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Integ Trans Syste 27634 pair #381738433
details
property
value
status
complete
benchmark
java_Break.c.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n013.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
2.35590100288 seconds
cpu usage
2.493861321
max memory
2.0463616E7
stage attributes
key
value
output-size
4524
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f2#(I4, I5, I6, I7) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5#(I24, I25, I26, I27) -> f2#(I24, I27, I26, 0) f2#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [1 + I28 <= I31] f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32] f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0) f2(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [1 + I28 <= I31] f2(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [I35 <= I32] f3(I36, I37, I38, I39) -> f4(I36, I39, I40, I41) [I41 = I40 /\ I40 = I40] f1(I42, I43, I44, I45) -> f2(I42, I45, I44, 1 + I45) The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 4 -> 9 5 -> 6 6 -> 7, 8 7 -> 8 -> 9 9 -> 7, 8 Where: 0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 1) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f2#(I4, I5, I6, I7) 3) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 4) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 5) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] 6) f5#(I24, I25, I26, I27) -> f2#(I24, I27, I26, 0) 7) f2#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [1 + I28 <= I31] 8) f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32] 9) f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) We have the following SCCs. { 8, 9 } DP problem for innermost termination. P = f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32] f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0) f2(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [1 + I28 <= I31] f2(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [I35 <= I32] f3(I36, I37, I38, I39) -> f4(I36, I39, I40, I41) [I41 = I40 /\ I40 = I40] f1(I42, I43, I44, I45) -> f2(I42, I45, I44, 1 + I45) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4)] = z1 + -1 * (1 + z4) NU[f2#(z1,z2,z3,z4)] = z1 + -1 * z4 This gives the following inequalities: I35 <= I32 ==> I32 + -1 * I35 > I32 + -1 * (1 + I35) with I32 + -1 * I35 >= 0 ==> I42 + -1 * (1 + I45) >= I42 + -1 * (1 + I45) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0)
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