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Integ Trans Syste 27634 pair #381738265
details
property
value
status
complete
benchmark
cfg.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n095.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
4.10290002823 seconds
cpu usage
4.324264354
max memory
2.461696E7
stage attributes
key
value
output-size
2242
starexec-result
MAYBE
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f4#(x1, x2, x3, x4, x5) -> f3#(x1, x2, x3, x4, x5) f3#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4] f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9] f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14] f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19] R = f4(x1, x2, x3, x4, x5) -> f3(x1, x2, x3, x4, x5) f3(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4] f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9] f2(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14] f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19] The dependency graph for this problem is: 0 -> 1, 2 1 -> 3 2 -> 4 3 -> 4 4 -> 3 Where: 0) f4#(x1, x2, x3, x4, x5) -> f3#(x1, x2, x3, x4, x5) 1) f3#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4] 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9] 3) f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14] 4) f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14] f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19] R = f4(x1, x2, x3, x4, x5) -> f3(x1, x2, x3, x4, x5) f3(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4] f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9] f2(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14] f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]
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