details

property	value
status	complete
benchmark	CountUpRound.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n130.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.25529 seconds
cpu usage	1.94092
user time	1.03811
system time	0.902812
max virtual memory	684240.0
max residence set size	8696.0

stage attributes

key	value
starexec-result	YES

output

1.89/2.25 YES 1.89/2.25 1.89/2.25 DP problem for innermost termination. 1.89/2.25 P = 1.89/2.25 init#(x1, x2) -> f1#(rnd1, rnd2) 1.89/2.25 f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 f3#(I5, I6) -> f2#(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7] 1.89/2.25 f2#(I8, I9) -> f3#(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1] 1.89/2.25 f1#(I11, I12) -> f2#(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1] 1.89/2.25 R = 1.89/2.25 init(x1, x2) -> f1(rnd1, rnd2) 1.89/2.25 f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7] 1.89/2.25 f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1] 1.89/2.25 f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1] 1.89/2.25 1.89/2.25 The dependency graph for this problem is: 1.89/2.25 0 -> 5 1.89/2.25 1 -> 2 1.89/2.25 2 -> 1 1.89/2.25 3 -> 2 1.89/2.25 4 -> 3 1.89/2.25 5 -> 2, 4 1.89/2.25 Where: 1.89/2.25 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1.89/2.25 1) f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 2) f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 3) f3#(I5, I6) -> f2#(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7] 1.89/2.25 4) f2#(I8, I9) -> f3#(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1] 1.89/2.25 5) f1#(I11, I12) -> f2#(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1] 1.89/2.25 1.89/2.25 We have the following SCCs. 1.89/2.25 { 1, 2 } 1.89/2.25 1.89/2.25 DP problem for innermost termination. 1.89/2.25 P = 1.89/2.25 f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 R = 1.89/2.25 init(x1, x2) -> f1(rnd1, rnd2) 1.89/2.25 f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7] 1.89/2.25 f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1] 1.89/2.25 f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1] 1.89/2.25 1.89/2.25 We use the reverse value criterion with the projection function NU: 1.89/2.25 NU[f2#(z1,z2)] = z1 - 1 + -1 * z2 1.89/2.25 NU[f3#(z1,z2)] = z1 - 1 + -1 * (z2 + 2) 1.89/2.25 1.89/2.25 This gives the following inequalities: 1.89/2.25 -1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1 ==> I0 - 1 + -1 * (I1 + 2) >= I0 - 1 + -1 * (I1 + 2) 1.89/2.25 -1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1 ==> I2 - 1 + -1 * I3 > I2 - 1 + -1 * (I3 + 2) with I2 - 1 + -1 * I3 >= 0 1.89/2.25 1.89/2.25 We remove all the strictly oriented dependency pairs. 1.89/2.25 1.89/2.25 DP problem for innermost termination. 1.89/2.25 P = 1.89/2.25 f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 R = 1.89/2.25 init(x1, x2) -> f1(rnd1, rnd2) 1.89/2.25 f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1] 1.89/2.25 f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7] 1.89/2.25 f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1] 1.89/2.25 f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1] 1.89/2.25 1.89/2.25 The dependency graph for this problem is: 1.89/2.25 1 -> 1.89/2.25 Where: 1.89/2.25 1) f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1] 1.89/2.25 1.89/2.25 We have the following SCCs. 1.89/2.25 1.89/5.23 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54