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Integ Trans Syste 27634 pair #381738517
details
property
value
status
complete
benchmark
fibcall.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
10.6439499855 seconds
cpu usage
11.336278743
max memory
2.8786688E7
stage attributes
key
value
output-size
3869
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 = f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) 2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 3) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5 NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5 This gives the following inequalities: ==> I14 + -1 * I13 >= I14 + -1 * I13 rnd2 = I18 /\ I22 <= I23 ==> I23 + -1 * I22 > I23 + -1 * (1 + I22) with I23 + -1 * I22 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] The dependency graph for this problem is: 2 -> Where: 2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) We have the following SCCs.
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