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Integ TRS Inner 43557 pair #381740714
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
n020.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.0676708221436 seconds
cpu usage
0.06601869
max memory
8663040.0
stage attributes
key
value
output-size
1142
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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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return to Integ TRS Inner 43557