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SRS Standard pair #516973991
details
property
value
status
complete
benchmark
size-12-alpha-3-num-532.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n066.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.62732791901 seconds
cpu usage
4.703001133
max memory
9.10503936E8
stage attributes
key
value
output-size
6689
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 }, it remains to prove termination of the 3-rule system { 0 0 1 ⟶ 2 , 0 2 ⟶ 1 2 0 0 , 1 2 ⟶ } The system was reversed. After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 3-rule system { 0 1 1 ⟶ 2 , 2 1 ⟶ 1 1 2 0 , 2 0 ⟶ } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (2,↑) ↦ 2, (0,↓) ↦ 3, (2,↓) ↦ 4 }, it remains to prove termination of the 6-rule system { 0 1 1 ⟶ 2 , 2 1 ⟶ 2 3 , 2 1 ⟶ 0 , 3 1 1 →= 4 , 4 1 →= 1 1 4 3 , 4 3 →= } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (5,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (5,2) ↦ 3, (2,1) ↦ 4, (1,4) ↦ 5, (2,4) ↦ 6, (1,6) ↦ 7, (2,6) ↦ 8, (2,3) ↦ 9, (3,1) ↦ 10, (3,4) ↦ 11, (3,6) ↦ 12, (0,4) ↦ 13, (0,6) ↦ 14, (4,1) ↦ 15, (4,4) ↦ 16, (4,6) ↦ 17, (4,3) ↦ 18, (5,4) ↦ 19, (5,1) ↦ 20, (5,6) ↦ 21 }, it remains to prove termination of the 51-rule system { 0 1 2 2 ⟶ 3 4 , 0 1 2 5 ⟶ 3 6 , 0 1 2 7 ⟶ 3 8 , 3 4 2 ⟶ 3 9 10 , 3 4 5 ⟶ 3 9 11 , 3 4 7 ⟶ 3 9 12 , 3 4 2 ⟶ 0 1 , 3 4 5 ⟶ 0 13 , 3 4 7 ⟶ 0 14 , 9 10 2 2 →= 6 15 , 9 10 2 5 →= 6 16 , 9 10 2 7 →= 6 17 , 18 10 2 2 →= 16 15 , 18 10 2 5 →= 16 16 , 18 10 2 7 →= 16 17 , 13 15 2 →= 1 2 5 18 10 , 13 15 5 →= 1 2 5 18 11 , 13 15 7 →= 1 2 5 18 12 , 5 15 2 →= 2 2 5 18 10 , 5 15 5 →= 2 2 5 18 11 , 5 15 7 →= 2 2 5 18 12 , 6 15 2 →= 4 2 5 18 10 , 6 15 5 →= 4 2 5 18 11 , 6 15 7 →= 4 2 5 18 12 , 11 15 2 →= 10 2 5 18 10 , 11 15 5 →= 10 2 5 18 11 , 11 15 7 →= 10 2 5 18 12 , 16 15 2 →= 15 2 5 18 10 , 16 15 5 →= 15 2 5 18 11 , 16 15 7 →= 15 2 5 18 12 , 19 15 2 →= 20 2 5 18 10 , 19 15 5 →= 20 2 5 18 11 , 19 15 7 →= 20 2 5 18 12 , 13 18 10 →= 1 , 13 18 11 →= 13 , 13 18 12 →= 14 , 5 18 10 →= 2 , 5 18 11 →= 5 , 5 18 12 →= 7 , 6 18 10 →= 4 , 6 18 11 →= 6 , 6 18 12 →= 8 , 11 18 10 →= 10 , 11 18 11 →= 11 , 11 18 12 →= 12 , 16 18 10 →= 15 , 16 18 11 →= 16 , 16 18 12 →= 17 , 19 18 10 →= 20 , 19 18 11 →= 19 , 19 18 12 →= 21 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟
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