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Integer_Transition_Systems 2019-03-29 01.54 pair #432274809
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
n160.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
10.3948 seconds
cpu usage
10.5713
user time
5.1452
system time
5.4261
max virtual memory
704560.0
max residence set size
10960.0
stage attributes
key
value
starexec-result
YES
output
10.50/10.39 YES 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 The dependency graph for this problem is: 10.50/10.39 0 -> 7 10.50/10.39 1 -> 2, 4, 6 10.50/10.39 2 -> 1 10.50/10.39 3 -> 2, 4, 6 10.50/10.39 4 -> 3 10.50/10.39 5 -> 10.50/10.39 6 -> 5 10.50/10.39 7 -> 2, 4, 6 10.50/10.39 Where: 10.50/10.39 0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 10.50/10.39 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 2) f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 4) f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 6) f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 7) f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 We have the following SCCs. 10.50/10.39 { 1, 2, 3, 4 } 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 We use the reverse value criterion with the projection function NU: 10.50/10.39 NU[f6#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 NU[f2#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 NU[f7#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 10.50/10.39 This gives the following inequalities: 10.50/10.39 ==> I3 + -1 * I2 >= I3 + -1 * I2 10.50/10.39 0 <= -1 - I6 + I7 ==> I7 + -1 * I6 > I7 + -1 * (1 + I6) with I7 + -1 * I6 >= 0 10.50/10.39 ==> I11 + -1 * I10 >= I11 + -1 * I10 10.50/10.39 I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0 ==> I15 + -1 * I14 > I15 + -1 * (1 + I14) with I15 + -1 * I14 >= 0 10.50/10.39 10.50/10.39 We remove all the strictly oriented dependency pairs. 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 The dependency graph for this problem is: 10.50/10.39 1 -> 10.50/10.39 3 -> 10.50/10.39 Where: 10.50/10.39 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 10.50/10.39 We have the following SCCs. 10.50/10.39 10.50/13.36 EOF
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