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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995091
details
property
value
status
complete
benchmark
13.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n042.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.180174 seconds
cpu usage
0.183368
user time
0.090964
system time
0.092404
max virtual memory
113176.0
max residence set size
8632.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.17 YES 0.00/0.17 0.00/0.17 DP problem for innermost termination. 0.00/0.17 P = 0.00/0.17 eval#(x, y) -> eval#(y, y) [x > 0 && x > y] 0.00/0.17 eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0] 0.00/0.17 R = 0.00/0.17 eval(x, y) -> eval(y, y) [x > 0 && x > y] 0.00/0.17 eval(I0, I1) -> eval(I0 - 1, I1) [I0 > 0 && I1 >= I0] 0.00/0.17 0.00/0.17 The dependency graph for this problem is: 0.00/0.17 0 -> 1 0.00/0.17 1 -> 1 0.00/0.17 Where: 0.00/0.17 0) eval#(x, y) -> eval#(y, y) [x > 0 && x > y] 0.00/0.17 1) eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0] 0.00/0.17 0.00/0.17 We have the following SCCs. 0.00/0.17 { 1 } 0.00/0.17 0.00/0.17 DP problem for innermost termination. 0.00/0.17 P = 0.00/0.17 eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0] 0.00/0.17 R = 0.00/0.17 eval(x, y) -> eval(y, y) [x > 0 && x > y] 0.00/0.17 eval(I0, I1) -> eval(I0 - 1, I1) [I0 > 0 && I1 >= I0] 0.00/0.17 0.00/0.17 We use the reverse value criterion with the projection function NU: 0.00/0.17 NU[eval#(z1,z2)] = z1 0.00/0.17 0.00/0.17 This gives the following inequalities: 0.00/0.17 I0 > 0 && I1 >= I0 ==> I0 > I0 - 1 with I0 >= 0 0.00/0.17 0.00/0.17 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.15 EOF
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