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SRS_Standard 2019-03-29 03.29 pair #432289182
details
property
value
status
complete
benchmark
matchbox2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n067.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
ttt2-1.19
configuration
ttt2
runtime (wallclock)
2.59272 seconds
cpu usage
8.89928
user time
7.25106
system time
1.64823
max virtual memory
3636216.0
max residence set size
74604.0
stage attributes
key
value
starexec-result
YES
output
8.67/2.57 YES 8.67/2.58 8.67/2.58 Problem: 8.67/2.58 t(o(x1)) -> m(a(x1)) 8.67/2.58 t(e(x1)) -> n(s(x1)) 8.67/2.58 a(l(x1)) -> a(t(x1)) 8.67/2.58 o(m(a(x1))) -> t(e(n(x1))) 8.67/2.58 s(a(x1)) -> l(a(t(o(m(a(t(e(x1)))))))) 8.67/2.58 n(s(x1)) -> a(l(a(t(x1)))) 8.67/2.58 8.67/2.58 Proof: 8.67/2.58 String Reversal Processor: 8.67/2.58 o(t(x1)) -> a(m(x1)) 8.67/2.58 e(t(x1)) -> s(n(x1)) 8.67/2.58 l(a(x1)) -> t(a(x1)) 8.67/2.58 a(m(o(x1))) -> n(e(t(x1))) 8.67/2.58 a(s(x1)) -> e(t(a(m(o(t(a(l(x1)))))))) 8.67/2.58 s(n(x1)) -> t(a(l(a(x1)))) 8.67/2.58 Matrix Interpretation Processor: dim=3 8.67/2.58 8.67/2.58 interpretation: 8.67/2.58 [1 1 0] 8.67/2.58 [l](x0) = [0 0 1]x0 8.67/2.58 [0 0 0] , 8.67/2.58 8.67/2.58 [1 0 0] 8.67/2.58 [n](x0) = [0 0 1]x0 8.67/2.58 [0 0 0] , 8.67/2.58 8.67/2.58 [1 1 0] [1] 8.67/2.58 [s](x0) = [0 0 0]x0 + [0] 8.67/2.58 [0 0 0] [1], 8.67/2.58 8.67/2.58 [1 1 0] [1] 8.67/2.58 [e](x0) = [0 0 0]x0 + [0] 8.67/2.58 [0 0 0] [1], 8.67/2.58 8.67/2.58 [1 0 0] 8.67/2.58 [m](x0) = [1 0 0]x0 8.67/2.58 [0 0 0] , 8.67/2.58 8.67/2.58 [1 0 1] [0] 8.67/2.58 [a](x0) = [0 0 0]x0 + [1] 8.67/2.58 [0 0 1] [0], 8.67/2.58 8.67/2.58 [1 0 0] 8.67/2.58 [t](x0) = [0 0 1]x0 8.67/2.58 [0 0 0] , 8.67/2.58 8.67/2.58 [1 0 1] [1] 8.67/2.58 [o](x0) = [0 0 0]x0 + [1] 8.67/2.58 [0 0 0] [0] 8.67/2.58 orientation: 8.67/2.58 [1 0 0] [1] [1 0 0] [0] 8.67/2.58 o(t(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(m(x1)) 8.67/2.58 [0 0 0] [0] [0 0 0] [0] 8.67/2.58 8.67/2.58 [1 0 1] [1] [1 0 1] [1] 8.67/2.58 e(t(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = s(n(x1)) 8.67/2.58 [0 0 0] [1] [0 0 0] [1] 8.67/2.58 8.67/2.58 [1 0 1] [1] [1 0 1] 8.67/2.58 l(a(x1)) = [0 0 1]x1 + [0] >= [0 0 1]x1 = t(a(x1)) 8.67/2.58 [0 0 0] [0] [0 0 0] 8.67/2.58 8.67/2.58 [1 0 1] [1] [1 0 1] [1] 8.67/2.58 a(m(o(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = n(e(t(x1))) 8.67/2.58 [0 0 0] [0] [0 0 0] [0] 8.67/2.58 8.67/2.58 [1 1 0] [2] [1 1 0] [2] 8.67/2.58 a(s(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = e(t(a(m(o(t(a(l(x1)))))))) 8.67/2.58 [0 0 0] [1] [0 0 0] [1] 8.67/2.58 8.67/2.58 [1 0 1] [1] [1 0 1] [1] 8.67/2.58 s(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = t(a(l(a(x1)))) 8.67/2.58 [0 0 0] [1] [0 0 0] [0] 8.67/2.58 problem: 8.67/2.58 e(t(x1)) -> s(n(x1)) 8.67/2.58 a(m(o(x1))) -> n(e(t(x1))) 8.67/2.58 a(s(x1)) -> e(t(a(m(o(t(a(l(x1)))))))) 8.67/2.58 s(n(x1)) -> t(a(l(a(x1)))) 8.67/2.58 Bounds Processor: 8.67/2.58 bound: 3 8.67/2.58 enrichment: match 8.67/2.58 automaton: 8.67/2.58 final states: {15,7,4,1} 8.67/2.58 transitions: 8.67/2.58 t1(47) -> 48* 8.67/2.58 t1(31) -> 32* 8.67/2.58 a1(44) -> 45* 8.67/2.58 a1(46) -> 47* 8.67/2.58 l1(45) -> 46* 8.67/2.58 n1(19) -> 20* 8.67/2.58 n1(33) -> 34* 8.67/2.58 n1(28) -> 29* 8.67/2.58 e1(32) -> 33* 8.67/2.58 s1(20) -> 21* 8.67/2.58 s1(29) -> 30* 8.67/2.58 t2(68) -> 69* 8.67/2.58 t2(63) -> 64*
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