details

property	value
status	complete
benchmark	dsa_test10.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	18.5970020294 seconds
cpu usage	19.897777183
max memory	1.9894272E7

stage attributes

key	value
output-size	6740
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 = f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) f6#(I0, I1, I2, I3, I4, I5) -> f1#(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] f2#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) The dependency graph for this problem is: 0 -> 1 1 -> 6 2 -> 6 3 -> 4 4 -> 5 5 -> 4 6 -> 2, 3 Where: 0) f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 1) f6#(I0, I1, I2, I3, I4, I5) -> f1#(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] 2) f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] 3) f2#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) 5) f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] 6) f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) We have the following SCCs. { 2, 6 } { 4, 5 } DP problem for innermost termination. P = f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) f3#(I24, I25, I26, I27, I28, I29) -> f5#(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3) NU[f5#(z1,z2,z3,z4,z5,z6)] = z1 + -1 * (1 + z3) This gives the following inequalities: ==> I18 + -1 * (1 + I20) >= I18 + -1 * (1 + I20) 1 + I26 <= I24 ==> I24 + -1 * (1 + I26) > I24 + -1 * (1 + (1 + I26)) with I24 + -1 * (1 + I26) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] f2(I6, I7, I8, I9, I10, I11) -> f1(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] f2(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, 0, I15, I16, rnd6) [rnd6 = rnd6 /\ I12 <= I13] f5(I18, I19, I20, I21, I22, I23) -> f3(I18, I19, I20, I21, I22, I23) f3(I24, I25, I26, I27, I28, I29) -> f5(I24, I25, 1 + I26, I27, I28, I29) [1 + I26 <= I24] f3(I30, I31, I32, I33, I34, I35) -> f4(I30, I31, I32, I33, I34, I35) [I30 <= I32] f1(I36, I37, I38, I39, I40, I41) -> f2(I36, I37, I38, I39, I40, I41) The dependency graph for this problem is: 4 -> Where: 4) f5#(I18, I19, I20, I21, I22, I23) -> f3#(I18, I19, I20, I21, I22, I23) We have the following SCCs. DP problem for innermost termination. P = f2#(I6, I7, I8, I9, I10, I11) -> f1#(I6, 1 + I7, I8, I9, I10, I11) [1 + I7 <= I6] f1#(I36, I37, I38, I39, I40, I41) -> f2#(I36, I37, I38, I39, I40, I41) R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f1(I0, 0, I2, rnd4, rnd5, I5) [rnd5 = rnd5 /\ rnd4 = rnd4] popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer Transition Systems