details

property	value
status	complete
benchmark	simple_control_on_input.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n109.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.822149991989 seconds
cpu usage	0.864710854
max memory	1.2599296E7

stage attributes

key	value
output-size	2680
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) -> f6#(x1, x2) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4#(I2, I3) -> f3#(I2, I3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) The dependency graph for this problem is: 0 -> 1 1 -> 3, 4 2 -> 5, 6 3 -> 2 4 -> 5 -> 2 6 -> Where: 0) f7#(x1, x2) -> f6#(x1, x2) 1) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 2) f4#(I2, I3) -> f3#(I2, I3) 3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5] 4) f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0] 5) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] 6) f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11] We have the following SCCs. { 2, 5 } DP problem for innermost termination. P = f4#(I2, I3) -> f3#(I2, I3) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8] R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2)] = z1 + -1 * (1 + z2) NU[f4#(z1,z2)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I2 + -1 * (1 + I3) >= I2 + -1 * (1 + I3) 1 + I9 <= I8 ==> I8 + -1 * (1 + I9) > I8 + -1 * (1 + (1 + I9)) with I8 + -1 * (1 + I9) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I2, I3) -> f3#(I2, I3) R = f7(x1, x2) -> f6(x1, x2) f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f4(I2, I3) -> f3(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 <= I5] f5(I6, I7) -> f1(I6, I7) [I7 <= 0] f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8] f3(I10, I11) -> f1(I10, I11) [I10 <= I11] f1(I12, I13) -> f2(I12, I13) The dependency graph for this problem is: 2 -> Where: 2) f4#(I2, I3) -> f3#(I2, I3) We have the following SCCs. popout

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job log

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actions

all output return to Integ Trans Syste 27634