details

property	value
status	complete
benchmark	n-6.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n103.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	11.8447980881 seconds
cpu usage	12.66114427
max memory	2.4551424E7

stage attributes

key	value
output-size	3591
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 = f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, 1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, 1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 5, 7 2 -> 5, 7 3 -> 2 4 -> 3 5 -> 4 6 -> 5, 7 7 -> 6 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 2) f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) 4) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] 5) f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] 6) f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) 7) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, 1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] We have the following SCCs. { 2, 3, 4, 5, 6, 7 } DP problem for innermost termination. P = f6#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f5#(I10, I11, I12, I13, I14) -> f6#(I10, I11, I12, 1 + I13, I14) f4#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) [I16 = I16] f1#(I20, I21, I22, I23, I24) -> f4#(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) f1#(I30, I31, I32, I33, I34) -> f3#(I30, I31, I35, I33, 1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f6(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, 1 + I13, I14) f4(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) [I16 = I16] f1(I20, I21, I22, I23, I24) -> f4(I20, I21, rnd3, I23, I24) [rnd3 = rnd3 /\ 0 <= -1 - I23 + I24] f3(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) f1(I30, I31, I32, I33, I34) -> f3(I30, I31, I35, I33, 1 + I34) [0 <= I35 /\ I35 <= 0 /\ I35 = I35 /\ 0 <= -1 - I33 + I34] f1(I36, I37, I38, I39, I40) -> f2(rnd1, I37, I38, I39, I40) [rnd1 = rnd1 /\ -1 * I39 + I40 <= 0] popout

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

job log

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actions

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