details

property	value
status	complete
benchmark	bubbleSort.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	24.6793 seconds
cpu usage	25.0876
user time	13.0352
system time	12.0524
max virtual memory	729200.0
max residence set size	14516.0

stage attributes

key	value
starexec-result	YES

output

25.02/24.67 YES 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 25.02/24.67 f10#(I0, I1, I2, I3, I4) -> f1#(4, 0, I2, I3, I4) 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I47, I48, I49) 25.02/24.67 f5#(I50, I51, I52, I53, I54) -> f3#(I50, I51, I52, I53, I54) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 The dependency graph for this problem is: 25.02/24.67 0 -> 1 25.02/24.67 1 -> 12 25.02/24.67 2 -> 8 25.02/24.67 3 -> 10, 11 25.02/24.67 4 -> 6, 7 25.02/24.67 5 -> 12 25.02/24.67 6 -> 9 25.02/24.67 7 -> 9 25.02/24.67 8 -> 4, 5 25.02/24.67 9 -> 8 25.02/24.67 10 -> 25.02/24.67 11 -> 25.02/24.67 12 -> 2, 3 25.02/24.67 Where: 25.02/24.67 0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 25.02/24.67 1) f10#(I0, I1, I2, I3, I4) -> f1#(4, 0, I2, I3, I4) 25.02/24.67 2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 3) f2#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 4) f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 5) f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 6) f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 7) f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 8) f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 10) f5#(I45, I46, I47, I48, I49) -> f3#(I45, I46, I47, I48, I49) 25.02/24.67 11) f5#(I50, I51, I52, I53, I54) -> f3#(I50, I51, I52, I53, I54) 25.02/24.67 12) f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We have the following SCCs. 25.02/24.67 { 2, 4, 5, 6, 7, 8, 9, 12 } 25.02/24.67 25.02/24.67 DP problem for innermost termination. 25.02/24.67 P = 25.02/24.67 f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f8#(I15, I16, I17, I18, I19) -> f9#(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8#(I20, I21, I22, I23, I24) -> f1#(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) 25.02/24.67 f9#(I30, I31, I32, I33, I34) -> f6#(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7#(I35, I36, I37, I38, I39) -> f8#(I35, I36, I37, I38, I39) 25.02/24.67 f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f1#(I60, I61, I62, I63, I64) -> f2#(I60, I61, I62, I63, I64) 25.02/24.67 R = 25.02/24.67 f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 25.02/24.67 f10(I0, I1, I2, I3, I4) -> f1(4, 0, I2, I3, I4) 25.02/24.67 f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [0 <= I5] 25.02/24.67 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [1 + I10 <= 0] 25.02/24.67 f8(I15, I16, I17, I18, I19) -> f9(I15, I16, I17, I18, I19) [1 + I16 <= I15] 25.02/24.67 f8(I20, I21, I22, I23, I24) -> f1(-1 + I20, I21, I22, I23, I24) [I20 <= I21] 25.02/24.67 f9(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) 25.02/24.67 f9(I30, I31, I32, I33, I34) -> f6(I30, I31, rnd3, I31, 1 + I31) [rnd3 = rnd3] 25.02/24.67 f7(I35, I36, I37, I38, I39) -> f8(I35, I36, I37, I38, I39) 25.02/24.67 f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) 25.02/24.67 f5(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) 25.02/24.67 f5(I50, I51, I52, I53, I54) -> f3(I50, I51, I52, I53, I54) 25.02/24.67 f3(I55, I56, I57, I58, I59) -> f4(I55, I56, I57, I58, I59) 25.02/24.67 f1(I60, I61, I62, I63, I64) -> f2(I60, I61, I62, I63, I64) 25.02/24.67 25.02/24.67 We use the extended value criterion with the projection function NU: 25.02/24.67 NU[f6#(x0,x1,x2,x3,x4)] = x0 - x1 - 2 25.02/24.67 NU[f1#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f9#(x0,x1,x2,x3,x4)] = x0 - x1 - 2 25.02/24.67 NU[f8#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f7#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 NU[f2#(x0,x1,x2,x3,x4)] = x0 - x1 - 1 25.02/24.67 25.02/24.67 This gives the following inequalities: 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