details

property	value
status	complete
benchmark	n-1d.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	20.1271870136 seconds
cpu usage	21.482144677
max memory	2.0168704E7

stage attributes

key	value
output-size	6805
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) -> f7#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13] f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0] f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 6, 8 2 -> 6, 8 3 -> 2 4 -> 3 5 -> 3 6 -> 4, 5 7 -> 6, 8 8 -> 7 Where: 0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] 5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] 6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] 7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13] f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0] f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3)] = -x2 + x3 + 1 NU[f4#(x0,x1,x2,x3)] = -x2 + x3 NU[f5#(x0,x1,x2,x3)] = -x2 + x3 NU[f1#(x0,x1,x2,x3)] = -x2 + x3 + 1 NU[f6#(x0,x1,x2,x3)] = -x2 + x3 + 1 This gives the following inequalities: ==> -I6 + I7 + 1 >= -I6 + I7 + 1 ==> -I10 + I11 >= -(1 + I10) + I11 + 1 1 <= I13 ==> -I14 + I15 >= -I14 + I15 1 + I17 <= 0 ==> -I18 + I19 >= -I18 + I19 rnd2 = rnd2 /\ 0 <= -1 - I22 + I23 ==> -I22 + I23 + 1 > -I22 + I23 with -I22 + I23 + 1 >= 0 ==> -I26 + I27 + 1 >= -I26 + I27 + 1 0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31 ==> -I30 + I31 + 1 >= -I30 + I31 + 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] popout

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

job log

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actions

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