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SRS Standard pair #516974933
details
property
value
status
complete
benchmark
size-12-alpha-3-num-95.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n120.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
2.40435409546 seconds
cpu usage
6.693337507
max memory
1.168883712E9
stage attributes
key
value
output-size
1397
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 0 ⟶ 1 2 , 1 ⟶ , 2 ⟶ , 2 1 ⟶ 1 0 2 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 2 ↦ 1, 1 ↦ 2 }, it remains to prove termination of the 5-rule system { 0 ⟶ , 0 0 ⟶ 1 2 , 2 ⟶ , 1 ⟶ , 2 1 ⟶ 1 0 2 } Loop of length 16 starting with a string of length 8 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .aa.bbbbbb rule aa-> bc at position 0 .bc.bbbbbb rule cb-> bac at position 1 .bbac.bbbbb rule cb-> bac at position 3 .bbabac.bbbb rule b-> at position 3 .bbaac.bbbb rule cb-> bac at position 4 .bbaabac.bbb rule cb-> bac at position 6 .bbaababac.bb rule b-> at position 6 .bbaabaac.bb rule aa-> bc at position 5 .bbaabbcc.bb rule cb-> bac at position 7 .bbaabbcbac.b rule cb-> bac at position 6 .bbaabbbacac.b rule c-> at position 8 .bbaabbbaac.b rule aa-> bc at position 7 .bbaabbbbcc.b rule cb-> bac at position 9 .bbaabbbbcbac. rule cb-> bac at position 8 .bbaabbbbbacac. rule c-> at position 10 .bbaabbbbbaac. rule aa-> bc at position 9 .bbaabbbbbbcc.
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