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SRS Standard pair #516971783
details
property
value
status
complete
benchmark
247020.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n005.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.10894608498 seconds
cpu usage
2.042414319
max memory
5.74124032E8
stage attributes
key
value
output-size
5635
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 3 ↦ 2, 4 ↦ 3, 5 ↦ 4, 2 ↦ 5 }, it remains to prove termination of the 5-rule system { 0 ⟶ 1 , 0 0 ⟶ 0 , 2 3 4 ⟶ 3 2 4 , 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ⟶ 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 1 0 0 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 0 0 1 0 1 1 1 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 1 0 0 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 1 0 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 0 1 1 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 , 1 0 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 1 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 1 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 1 0 1 0 ⟶ 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 } Applying sparse untiling TRFCU(2) after reversal [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 4-rule system { 0 ⟶ 1 , 0 0 ⟶ 0 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ⟶ 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 1 0 0 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 0 0 1 0 1 1 1 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 1 0 0 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 1 0 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 0 1 1 0 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 , 1 0 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 1 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 1 0 1 1 0 1 1 1 1 0 1 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 1 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 1 0 1 0 ⟶ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 15 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 34 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { }, it remains to prove termination of the 0-rule system { } The system is trivially terminating.
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