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Compl Integ Trans Syste 26843 pair #381744341
details
property
value
status
complete
benchmark
randomwalk.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n042.star.cs.uiowa.edu
space
T2
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
5.50632810593 seconds
cpu usage
14.267688602
max memory
3.75988224E8
stage attributes
key
value
output-size
7536
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox/benchmark/theBenchmark.koat # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, INF). (0) CpxIntTrs (1) Loat Proof [FINISHED, 3846 ms] (2) BOUNDS(1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D, E, F, G) -> Com_1(f10(1, 1, H, 0, 2, 1, G)) :|: H >= 0 f10(A, B, C, D, E, F, G) -> Com_1(f21(A, B - 1, C, D, E, F, 0)) :|: F >= 1 && E >= F && B >= 1 && 0 >= C f10(A, B, C, D, E, F, G) -> Com_1(f21(A + 1, A + 1, H, D, E, F, I)) :|: I >= 0 && 1 >= I && H >= 0 && F >= 1 && E >= F && 0 >= B && 0 >= C f10(A, B, C, D, E, F, G) -> Com_1(f21(A, B, C - 1, D, E, F, H)) :|: H >= 0 && 1 >= H && F >= 1 && C >= 1 && E >= F f21(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, D, E, F - 1, G)) :|: 0 >= G f21(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, D, E, F + 1, G)) :|: G >= 1 f10(A, B, C, D, E, F, G) -> Com_1(f31(A, B, C, D, E, F, G)) :|: 0 >= F && E >= F f10(A, B, C, D, E, F, G) -> Com_1(f31(A, B, C, D, E, F, G)) :|: F >= 1 + E The start-symbols are:[f0_7] ---------------------------------------- (1) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f0 0: f0 -> f10 : A'=1, B'=1, C'=free, D'=0, E'=2, F'=1, [ free>=0 ], cost: 1 1: f10 -> f21 : B'=-1+B, G'=0, [ F>=1 && E>=F && B>=1 && 0>=C ], cost: 1 2: f10 -> f21 : A'=1+A, B'=1+A, C'=free_2, G'=free_1, [ free_1>=0 && 1>=free_1 && free_2>=0 && F>=1 && E>=F && 0>=B && 0>=C ], cost: 1 3: f10 -> f21 : C'=-1+C, G'=free_3, [ free_3>=0 && 1>=free_3 && F>=1 && C>=1 && E>=F ], cost: 1 6: f10 -> f31 : [ 0>=F && E>=F ], cost: 1 7: f10 -> f31 : [ F>=1+E ], cost: 1 4: f21 -> f10 : F'=-1+F, [ 0>=G ], cost: 1 5: f21 -> f10 : F'=1+F, [ G>=1 ], cost: 1 Removed unreachable and leaf rules: Start location: f0 0: f0 -> f10 : A'=1, B'=1, C'=free, D'=0, E'=2, F'=1, [ free>=0 ], cost: 1 1: f10 -> f21 : B'=-1+B, G'=0, [ F>=1 && E>=F && B>=1 && 0>=C ], cost: 1 2: f10 -> f21 : A'=1+A, B'=1+A, C'=free_2, G'=free_1, [ free_1>=0 && 1>=free_1 && free_2>=0 && F>=1 && E>=F && 0>=B && 0>=C ], cost: 1 3: f10 -> f21 : C'=-1+C, G'=free_3, [ free_3>=0 && 1>=free_3 && F>=1 && C>=1 && E>=F ], cost: 1 4: f21 -> f10 : F'=-1+F, [ 0>=G ], cost: 1 5: f21 -> f10 : F'=1+F, [ G>=1 ], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on tree-shaped paths): Start location: f0 0: f0 -> f10 : A'=1, B'=1, C'=free, D'=0, E'=2, F'=1, [ free>=0 ], cost: 1 8: f10 -> f10 : B'=-1+B, F'=-1+F, G'=0, [ F>=1 && E>=F && B>=1 && 0>=C ], cost: 2 9: f10 -> f10 : A'=1+A, B'=1+A, C'=free_2, F'=-1+F, G'=free_1, [ free_1>=0 && free_2>=0 && F>=1 && E>=F && 0>=B && 0>=C && 0>=free_1 ], cost: 2 10: f10 -> f10 : A'=1+A, B'=1+A, C'=free_2, F'=1+F, G'=free_1, [ 1>=free_1 && free_2>=0 && F>=1 && E>=F && 0>=B && 0>=C && free_1>=1 ], cost: 2
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