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SRS_Standard 2019-03-29 03.29 pair #432292350
details
property
value
status
complete
benchmark
z110.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n135.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
ttt2-1.19
configuration
ttt2
runtime (wallclock)
1.83644 seconds
cpu usage
6.03463
user time
4.60429
system time
1.43034
max virtual memory
3529704.0
max residence set size
68304.0
stage attributes
key
value
starexec-result
YES
output
4.79/1.82 YES 4.79/1.83 4.79/1.83 Problem: 4.79/1.83 a(a(x1)) -> b(x1) 4.79/1.83 b(a(x1)) -> a(b(x1)) 4.79/1.83 b(b(c(x1))) -> c(a(x1)) 4.79/1.83 b(b(x1)) -> a(a(a(x1))) 4.79/1.83 c(a(x1)) -> b(a(c(x1))) 4.79/1.83 4.79/1.83 Proof: 4.79/1.83 Matrix Interpretation Processor: dim=1 4.79/1.83 4.79/1.83 interpretation: 4.79/1.83 [c](x0) = 3x0, 4.79/1.83 4.79/1.83 [b](x0) = x0 + 8, 4.79/1.83 4.79/1.83 [a](x0) = x0 + 4 4.79/1.83 orientation: 4.79/1.83 a(a(x1)) = x1 + 8 >= x1 + 8 = b(x1) 4.79/1.83 4.79/1.83 b(a(x1)) = x1 + 12 >= x1 + 12 = a(b(x1)) 4.79/1.83 4.79/1.83 b(b(c(x1))) = 3x1 + 16 >= 3x1 + 12 = c(a(x1)) 4.79/1.83 4.79/1.83 b(b(x1)) = x1 + 16 >= x1 + 12 = a(a(a(x1))) 4.79/1.83 4.79/1.83 c(a(x1)) = 3x1 + 12 >= 3x1 + 12 = b(a(c(x1))) 4.79/1.83 problem: 4.79/1.83 a(a(x1)) -> b(x1) 4.79/1.83 b(a(x1)) -> a(b(x1)) 4.79/1.83 c(a(x1)) -> b(a(c(x1))) 4.79/1.83 Matrix Interpretation Processor: dim=3 4.79/1.83 4.79/1.83 interpretation: 4.79/1.83 [1 0 0] [1] 4.79/1.83 [c](x0) = [0 0 0]x0 + [0] 4.79/1.83 [1 0 0] [1], 4.79/1.83 4.79/1.83 [1 0 0] 4.79/1.83 [b](x0) = [0 0 0]x0 4.79/1.83 [0 0 0] , 4.79/1.83 4.79/1.83 [1 1 0] [1] 4.79/1.83 [a](x0) = [0 0 0]x0 + [0] 4.79/1.83 [0 0 0] [0] 4.79/1.83 orientation: 4.79/1.83 [1 1 0] [2] [1 0 0] 4.79/1.83 a(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = b(x1) 4.79/1.83 [0 0 0] [0] [0 0 0] 4.79/1.83 4.79/1.83 [1 1 0] [1] [1 0 0] [1] 4.79/1.83 b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(b(x1)) 4.79/1.83 [0 0 0] [0] [0 0 0] [0] 4.79/1.83 4.79/1.83 [1 1 0] [2] [1 0 0] [2] 4.79/1.83 c(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(c(x1))) 4.79/1.83 [1 1 0] [2] [0 0 0] [0] 4.79/1.83 problem: 4.79/1.83 b(a(x1)) -> a(b(x1)) 4.79/1.83 c(a(x1)) -> b(a(c(x1))) 4.79/1.83 String Reversal Processor: 4.79/1.83 a(b(x1)) -> b(a(x1)) 4.79/1.83 a(c(x1)) -> c(a(b(x1))) 4.79/1.83 Bounds Processor: 4.79/1.83 bound: 1 4.79/1.83 enrichment: match 4.79/1.83 automaton: 4.79/1.83 final states: {4,1} 4.79/1.83 transitions: 4.79/1.83 b1(8) -> 9* 4.79/1.83 a1(7) -> 8* 4.79/1.83 f30() -> 2* 4.79/1.83 b0(2) -> 5* 4.79/1.83 b0(3) -> 1* 4.79/1.83 a0(5) -> 6* 4.79/1.83 a0(2) -> 3* 4.79/1.83 c0(6) -> 4* 4.79/1.83 1 -> 8,3 4.79/1.83 2 -> 7* 4.79/1.83 4 -> 8,3 4.79/1.83 9 -> 6* 4.79/1.83 problem: 4.79/1.83 4.79/1.83 Qed 4.79/1.83 EOF
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