details

property	value
status	complete
benchmark	ex20.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.29282 seconds
cpu usage	6.32462
user time	3.30321
system time	3.02141
max virtual memory	700652.0
max residence set size	8756.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f11#(x1, x2, x3, x4, x5, x6) -> f10#(x1, x2, x3, x4, x5, x6) f10#(I0, I1, I2, I3, I4, I5) -> f6#(I0, I1, I2, I3, I4, rnd6) [rnd6 = rnd6] f2#(I6, I7, I8, I9, I10, I11) -> f9#(I6, I7, I8, I9, I10, I11) [0 <= I7] f2#(I12, I13, I14, I15, I16, I17) -> f7#(I12, I13, I14, I15, I16, I17) [1 + I13 <= 0] f9#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I21, I22, I23) [1 + I19 <= I18] f9#(I24, I25, I26, I27, I28, I29) -> f7#(I24, I25, I26, I27, I28, I29) [I24 <= I25] f3#(I36, I37, I38, I39, I40, I41) -> f1#(I36, I37, I38, I39, I40, I41) f6#(I42, I43, I44, I45, I46, I47) -> f5#(I42, I43, I44, I45, I46, I47) [1 + I47 <= 0] f6#(I48, I49, I50, I51, I52, I53) -> f5#(I48, I49, I50, I51, I52, I53) [1 <= I53] f6#(I54, I55, I56, I57, I58, I59) -> f4#(I54, I55, I56, I57, I54, I59) [0 <= I59 /\ I59 <= 0] f5#(I60, I61, I62, I63, I64, I65) -> f4#(I60, I61, I62, I63, 0, I65) f4#(I66, I67, I68, I69, I70, I71) -> f3#(I66, I67, 0, I69, I70, I71) f1#(I72, I73, I74, I75, I76, I77) -> f3#(I72, I73, 1 + I74, 2 + I75, I76, I77) [I74 <= I76] f1#(I78, I79, I80, I81, I82, I83) -> f2#(I78, I79, I80, I81, I82, I83) [1 + I82 <= I80] R = f11(x1, x2, x3, x4, x5, x6) -> f10(x1, x2, x3, x4, x5, x6) f10(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, rnd6) [rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f9(I6, I7, I8, I9, I10, I11) [0 <= I7] f2(I12, I13, I14, I15, I16, I17) -> f7(I12, I13, I14, I15, I16, I17) [1 + I13 <= 0] f9(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I21, I22, I23) [1 + I19 <= I18] f9(I24, I25, I26, I27, I28, I29) -> f7(I24, I25, I26, I27, I28, I29) [I24 <= I25] f7(I30, I31, I32, I33, I34, I35) -> f8(I30, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f1(I36, I37, I38, I39, I40, I41) f6(I42, I43, I44, I45, I46, I47) -> f5(I42, I43, I44, I45, I46, I47) [1 + I47 <= 0] f6(I48, I49, I50, I51, I52, I53) -> f5(I48, I49, I50, I51, I52, I53) [1 <= I53] f6(I54, I55, I56, I57, I58, I59) -> f4(I54, I55, I56, I57, I54, I59) [0 <= I59 /\ I59 <= 0] f5(I60, I61, I62, I63, I64, I65) -> f4(I60, I61, I62, I63, 0, I65) f4(I66, I67, I68, I69, I70, I71) -> f3(I66, I67, 0, I69, I70, I71) f1(I72, I73, I74, I75, I76, I77) -> f3(I72, I73, 1 + I74, 2 + I75, I76, I77) [I74 <= I76] f1(I78, I79, I80, I81, I82, I83) -> f2(I78, I79, I80, I81, I82, I83) [1 + I82 <= I80] The dependency graph for this problem is: 0 -> 1 1 -> 7, 8, 9 2 -> 4, 5 3 -> 4 -> 5 -> 6 -> 12, 13 7 -> 10 8 -> 10 9 -> 11 10 -> 11 11 -> 6 12 -> 6 13 -> 2, 3 Where: 0) f11#(x1, x2, x3, x4, x5, x6) -> f10#(x1, x2, x3, x4, x5, x6) 1) f10#(I0, I1, I2, I3, I4, I5) -> f6#(I0, I1, I2, I3, I4, rnd6) [rnd6 = rnd6] 2) f2#(I6, I7, I8, I9, I10, I11) -> f9#(I6, I7, I8, I9, I10, I11) [0 <= I7] 3) f2#(I12, I13, I14, I15, I16, I17) -> f7#(I12, I13, I14, I15, I16, I17) [1 + I13 <= 0] 4) f9#(I18, I19, I20, I21, I22, I23) -> f7#(I18, I19, I20, I21, I22, I23) [1 + I19 <= I18] 5) f9#(I24, I25, I26, I27, I28, I29) -> f7#(I24, I25, I26, I27, I28, I29) [I24 <= I25] 6) f3#(I36, I37, I38, I39, I40, I41) -> f1#(I36, I37, I38, I39, I40, I41) 7) f6#(I42, I43, I44, I45, I46, I47) -> f5#(I42, I43, I44, I45, I46, I47) [1 + I47 <= 0] 8) f6#(I48, I49, I50, I51, I52, I53) -> f5#(I48, I49, I50, I51, I52, I53) [1 <= I53] 9) f6#(I54, I55, I56, I57, I58, I59) -> f4#(I54, I55, I56, I57, I54, I59) [0 <= I59 /\ I59 <= 0] 10) f5#(I60, I61, I62, I63, I64, I65) -> f4#(I60, I61, I62, I63, 0, I65) 11) f4#(I66, I67, I68, I69, I70, I71) -> f3#(I66, I67, 0, I69, I70, I71) 12) f1#(I72, I73, I74, I75, I76, I77) -> f3#(I72, I73, 1 + I74, 2 + I75, I76, I77) [I74 <= I76] 13) f1#(I78, I79, I80, I81, I82, I83) -> f2#(I78, I79, I80, I81, I82, I83) [1 + I82 <= I80] We have the following SCCs. { 6, 12 } DP problem for innermost termination. P = f3#(I36, I37, I38, I39, I40, I41) -> f1#(I36, I37, I38, I39, I40, I41) f1#(I72, I73, I74, I75, I76, I77) -> f3#(I72, I73, 1 + I74, 2 + I75, I76, I77) [I74 <= I76] R = f11(x1, x2, x3, x4, x5, x6) -> f10(x1, x2, x3, x4, x5, x6) f10(I0, I1, I2, I3, I4, I5) -> f6(I0, I1, I2, I3, I4, rnd6) [rnd6 = rnd6] f2(I6, I7, I8, I9, I10, I11) -> f9(I6, I7, I8, I9, I10, I11) [0 <= I7] f2(I12, I13, I14, I15, I16, I17) -> f7(I12, I13, I14, I15, I16, I17) [1 + I13 <= 0] f9(I18, I19, I20, I21, I22, I23) -> f7(I18, I19, I20, I21, I22, I23) [1 + I19 <= I18] f9(I24, I25, I26, I27, I28, I29) -> f7(I24, I25, I26, I27, I28, I29) [I24 <= I25] f7(I30, I31, I32, I33, I34, I35) -> f8(I30, I31, I32, I33, I34, I35) f3(I36, I37, I38, I39, I40, I41) -> f1(I36, I37, I38, I39, I40, I41) f6(I42, I43, I44, I45, I46, I47) -> f5(I42, I43, I44, I45, I46, I47) [1 + I47 <= 0] f6(I48, I49, I50, I51, I52, I53) -> f5(I48, I49, I50, I51, I52, I53) [1 <= I53] f6(I54, I55, I56, I57, I58, I59) -> f4(I54, I55, I56, I57, I54, I59) [0 <= I59 /\ I59 <= 0] f5(I60, I61, I62, I63, I64, I65) -> f4(I60, I61, I62, I63, 0, I65) f4(I66, I67, I68, I69, I70, I71) -> f3(I66, I67, 0, I69, I70, I71) f1(I72, I73, I74, I75, I76, I77) -> f3(I72, I73, 1 + I74, 2 + I75, I76, I77) [I74 <= I76] f1(I78, I79, I80, I81, I82, I83) -> f2(I78, I79, I80, I81, I82, I83) [1 + I82 <= I80] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6)] = z5 + -1 * z3 NU[f3#(z1,z2,z3,z4,z5,z6)] = z5 + -1 * z3 This gives the following inequalities: ==> I40 + -1 * I38 >= I40 + -1 * I38 I74 <= I76 ==> I76 + -1 * I74 > I76 + -1 * (1 + I74) with I76 + -1 * I74 >= 0 We remove all the strictly oriented dependency pairs. popout

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

job log

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actions

all output return to ITS