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ITS pair #487098835
details
property
value
status
complete
benchmark
p-3.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)
16.2416 seconds
cpu usage
16.3777
user time
8.65691
system time
7.72077
max virtual memory
744660.0
max residence set size
10952.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) The dependency graph for this problem is: 0 -> 7 1 -> 2, 4, 6 2 -> 1 3 -> 2, 4, 6 4 -> 3 5 -> 6 -> 5 7 -> 2, 4, 6 Where: 0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 2) f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 4) f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] 6) f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 7) f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) We use the reverse value criterion with the projection function NU: NU[f6#(z1,z2,z3,z4)] = z4 + -1 * z3 NU[f2#(z1,z2,z3,z4)] = z4 + -1 * z3 NU[f7#(z1,z2,z3,z4)] = z4 + -1 * z3 This gives the following inequalities: ==> I3 + -1 * I2 >= I3 + -1 * I2 0 <= -1 - I6 + I7 ==> I7 + -1 * I6 > I7 + -1 * (1 + I6) with I7 + -1 * I6 >= 0 ==> I11 + -1 * I10 >= I11 + -1 * I10 I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0 ==> I15 + -1 * I14 > I15 + -1 * (1 + I14) with I15 + -1 * I14 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) R = f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) The dependency graph for this problem is: 1 -> 3 -> Where: 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) We have the following SCCs.
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