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SRS Stand 10685 pair #381717699
details
property
value
status
complete
benchmark
z111.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
4.77277493477 seconds
cpu usage
17.654657205
max memory
7.47331584E8
stage attributes
key
value
output-size
3819
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) b(b(x1)) -> c(c(c(x1))) c(c(x1)) -> d(d(d(x1))) c(d(d(x1))) -> a(x1) Proof: String Reversal Processor: a(a(x1)) -> c(c(c(b(x1)))) c(b(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) b(b(x1)) -> c(c(c(x1))) c(c(x1)) -> d(d(d(x1))) d(d(c(x1))) -> a(x1) Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 2, [b](x0) = x0 + 5, [c](x0) = x0 + 3, [a](x0) = x0 + 7 orientation: a(a(x1)) = x1 + 14 >= x1 + 14 = c(c(c(b(x1)))) c(b(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1)))) a(x1) = x1 + 7 >= x1 + 7 = d(c(d(x1))) b(b(x1)) = x1 + 10 >= x1 + 9 = c(c(c(x1))) c(c(x1)) = x1 + 6 >= x1 + 6 = d(d(d(x1))) d(d(c(x1))) = x1 + 7 >= x1 + 7 = a(x1) problem: a(a(x1)) -> c(c(c(b(x1)))) c(b(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(c(x1)) -> d(d(d(x1))) d(d(c(x1))) -> a(x1) String Reversal Processor: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(c(x1)) -> d(d(d(x1))) c(d(d(x1))) -> a(x1) Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 2, [b](x0) = x0, [c](x0) = x0 + 8, [a](x0) = x0 + 12 orientation: a(a(x1)) = x1 + 24 >= x1 + 24 = b(c(c(c(x1)))) b(c(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1)))) a(x1) = x1 + 12 >= x1 + 12 = d(c(d(x1))) c(c(x1)) = x1 + 16 >= x1 + 6 = d(d(d(x1))) c(d(d(x1))) = x1 + 12 >= x1 + 12 = a(x1) problem: a(a(x1)) -> b(c(c(c(x1)))) b(c(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) c(d(d(x1))) -> a(x1) String Reversal Processor: a(a(x1)) -> c(c(c(b(x1)))) c(b(x1)) -> d(d(d(d(x1)))) a(x1) -> d(c(d(x1))) d(d(c(x1))) -> a(x1) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [d](x0) = [0 0 0]x0 [1 0 0] , [3 0 1] [3] [b](x0) = [2 0 2]x0 + [0] [3 1 2] [1], [1 1 0] [1]
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