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Integ Trans Syste 27634 pair #381738248
details
property
value
status
complete
benchmark
Round3.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n100.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.940151929855 seconds
cpu usage
0.989506047
max memory
1.273856E7
stage attributes
key
value
output-size
2301
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 = init#(x1, x2) -> f1#(rnd1, rnd2) f4#(I0, I1) -> f3#(I0 + 1, I2) [0 <= I1 - 1 /\ I0 + 1 - 3 * y1 <= 2 /\ 0 <= I0 + 1 - 3 * y1 /\ I0 + 1 - 3 * y1 = I2] f3#(I3, I4) -> f4#(I3, I4) [0 <= I4 - 1] f2#(I5, I6) -> f3#(I7, I8) [-1 <= I6 - 1 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 - 3 * I9 <= 2 /\ 0 <= I7 - 3 * I9 /\ I7 - 3 * I9 = I8] f1#(I10, I11) -> f2#(I10, I11) [-1 <= I11 - 1 /\ -1 <= I12 - 1 /\ 0 <= I10 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f4(I0, I1) -> f3(I0 + 1, I2) [0 <= I1 - 1 /\ I0 + 1 - 3 * y1 <= 2 /\ 0 <= I0 + 1 - 3 * y1 /\ I0 + 1 - 3 * y1 = I2] f3(I3, I4) -> f4(I3, I4) [0 <= I4 - 1] f2(I5, I6) -> f3(I7, I8) [-1 <= I6 - 1 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 - 3 * I9 <= 2 /\ 0 <= I7 - 3 * I9 /\ I7 - 3 * I9 = I8] f1(I10, I11) -> f2(I10, I11) [-1 <= I11 - 1 /\ -1 <= I12 - 1 /\ 0 <= I10 - 1] The dependency graph for this problem is: 0 -> 4 1 -> 2 2 -> 1 3 -> 2 4 -> 3 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f4#(I0, I1) -> f3#(I0 + 1, I2) [0 <= I1 - 1 /\ I0 + 1 - 3 * y1 <= 2 /\ 0 <= I0 + 1 - 3 * y1 /\ I0 + 1 - 3 * y1 = I2] 2) f3#(I3, I4) -> f4#(I3, I4) [0 <= I4 - 1] 3) f2#(I5, I6) -> f3#(I7, I8) [-1 <= I6 - 1 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 - 3 * I9 <= 2 /\ 0 <= I7 - 3 * I9 /\ I7 - 3 * I9 = I8] 4) f1#(I10, I11) -> f2#(I10, I11) [-1 <= I11 - 1 /\ -1 <= I12 - 1 /\ 0 <= I10 - 1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f4#(I0, I1) -> f3#(I0 + 1, I2) [0 <= I1 - 1 /\ I0 + 1 - 3 * y1 <= 2 /\ 0 <= I0 + 1 - 3 * y1 /\ I0 + 1 - 3 * y1 = I2] f3#(I3, I4) -> f4#(I3, I4) [0 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f4(I0, I1) -> f3(I0 + 1, I2) [0 <= I1 - 1 /\ I0 + 1 - 3 * y1 <= 2 /\ 0 <= I0 + 1 - 3 * y1 /\ I0 + 1 - 3 * y1 = I2] f3(I3, I4) -> f4(I3, I4) [0 <= I4 - 1] f2(I5, I6) -> f3(I7, I8) [-1 <= I6 - 1 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 - 3 * I9 <= 2 /\ 0 <= I7 - 3 * I9 /\ I7 - 3 * I9 = I8] f1(I10, I11) -> f2(I10, I11) [-1 <= I11 - 1 /\ -1 <= I12 - 1 /\ 0 <= I10 - 1]
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