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Integ Trans Syste 27634 pair #381737641
details
property
value
status
complete
benchmark
p-61.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n096.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.91784000397 seconds
cpu usage
2.02034355
max memory
1.7616896E7
stage attributes
key
value
output-size
3164
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, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19] f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 1) f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19] f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z4 NU[f3#(z1,z2,z3,z4,z5)] = z4 This gives the following inequalities: ==> I8 (>! \union =) I8 0 <= -101 + I13 /\ 0 <= 9 - I14 ==> I13 >! -1 + I13 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14] f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19] f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0] The dependency graph for this problem is: 2 -> Where: 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) We have the following SCCs.
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