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Integ Trans Syste 27634 pair #381739075
details
property
value
status
complete
benchmark
array_init_assign.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n106.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.61338591576 seconds
cpu usage
1.701361361
max memory
1.6007168E7
stage attributes
key
value
output-size
4604
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 = f9#(x1, x2) -> f8#(x1, x2) f8#(I0, I1) -> f3#(0, I1) f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2] f4#(I4, I5) -> f7#(I4, I5) [2 <= I5] f7#(I6, I7) -> f5#(I6, I7) f7#(I8, I9) -> f5#(I8, I9) f2#(I12, I13) -> f4#(I12, I13) f3#(I14, I15) -> f1#(I14, I15) f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2] f1#(I18, I19) -> f2#(I18, 0) [2 <= I18] R = f9(x1, x2) -> f8(x1, x2) f8(I0, I1) -> f3(0, I1) f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2] f4(I4, I5) -> f7(I4, I5) [2 <= I5] f7(I6, I7) -> f5(I6, I7) f7(I8, I9) -> f5(I8, I9) f5(I10, I11) -> f6(I10, I11) f2(I12, I13) -> f4(I12, I13) f3(I14, I15) -> f1(I14, I15) f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2] f1(I18, I19) -> f2(I18, 0) [2 <= I18] The dependency graph for this problem is: 0 -> 1 1 -> 7 2 -> 6 3 -> 4, 5 4 -> 5 -> 6 -> 2, 3 7 -> 8, 9 8 -> 7 9 -> 6 Where: 0) f9#(x1, x2) -> f8#(x1, x2) 1) f8#(I0, I1) -> f3#(0, I1) 2) f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2] 3) f4#(I4, I5) -> f7#(I4, I5) [2 <= I5] 4) f7#(I6, I7) -> f5#(I6, I7) 5) f7#(I8, I9) -> f5#(I8, I9) 6) f2#(I12, I13) -> f4#(I12, I13) 7) f3#(I14, I15) -> f1#(I14, I15) 8) f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2] 9) f1#(I18, I19) -> f2#(I18, 0) [2 <= I18] We have the following SCCs. { 7, 8 } { 2, 6 } DP problem for innermost termination. P = f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2] f2#(I12, I13) -> f4#(I12, I13) R = f9(x1, x2) -> f8(x1, x2) f8(I0, I1) -> f3(0, I1) f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2] f4(I4, I5) -> f7(I4, I5) [2 <= I5] f7(I6, I7) -> f5(I6, I7) f7(I8, I9) -> f5(I8, I9) f5(I10, I11) -> f6(I10, I11) f2(I12, I13) -> f4(I12, I13) f3(I14, I15) -> f1(I14, I15) f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2] f1(I18, I19) -> f2(I18, 0) [2 <= I18] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = 2 + -1 * (1 + z2) NU[f4#(z1,z2)] = 2 + -1 * (1 + z2) This gives the following inequalities: 1 + I3 <= 2 ==> 2 + -1 * (1 + I3) > 2 + -1 * (1 + (1 + I3)) with 2 + -1 * (1 + I3) >= 0 ==> 2 + -1 * (1 + I13) >= 2 + -1 * (1 + I13) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I12, I13) -> f4#(I12, I13) R = f9(x1, x2) -> f8(x1, x2) f8(I0, I1) -> f3(0, I1) f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2] f4(I4, I5) -> f7(I4, I5) [2 <= I5] f7(I6, I7) -> f5(I6, I7) f7(I8, I9) -> f5(I8, I9) f5(I10, I11) -> f6(I10, I11) f2(I12, I13) -> f4(I12, I13) f3(I14, I15) -> f1(I14, I15)
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