1) f2(I2, I3) -> f4(I2, rnd2) [rnd2 = rnd2] f4(I4, I5) -> f3(I4, I5) [1 + I5 <= 0] f4(I6, I7) -> f3(I6, I7) [1 <= I7] f4(I8, I9) -> f5(I8, I9) [0 <= I9 /\ I9 <= 0] f3(I10, I11) -> f1(0, I11) [4 <= I10 /\ I10 <= 4] f3(I12, I13) -> f1(I12, I13) [1 + I12 <= 4] f3(I14, I15) -> f1(I14, I15) [5 <= I14] f1(I16, I17) -> f2(1 + I16, I17)
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