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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995137
details
property
value
status
complete
benchmark
thousand.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n016.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.131367 seconds
cpu usage
0.133962
user time
0.078908
system time
0.055054
max virtual memory
113176.0
max residence set size
7776.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.13 YES 0.00/0.13 0.00/0.13 DP problem for innermost termination. 0.00/0.13 P = 0.00/0.13 f#(true, x) -> f#(1000 >= x, x + 1) 0.00/0.13 R = 0.00/0.13 f(true, x) -> f(1000 >= x, x + 1) 0.00/0.13 0.00/0.13 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.13 0.00/0.13 DP problem for innermost termination. 0.00/0.13 P = 0.00/0.13 f#(true, x) -> f_1#(x) 0.00/0.13 f_1#(x) -> f#(1000 >= x, x + 1) 0.00/0.13 f_1#(x) -> f_1#(x + 1) [1000 >= x] 0.00/0.13 R = 0.00/0.13 f(true, x) -> f(1000 >= x, x + 1) 0.00/0.13 0.00/0.13 The dependency graph for this problem is: 0.00/0.13 0 -> 2, 1 0.00/0.13 1 -> 0.00/0.13 2 -> 2, 1 0.00/0.13 Where: 0.00/0.13 0) f#(true, x) -> f_1#(x) 0.00/0.13 1) f_1#(x) -> f#(1000 >= x, x + 1) 0.00/0.13 2) f_1#(x) -> f_1#(x + 1) [1000 >= x] 0.00/0.13 0.00/0.13 We have the following SCCs. 0.00/0.13 { 2 } 0.00/0.13 0.00/0.13 DP problem for innermost termination. 0.00/0.13 P = 0.00/0.13 f_1#(x) -> f_1#(x + 1) [1000 >= x] 0.00/0.13 R = 0.00/0.13 f(true, x) -> f(1000 >= x, x + 1) 0.00/0.13 0.00/0.13 We use the reverse value criterion with the projection function NU: 0.00/0.13 NU[f_1#(z1)] = 1000 + -1 * z1 0.00/0.13 0.00/0.13 This gives the following inequalities: 0.00/0.13 1000 >= x ==> 1000 + -1 * x > 1000 + -1 * (x + 1) with 1000 + -1 * x >= 0 0.00/0.13 0.00/0.13 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.11 EOF
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