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Integer_Transition_Systems 2019-03-29 01.54 pair #432275643
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
n078.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.490693 seconds
cpu usage
0.499871
user time
0.243708
system time
0.256163
max virtual memory
113176.0
max residence set size
8004.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.49 YES 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 f7#(x1) -> f6#(x1) 0.00/0.49 f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1] 0.00/0.49 f4#(I1) -> f3#(I1) 0.00/0.49 f5#(I2) -> f4#(I2) [1 <= I2] 0.00/0.49 f5#(I3) -> f1#(I3) [I3 <= 0] 0.00/0.49 f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] 0.00/0.49 f3#(I5) -> f1#(I5) [20 <= I5] 0.00/0.49 R = 0.00/0.49 f7(x1) -> f6(x1) 0.00/0.49 f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] 0.00/0.49 f4(I1) -> f3(I1) 0.00/0.49 f5(I2) -> f4(I2) [1 <= I2] 0.00/0.49 f5(I3) -> f1(I3) [I3 <= 0] 0.00/0.49 f3(I4) -> f4(1 + I4) [1 + I4 <= 20] 0.00/0.49 f3(I5) -> f1(I5) [20 <= I5] 0.00/0.49 f1(I6) -> f2(I6) 0.00/0.49 0.00/0.49 The dependency graph for this problem is: 0.00/0.49 0 -> 1 0.00/0.49 1 -> 3, 4 0.00/0.49 2 -> 5, 6 0.00/0.49 3 -> 2 0.00/0.49 4 -> 0.00/0.49 5 -> 2 0.00/0.49 6 -> 0.00/0.49 Where: 0.00/0.49 0) f7#(x1) -> f6#(x1) 0.00/0.49 1) f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1] 0.00/0.49 2) f4#(I1) -> f3#(I1) 0.00/0.49 3) f5#(I2) -> f4#(I2) [1 <= I2] 0.00/0.49 4) f5#(I3) -> f1#(I3) [I3 <= 0] 0.00/0.49 5) f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] 0.00/0.49 6) f3#(I5) -> f1#(I5) [20 <= I5] 0.00/0.49 0.00/0.49 We have the following SCCs. 0.00/0.49 { 2, 5 } 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 f4#(I1) -> f3#(I1) 0.00/0.49 f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20] 0.00/0.49 R = 0.00/0.49 f7(x1) -> f6(x1) 0.00/0.49 f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] 0.00/0.49 f4(I1) -> f3(I1) 0.00/0.49 f5(I2) -> f4(I2) [1 <= I2] 0.00/0.49 f5(I3) -> f1(I3) [I3 <= 0] 0.00/0.49 f3(I4) -> f4(1 + I4) [1 + I4 <= 20] 0.00/0.49 f3(I5) -> f1(I5) [20 <= I5] 0.00/0.49 f1(I6) -> f2(I6) 0.00/0.49 0.00/0.49 We use the reverse value criterion with the projection function NU: 0.00/0.49 NU[f3#(z1)] = 20 + -1 * (1 + z1) 0.00/0.49 NU[f4#(z1)] = 20 + -1 * (1 + z1) 0.00/0.49 0.00/0.49 This gives the following inequalities: 0.00/0.49 ==> 20 + -1 * (1 + I1) >= 20 + -1 * (1 + I1) 0.00/0.49 1 + I4 <= 20 ==> 20 + -1 * (1 + I4) > 20 + -1 * (1 + (1 + I4)) with 20 + -1 * (1 + I4) >= 0 0.00/0.49 0.00/0.49 We remove all the strictly oriented dependency pairs. 0.00/0.49 0.00/0.49 DP problem for innermost termination. 0.00/0.49 P = 0.00/0.49 f4#(I1) -> f3#(I1) 0.00/0.49 R = 0.00/0.49 f7(x1) -> f6(x1) 0.00/0.49 f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1] 0.00/0.49 f4(I1) -> f3(I1) 0.00/0.49 f5(I2) -> f4(I2) [1 <= I2] 0.00/0.49 f5(I3) -> f1(I3) [I3 <= 0] 0.00/0.49 f3(I4) -> f4(1 + I4) [1 + I4 <= 20] 0.00/0.49 f3(I5) -> f1(I5) [20 <= I5] 0.00/0.49 f1(I6) -> f2(I6) 0.00/0.49 0.00/0.49 The dependency graph for this problem is: 0.00/0.49 2 -> 0.00/0.49 Where: 0.00/0.49 2) f4#(I1) -> f3#(I1) 0.00/0.49 0.00/0.49 We have the following SCCs. 0.00/0.49 0.00/3.47 EOF
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