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Integer_Transition_Systems 2019-03-29 01.54 pair #432273147
details
property
value
status
complete
benchmark
upAndDown_rec.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n042.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.69361 seconds
cpu usage
1.7102
user time
0.852633
system time
0.857569
max virtual memory
684948.0
max residence set size
8316.0
stage attributes
key
value
starexec-result
MAYBE
output
1.59/1.69 MAYBE 1.59/1.69 1.59/1.69 DP problem for innermost termination. 1.59/1.69 P = 1.59/1.69 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1.59/1.69 f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1] 1.59/1.69 f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10] 1.59/1.69 f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1] 1.59/1.69 f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9] 1.59/1.69 f1#(I12, I13, I14) -> f2#(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 1.59/1.69 R = 1.59/1.69 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 1.59/1.69 f2(I0, I1, I2) -> f2(1, 1, 1) [0 = I2 /\ I0 <= 1] 1.59/1.69 f2(I3, I4, I5) -> f2(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10] 1.59/1.69 f2(I6, I7, I8) -> f2(0, 0, 9) [10 = I8 /\ I6 <= 1] 1.59/1.69 f2(I9, I10, I11) -> f2(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9] 1.59/1.69 f1(I12, I13, I14) -> f2(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 1.59/1.69 1.59/1.69 The dependency graph for this problem is: 1.59/1.69 0 -> 5 1.59/1.69 1 -> 2 1.59/1.69 2 -> 2, 3 1.59/1.69 3 -> 4 1.59/1.69 4 -> 1, 4 1.59/1.69 5 -> 1, 3, 4 1.59/1.69 Where: 1.59/1.69 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1.59/1.69 1) f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1] 1.59/1.69 2) f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10] 1.59/1.69 3) f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1] 1.59/1.69 4) f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9] 1.59/1.69 5) f1#(I12, I13, I14) -> f2#(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 1.59/1.69 1.59/1.69 We have the following SCCs. 1.59/1.69 { 1, 2, 3, 4 } 1.59/1.69 1.59/1.69 DP problem for innermost termination. 1.59/1.69 P = 1.59/1.69 f2#(I0, I1, I2) -> f2#(1, 1, 1) [0 = I2 /\ I0 <= 1] 1.59/1.69 f2#(I3, I4, I5) -> f2#(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10] 1.59/1.69 f2#(I6, I7, I8) -> f2#(0, 0, 9) [10 = I8 /\ I6 <= 1] 1.59/1.69 f2#(I9, I10, I11) -> f2#(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9] 1.59/1.69 R = 1.59/1.69 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 1.59/1.69 f2(I0, I1, I2) -> f2(1, 1, 1) [0 = I2 /\ I0 <= 1] 1.59/1.69 f2(I3, I4, I5) -> f2(1, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10] 1.59/1.69 f2(I6, I7, I8) -> f2(0, 0, 9) [10 = I8 /\ I6 <= 1] 1.59/1.69 f2(I9, I10, I11) -> f2(0, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9] 1.59/1.69 f1(I12, I13, I14) -> f2(0, 0, I13) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 1.59/1.69 1.59/4.67 EOF
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