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ITS pair #487098016
details
property
value
status
complete
benchmark
java_AG313.c.t2.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)
14.9276 seconds
cpu usage
14.5771
user time
7.64126
system time
6.93588
max virtual memory
732288.0
max residence set size
13420.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f2#(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f3#(I60, I61, I62, I63, I58, I59, I60, I61, I64) [I64 = I63 /\ 0 <= I60 /\ I60 <= 0 /\ I63 = I63] f5#(I65, I66, I67, I68, I69, I70, I71, I72, I73) -> f2#(I71, I72, I73, I68, I69, I70, I71, I72, I71) [1 + I71 <= 0] f5#(I74, I75, I76, I77, I78, I79, I80, I81, I82) -> f2#(I80, I81, I82, I77, I78, I79, I80, I81, I80) [1 <= I80] f2#(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f1#(I100, I101, I102, I97, I98, I99, I100, I101, I102) [1 <= I101 /\ 1 <= I102] f2#(I103, I104, I105, I106, I107, I108, I109, I110, I111) -> f3#(I109, I110, I111, I112, I107, I108, I109, I110, I113) [I113 = I112 /\ I111 <= 0 /\ I112 = I112] f2#(I114, I115, I116, I117, I118, I119, I120, I121, I122) -> f3#(I120, I121, I122, I123, I118, I119, I120, I121, I124) [I124 = I123 /\ I121 <= 0 /\ I123 = I123] f1#(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2#(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f2(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f4(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f3(I60, I61, I62, I63, I58, I59, I60, I61, I64) [I64 = I63 /\ 0 <= I60 /\ I60 <= 0 /\ I63 = I63] f5(I65, I66, I67, I68, I69, I70, I71, I72, I73) -> f2(I71, I72, I73, I68, I69, I70, I71, I72, I71) [1 + I71 <= 0] f5(I74, I75, I76, I77, I78, I79, I80, I81, I82) -> f2(I80, I81, I82, I77, I78, I79, I80, I81, I80) [1 <= I80] f3(I83, I84, I85, I86, I87, I88, I89, I90, I91) -> f4(I89, I90, I91, I92, rnd5, rnd6, rnd7, rnd8, I93) [I93 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = I92 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ I92 = I92] f2(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f1(I100, I101, I102, I97, I98, I99, I100, I101, I102) [1 <= I101 /\ 1 <= I102] f2(I103, I104, I105, I106, I107, I108, I109, I110, I111) -> f3(I109, I110, I111, I112, I107, I108, I109, I110, I113) [I113 = I112 /\ I111 <= 0 /\ I112 = I112] f2(I114, I115, I116, I117, I118, I119, I120, I121, I122) -> f3(I120, I121, I122, I123, I118, I119, I120, I121, I124) [I124 = I123 /\ I121 <= 0 /\ I123 = I123] f1(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133) The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6, 7, 8 2 -> 3 -> 9, 10, 11 4 -> 12 5 -> 6, 7, 8 6 -> 7 -> 10, 11 8 -> 9, 11 9 -> 12 10 -> 11 -> 12 -> 9, 10, 11 Where: 0) f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 1) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 2) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 3) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f2#(I18, I19, I20, I21, I22, I23, I24, I25, I26) 4) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) 5) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] 6) f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f3#(I60, I61, I62, I63, I58, I59, I60, I61, I64) [I64 = I63 /\ 0 <= I60 /\ I60 <= 0 /\ I63 = I63] 7) f5#(I65, I66, I67, I68, I69, I70, I71, I72, I73) -> f2#(I71, I72, I73, I68, I69, I70, I71, I72, I71) [1 + I71 <= 0] 8) f5#(I74, I75, I76, I77, I78, I79, I80, I81, I82) -> f2#(I80, I81, I82, I77, I78, I79, I80, I81, I80) [1 <= I80] 9) f2#(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f1#(I100, I101, I102, I97, I98, I99, I100, I101, I102) [1 <= I101 /\ 1 <= I102] 10) f2#(I103, I104, I105, I106, I107, I108, I109, I110, I111) -> f3#(I109, I110, I111, I112, I107, I108, I109, I110, I113) [I113 = I112 /\ I111 <= 0 /\ I112 = I112] 11) f2#(I114, I115, I116, I117, I118, I119, I120, I121, I122) -> f3#(I120, I121, I122, I123, I118, I119, I120, I121, I124) [I124 = I123 /\ I121 <= 0 /\ I123 = I123] 12) f1#(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2#(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133) We have the following SCCs. { 9, 12 } DP problem for innermost termination. P = f2#(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f1#(I100, I101, I102, I97, I98, I99, I100, I101, I102) [1 <= I101 /\ 1 <= I102] f1#(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2#(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f2(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f4(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, I49, I50, I51, I52, rnd9) [rnd9 = rnd4 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f3(I60, I61, I62, I63, I58, I59, I60, I61, I64) [I64 = I63 /\ 0 <= I60 /\ I60 <= 0 /\ I63 = I63] f5(I65, I66, I67, I68, I69, I70, I71, I72, I73) -> f2(I71, I72, I73, I68, I69, I70, I71, I72, I71) [1 + I71 <= 0] f5(I74, I75, I76, I77, I78, I79, I80, I81, I82) -> f2(I80, I81, I82, I77, I78, I79, I80, I81, I80) [1 <= I80] f3(I83, I84, I85, I86, I87, I88, I89, I90, I91) -> f4(I89, I90, I91, I92, rnd5, rnd6, rnd7, rnd8, I93) [I93 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = I92 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ I92 = I92] f2(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f1(I100, I101, I102, I97, I98, I99, I100, I101, I102) [1 <= I101 /\ 1 <= I102] f2(I103, I104, I105, I106, I107, I108, I109, I110, I111) -> f3(I109, I110, I111, I112, I107, I108, I109, I110, I113) [I113 = I112 /\ I111 <= 0 /\ I112 = I112] f2(I114, I115, I116, I117, I118, I119, I120, I121, I122) -> f3(I120, I121, I122, I123, I118, I119, I120, I121, I124) [I124 = I123 /\ I121 <= 0 /\ I123 = I123] f1(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = -1 * z8 + z9 + -1 * 1 NU[f2#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z9 + -1 * 1 This gives the following inequalities: 1 <= I101 /\ 1 <= I102 ==> I102 + -1 * 1 > -1 * I101 + I102 + -1 * 1 with I102 + -1 * 1 >= 0 ==> -1 * I132 + I133 + -1 * 1 >= -1 * I132 + I133 + -1 * 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I125, I126, I127, I128, I129, I130, I131, I132, I133) -> f2#(I131, I132, I133, I128, I129, I130, I131, I132, -1 * I132 + I133)
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