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Integer_Transition_Systems 2019-03-29 01.54 pair #432275529
details
property
value
status
complete
benchmark
heidy10.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.46649 seconds
cpu usage
1.00696
user time
0.494916
system time
0.512039
max virtual memory
233428.0
max residence set size
8384.0
stage attributes
key
value
starexec-result
YES
output
0.98/1.46 YES 0.98/1.46 0.98/1.46 DP problem for innermost termination. 0.98/1.46 P = 0.98/1.46 f5#(x1, x2) -> f4#(x1, x2) 0.98/1.46 f4#(I0, I1) -> f3#(I0, I1) 0.98/1.46 f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 0.98/1.46 f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 R = 0.98/1.46 f5(x1, x2) -> f4(x1, x2) 0.98/1.46 f4(I0, I1) -> f3(I0, I1) 0.98/1.46 f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1(I4, I5) -> f3(I4, I5) [I5 <= 0] 0.98/1.46 f2(I6, I7) -> f1(I6, I7) 0.98/1.46 f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] 0.98/1.46 0.98/1.46 The dependency graph for this problem is: 0.98/1.46 0 -> 1 0.98/1.46 1 -> 2 0.98/1.46 2 -> 3, 5 0.98/1.46 3 -> 2 0.98/1.46 4 -> 3, 5 0.98/1.46 5 -> 4 0.98/1.46 Where: 0.98/1.46 0) f5#(x1, x2) -> f4#(x1, x2) 0.98/1.46 1) f4#(I0, I1) -> f3#(I0, I1) 0.98/1.46 2) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 0.98/1.46 4) f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 0.98/1.46 We have the following SCCs. 0.98/1.46 { 2, 3, 4, 5 } 0.98/1.46 0.98/1.46 DP problem for innermost termination. 0.98/1.46 P = 0.98/1.46 f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 0.98/1.46 f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 R = 0.98/1.46 f5(x1, x2) -> f4(x1, x2) 0.98/1.46 f4(I0, I1) -> f3(I0, I1) 0.98/1.46 f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1(I4, I5) -> f3(I4, I5) [I5 <= 0] 0.98/1.46 f2(I6, I7) -> f1(I6, I7) 0.98/1.46 f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] 0.98/1.46 0.98/1.46 We use the basic value criterion with the projection function NU: 0.98/1.46 NU[f2#(z1,z2)] = z1 0.98/1.46 NU[f1#(z1,z2)] = z1 0.98/1.46 NU[f3#(z1,z2)] = z1 0.98/1.46 0.98/1.46 This gives the following inequalities: 0.98/1.46 rnd2 = rnd2 /\ 1 <= I2 ==> I2 >! -1 + I2 0.98/1.46 I5 <= 0 ==> I4 (>! \union =) I4 0.98/1.46 ==> I6 (>! \union =) I6 0.98/1.46 1 <= I9 ==> I8 (>! \union =) I8 0.98/1.46 0.98/1.46 We remove all the strictly oriented dependency pairs. 0.98/1.46 0.98/1.46 DP problem for innermost termination. 0.98/1.46 P = 0.98/1.46 f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 0.98/1.46 f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 R = 0.98/1.46 f5(x1, x2) -> f4(x1, x2) 0.98/1.46 f4(I0, I1) -> f3(I0, I1) 0.98/1.46 f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1(I4, I5) -> f3(I4, I5) [I5 <= 0] 0.98/1.46 f2(I6, I7) -> f1(I6, I7) 0.98/1.46 f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] 0.98/1.46 0.98/1.46 The dependency graph for this problem is: 0.98/1.46 3 -> 0.98/1.46 4 -> 3, 5 0.98/1.46 5 -> 4 0.98/1.46 Where: 0.98/1.46 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 0.98/1.46 4) f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 0.98/1.46 We have the following SCCs. 0.98/1.46 { 4, 5 } 0.98/1.46 0.98/1.46 DP problem for innermost termination. 0.98/1.46 P = 0.98/1.46 f2#(I6, I7) -> f1#(I6, I7) 0.98/1.46 f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] 0.98/1.46 R = 0.98/1.46 f5(x1, x2) -> f4(x1, x2) 0.98/1.46 f4(I0, I1) -> f3(I0, I1) 0.98/1.46 f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 0.98/1.46 f1(I4, I5) -> f3(I4, I5) [I5 <= 0] 0.98/1.46 f2(I6, I7) -> f1(I6, I7) 0.98/1.46 f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] 0.98/1.46
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