details

property	value
status	complete
benchmark	java_Break.c.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.66538619995 seconds
cpu usage	1.783113745
max memory	1.5921152E7

stage attributes

key	value
output-size	3888
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) -> f6#(x1, x2, x3) f6#(I0, I1, I2) -> f5#(I0, I1, I2) f6#(I3, I4, I5) -> f2#(I3, I4, I5) f6#(I6, I7, I8) -> f3#(I6, I7, I8) f6#(I9, I10, I11) -> f1#(I9, I10, I11) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f5#(I18, I19, I20) -> f2#(I20, I19, 0) f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23] f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f5(I0, I1, I2) f6(I3, I4, I5) -> f2(I3, I4, I5) f6(I6, I7, I8) -> f3(I6, I7, I8) f6(I9, I10, I11) -> f1(I9, I10, I11) f6(I12, I13, I14) -> f4(I12, I13, I14) f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f5(I18, I19, I20) -> f2(I20, I19, 0) f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23] f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10] f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30] f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 4 -> 9 5 -> 6 6 -> 8 7 -> 8 -> 9 9 -> 7, 8 Where: 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 1) f6#(I0, I1, I2) -> f5#(I0, I1, I2) 2) f6#(I3, I4, I5) -> f2#(I3, I4, I5) 3) f6#(I6, I7, I8) -> f3#(I6, I7, I8) 4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 6) f5#(I18, I19, I20) -> f2#(I20, I19, 0) 7) f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23] 8) f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] 9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) We have the following SCCs. { 8, 9 } DP problem for innermost termination. P = f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f5(I0, I1, I2) f6(I3, I4, I5) -> f2(I3, I4, I5) f6(I6, I7, I8) -> f3(I6, I7, I8) f6(I9, I10, I11) -> f1(I9, I10, I11) f6(I12, I13, I14) -> f4(I12, I13, I14) f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f5(I18, I19, I20) -> f2(I20, I19, 0) f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23] f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10] f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30] f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z3) NU[f2#(z1,z2,z3)] = 10 + -1 * z3 This gives the following inequalities: I26 <= 10 ==> 10 + -1 * I26 > 10 + -1 * (1 + I26) with 10 + -1 * I26 >= 0 ==> 10 + -1 * (1 + I34) >= 10 + -1 * (1 + I34) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f5(I0, I1, I2) f6(I3, I4, I5) -> f2(I3, I4, I5) f6(I6, I7, I8) -> f3(I6, I7, I8) f6(I9, I10, I11) -> f1(I9, I10, I11) f6(I12, I13, I14) -> f4(I12, I13, I14) f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f5(I18, I19, I20) -> f2(I20, I19, 0) popout

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job log

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