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Integer_Transition_Systems 2019-03-29 01.54 pair #432274821
details
property
value
status
complete
benchmark
ex14.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.82522 seconds
cpu usage
1.84232
user time
0.893326
system time
0.948995
max virtual memory
130000.0
max residence set size
8668.0
stage attributes
key
value
starexec-result
YES
output
1.76/1.82 YES 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1.76/1.82 f4#(I0, I1, I2) -> f3#(I0, 1, I2) 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 The dependency graph for this problem is: 1.76/1.82 0 -> 1 1.76/1.82 1 -> 2 1.76/1.82 2 -> 3 1.76/1.82 3 -> 2 1.76/1.82 Where: 1.76/1.82 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1.76/1.82 1) f4#(I0, I1, I2) -> f3#(I0, 1, I2) 1.76/1.82 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 1.76/1.82 We have the following SCCs. 1.76/1.82 { 2, 3 } 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 We use the reverse value criterion with the projection function NU: 1.76/1.82 NU[f1#(z1,z2,z3)] = z1 + -1 * z2 1.76/1.82 NU[f3#(z1,z2,z3)] = z1 + -1 * z2 1.76/1.82 1.76/1.82 This gives the following inequalities: 1.76/1.82 ==> I3 + -1 * I4 >= I3 + -1 * I4 1.76/1.82 I7 <= I6 ==> I6 + -1 * I7 > I6 + -1 * (1 + I7) with I6 + -1 * I7 >= 0 1.76/1.82 1.76/1.82 We remove all the strictly oriented dependency pairs. 1.76/1.82 1.76/1.82 DP problem for innermost termination. 1.76/1.82 P = 1.76/1.82 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 R = 1.76/1.82 f5(x1, x2, x3) -> f4(x1, x2, x3) 1.76/1.82 f4(I0, I1, I2) -> f3(I0, 1, I2) 1.76/1.82 f3(I3, I4, I5) -> f1(I3, I4, I5) 1.76/1.82 f1(I6, I7, I8) -> f3(I6, 1 + I7, I6 - I7) [I7 <= I6] 1.76/1.82 f1(I9, I10, I11) -> f2(I9, I10, I11) [1 + I9 <= I10] 1.76/1.82 1.76/1.82 The dependency graph for this problem is: 1.76/1.82 2 -> 1.76/1.82 Where: 1.76/1.82 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 1.76/1.82 1.76/1.82 We have the following SCCs. 1.76/1.82 1.76/4.80 EOF
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