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SRS Standard pair #516969167
details
property
value
status
complete
benchmark
uni-6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n175.star.cs.uiowa.edu
space
Waldmann_06_SRS
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
0.924440860748 seconds
cpu usage
0.983196432
max memory
4.50449408E8
stage attributes
key
value
output-size
1300
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO After renaming modulo the bijection { a ↦ 0, c ↦ 1, b ↦ 2 }, it remains to prove termination of the 3-rule system { 0 0 0 ⟶ 1 1 2 , 2 2 2 ⟶ 1 1 1 , 1 1 1 ⟶ 0 2 2 } Loop of length 19 starting with a string of length 7 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .aaa.cccc rule aaa-> bbc at position 0 .bbc.cccc rule ccc-> bbb at position 2 .bbbbb.cc rule bbb-> acc at position 2 .bbacc.cc rule ccc-> bbb at position 3 .bbabbb.c rule bbb-> acc at position 3 .bbaacc.c rule ccc-> bbb at position 4 .bbaabbb. rule bbb-> acc at position 4 .bbaaacc. rule aaa-> bbc at position 2 .bbbbccc. rule bbb-> acc at position 1 .baccccc. rule ccc-> bbb at position 2 .babbbcc. rule bbb-> acc at position 2 .baacccc. rule ccc-> bbb at position 3 .baabbbc. rule bbb-> acc at position 3 .baaaccc. rule aaa-> bbc at position 1 .bbbcccc. rule bbb-> acc at position 0 .acccccc. rule ccc-> bbb at position 1 .abbbccc. rule bbb-> acc at position 1 .aaccccc. rule ccc-> bbb at position 2 .aabbbcc. rule bbb-> acc at position 2 .aaacccc.
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