details

property	value
status	complete
benchmark	ex3.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n013.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.5205039978 seconds
cpu usage	2.667435587
max memory	2.039808E7

stage attributes

key	value
output-size	4025
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 = f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6] f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11] f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27] f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6] f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11] f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1] f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27] f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33] The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 3 -> 4 -> 6, 7 5 -> 4 6 -> 4 7 -> 2, 3 Where: 0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) 2) f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6] 3) f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11] 4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 5) f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27] 6) f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] 7) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33] We have the following SCCs. { 4, 6 } DP problem for innermost termination. P = f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6] f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11] f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1] f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27] f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2) NU[f3#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2) This gives the following inequalities: ==> I19 + -1 * (1 + I17) >= I19 + -1 * (1 + I17) 1 + I29 <= I31 ==> I31 + -1 * (1 + I29) > I31 + -1 * (1 + (1 + I29)) with I31 + -1 * (1 + I29) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6] f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11] f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1] f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27] f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31] f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33] The dependency graph for this problem is: 4 -> Where: 4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) We have the following SCCs. popout

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job log

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actions

all output return to Integ Trans Syste 27634