details

property	value
status	complete
benchmark	loop_on_input.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n090.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.15201 seconds
cpu usage	1.1753
user time	0.616715
system time	0.55858
max virtual memory	684948.0
max residence set size	8560.0

stage attributes

key	value
starexec-result	YES

output

1.08/1.15 YES 1.08/1.15 1.08/1.15 DP problem for innermost termination. 1.08/1.15 P = 1.08/1.15 f6#(x1) -> f5#(x1) 1.08/1.15 f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1] 1.08/1.15 f2#(I1) -> f3#(I1) 1.08/1.15 f3#(I2) -> f1#(I2) [1 + I2 <= 4] 1.08/1.15 f1#(I4) -> f2#(1 + I4) [1 <= I4] 1.08/1.15 f1#(I5) -> f2#(1) [I5 <= 0] 1.08/1.15 R = 1.08/1.15 f6(x1) -> f5(x1) 1.08/1.15 f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1] 1.08/1.15 f2(I1) -> f3(I1) 1.08/1.15 f3(I2) -> f1(I2) [1 + I2 <= 4] 1.08/1.15 f3(I3) -> f4(I3) [4 <= I3] 1.08/1.15 f1(I4) -> f2(1 + I4) [1 <= I4] 1.08/1.15 f1(I5) -> f2(1) [I5 <= 0] 1.08/1.15 1.08/1.15 The dependency graph for this problem is: 1.08/1.15 0 -> 1 1.08/1.15 1 -> 2 1.08/1.15 2 -> 3 1.08/1.15 3 -> 4, 5 1.08/1.15 4 -> 2 1.08/1.15 5 -> 2 1.08/1.15 Where: 1.08/1.15 0) f6#(x1) -> f5#(x1) 1.08/1.15 1) f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1] 1.08/1.15 2) f2#(I1) -> f3#(I1) 1.08/1.15 3) f3#(I2) -> f1#(I2) [1 + I2 <= 4] 1.08/1.15 4) f1#(I4) -> f2#(1 + I4) [1 <= I4] 1.08/1.15 5) f1#(I5) -> f2#(1) [I5 <= 0] 1.08/1.15 1.08/1.15 We have the following SCCs. 1.08/1.15 { 2, 3, 4, 5 } 1.08/1.15 1.08/1.15 DP problem for innermost termination. 1.08/1.15 P = 1.08/1.15 f2#(I1) -> f3#(I1) 1.08/1.15 f3#(I2) -> f1#(I2) [1 + I2 <= 4] 1.08/1.15 f1#(I4) -> f2#(1 + I4) [1 <= I4] 1.08/1.15 f1#(I5) -> f2#(1) [I5 <= 0] 1.08/1.15 R = 1.08/1.15 f6(x1) -> f5(x1) 1.08/1.15 f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1] 1.08/1.15 f2(I1) -> f3(I1) 1.08/1.15 f3(I2) -> f1(I2) [1 + I2 <= 4] 1.08/1.15 f3(I3) -> f4(I3) [4 <= I3] 1.08/1.15 f1(I4) -> f2(1 + I4) [1 <= I4] 1.08/1.15 f1(I5) -> f2(1) [I5 <= 0] 1.08/1.15 1.08/1.15 We use the extended value criterion with the projection function NU: 1.08/1.15 NU[f1#(x0)] = -x0 + 2 1.08/1.15 NU[f3#(x0)] = -x0 + 3 1.08/1.15 NU[f2#(x0)] = -x0 + 3 1.08/1.15 1.08/1.15 This gives the following inequalities: 1.08/1.15 ==> -I1 + 3 >= -I1 + 3 1.08/1.15 1 + I2 <= 4 ==> -I2 + 3 > -I2 + 2 with -I2 + 3 >= 0 1.08/1.15 1 <= I4 ==> -I4 + 2 >= -(1 + I4) + 3 1.08/1.15 I5 <= 0 ==> -I5 + 2 >= -1 + 3 1.08/1.15 1.08/1.15 We remove all the strictly oriented dependency pairs. 1.08/1.15 1.08/1.15 DP problem for innermost termination. 1.08/1.15 P = 1.08/1.15 f2#(I1) -> f3#(I1) 1.08/1.15 f1#(I4) -> f2#(1 + I4) [1 <= I4] 1.08/1.15 f1#(I5) -> f2#(1) [I5 <= 0] 1.08/1.15 R = 1.08/1.15 f6(x1) -> f5(x1) 1.08/1.15 f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1] 1.08/1.15 f2(I1) -> f3(I1) 1.08/1.15 f3(I2) -> f1(I2) [1 + I2 <= 4] 1.08/1.15 f3(I3) -> f4(I3) [4 <= I3] 1.08/1.15 f1(I4) -> f2(1 + I4) [1 <= I4] 1.08/1.15 f1(I5) -> f2(1) [I5 <= 0] 1.08/1.15 1.08/1.15 The dependency graph for this problem is: 1.08/1.15 2 -> 1.08/1.15 4 -> 2 1.08/1.15 5 -> 2 1.08/1.15 Where: 1.08/1.15 2) f2#(I1) -> f3#(I1) 1.08/1.15 4) f1#(I4) -> f2#(1 + I4) [1 <= I4] 1.08/1.15 5) f1#(I5) -> f2#(1) [I5 <= 0] 1.08/1.15 1.08/1.15 We have the following SCCs. 1.08/1.15 1.08/4.12 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54