details

property	value
status	complete
benchmark	loop_on_input.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.85725688934 seconds
cpu usage	1.962643405
max memory	1.8325504E7

stage attributes

key	value
output-size	2774
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 = f6#(x1, x2) -> f5#(x1, x2) f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2] f2#(I2, I3) -> f3#(I2, I3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4] f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9] f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0] R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2] f2(I2, I3) -> f3(I2, I3) f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4] f3(I6, I7) -> f4(I6, I7) [4 <= I7] f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9] f1(I10, I11) -> f2(I10, 1) [I11 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 4, 5 4 -> 2 5 -> 2 Where: 0) f6#(x1, x2) -> f5#(x1, x2) 1) f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2] 2) f2#(I2, I3) -> f3#(I2, I3) 3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4] 4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9] 5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f2#(I2, I3) -> f3#(I2, I3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4] f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9] f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0] R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2] f2(I2, I3) -> f3(I2, I3) f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4] f3(I6, I7) -> f4(I6, I7) [4 <= I7] f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9] f1(I10, I11) -> f2(I10, 1) [I11 <= 0] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1)] = -x1 + 2 NU[f3#(x0,x1)] = -x1 + 3 NU[f2#(x0,x1)] = -x1 + 3 This gives the following inequalities: ==> -I3 + 3 >= -I3 + 3 1 + I5 <= 4 ==> -I5 + 3 > -I5 + 2 with -I5 + 3 >= 0 1 <= I9 ==> -I9 + 2 >= -(1 + I9) + 3 I11 <= 0 ==> -I11 + 2 >= -1 + 3 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I2, I3) -> f3#(I2, I3) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9] f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0] R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2] f2(I2, I3) -> f3(I2, I3) f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4] f3(I6, I7) -> f4(I6, I7) [4 <= I7] f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9] f1(I10, I11) -> f2(I10, 1) [I11 <= 0] The dependency graph for this problem is: 2 -> 4 -> 2 5 -> 2 Where: 2) f2#(I2, I3) -> f3#(I2, I3) 4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9] 5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0] 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