details

property	value
status	complete
benchmark	ex18.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n171.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	17.8531 seconds
cpu usage	18.1448
user time	9.7574
system time	8.38737
max virtual memory	164388.0
max residence set size	10032.0

stage attributes

key	value
starexec-result	MAYBE

output

18.05/17.85 MAYBE 18.05/17.85 18.05/17.85 DP problem for innermost termination. 18.05/17.85 P = 18.05/17.85 f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 18.05/17.85 f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 18.05/17.85 f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7] 18.05/17.85 f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] 18.05/17.85 f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] 18.05/17.85 f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.05/17.85 f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 18.05/17.85 f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 18.05/17.85 f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) 18.05/17.85 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 18.05/17.85 f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 18.05/17.85 f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62] 18.05/17.85 f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] 18.05/17.85 R = 18.05/17.85 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 18.05/17.85 f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 18.05/17.85 f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] 18.05/17.85 f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] 18.05/17.85 f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] 18.05/17.85 f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.05/17.85 f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 18.05/17.85 f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 18.05/17.85 f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 18.05/17.85 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 18.05/17.85 f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] 18.05/17.85 f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 18.05/17.85 f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] 18.05/17.85 f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] 18.05/17.85 f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] 18.05/17.85 f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] 18.05/17.85 18.05/17.85 The dependency graph for this problem is: 18.05/17.85 0 -> 1 18.05/17.85 1 -> 11 18.05/17.85 2 -> 5, 6 18.05/17.85 3 -> 5, 6 18.05/17.85 4 -> 8 18.05/17.85 5 -> 8 18.05/17.85 6 -> 10 18.05/17.85 7 -> 9 18.05/17.85 8 -> 7 18.05/17.85 9 -> 7 18.05/17.85 10 -> 2, 3, 4 18.05/17.85 11 -> 12 18.05/17.85 12 -> 10 18.05/17.85 Where: 18.05/17.85 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 18.05/17.85 1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 18.05/17.85 2) f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7] 18.05/17.85 3) f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10] 18.05/17.85 4) f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] 18.05/17.85 5) f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 18.05/17.85 6) f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 18.05/17.85 7) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 18.05/17.85 8) f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) 18.05/17.85 9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 18.05/17.85 10) f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 18.05/17.85 11) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62] 18.05/17.85 12) f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] 18.05/17.85 18.05/17.85 We have the following SCCs. 18.05/17.85 { 2, 3, 6, 10 } 18.05/17.85 { 7, 9 } 18.05/17.85 18.05/17.85 DP problem for innermost termination. 18.05/17.85 P = 18.05/17.85 f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 18.05/17.85 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 18.05/17.85 R = 18.05/17.85 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 18.05/17.85 f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 18.05/17.85 f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7] 18.05/17.85 f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10] 18.05/17.85 f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17] 18.05/17.85 f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 18.05/17.85 f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 18.05/17.85 f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 18.05/17.85 f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 18.05/17.85 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40] 18.05/17.85 f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46] 18.05/17.85 f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 18.05/17.85 f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0] 18.05/17.85 f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62] 18.05/17.85 f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67] 18.05/17.85 f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100] 18.05/17.85 18.05/17.85 We use the reverse value criterion with the projection function NU: 18.05/17.85 NU[f6#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2) 18.05/17.85 NU[f7#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2) 18.05/17.85 18.05/17.85 This gives the following inequalities: 18.05/17.85 ==> I30 + -1 * (1 + I31) >= I30 + -1 * (1 + I31) 18.05/17.85 1 + I41 <= I40 ==> I40 + -1 * (1 + I41) > I40 + -1 * (1 + (1 + I41)) with I40 + -1 * (1 + I41) >= 0 18.05/17.85 18.05/17.85 We remove all the strictly oriented dependency pairs. 18.05/17.85 popout

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job log

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54