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SRS Standard pair #516973391
details
property
value
status
complete
benchmark
03.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n030.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
0.67756986618 seconds
cpu usage
1.081863194
max memory
4.7581184E8
stage attributes
key
value
output-size
1234
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 { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 2-rule system { 0 0 0 0 ⟶ 0 0 1 1 , 1 0 0 1 ⟶ 0 0 1 0 } Loop of length 15 starting with a string of length 12 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .aaaa.aabababa rule aaaa-> aabb at position 0 .aabb.aabababa rule baab-> aaba at position 3 .aabaaba.ababa rule baab-> aaba at position 5 .aabaaaaba.aba rule aaaa-> aabb at position 3 .aabaabbba.aba rule baab-> aaba at position 7 .aabaabbaaba.a rule baab-> aaba at position 6 .aabaabaabaa.a rule baab-> aaba at position 5 .aabaaaabaaa.a rule aaaa-> aabb at position 3 .aabaabbbaaa.a rule baab-> aaba at position 2 .aaaababbaaa.a rule aaaa-> aabb at position 8 .aaaababbaabb. rule baab-> aaba at position 7 .aaaababaabab. rule baab-> aaba at position 6 .aaaabaaabaab. rule baab-> aaba at position 8 .aaaabaaaaaba. rule aaaa-> aabb at position 5 .aaaabaabbaba. rule baab-> aaba at position 4 .aaaaaabababa.
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