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SRS Stand 10685 pair #381719608
details
property
value
status
complete
benchmark
z121.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n052.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
2.28557896614 seconds
cpu usage
7.880005254
max memory
8.4998144E8
stage attributes
key
value
output-size
2465
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: f(f(x1)) -> b(b(b(x1))) a(f(x1)) -> f(a(a(x1))) b(b(x1)) -> c(c(a(c(x1)))) d(b(x1)) -> d(a(b(x1))) c(c(x1)) -> d(d(d(x1))) b(d(x1)) -> d(b(x1)) c(d(d(x1))) -> f(x1) Proof: String Reversal Processor: f(f(x1)) -> b(b(b(x1))) f(a(x1)) -> a(a(f(x1))) b(b(x1)) -> c(a(c(c(x1)))) b(d(x1)) -> b(a(d(x1))) c(c(x1)) -> d(d(d(x1))) d(b(x1)) -> b(d(x1)) d(d(c(x1))) -> f(x1) Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 4, [c](x0) = x0 + 6, [a](x0) = x0, [b](x0) = x0 + 9, [f](x0) = x0 + 14 orientation: f(f(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1))) f(a(x1)) = x1 + 14 >= x1 + 14 = a(a(f(x1))) b(b(x1)) = x1 + 18 >= x1 + 18 = c(a(c(c(x1)))) b(d(x1)) = x1 + 13 >= x1 + 13 = b(a(d(x1))) c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1))) d(b(x1)) = x1 + 13 >= x1 + 13 = b(d(x1)) d(d(c(x1))) = x1 + 14 >= x1 + 14 = f(x1) problem: f(a(x1)) -> a(a(f(x1))) b(b(x1)) -> c(a(c(c(x1)))) b(d(x1)) -> b(a(d(x1))) c(c(x1)) -> d(d(d(x1))) d(b(x1)) -> b(d(x1)) d(d(c(x1))) -> f(x1) Matrix Interpretation Processor: dim=1 interpretation: [d](x0) = x0 + 4, [c](x0) = x0 + 8, [a](x0) = x0, [b](x0) = x0 + 14, [f](x0) = x0 + 4 orientation: f(a(x1)) = x1 + 4 >= x1 + 4 = a(a(f(x1))) b(b(x1)) = x1 + 28 >= x1 + 24 = c(a(c(c(x1)))) b(d(x1)) = x1 + 18 >= x1 + 18 = b(a(d(x1))) c(c(x1)) = x1 + 16 >= x1 + 12 = d(d(d(x1))) d(b(x1)) = x1 + 18 >= x1 + 18 = b(d(x1)) d(d(c(x1))) = x1 + 16 >= x1 + 4 = f(x1) problem: f(a(x1)) -> a(a(f(x1))) b(d(x1)) -> b(a(d(x1))) d(b(x1)) -> b(d(x1)) Bounds Processor: bound: 1 enrichment: match automaton: final states: {8,5,1} transitions: f50() -> 2* a0(4) -> 1* a0(6) -> 7* a0(3) -> 4* f0(2) -> 3* b0(7) -> 5* b0(6) -> 8*
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