details

property	value
status	complete
benchmark	p-55.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n108.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	11.7599840164 seconds
cpu usage	11.396916915
max memory	2.619392E7

stage attributes

key	value
output-size	7347
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 = f8#(x1, x2, x3) -> f7#(x1, x2, x3) f7#(I0, I1, I2) -> f1#(I0, I1, I2) f6#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] f5#(I9, I10, I11) -> f1#(I9, I10, I11) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] R = f8(x1, x2, x3) -> f7(x1, x2, x3) f7(I0, I1, I2) -> f1(I0, I1, I2) f6(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] f5(I9, I10, I11) -> f1(I9, I10, I11) f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17] f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1] f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22] f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3, 8 2 -> 3, 8 3 -> 2 4 -> 3, 8 5 -> 4 6 -> 5 7 -> 5 8 -> 6, 7 Where: 0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 1) f7#(I0, I1, I2) -> f1#(I0, I1, I2) 2) f6#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 6) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] 7) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] 8) f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = f6#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] f5#(I9, I10, I11) -> f1#(I9, I10, I11) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] R = f8(x1, x2, x3) -> f7(x1, x2, x3) f7(I0, I1, I2) -> f1(I0, I1, I2) f6(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] f5(I9, I10, I11) -> f1(I9, I10, I11) f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17] f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1] f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22] f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2)] = -x2 - 2 NU[f4#(x0,x1,x2)] = -x2 - 2 NU[f5#(x0,x1,x2)] = -x2 - 1 NU[f1#(x0,x1,x2)] = -x2 - 1 NU[f6#(x0,x1,x2)] = -x2 - 1 This gives the following inequalities: ==> -I5 - 1 >= -I5 - 1 -1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7 ==> -I8 - 1 > -(1 + I8) - 1 with -I8 - 1 >= 0 ==> -I11 - 1 >= -I11 - 1 ==> -I14 - 2 >= -(1 + I14) - 1 0 <= I17 ==> -I17 - 2 >= -I17 - 2 1 + I20 <= -1 ==> -I20 - 2 >= -I20 - 2 0 <= -1 - I22 ==> -I23 - 1 >= -I23 - 2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I3, I4, I5) -> f1#(I3, I4, I5) f5#(I9, I10, I11) -> f1#(I9, I10, I11) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] popout

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all output return to Integ Trans Syste 27634