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SRS Standard pair #487519479
details
property
value
status
complete
benchmark
z112.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n043.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
1.49194407463 seconds
cpu usage
4.565164536
max memory
6.44079616E8
stage attributes
key
value
output-size
3718
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: a(a(x1)) -> b(c(x1)) b(b(x1)) -> c(d(x1)) b(x1) -> a(x1) c(c(x1)) -> d(f(x1)) d(d(x1)) -> f(f(f(x1))) d(x1) -> b(x1) f(f(x1)) -> g(a(x1)) g(g(x1)) -> a(x1) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [0] [f](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [1 0 0] [0] [c](x0) = [0 0 0]x0 + [1] [0 0 0] [0], [1 0 0] [0] [d](x0) = [0 0 0]x0 + [1] [0 1 0] [0], [1 0 1] [0] [g](x0) = [0 0 1]x0 + [0] [0 0 0] [1], [1 0 0] [0] [a](x0) = [0 0 0]x0 + [1] [0 0 0] [0], [1 0 0] [0] [b](x0) = [0 0 0]x0 + [1] [0 0 0] [0] orientation: [1 0 0] [0] [1 0 0] [0] a(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(c(x1)) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] b(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(d(x1)) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] b(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] c(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(f(x1)) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] d(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = f(f(f(x1))) [0 0 0] [1] [0 0 0] [1] [1 0 0] [0] [1 0 0] [0] d(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(x1) [0 1 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] f(f(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = g(a(x1)) [0 0 0] [1] [0 0 0] [1] [1 0 1] [1] [1 0 0] [0] g(g(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1) [0 0 0] [1] [0 0 0] [0] problem: a(a(x1)) -> b(c(x1)) b(b(x1)) -> c(d(x1)) b(x1) -> a(x1) c(c(x1)) -> d(f(x1)) d(d(x1)) -> f(f(f(x1))) d(x1) -> b(x1) f(f(x1)) -> g(a(x1)) String Reversal Processor: a(a(x1)) -> c(b(x1)) b(b(x1)) -> d(c(x1)) b(x1) -> a(x1) c(c(x1)) -> f(d(x1)) d(d(x1)) -> f(f(f(x1))) d(x1) -> b(x1) f(f(x1)) -> a(g(x1)) Matrix Interpretation Processor: dim=1 interpretation: [f](x0) = x0 + 6, [c](x0) = x0 + 9,
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