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SRS Standard pair #516973277
details
property
value
status
complete
benchmark
num-515.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n166.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
5.2555949688 seconds
cpu usage
18.71305958
max memory
3.676291072E9
stage attributes
key
value
output-size
1513
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, b ↦ 1, c ↦ 2 }, it remains to prove termination of the 2-rule system { 0 0 0 ⟶ 1 , 1 2 ⟶ 2 2 0 0 0 0 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 2-rule system { 0 0 0 ⟶ 1 , 2 1 ⟶ 0 0 0 0 2 2 } Loop of length 16 starting with a string of length 5 using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }: .cb.bbb rule cb-> aaaacc at position 0 .aaaacc.bbb rule cb-> aaaacc at position 5 .aaaacaaaacc.bb rule aaa-> b at position 5 .aaaacbacc.bb rule cb-> aaaacc at position 8 .aaaacbacaaaacc.b rule aaa-> b at position 8 .aaaacbacbacc.b rule cb-> aaaacc at position 7 .aaaacbaaaaaccacc.b rule aaa-> b at position 6 .aaaacbbaaccacc.b rule cb-> aaaacc at position 13 .aaaacbbaaccacaaaacc. rule aaa-> b at position 13 .aaaacbbaaccacbacc. rule cb-> aaaacc at position 12 .aaaacbbaaccaaaaaccacc. rule aaa-> b at position 11 .aaaacbbaaccbaaccacc. rule cb-> aaaacc at position 10 .aaaacbbaacaaaaccaaccacc. rule aaa-> b at position 10 .aaaacbbaacbaccaaccacc. rule cb-> aaaacc at position 9 .aaaacbbaaaaaaccaccaaccacc. rule aaa-> b at position 7 .aaaacbbbaaaccaccaaccacc. rule aaa-> b at position 8 .aaaacbbbbccaccaaccacc.
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