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SRS Standard pair #516975053
details
property
value
status
complete
benchmark
size-12-alpha-3-num-3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.11630702019 seconds
cpu usage
2.78716371
max memory
6.43760128E8
stage attributes
key
value
output-size
1375
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 }, it remains to prove termination of the 5-rule system { 0 ⟶ , 0 ⟶ 1 , 0 1 ⟶ 1 0 2 0 , 1 ⟶ , 2 2 ⟶ } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 5-rule system { 0 ⟶ , 0 ⟶ 1 , 1 0 ⟶ 0 2 0 1 , 1 ⟶ , 2 2 ⟶ } Loop of length 17 starting with a string of length 4 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .a.aaa rule a-> b at position 0 .b.aaa rule ba-> acab at position 0 .acab.aa rule a-> b at position 2 .acbb.aa rule ba-> acab at position 3 .acbacab.a rule ba-> acab at position 2 .acacabcab.a rule a-> at position 2 .accabcab.a rule cc-> at position 1 .aabcab.a rule a-> b at position 4 .aabcbb.a rule ba-> acab at position 5 .aabcbacab. rule ba-> acab at position 4 .aabcacabcab. rule a-> at position 4 .aabccabcab. rule cc-> at position 3 .aababcab. rule ba-> acab at position 2 .aaacabbcab. rule a-> at position 4 .aaacbbcab. rule b-> at position 4 .aaacbcab. rule b-> at position 4 .aaaccab. rule cc-> at position 3 .aaaab.
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