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Integer TRS Innermost pair #487100443
details
property
value
status
complete
benchmark
countUpNo.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.06792 seconds
cpu usage
0.068938
user time
0.037686
system time
0.031252
max virtual memory
113188.0
max residence set size
8304.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE DP problem for innermost termination. P = cu#(true, I0) -> cu#(I0 < exp(I0), I0 + 1) cu#(true, I0) -> exp#(I0) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cu#(true, I0) -> cu_1#(I0) cu#(true, I0) -> exp#(I0) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) The dependency graph for this problem is: 0 -> 2 1 -> 2 -> 1, 0 Where: 0) cu#(true, I0) -> cu_1#(I0) 1) cu#(true, I0) -> exp#(I0) 2) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) We have the following SCCs. { 0, 2 } DP problem for innermost termination. P = cu#(true, I0) -> cu_1#(I0) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) R = exp(x) -> 2 * x cu(true, I0) -> cu(I0 < exp(I0), I0 + 1)
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