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SRS Standard pair #487519293
details
property
value
status
complete
benchmark
secr10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n129.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
44.9503219128 seconds
cpu usage
177.700791579
max memory
1.878982656E9
stage attributes
key
value
output-size
13150
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)) -> c(b(a(b(a(x1))))) b(a(b(x1))) -> b(x1) a(a(a(x1))) -> c(c(a(x1))) c(c(x1)) -> a(b(c(b(a(x1))))) a(c(a(x1))) -> c(c(a(x1))) c(a(c(x1))) -> a(a(c(x1))) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [b](x0) = [0 0 0]x0 [0 1 0] , [1 1 0] [0] [a](x0) = [0 1 1]x0 + [1] [0 0 0] [0], [1 1 0] [0] [c](x0) = [0 1 0]x0 + [1] [0 0 0] [0] orientation: [1 2 1] [1] [1 1 0] [0] a(a(x1)) = [0 1 1]x1 + [2] >= [0 0 0]x1 + [1] = c(b(a(b(a(x1))))) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] b(a(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 = b(x1) [0 1 0] [1] [0 1 0] [1 3 2] [3] [1 3 2] [3] a(a(a(x1))) = [0 1 1]x1 + [3] >= [0 1 1]x1 + [3] = c(c(a(x1))) [0 0 0] [0] [0 0 0] [0] [1 2 0] [1] [1 1 0] [0] c(c(x1)) = [0 1 0]x1 + [2] >= [0 0 0]x1 + [2] = a(b(c(b(a(x1))))) [0 0 0] [0] [0 0 0] [0] [1 3 2] [3] [1 3 2] [3] a(c(a(x1))) = [0 1 1]x1 + [3] >= [0 1 1]x1 + [3] = c(c(a(x1))) [0 0 0] [0] [0 0 0] [0] [1 3 0] [3] [1 3 0] [3] c(a(c(x1))) = [0 1 0]x1 + [3] >= [0 1 0]x1 + [3] = a(a(c(x1))) [0 0 0] [0] [0 0 0] [0] problem: b(a(b(x1))) -> b(x1) a(a(a(x1))) -> c(c(a(x1))) a(c(a(x1))) -> c(c(a(x1))) c(a(c(x1))) -> a(a(c(x1))) Matrix Interpretation Processor: dim=1 interpretation: [b](x0) = 2x0 + 2, [a](x0) = 2x0 + 2, [c](x0) = 2x0 + 2 orientation: b(a(b(x1))) = 8x1 + 14 >= 2x1 + 2 = b(x1) a(a(a(x1))) = 8x1 + 14 >= 8x1 + 14 = c(c(a(x1))) a(c(a(x1))) = 8x1 + 14 >= 8x1 + 14 = c(c(a(x1))) c(a(c(x1))) = 8x1 + 14 >= 8x1 + 14 = a(a(c(x1))) problem: a(a(a(x1))) -> c(c(a(x1))) a(c(a(x1))) -> c(c(a(x1))) c(a(c(x1))) -> a(a(c(x1))) String Reversal Processor: a(a(a(x1))) -> a(c(c(x1))) a(c(a(x1))) -> a(c(c(x1))) c(a(c(x1))) -> c(a(a(x1))) DP Processor: DPs: a#(a(a(x1))) -> c#(x1) a#(a(a(x1))) -> c#(c(x1)) a#(a(a(x1))) -> a#(c(c(x1))) a#(c(a(x1))) -> c#(x1) a#(c(a(x1))) -> c#(c(x1)) a#(c(a(x1))) -> a#(c(c(x1))) c#(a(c(x1))) -> a#(x1) c#(a(c(x1))) -> a#(a(x1)) c#(a(c(x1))) -> c#(a(a(x1))) TRS: a(a(a(x1))) -> a(c(c(x1))) a(c(a(x1))) -> a(c(c(x1))) c(a(c(x1))) -> c(a(a(x1))) TDG Processor:
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