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Integ Trans Syste 27634 pair #381737335
details
property
value
status
complete
benchmark
java_FibRecursive.c.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n009.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
3.33052802086 seconds
cpu usage
3.544258536
max memory
2.4051712E7
stage attributes
key
value
output-size
5749
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f8#(x1, x2, x3) -> f7#(x1, x2, x3) f7#(I0, I1, I2) -> f3#(I0, I1, I2) f7#(I3, I4, I5) -> f6#(I3, I4, I5) f7#(I6, I7, I8) -> f5#(I6, I7, I8) f7#(I9, I10, I11) -> f4#(I9, I10, I11) f7#(I12, I13, I14) -> f1#(I12, I13, I14) f3#(I18, I19, I20) -> f6#(I20, I19, I20) f6#(I21, I22, I23) -> f5#(I23, I22, I23) [I23 <= 0] f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26] f4#(I30, I31, I32) -> f5#(I32, I31, I32) [1 <= I32 /\ I32 <= 1] f4#(I33, I34, I35) -> f1#(I35, I34, I35) [1 + I35 <= 1] f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38] f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) R = f8(x1, x2, x3) -> f7(x1, x2, x3) f7(I0, I1, I2) -> f3(I0, I1, I2) f7(I3, I4, I5) -> f6(I3, I4, I5) f7(I6, I7, I8) -> f5(I6, I7, I8) f7(I9, I10, I11) -> f4(I9, I10, I11) f7(I12, I13, I14) -> f1(I12, I13, I14) f7(I15, I16, I17) -> f2(I15, I16, I17) f3(I18, I19, I20) -> f6(I20, I19, I20) f6(I21, I22, I23) -> f5(I23, I22, I23) [I23 <= 0] f6(I24, I25, I26) -> f4(I26, I25, I26) [1 <= I26] f5(I27, I28, I29) -> f2(I29, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f4(I30, I31, I32) -> f5(I32, I31, I32) [1 <= I32 /\ I32 <= 1] f4(I33, I34, I35) -> f1(I35, I34, I35) [1 + I35 <= 1] f4(I36, I37, I38) -> f1(I38, I37, I38) [2 <= I38] f1(I39, I40, I41) -> f3(I41, I40, -1 + I41) f1(I42, I43, I44) -> f3(I44, I43, -2 + I44) f1(I45, I46, I47) -> f2(I47, I48, I49) [I49 = I48 /\ I48 = I48] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 4 -> 9, 10, 11 5 -> 12, 13 6 -> 7, 8 7 -> 8 -> 9, 11 9 -> 10 -> 12, 13 11 -> 12, 13 12 -> 6 13 -> 6 Where: 0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 1) f7#(I0, I1, I2) -> f3#(I0, I1, I2) 2) f7#(I3, I4, I5) -> f6#(I3, I4, I5) 3) f7#(I6, I7, I8) -> f5#(I6, I7, I8) 4) f7#(I9, I10, I11) -> f4#(I9, I10, I11) 5) f7#(I12, I13, I14) -> f1#(I12, I13, I14) 6) f3#(I18, I19, I20) -> f6#(I20, I19, I20) 7) f6#(I21, I22, I23) -> f5#(I23, I22, I23) [I23 <= 0] 8) f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26] 9) f4#(I30, I31, I32) -> f5#(I32, I31, I32) [1 <= I32 /\ I32 <= 1] 10) f4#(I33, I34, I35) -> f1#(I35, I34, I35) [1 + I35 <= 1] 11) f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38] 12) f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) 13) f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) We have the following SCCs. { 6, 8, 11, 12, 13 } DP problem for innermost termination. P = f3#(I18, I19, I20) -> f6#(I20, I19, I20) f6#(I24, I25, I26) -> f4#(I26, I25, I26) [1 <= I26] f4#(I36, I37, I38) -> f1#(I38, I37, I38) [2 <= I38] f1#(I39, I40, I41) -> f3#(I41, I40, -1 + I41) f1#(I42, I43, I44) -> f3#(I44, I43, -2 + I44) R = f8(x1, x2, x3) -> f7(x1, x2, x3) f7(I0, I1, I2) -> f3(I0, I1, I2) f7(I3, I4, I5) -> f6(I3, I4, I5) f7(I6, I7, I8) -> f5(I6, I7, I8) f7(I9, I10, I11) -> f4(I9, I10, I11) f7(I12, I13, I14) -> f1(I12, I13, I14) f7(I15, I16, I17) -> f2(I15, I16, I17) f3(I18, I19, I20) -> f6(I20, I19, I20) f6(I21, I22, I23) -> f5(I23, I22, I23) [I23 <= 0] f6(I24, I25, I26) -> f4(I26, I25, I26) [1 <= I26] f5(I27, I28, I29) -> f2(I29, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] f4(I30, I31, I32) -> f5(I32, I31, I32) [1 <= I32 /\ I32 <= 1] f4(I33, I34, I35) -> f1(I35, I34, I35) [1 + I35 <= 1] f4(I36, I37, I38) -> f1(I38, I37, I38) [2 <= I38] f1(I39, I40, I41) -> f3(I41, I40, -1 + I41)
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