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SRS Stand 10685 pair #381718118
details
property
value
status
complete
benchmark
beans1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n034.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
ttt2-1.17+nonreach
configuration
ttt2-1.17+nonreach
runtime (wallclock)
2.88939213753 seconds
cpu usage
10.075028659
max memory
1.007579136E9
stage attributes
key
value
output-size
4292
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: 1(2(1(x1))) -> 2(0(2(x1))) 0(2(1(x1))) -> 1(0(2(x1))) L(2(1(x1))) -> L(1(0(2(x1)))) 1(2(0(x1))) -> 2(0(1(x1))) 1(2(R(x1))) -> 2(0(1(R(x1)))) 0(2(0(x1))) -> 1(0(1(x1))) L(2(0(x1))) -> L(1(0(1(x1)))) 0(2(R(x1))) -> 1(0(1(R(x1)))) Proof: String Reversal Processor: 1(2(1(x1))) -> 2(0(2(x1))) 1(2(0(x1))) -> 2(0(1(x1))) 1(2(L(x1))) -> 2(0(1(L(x1)))) 0(2(1(x1))) -> 1(0(2(x1))) R(2(1(x1))) -> R(1(0(2(x1)))) 0(2(0(x1))) -> 1(0(1(x1))) 0(2(L(x1))) -> 1(0(1(L(x1)))) R(2(0(x1))) -> R(1(0(1(x1)))) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [R](x0) = [0 0 1]x0 [1 0 0] , [1 0 0] [L](x0) = [0 0 0]x0 [0 0 0] , [1 0 0] [0](x0) = [0 0 0]x0 [0 0 0] , [1 0 0] [0] [2](x0) = [0 0 0]x0 + [0] [0 0 0] [1], [1 0 0] [1](x0) = [0 0 0]x0 [0 0 1] orientation: [1 0 0] [0] [1 0 0] [0] 1(2(1(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(2(x1))) [0 0 0] [1] [0 0 0] [1] [1 0 0] [0] [1 0 0] [0] 1(2(0(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(1(x1))) [0 0 0] [1] [0 0 0] [1] [1 0 0] [0] [1 0 0] [0] 1(2(L(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 2(0(1(L(x1)))) [0 0 0] [1] [0 0 0] [1] [1 0 0] [1 0 0] 0(2(1(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(2(x1))) [0 0 0] [0 0 0] [1 0 0] [1] [1 0 0] R(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(2(x1)))) [1 0 0] [0] [1 0 0] [1 0 0] [1 0 0] 0(2(0(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(1(x1))) [0 0 0] [0 0 0] [1 0 0] [1 0 0] 0(2(L(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(1(L(x1)))) [0 0 0] [0 0 0] [1 0 0] [1] [1 0 0] R(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(1(x1)))) [1 0 0] [0] [1 0 0] problem: 1(2(1(x1))) -> 2(0(2(x1))) 1(2(0(x1))) -> 2(0(1(x1))) 1(2(L(x1))) -> 2(0(1(L(x1)))) 0(2(1(x1))) -> 1(0(2(x1))) 0(2(0(x1))) -> 1(0(1(x1))) 0(2(L(x1))) -> 1(0(1(L(x1)))) String Reversal Processor: 1(2(1(x1))) -> 2(0(2(x1))) 0(2(1(x1))) -> 1(0(2(x1))) L(2(1(x1))) -> L(1(0(2(x1)))) 1(2(0(x1))) -> 2(0(1(x1))) 0(2(0(x1))) -> 1(0(1(x1))) L(2(0(x1))) -> L(1(0(1(x1)))) Matrix Interpretation Processor: dim=1 interpretation: [L](x0) = 2x0 + 8,
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