details

property	value
status	complete
benchmark	array3.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	5.64468097687 seconds
cpu usage	5.871280089
max memory	2.1454848E7

stage attributes

key	value
output-size	5342
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) f6#(I0, I1, I2, I3) -> f3#(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4] f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20] f1#(I24, I25, I26, I27) -> f2#(I24, I25, 0, I27) [I24 <= I25] R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f3(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10] f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20] f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25] The dependency graph for this problem is: 0 -> 1 1 -> 4 2 -> 3 3 -> 2 4 -> 5, 6 5 -> 4 6 -> 3 Where: 0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 1) f6#(I0, I1, I2, I3) -> f3#(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] 2) f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4] 3) f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 5) f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20] 6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, 0, I27) [I24 <= I25] We have the following SCCs. { 4, 5 } { 2, 3 } DP problem for innermost termination. P = f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4] f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f3(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10] f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20] f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3,z4)] = z1 + -1 * (1 + z3) NU[f4#(z1,z2,z3,z4)] = z1 + -1 * (1 + z3) This gives the following inequalities: 1 + I6 <= I4 ==> I4 + -1 * (1 + I6) > I4 + -1 * (1 + (1 + I6)) with I4 + -1 * (1 + I6) >= 0 ==> I12 + -1 * (1 + I14) >= I12 + -1 * (1 + I14) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f3(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10] f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20] f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25] The dependency graph for this problem is: 3 -> Where: 3) f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) We have the following SCCs. DP problem for innermost termination. P = f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20] R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f3(I0, 0, 0, rnd4) [rnd4 = rnd4 /\ y1 = 0] popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer Transition Systems