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Integer_Transition_Systems 2019-03-29 01.54 pair #432274329
details
property
value
status
complete
benchmark
232.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n181.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
2.93334 seconds
cpu usage
2.97486
user time
1.5087
system time
1.46617
max virtual memory
723228.0
max residence set size
8576.0
stage attributes
key
value
starexec-result
YES
output
2.92/2.93 YES 2.92/2.93 2.92/2.93 DP problem for innermost termination. 2.92/2.93 P = 2.92/2.93 f5#(x1, x2) -> f4#(x1, x2) 2.92/2.93 f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] 2.92/2.93 f4#(I2, I3) -> f2#(I2, I3) 2.92/2.93 f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0] 2.92/2.93 R = 2.92/2.93 f5(x1, x2) -> f4(x1, x2) 2.92/2.93 f2(I0, I1) -> f1(I0, I0) [1 <= I0] 2.92/2.93 f4(I2, I3) -> f2(I2, I3) 2.92/2.93 f3(I4, I5) -> f1(I4, I5) 2.92/2.93 f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0] 2.92/2.93 2.92/2.93 The dependency graph for this problem is: 2.92/2.93 0 -> 2 2.92/2.93 1 -> 4 2.92/2.93 2 -> 1 2.92/2.93 3 -> 4, 5 2.92/2.93 4 -> 3 2.92/2.93 5 -> 1 2.92/2.93 Where: 2.92/2.93 0) f5#(x1, x2) -> f4#(x1, x2) 2.92/2.93 1) f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] 2.92/2.93 2) f4#(I2, I3) -> f2#(I2, I3) 2.92/2.93 3) f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0] 2.92/2.93 2.92/2.93 We have the following SCCs. 2.92/2.93 { 1, 3, 4, 5 } 2.92/2.93 2.92/2.93 DP problem for innermost termination. 2.92/2.93 P = 2.92/2.93 f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] 2.92/2.93 f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0] 2.92/2.93 R = 2.92/2.93 f5(x1, x2) -> f4(x1, x2) 2.92/2.93 f2(I0, I1) -> f1(I0, I0) [1 <= I0] 2.92/2.93 f4(I2, I3) -> f2(I2, I3) 2.92/2.93 f3(I4, I5) -> f1(I4, I5) 2.92/2.93 f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0] 2.92/2.93 2.92/2.93 We use the extended value criterion with the projection function NU: 2.92/2.93 NU[f3#(x0,x1)] = x0 - 2 2.92/2.93 NU[f1#(x0,x1)] = x0 - 2 2.92/2.93 NU[f2#(x0,x1)] = x0 - 1 2.92/2.93 2.92/2.93 This gives the following inequalities: 2.92/2.93 1 <= I0 ==> I0 - 1 > I0 - 2 with I0 - 1 >= 0 2.92/2.93 ==> I4 - 2 >= I4 - 2 2.92/2.93 1 <= I7 ==> I6 - 2 >= I6 - 2 2.92/2.93 I9 <= 0 ==> I8 - 2 >= (-1 + I8) - 1 2.92/2.93 2.92/2.93 We remove all the strictly oriented dependency pairs. 2.92/2.93 2.92/2.93 DP problem for innermost termination. 2.92/2.93 P = 2.92/2.93 f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0] 2.92/2.93 R = 2.92/2.93 f5(x1, x2) -> f4(x1, x2) 2.92/2.93 f2(I0, I1) -> f1(I0, I0) [1 <= I0] 2.92/2.93 f4(I2, I3) -> f2(I2, I3) 2.92/2.93 f3(I4, I5) -> f1(I4, I5) 2.92/2.93 f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0] 2.92/2.93 2.92/2.93 The dependency graph for this problem is: 2.92/2.93 3 -> 4, 5 2.92/2.93 4 -> 3 2.92/2.93 5 -> 2.92/2.93 Where: 2.92/2.93 3) f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 5) f1#(I8, I9) -> f2#(-1 + I8, I9) [I9 <= 0] 2.92/2.93 2.92/2.93 We have the following SCCs. 2.92/2.93 { 3, 4 } 2.92/2.93 2.92/2.93 DP problem for innermost termination. 2.92/2.93 P = 2.92/2.93 f3#(I4, I5) -> f1#(I4, I5) 2.92/2.93 f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 2.92/2.93 R = 2.92/2.93 f5(x1, x2) -> f4(x1, x2) 2.92/2.93 f2(I0, I1) -> f1(I0, I0) [1 <= I0] 2.92/2.93 f4(I2, I3) -> f2(I2, I3) 2.92/2.93 f3(I4, I5) -> f1(I4, I5) 2.92/2.93 f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] 2.92/2.93 f1(I8, I9) -> f2(-1 + I8, I9) [I9 <= 0] 2.92/2.93
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