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SRS Standard pair #516970757
details
property
value
status
complete
benchmark
212534.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
2.82941603661 seconds
cpu usage
9.220065904
max memory
2.192523264E9
stage attributes
key
value
output-size
180362
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 48-rule system { 0 1 2 ⟶ 2 1 0 3 , 0 1 2 ⟶ 2 1 3 0 , 0 1 2 ⟶ 2 1 3 0 4 , 0 1 2 ⟶ 2 1 3 4 0 , 0 1 2 ⟶ 2 1 3 4 0 4 , 0 0 1 2 ⟶ 0 2 1 0 3 , 0 1 1 2 ⟶ 1 0 4 1 2 , 0 1 1 2 ⟶ 1 2 1 0 3 , 0 1 2 0 ⟶ 2 1 0 3 0 3 , 0 1 2 5 ⟶ 2 1 0 3 5 , 0 1 2 5 ⟶ 5 1 0 3 2 , 0 1 4 2 ⟶ 2 1 3 4 0 , 0 1 4 2 ⟶ 2 4 1 0 3 , 0 1 4 2 ⟶ 0 2 1 0 3 4 , 2 1 1 5 ⟶ 2 5 1 4 1 , 2 1 1 5 ⟶ 5 1 3 2 1 , 2 1 2 0 ⟶ 2 2 1 0 3 , 2 3 5 5 ⟶ 5 1 0 3 2 5 , 2 5 3 2 ⟶ 5 1 3 2 2 , 5 1 2 0 ⟶ 2 5 1 0 3 , 0 1 0 1 2 ⟶ 0 2 1 1 3 0 , 0 1 1 2 5 ⟶ 0 1 3 5 1 2 , 0 1 1 4 2 ⟶ 4 1 0 3 2 1 , 0 1 1 5 0 ⟶ 0 1 0 5 4 1 , 0 1 1 5 4 ⟶ 1 4 1 0 3 5 , 0 1 2 3 5 ⟶ 0 1 3 4 5 2 , 0 1 2 3 5 ⟶ 0 2 5 1 0 3 , 0 1 2 3 5 ⟶ 1 0 3 0 2 5 , 0 1 2 3 5 ⟶ 5 1 3 0 2 0 , 0 1 4 0 2 ⟶ 4 0 1 3 0 2 , 0 1 4 0 2 ⟶ 4 0 2 1 3 0 , 0 4 5 3 5 ⟶ 5 1 0 3 4 5 , 2 1 0 1 5 ⟶ 1 0 3 2 5 1 , 2 1 4 3 5 ⟶ 1 3 5 4 4 2 , 2 1 4 3 5 ⟶ 5 4 4 1 3 2 , 2 1 4 5 0 ⟶ 2 4 1 0 3 5 , 2 2 3 5 0 ⟶ 5 2 4 2 0 3 , 2 2 4 3 5 ⟶ 5 1 3 4 2 2 , 2 3 1 1 2 ⟶ 1 3 2 1 2 0 , 2 3 1 1 2 ⟶ 4 1 2 2 1 3 , 2 3 2 0 5 ⟶ 2 2 1 0 3 5 , 2 3 3 1 5 ⟶ 1 3 5 1 3 2 , 2 5 0 3 5 ⟶ 5 2 1 3 0 5 , 4 2 0 1 2 ⟶ 4 2 2 1 3 0 , 5 0 1 2 2 ⟶ 4 1 5 2 0 2 , 5 1 4 2 2 ⟶ 5 1 3 2 4 2 , 5 1 4 3 2 ⟶ 4 5 1 3 4 2 , 5 5 4 3 2 ⟶ 5 1 3 4 5 2 } The system was reversed. After renaming modulo the bijection { 2 ↦ 0, 1 ↦ 1, 0 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 48-rule system { 0 1 2 ⟶ 3 2 1 0 , 0 1 2 ⟶ 2 3 1 0 , 0 1 2 ⟶ 4 2 3 1 0 , 0 1 2 ⟶ 2 4 3 1 0 , 0 1 2 ⟶ 4 2 4 3 1 0 , 0 1 2 2 ⟶ 3 2 1 0 2 , 0 1 1 2 ⟶ 0 1 4 2 1 , 0 1 1 2 ⟶ 3 2 1 0 1 , 2 0 1 2 ⟶ 3 2 3 2 1 0 , 5 0 1 2 ⟶ 5 3 2 1 0 , 5 0 1 2 ⟶ 0 3 2 1 5 , 0 4 1 2 ⟶ 2 4 3 1 0 , 0 4 1 2 ⟶ 3 2 1 4 0 , 0 4 1 2 ⟶ 4 3 2 1 0 2 , 5 1 1 0 ⟶ 1 4 1 5 0 , 5 1 1 0 ⟶ 1 0 3 1 5 , 2 0 1 0 ⟶ 3 2 1 0 0 , 5 5 3 0 ⟶ 5 0 3 2 1 5 , 0 3 5 0 ⟶ 0 0 3 1 5 , 2 0 1 5 ⟶ 3 2 1 5 0 , 0 1 2 1 2 ⟶ 2 3 1 1 0 2 , 5 0 1 1 2 ⟶ 0 1 5 3 1 2 , 0 4 1 1 2 ⟶ 1 0 3 2 1 4 , 2 5 1 1 2 ⟶ 1 4 5 2 1 2 , 4 5 1 1 2 ⟶ 5 3 2 1 4 1 , 5 3 0 1 2 ⟶ 0 5 4 3 1 2 , 5 3 0 1 2 ⟶ 3 2 1 5 0 2 , 5 3 0 1 2 ⟶ 5 0 2 3 2 1 , 5 3 0 1 2 ⟶ 2 0 2 3 1 5 , 0 2 4 1 2 ⟶ 0 2 3 1 2 4 , 0 2 4 1 2 ⟶ 2 3 1 0 2 4 , 5 3 5 4 2 ⟶ 5 4 3 2 1 5 , 5 1 2 1 0 ⟶ 1 5 0 3 2 1 , 5 3 4 1 0 ⟶ 0 4 4 5 3 1 , 5 3 4 1 0 ⟶ 0 3 1 4 4 5 , 2 5 4 1 0 ⟶ 5 3 2 1 4 0 , 2 5 3 0 0 ⟶ 3 2 0 4 0 5 ,
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