details

property	value
status	complete
benchmark	heidy10.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n069.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.798520088196 seconds
cpu usage	0.844082927
max memory	1.2648448E7

stage attributes

key	value
output-size	3633
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 = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f3#(I0, I1) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f3#(I0, I1) 2) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 4) f2#(I6, I7) -> f1#(I6, I7) 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f1#(z1,z2)] = z1 NU[f3#(z1,z2)] = z1 This gives the following inequalities: rnd2 = rnd2 /\ 1 <= I2 ==> I2 >! -1 + I2 I5 <= 0 ==> I4 (>! \union =) I4 ==> I6 (>! \union =) I6 1 <= I9 ==> I8 (>! \union =) I8 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] The dependency graph for this problem is: 3 -> 4 -> 3, 5 5 -> 4 Where: 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 4) f2#(I6, I7) -> f1#(I6, I7) 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) popout

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all output return to Integ Trans Syste 27634