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ITS pair #487098676
details
property
value
status
complete
benchmark
java_Sequence.c.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)
20.0011 seconds
cpu usage
20.1293
user time
10.7581
system time
9.37123
max virtual memory
685496.0
max residence set size
8776.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68] f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89] f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5, 6, 7 1 -> 8 2 -> 9, 10 3 -> 11 4 -> 12, 13 5 -> 14 6 -> 7 -> 8 8 -> 9 9 -> 11 10 -> 12 11 -> 9, 10 12 -> 14 13 -> 14 -> 12, 13 Where: 0) f9#(x1, x2, x3, x4, x5, x6) -> f8#(x1, x2, x3, x4, x5, x6) 1) f8#(I0, I1, I2, I3, I4, I5) -> f7#(I0, I1, I2, I3, I4, I5) 2) f8#(I6, I7, I8, I9, I10, I11) -> f6#(I6, I7, I8, I9, I10, I11) 3) f8#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, I16, I17) 4) f8#(I18, I19, I20, I21, I22, I23) -> f4#(I18, I19, I20, I21, I22, I23) 5) f8#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, I28, I29) 6) f8#(I30, I31, I32, I33, I34, I35) -> f1#(I30, I31, I32, I33, I34, I35) 7) f8#(I42, I43, I44, I45, I46, I47) -> f7#(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] 8) f7#(I48, I49, I50, I51, I52, I53) -> f6#(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] 9) f6#(I56, I57, I58, I59, I60, I61) -> f5#(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] 10) f6#(I64, I65, I66, I67, I68, I69) -> f4#(I68, I69, I66, I67, I68, 5) [100 <= I68] 11) f5#(I70, I71, I72, I73, I74, I75) -> f6#(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] 12) f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] 13) f4#(I84, I85, I86, I87, I88, I89) -> f1#(I88, I89, I86, I87, I88, I89) [21 <= I89] 14) f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) We have the following SCCs. { 9, 11 } { 12, 14 } DP problem for innermost termination. P = f4#(I78, I79, I80, I81, I82, I83) -> f3#(I82, I83, I80, I81, I82, I83) [I83 <= 20] f3#(I90, I91, I92, I93, I94, I95) -> f4#(I94, I95, I92, I93, I94, 3 + I95) R = f9(x1, x2, x3, x4, x5, x6) -> f8(x1, x2, x3, x4, x5, x6) f8(I0, I1, I2, I3, I4, I5) -> f7(I0, I1, I2, I3, I4, I5) f8(I6, I7, I8, I9, I10, I11) -> f6(I6, I7, I8, I9, I10, I11) f8(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, I16, I17) f8(I18, I19, I20, I21, I22, I23) -> f4(I18, I19, I20, I21, I22, I23) f8(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, I28, I29) f8(I30, I31, I32, I33, I34, I35) -> f1(I30, I31, I32, I33, I34, I35) f8(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) f8(I42, I43, I44, I45, I46, I47) -> f7(I46, I47, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f7(I48, I49, I50, I51, I52, I53) -> f6(I52, I53, I54, I51, 0, I55) [I55 = I54 /\ I54 = I54] f6(I56, I57, I58, I59, I60, I61) -> f5(I60, I61, I62, I59, I60, I63) [I63 = I62 /\ I60 <= 99 /\ I62 = I62] f6(I64, I65, I66, I67, I68, I69) -> f4(I68, I69, I66, I67, I68, 5) [100 <= I68] f5(I70, I71, I72, I73, I74, I75) -> f6(I74, I75, I76, I73, 1 + I74, I77) [I77 = I76 /\ I76 = I76] f4(I78, I79, I80, I81, I82, I83) -> f3(I82, I83, I80, I81, I82, I83) [I83 <= 20] f4(I84, I85, I86, I87, I88, I89) -> f1(I88, I89, I86, I87, I88, I89) [21 <= I89] f3(I90, I91, I92, I93, I94, I95) -> f4(I94, I95, I92, I93, I94, 3 + I95) f1(I96, I97, I98, I99, I100, I101) -> f2(I100, I101, I102, I103, I104, I105) [I105 = I103 /\ I104 = I102 /\ I103 = I103 /\ I102 = I102] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = 20 + -1 * (3 + z6)
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