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Integ Trans Syste 27634 pair #381739858
details
property
value
status
complete
benchmark
array2.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.73545384407 seconds
cpu usage
0.776929348
max memory
1.2197888E7
stage attributes
key
value
output-size
2051
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 = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f3#(0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f3#(I2, I3) -> f1#(I2, I3) f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f3(I2, I3) -> f1(I2, I3) f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50] f1(I6, I7) -> f2(I6, I7) [50 <= I6] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f3#(0, rnd2) [y1 = 0 /\ rnd2 = rnd2] 2) f3#(I2, I3) -> f1#(I2, I3) 3) f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I2, I3) -> f1#(I2, I3) f1#(I4, I5) -> f3#(1 + I4, I5) [1 + I4 <= 50] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f3(I2, I3) -> f1(I2, I3) f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50] f1(I6, I7) -> f2(I6, I7) [50 <= I6] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2)] = 50 + -1 * (1 + z1) NU[f3#(z1,z2)] = 50 + -1 * (1 + z1) This gives the following inequalities: ==> 50 + -1 * (1 + I2) >= 50 + -1 * (1 + I2) 1 + I4 <= 50 ==> 50 + -1 * (1 + I4) > 50 + -1 * (1 + (1 + I4)) with 50 + -1 * (1 + I4) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I2, I3) -> f1#(I2, I3) R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(0, rnd2) [y1 = 0 /\ rnd2 = rnd2] f3(I2, I3) -> f1(I2, I3) f1(I4, I5) -> f3(1 + I4, I5) [1 + I4 <= 50] f1(I6, I7) -> f2(I6, I7) [50 <= I6] The dependency graph for this problem is: 2 -> Where: 2) f3#(I2, I3) -> f1#(I2, I3) We have the following SCCs.
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