details

property	value
status	complete
benchmark	simple_control_on_input.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.49195 seconds
cpu usage	0.501432
user time	0.258608
system time	0.242824
max virtual memory	113188.0
max residence set size	8004.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f7#(x1) -> f6#(x1) f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1] f4#(I1) -> f3#(I1) f5#(I2) -> f4#(I2) [1 <= I2] f5#(I3) -> f1#(I3) [I3 <= 0] f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] f3#(I5) -> f1#(I5) [20 <= I5] R = f7(x1) -> f6(x1) f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] f4(I1) -> f3(I1) f5(I2) -> f4(I2) [1 <= I2] f5(I3) -> f1(I3) [I3 <= 0] f3(I4) -> f4(1 + I4) [1 + I4 <= 20] f3(I5) -> f1(I5) [20 <= I5] f1(I6) -> f2(I6) The dependency graph for this problem is: 0 -> 1 1 -> 3, 4 2 -> 5, 6 3 -> 2 4 -> 5 -> 2 6 -> Where: 0) f7#(x1) -> f6#(x1) 1) f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1] 2) f4#(I1) -> f3#(I1) 3) f5#(I2) -> f4#(I2) [1 <= I2] 4) f5#(I3) -> f1#(I3) [I3 <= 0] 5) f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] 6) f3#(I5) -> f1#(I5) [20 <= I5] We have the following SCCs. { 2, 5 } DP problem for innermost termination. P = f4#(I1) -> f3#(I1) f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] R = f7(x1) -> f6(x1) f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] f4(I1) -> f3(I1) f5(I2) -> f4(I2) [1 <= I2] f5(I3) -> f1(I3) [I3 <= 0] f3(I4) -> f4(1 + I4) [1 + I4 <= 20] f3(I5) -> f1(I5) [20 <= I5] f1(I6) -> f2(I6) We use the reverse value criterion with the projection function NU: NU[f3#(z1)] = 20 + -1 * (1 + z1) NU[f4#(z1)] = 20 + -1 * (1 + z1) This gives the following inequalities: ==> 20 + -1 * (1 + I1) >= 20 + -1 * (1 + I1) 1 + I4 <= 20 ==> 20 + -1 * (1 + I4) > 20 + -1 * (1 + (1 + I4)) with 20 + -1 * (1 + I4) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I1) -> f3#(I1) R = f7(x1) -> f6(x1) f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] f4(I1) -> f3(I1) f5(I2) -> f4(I2) [1 <= I2] f5(I3) -> f1(I3) [I3 <= 0] f3(I4) -> f4(1 + I4) [1 + I4 <= 20] f3(I5) -> f1(I5) [20 <= I5] f1(I6) -> f2(I6) The dependency graph for this problem is: 2 -> Where: 2) f4#(I1) -> f3#(I1) We have the following SCCs. popout

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