details

property	value
status	complete
benchmark	ase_example.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)	16.6902 seconds
cpu usage	16.5299
user time	9.15418
system time	7.37569
max virtual memory	745484.0
max residence set size	11860.0

stage attributes

key	value
starexec-result	YES

output

16.48/16.68 YES 16.48/16.68 16.48/16.68 DP problem for innermost termination. 16.48/16.68 P = 16.48/16.68 f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 16.48/16.68 f8#(I0, I1, I2, I3, I4, I5) -> f1#(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 16.48/16.68 f2#(I6, I7, I8, I9, I10, I11) -> f1#(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 16.48/16.68 f2#(I12, I13, I14, I15, I16, I17) -> f6#(0, I13, I14, I15, I16, I17) [I14 <= I12] 16.48/16.68 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 16.48/16.68 f7#(I24, I25, I26, I27, I28, I29) -> f6#(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 16.48/16.68 f7#(I30, I31, I32, I33, I34, I35) -> f5#(0, I31, I32, I33, I34, I35) [I32 <= I30] 16.48/16.68 f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) 16.48/16.68 f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) 16.48/16.68 R = 16.48/16.68 f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 16.48/16.68 f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 16.48/16.68 f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 16.48/16.68 f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] 16.48/16.68 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 16.48/16.68 f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 16.48/16.68 f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] 16.48/16.68 f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) 16.48/16.68 f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] 16.48/16.68 f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) 16.48/16.68 16.48/16.68 The dependency graph for this problem is: 16.48/16.68 0 -> 1 16.48/16.68 1 -> 9 16.48/16.68 2 -> 9 16.48/16.68 3 -> 7 16.48/16.68 4 -> 8 16.48/16.68 5 -> 7 16.48/16.68 6 -> 4 16.48/16.68 7 -> 5, 6 16.48/16.68 8 -> 4 16.48/16.68 9 -> 2, 3 16.48/16.68 Where: 16.48/16.68 0) f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 16.48/16.68 1) f8#(I0, I1, I2, I3, I4, I5) -> f1#(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 16.48/16.68 2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 16.48/16.68 3) f2#(I12, I13, I14, I15, I16, I17) -> f6#(0, I13, I14, I15, I16, I17) [I14 <= I12] 16.48/16.68 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 16.48/16.68 5) f7#(I24, I25, I26, I27, I28, I29) -> f6#(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 16.48/16.68 6) f7#(I30, I31, I32, I33, I34, I35) -> f5#(0, I31, I32, I33, I34, I35) [I32 <= I30] 16.48/16.68 7) f6#(I36, I37, I38, I39, I40, I41) -> f7#(I36, I37, I38, I39, I40, I41) 16.48/16.68 8) f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 9) f1#(I54, I55, I56, I57, I58, I59) -> f2#(I54, I55, I56, I57, I58, I59) 16.48/16.68 16.48/16.68 We have the following SCCs. 16.48/16.68 { 2, 9 } 16.48/16.68 { 5, 7 } 16.48/16.68 { 4, 8 } 16.48/16.68 16.48/16.68 DP problem for innermost termination. 16.48/16.68 P = 16.48/16.68 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 16.48/16.68 f3#(I42, I43, I44, I45, I46, I47) -> f5#(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 R = 16.48/16.68 f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 16.48/16.68 f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 16.48/16.68 f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 16.48/16.68 f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] 16.48/16.68 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 16.48/16.68 f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 16.48/16.68 f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] 16.48/16.68 f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) 16.48/16.68 f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] 16.48/16.68 f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) 16.48/16.68 16.48/16.68 We use the reverse value criterion with the projection function NU: 16.48/16.68 NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) 16.48/16.68 NU[f5#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z1) 16.48/16.68 16.48/16.68 This gives the following inequalities: 16.48/16.68 ==> I20 + -1 * (1 + I18) >= I20 + -1 * (1 + I18) 16.48/16.68 1 + I42 <= I44 ==> I44 + -1 * (1 + I42) > I44 + -1 * (1 + (1 + I42)) with I44 + -1 * (1 + I42) >= 0 16.48/16.68 16.48/16.68 We remove all the strictly oriented dependency pairs. 16.48/16.68 16.48/16.68 DP problem for innermost termination. 16.48/16.68 P = 16.48/16.68 f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 16.48/16.68 R = 16.48/16.68 f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) 16.48/16.68 f8(I0, I1, I2, I3, I4, I5) -> f1(0, rnd2, 10, 10, rnd5, rnd6) [rnd5 = rnd5 /\ rnd2 = 10 /\ rnd6 = rnd6] 16.48/16.68 f2(I6, I7, I8, I9, I10, I11) -> f1(1 + I6, I7, I8, I9, I10, I11) [1 + I6 <= I8] 16.48/16.68 f2(I12, I13, I14, I15, I16, I17) -> f6(0, I13, I14, I15, I16, I17) [I14 <= I12] 16.48/16.68 f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) 16.48/16.68 f7(I24, I25, I26, I27, I28, I29) -> f6(1 + I24, I25, I26, I27, I28, I29) [1 + I24 <= I26] 16.48/16.68 f7(I30, I31, I32, I33, I34, I35) -> f5(0, I31, I32, I33, I34, I35) [I32 <= I30] 16.48/16.68 f6(I36, I37, I38, I39, I40, I41) -> f7(I36, I37, I38, I39, I40, I41) 16.48/16.68 f3(I42, I43, I44, I45, I46, I47) -> f5(1 + I42, I43, I44, I45, I46, I47) [1 + I42 <= I44] 16.48/16.68 f3(I48, I49, I50, I51, I52, I53) -> f4(I48, I49, I50, I51, I52, I53) [I50 <= I48] 16.48/16.68 f1(I54, I55, I56, I57, I58, I59) -> f2(I54, I55, I56, I57, I58, I59) 16.48/16.68 16.48/16.68 The dependency graph for this problem is: 16.48/16.68 4 -> popout

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

job log

popout

actions

all output return to Integer_Transition_Systems 2019-03-29 01.54