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SRS Standard pair #516968618
details
property
value
status
complete
benchmark
z083.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
Zantema_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
2.95791316032 seconds
cpu usage
10.194630075
max memory
7.66803968E8
stage attributes
key
value
output-size
4907
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: b(d(b(x1))) -> c(d(b(x1))) b(a(c(x1))) -> b(c(x1)) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(x1)) Proof: Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0, [a](x0) = x0 + 8, [b](x0) = x0 + 8, [c](x0) = x0 + 8 orientation: b(d(b(x1))) = x1 + 16 >= x1 + 16 = c(d(b(x1))) b(a(c(x1))) = x1 + 24 >= x1 + 16 = b(c(x1)) a(d(x1)) = x1 + 8 >= x1 + 8 = d(c(x1)) b(b(b(x1))) = x1 + 24 >= x1 + 24 = a(b(c(x1))) d(c(x1)) = x1 + 8 >= x1 + 8 = b(d(x1)) d(c(x1)) = x1 + 8 >= x1 + 8 = d(b(d(x1))) d(a(c(x1))) = x1 + 16 >= x1 + 16 = b(b(x1)) problem: b(d(b(x1))) -> c(d(b(x1))) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(x1)) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [0] [d](x0) = [0 0 0]x0 + [1] [0 0 1] [0], [1 0 1] [a](x0) = [0 1 1]x0 [0 1 0] , [1 0 0] [b](x0) = [0 0 0]x0 [0 1 0] , [1 0 0] [0] [c](x0) = [0 0 0]x0 + [0] [0 0 0] [1] orientation: [1 0 0] [0] [1 0 0] [0] b(d(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(d(b(x1))) [0 0 0] [1] [0 0 0] [1] [1 0 1] [0] [1 0 0] [0] a(d(x1)) = [0 0 1]x1 + [1] >= [0 0 0]x1 + [1] = d(c(x1)) [0 0 0] [1] [0 0 0] [1] [1 0 0] [1 0 0] b(b(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = a(b(c(x1))) [0 0 0] [0 0 0] [1 0 0] [0] [1 0 0] [0] d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = b(d(x1)) [0 0 0] [1] [0 0 0] [1] [1 0 0] [0] [1 0 0] [0] d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(b(d(x1))) [0 0 0] [1] [0 0 0] [1] [1 0 0] [1] [1 0 0] d(a(c(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = b(b(x1)) [0 0 0] [0] [0 0 0] problem: b(d(b(x1))) -> c(d(b(x1))) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) Matrix Interpretation Processor: dim=3 interpretation:
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