details

property	value
status	complete
benchmark	dsa_test6.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n097.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.36818480492 seconds
cpu usage	1.429821098
max memory	1.5466496E7

stage attributes

key	value
output-size	2235
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, x3) -> f4#(x1, x2, x3) f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [rnd3 = rnd3] f3#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3] f3(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6] f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1) f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [rnd3 = rnd3] 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3] f3(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6] f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = z1 + -1 * (1 + z2) NU[f3#(z1,z2,z3)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I3 + -1 * (1 + I4) >= I3 + -1 * (1 + I4) 1 + I7 <= I6 ==> I6 + -1 * (1 + I7) > I6 + -1 * (1 + (1 + I7)) with I6 + -1 * (1 + I7) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I3, I4, I5) -> f1#(I3, I4, I5) R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f3(I0, 0, rnd3) [rnd3 = rnd3] f3(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6] f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10] The dependency graph for this problem is: 2 -> Where: 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) We have the following SCCs. popout

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all output return to Integ Trans Syste 27634