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SRS Standard pair #487088192
details
property
value
status
complete
benchmark
212892.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
50.023 seconds
cpu usage
194.235
user time
189.068
system time
5.16692
max virtual memory
1.246966E7
max residence set size
235484.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE Problem: 0(0(x1)) -> 1(0(2(0(2(x1))))) 0(0(x1)) -> 0(2(3(4(0(2(x1)))))) 0(3(x1)) -> 0(2(3(2(x1)))) 0(3(x1)) -> 2(0(2(1(3(x1))))) 0(3(x1)) -> 2(3(0(2(2(x1))))) 0(3(x1)) -> 0(2(2(2(3(2(x1)))))) 0(0(0(x1))) -> 0(2(0(0(x1)))) 0(0(4(x1))) -> 2(4(0(0(3(2(x1)))))) 0(0(5(x1))) -> 0(5(2(0(x1)))) 0(1(3(x1))) -> 3(2(2(1(0(x1))))) 0(3(0(x1))) -> 2(0(2(3(4(0(x1)))))) 0(4(3(x1))) -> 3(0(2(4(x1)))) 0(4(3(x1))) -> 1(0(2(1(3(4(x1)))))) 0(4(3(x1))) -> 2(2(4(3(4(0(x1)))))) 0(4(5(x1))) -> 2(4(0(5(2(5(x1)))))) 0(4(5(x1))) -> 5(3(2(4(4(0(x1)))))) 0(5(0(x1))) -> 0(5(2(4(0(x1))))) 0(5(0(x1))) -> 5(0(2(3(2(0(x1)))))) 0(5(3(x1))) -> 3(2(5(3(2(0(x1)))))) 1(0(0(x1))) -> 1(0(2(5(0(2(x1)))))) 1(0(3(x1))) -> 2(1(3(0(5(2(x1)))))) 1(0(4(x1))) -> 3(0(2(1(4(x1))))) 1(0(4(x1))) -> 2(1(2(1(4(0(x1)))))) 3(0(0(x1))) -> 3(0(2(0(x1)))) 3(0(3(x1))) -> 3(0(2(3(x1)))) 3(0(3(x1))) -> 3(3(2(2(0(x1))))) 3(0(4(x1))) -> 2(2(3(4(0(x1))))) 3(0(4(x1))) -> 2(3(4(2(0(3(x1)))))) 3(0(4(x1))) -> 5(5(2(4(0(3(x1)))))) 3(3(4(x1))) -> 3(2(1(3(2(4(x1)))))) 3(5(0(x1))) -> 3(5(1(0(2(1(x1)))))) 4(0(4(x1))) -> 0(2(4(4(0(x1))))) 4(0(4(x1))) -> 1(4(0(2(4(0(x1)))))) 0(0(4(3(x1)))) -> 3(0(0(2(1(4(x1)))))) 0(0(5(3(x1)))) -> 0(5(2(5(0(3(x1)))))) 0(1(0(3(x1)))) -> 2(1(0(3(0(x1))))) 0(1(1(0(x1)))) -> 2(1(0(0(1(2(x1)))))) 0(4(0(0(x1)))) -> 0(4(0(2(1(0(x1)))))) 0(4(1(3(x1)))) -> 2(1(4(2(0(3(x1)))))) 0(4(5(3(x1)))) -> 4(3(2(5(0(x1))))) 0(4(5(3(x1)))) -> 4(3(5(0(2(4(x1)))))) 0(5(0(4(x1)))) -> 5(0(0(2(4(x1))))) 0(5(5(3(x1)))) -> 2(3(5(5(0(x1))))) 1(3(0(4(x1)))) -> 4(0(1(2(3(x1))))) 4(0(3(5(x1)))) -> 5(2(3(4(2(0(x1)))))) 0(0(1(0(4(x1))))) -> 0(0(0(1(4(2(x1)))))) 0(0(5(5(3(x1))))) -> 0(5(0(3(2(5(x1)))))) 3(0(4(3(4(x1))))) -> 3(0(3(4(2(4(x1)))))) 3(3(0(4(5(x1))))) -> 1(4(5(3(0(3(x1)))))) 3(4(2(0(3(x1))))) -> 2(3(0(2(3(4(x1)))))) Proof: Open
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