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	n060.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	2.62188887596 seconds
cpu usage	9.138477416
max memory	1.147744256E9

stage attributes

key	value
output-size	3518
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: 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 0] [R](x0) = [0 1 0]x0 [0 0 0] , [1 1 0] [0] [2](x0) = [0 0 0]x0 + [1] [0 0 0] [0], [1 1 0] [1] [L](x0) = [1 1 1]x0 + [1] [0 0 0] [0], [1 0 0] [1](x0) = [0 1 0]x0 [0 1 0] , [1 0 0] [0](x0) = [0 0 0]x0 [0 0 0] orientation: [1 1 0] [0] [1 1 0] [0] 1(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(2(x1))) [0 0 0] [1] [0 0 0] [0] [1 0 0] [0] [1 0 0] [0] 1(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(1(x1))) [0 0 0] [1] [0 0 0] [0] [2 2 1] [2] [1 1 0] [1] 1(2(L(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = 2(0(1(L(x1)))) [0 0 0] [1] [0 0 0] [0] [1 1 0] [1 1 0] 0(2(1(x1))) = [0 0 0]x1 >= [0 0 0]x1 = 1(0(2(x1))) [0 0 0] [0 0 0] [1 1 0] [0] [1 1 0] R(2(1(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(2(x1)))) [0 0 0] [0] [0 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] [2 2 1] [2] [1 1 0] [1] 0(2(L(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 1(0(1(L(x1)))) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] R(2(0(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 = R(1(0(1(x1)))) [0 0 0] [0] [0 0 0] problem: 1(2(1(x1))) -> 2(0(2(x1))) 1(2(0(x1))) -> 2(0(1(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))) R(2(0(x1))) -> R(1(0(1(x1)))) String Reversal Processor: 1(2(1(x1))) -> 2(0(2(x1))) 0(2(1(x1))) -> 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))) 0(2(R(x1))) -> 1(0(1(R(x1)))) WPO Processor: algebra: Sum weight function: w0 = 0 popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Standard