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Integer TRS Innermost pair #487528898
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
n022.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.186667203903 seconds
cpu usage
0.146550278
max memory
8679424.0
stage attributes
key
value
output-size
1304
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f#(true, x) -> f#(1000 >= x, x + 1) R = f(true, x) -> f(1000 >= x, x + 1) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = f#(true, x) -> f_1#(x) f_1#(x) -> f#(1000 >= x, x + 1) f_1#(x) -> f_1#(x + 1) [1000 >= x] R = f(true, x) -> f(1000 >= x, x + 1) The dependency graph for this problem is: 0 -> 2, 1 1 -> 2 -> 2, 1 Where: 0) f#(true, x) -> f_1#(x) 1) f_1#(x) -> f#(1000 >= x, x + 1) 2) f_1#(x) -> f_1#(x + 1) [1000 >= x] We have the following SCCs. { 2 } DP problem for innermost termination. P = f_1#(x) -> f_1#(x + 1) [1000 >= x] R = f(true, x) -> f(1000 >= x, x + 1) We use the reverse value criterion with the projection function NU: NU[f_1#(z1)] = 1000 + -1 * z1 This gives the following inequalities: 1000 >= x ==> 1000 + -1 * x > 1000 + -1 * (x + 1) with 1000 + -1 * x >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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