details

property	value
status	complete
benchmark	spiral.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n106.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	27.5738050938 seconds
cpu usage	29.514060797
max memory	2.6710016E7

stage attributes

key	value
output-size	3061
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f2#(I0, I1) f3#(I2, I3) -> f1#(I2, I3) [1 <= I3] f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 0] f2#(I6, I7) -> f3#(I6, I7) [1 <= I6] f2#(I8, I9) -> f3#(I8, I9) [1 + I8 <= 0] f1#(I10, I11) -> f2#(I10, -1 + I11) [2 - I11 <= I10 /\ I11 <= I10] f1#(I12, I13) -> f2#(-1 + I12, I13) [1 + I13 <= I12 /\ I13 <= 1 - I12] f1#(I14, I15) -> f2#(I14, 1 + I15) [1 + I14 <= -1 * I15 /\ I14 <= I15] f1#(I16, I17) -> f2#(1 + I16, I17) [1 + I16 <= I17 /\ -1 * I16 <= I17] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f2(I0, I1) f3(I2, I3) -> f1(I2, I3) [1 <= I3] f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 0] f2(I6, I7) -> f3(I6, I7) [1 <= I6] f2(I8, I9) -> f3(I8, I9) [1 + I8 <= 0] f1(I10, I11) -> f2(I10, -1 + I11) [2 - I11 <= I10 /\ I11 <= I10] f1(I12, I13) -> f2(-1 + I12, I13) [1 + I13 <= I12 /\ I13 <= 1 - I12] f1(I14, I15) -> f2(I14, 1 + I15) [1 + I14 <= -1 * I15 /\ I14 <= I15] f1(I16, I17) -> f2(1 + I16, I17) [1 + I16 <= I17 /\ -1 * I16 <= I17] The dependency graph for this problem is: 0 -> 1 1 -> 4, 5 2 -> 6, 8, 9 3 -> 6, 7, 8 4 -> 2, 3 5 -> 2, 3 6 -> 4 7 -> 4, 5 8 -> 5 9 -> 4, 5 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f2#(I0, I1) 2) f3#(I2, I3) -> f1#(I2, I3) [1 <= I3] 3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 0] 4) f2#(I6, I7) -> f3#(I6, I7) [1 <= I6] 5) f2#(I8, I9) -> f3#(I8, I9) [1 + I8 <= 0] 6) f1#(I10, I11) -> f2#(I10, -1 + I11) [2 - I11 <= I10 /\ I11 <= I10] 7) f1#(I12, I13) -> f2#(-1 + I12, I13) [1 + I13 <= I12 /\ I13 <= 1 - I12] 8) f1#(I14, I15) -> f2#(I14, 1 + I15) [1 + I14 <= -1 * I15 /\ I14 <= I15] 9) f1#(I16, I17) -> f2#(1 + I16, I17) [1 + I16 <= I17 /\ -1 * I16 <= I17] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9 } DP problem for innermost termination. P = f3#(I2, I3) -> f1#(I2, I3) [1 <= I3] f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 0] f2#(I6, I7) -> f3#(I6, I7) [1 <= I6] f2#(I8, I9) -> f3#(I8, I9) [1 + I8 <= 0] f1#(I10, I11) -> f2#(I10, -1 + I11) [2 - I11 <= I10 /\ I11 <= I10] f1#(I12, I13) -> f2#(-1 + I12, I13) [1 + I13 <= I12 /\ I13 <= 1 - I12] f1#(I14, I15) -> f2#(I14, 1 + I15) [1 + I14 <= -1 * I15 /\ I14 <= I15] f1#(I16, I17) -> f2#(1 + I16, I17) [1 + I16 <= I17 /\ -1 * I16 <= I17] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f2(I0, I1) f3(I2, I3) -> f1(I2, I3) [1 <= I3] f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 0] f2(I6, I7) -> f3(I6, I7) [1 <= I6] f2(I8, I9) -> f3(I8, I9) [1 + I8 <= 0] f1(I10, I11) -> f2(I10, -1 + I11) [2 - I11 <= I10 /\ I11 <= I10] f1(I12, I13) -> f2(-1 + I12, I13) [1 + I13 <= I12 /\ I13 <= 1 - I12] f1(I14, I15) -> f2(I14, 1 + I15) [1 + I14 <= -1 * I15 /\ I14 <= I15] f1(I16, I17) -> f2(1 + I16, I17) [1 + I16 <= I17 /\ -1 * I16 <= I17] popout

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

job log

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actions

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