details

property	value
status	complete
benchmark	n-15a.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	3.57886695862 seconds
cpu usage	3.807041603
max memory	2.0627456E7

stage attributes

key	value
output-size	3426
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f10#(x1, x2) -> f1#(x1, x2) f9#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0] f6#(I11, I12) -> f7#(I11, I12) [14 <= I12] f6#(I13, I14) -> f7#(I13, I14) [1 + I14 <= 13] f2#(I15, I16) -> f6#(I15, -1 + I16) f5#(I17, I18) -> f2#(I17, I18) f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0] f3#(I21, I22) -> f4#(I21, I22) [14 <= I22] f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13] f2#(I25, I26) -> f3#(I25, -1 + I26) f1#(I27, I28) -> f2#(I27, I28) R = f10(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f8(rnd1, rnd2) [y1 = -1 + I1 /\ y1 <= 13 /\ 13 <= y1 /\ rnd2 = rnd2 /\ 0 <= -1 - rnd2 /\ rnd1 = rnd1] f9(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f9(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0] f7(I8, I9) -> f8(I10, I9) [I10 = I10 /\ 0 <= -1 - I9] f6(I11, I12) -> f7(I11, I12) [14 <= I12] f6(I13, I14) -> f7(I13, I14) [1 + I14 <= 13] f2(I15, I16) -> f6(I15, -1 + I16) f5(I17, I18) -> f2(I17, I18) f4(I19, I20) -> f5(I19, I20) [-1 * I20 <= 0] f3(I21, I22) -> f4(I21, I22) [14 <= I22] f3(I23, I24) -> f4(I23, I24) [1 + I24 <= 13] f2(I25, I26) -> f3(I25, -1 + I26) f1(I27, I28) -> f2(I27, I28) The dependency graph for this problem is: 0 -> 11 1 -> 2, 5, 10 2 -> 1 3 -> 4 -> 5 -> 3, 4 6 -> 2, 5, 10 7 -> 6 8 -> 7 9 -> 7 10 -> 8, 9 11 -> 2, 5, 10 Where: 0) f10#(x1, x2) -> f1#(x1, x2) 1) f9#(I2, I3) -> f2#(I2, I3) 2) f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0] 3) f6#(I11, I12) -> f7#(I11, I12) [14 <= I12] 4) f6#(I13, I14) -> f7#(I13, I14) [1 + I14 <= 13] 5) f2#(I15, I16) -> f6#(I15, -1 + I16) 6) f5#(I17, I18) -> f2#(I17, I18) 7) f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0] 8) f3#(I21, I22) -> f4#(I21, I22) [14 <= I22] 9) f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13] 10) f2#(I25, I26) -> f3#(I25, -1 + I26) 11) f1#(I27, I28) -> f2#(I27, I28) We have the following SCCs. { 1, 2, 6, 7, 8, 9, 10 } DP problem for innermost termination. P = f9#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0] f5#(I17, I18) -> f2#(I17, I18) f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0] f3#(I21, I22) -> f4#(I21, I22) [14 <= I22] f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13] f2#(I25, I26) -> f3#(I25, -1 + I26) R = f10(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f8(rnd1, rnd2) [y1 = -1 + I1 /\ y1 <= 13 /\ 13 <= y1 /\ rnd2 = rnd2 /\ 0 <= -1 - rnd2 /\ rnd1 = rnd1] f9(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f9(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0] f7(I8, I9) -> f8(I10, I9) [I10 = I10 /\ 0 <= -1 - I9] f6(I11, I12) -> f7(I11, I12) [14 <= I12] f6(I13, I14) -> f7(I13, I14) [1 + I14 <= 13] f2(I15, I16) -> f6(I15, -1 + I16) f5(I17, I18) -> f2(I17, I18) f4(I19, I20) -> f5(I19, I20) [-1 * I20 <= 0] f3(I21, I22) -> f4(I21, I22) [14 <= I22] f3(I23, I24) -> f4(I23, I24) [1 + I24 <= 13] f2(I25, I26) -> f3(I25, -1 + I26) f1(I27, I28) -> f2(I27, I28) popout

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

job log

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actions

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