details

property	value
status	complete
benchmark	flip2_rec.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n018.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.983525 seconds
cpu usage	1.00109
user time	0.523842
system time	0.477245
max virtual memory	126024.0
max residence set size	8616.0

stage attributes

key	value
starexec-result	MAYBE

output

0.92/0.98 MAYBE 0.92/0.98 0.92/0.98 DP problem for innermost termination. 0.92/0.98 P = 0.92/0.98 init#(x1, x2) -> f1#(rnd1, rnd2) 0.92/0.98 f2#(I0, I1) -> f2#(I1, I0) [1 <= I0 - 1 /\ I1 <= I0 - 1 /\ 0 <= I1 - 1] 0.92/0.98 f2#(I2, I3) -> f2#(I3, I3) [1 <= I3 - 1 /\ 0 <= I2 - 1 /\ I2 <= I3 - 1] 0.92/0.98 f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ I4 - 1 <= I4 - 1 /\ 0 <= I4 - 1] 0.92/0.98 f1#(I6, I7) -> f2#(I7 + 5, I7) [0 <= I6 - 1 /\ -1 <= I7 - 1 /\ I7 <= I7 + 5 - 1] 0.92/0.98 R = 0.92/0.98 init(x1, x2) -> f1(rnd1, rnd2) 0.92/0.98 f2(I0, I1) -> f2(I1, I0) [1 <= I0 - 1 /\ I1 <= I0 - 1 /\ 0 <= I1 - 1] 0.92/0.98 f2(I2, I3) -> f2(I3, I3) [1 <= I3 - 1 /\ 0 <= I2 - 1 /\ I2 <= I3 - 1] 0.92/0.98 f2(I4, I5) -> f2(I4, I4 - 1) [I4 = I5 /\ I4 - 1 <= I4 - 1 /\ 0 <= I4 - 1] 0.92/0.98 f1(I6, I7) -> f2(I7 + 5, I7) [0 <= I6 - 1 /\ -1 <= I7 - 1 /\ I7 <= I7 + 5 - 1] 0.92/0.98 0.92/0.98 The dependency graph for this problem is: 0.92/0.98 0 -> 4 0.92/0.98 1 -> 2 0.92/0.98 2 -> 3 0.92/0.98 3 -> 1 0.92/0.98 4 -> 1 0.92/0.98 Where: 0.92/0.98 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.92/0.98 1) f2#(I0, I1) -> f2#(I1, I0) [1 <= I0 - 1 /\ I1 <= I0 - 1 /\ 0 <= I1 - 1] 0.92/0.98 2) f2#(I2, I3) -> f2#(I3, I3) [1 <= I3 - 1 /\ 0 <= I2 - 1 /\ I2 <= I3 - 1] 0.92/0.98 3) f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ I4 - 1 <= I4 - 1 /\ 0 <= I4 - 1] 0.92/0.98 4) f1#(I6, I7) -> f2#(I7 + 5, I7) [0 <= I6 - 1 /\ -1 <= I7 - 1 /\ I7 <= I7 + 5 - 1] 0.92/0.98 0.92/0.98 We have the following SCCs. 0.92/0.98 { 1, 2, 3 } 0.92/0.98 0.92/0.98 DP problem for innermost termination. 0.92/0.98 P = 0.92/0.98 f2#(I0, I1) -> f2#(I1, I0) [1 <= I0 - 1 /\ I1 <= I0 - 1 /\ 0 <= I1 - 1] 0.92/0.98 f2#(I2, I3) -> f2#(I3, I3) [1 <= I3 - 1 /\ 0 <= I2 - 1 /\ I2 <= I3 - 1] 0.92/0.98 f2#(I4, I5) -> f2#(I4, I4 - 1) [I4 = I5 /\ I4 - 1 <= I4 - 1 /\ 0 <= I4 - 1] 0.92/0.98 R = 0.92/0.98 init(x1, x2) -> f1(rnd1, rnd2) 0.92/0.98 f2(I0, I1) -> f2(I1, I0) [1 <= I0 - 1 /\ I1 <= I0 - 1 /\ 0 <= I1 - 1] 0.92/0.98 f2(I2, I3) -> f2(I3, I3) [1 <= I3 - 1 /\ 0 <= I2 - 1 /\ I2 <= I3 - 1] 0.92/0.98 f2(I4, I5) -> f2(I4, I4 - 1) [I4 = I5 /\ I4 - 1 <= I4 - 1 /\ 0 <= I4 - 1] 0.92/0.98 f1(I6, I7) -> f2(I7 + 5, I7) [0 <= I6 - 1 /\ -1 <= I7 - 1 /\ I7 <= I7 + 5 - 1] 0.92/0.98 0.92/3.96 EOF popout

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

job log

popout

actions

all output return to Integer_Transition_Systems 2019-03-29 01.54