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ITS pair #487097698
details
property
value
status
complete
benchmark
ex20.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
5.03469 seconds
cpu usage
4.56469
user time
2.39537
system time
2.16932
max virtual memory
735168.0
max residence set size
8728.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5] f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20] f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44] f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5] f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20] f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44] f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66] 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) -> f10#(x1, x2, x3, x4, x5) 1) f10#(I0, I1, I2, I3, I4) -> f6#(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] 2) f2#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [0 <= I5] 3) f2#(I10, I11, I12, I13, I14) -> f7#(I10, I11, I12, I13, I14) [1 + I10 <= 0] 4) f9#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, I18, I19) [1 + I15 <= 1023] 5) f9#(I20, I21, I22, I23, I24) -> f7#(I20, I21, I22, I23, I24) [1023 <= I20] 6) f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) 7) f6#(I35, I36, I37, I38, I39) -> f5#(I35, I36, I37, I38, I39) [1 + I39 <= 0] 8) f6#(I40, I41, I42, I43, I44) -> f5#(I40, I41, I42, I43, I44) [1 <= I44] 9) f6#(I45, I46, I47, I48, I49) -> f4#(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] 10) f5#(I50, I51, I52, I53, I54) -> f4#(I50, I51, I52, 0, I54) 11) f4#(I55, I56, I57, I58, I59) -> f3#(I55, 0, I57, I58, I59) 12) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] 13) f1#(I65, I66, I67, I68, I69) -> f2#(I65, I66, I67, I68, I69) [1 + I68 <= I66] We have the following SCCs. { 6, 12 } DP problem for innermost termination. P = f3#(I30, I31, I32, I33, I34) -> f1#(I30, I31, I32, I33, I34) f1#(I60, I61, I62, I63, I64) -> f3#(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] R = f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) f10(I0, I1, I2, I3, I4) -> f6(I0, I1, I2, I3, rnd5) [rnd5 = rnd5] f2(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [0 <= I5] f2(I10, I11, I12, I13, I14) -> f7(I10, I11, I12, I13, I14) [1 + I10 <= 0] f9(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, I18, I19) [1 + I15 <= 1023] f9(I20, I21, I22, I23, I24) -> f7(I20, I21, I22, I23, I24) [1023 <= I20] f7(I25, I26, I27, I28, I29) -> f8(I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34) -> f1(I30, I31, I32, I33, I34) f6(I35, I36, I37, I38, I39) -> f5(I35, I36, I37, I38, I39) [1 + I39 <= 0] f6(I40, I41, I42, I43, I44) -> f5(I40, I41, I42, I43, I44) [1 <= I44] f6(I45, I46, I47, I48, I49) -> f4(I45, I46, I47, 1023, I49) [0 <= I49 /\ I49 <= 0] f5(I50, I51, I52, I53, I54) -> f4(I50, I51, I52, 0, I54) f4(I55, I56, I57, I58, I59) -> f3(I55, 0, I57, I58, I59) f1(I60, I61, I62, I63, I64) -> f3(I60, 1 + I61, 2 + I62, I63, I64) [I61 <= I63] f1(I65, I66, I67, I68, I69) -> f2(I65, I66, I67, I68, I69) [1 + I68 <= I66] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z4 + -1 * z2 NU[f3#(z1,z2,z3,z4,z5)] = z4 + -1 * z2 This gives the following inequalities: ==> I33 + -1 * I31 >= I33 + -1 * I31 I61 <= I63 ==> I63 + -1 * I61 > I63 + -1 * (1 + I61) with I63 + -1 * I61 >= 0 We remove all the strictly oriented dependency pairs.
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