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Runti Compl Full Rewri 10127 pair #381902989
details
property
value
status
complete
benchmark
thiemann38.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
291.448531151 seconds
cpu usage
304.917856784
max memory
5.461876736E9
stage attributes
key
value
output-size
4493
starexec-result
WORST_CASE(Omega(n^1), ?)
output
/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). (0) CpxTRS (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (2) TRS for Loop Detection (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (4) BEST (5) proven lower bound (6) LowerBoundPropagationProof [FINISHED, 0 ms] (7) BOUNDS(n^1, INF) (8) TRS for Loop Detection ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: length(nil) -> 0 length(cons(x, l)) -> s(length(l)) lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) head(cons(x, l)) -> x head(nil) -> undefined tail(nil) -> nil tail(cons(x, l)) -> l reverse(l) -> rev(0, l, nil, l) rev(x, l, accu, orig) -> if(lt(x, length(orig)), x, l, accu, orig) if(true, x, l, accu, orig) -> rev(s(x), tail(l), cons(head(l), accu), orig) if(false, x, l, accu, orig) -> accu S is empty. Rewrite Strategy: FULL ---------------------------------------- (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (2) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF). The TRS R consists of the following rules: length(nil) -> 0 length(cons(x, l)) -> s(length(l)) lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) head(cons(x, l)) -> x head(nil) -> undefined tail(nil) -> nil tail(cons(x, l)) -> l reverse(l) -> rev(0, l, nil, l) rev(x, l, accu, orig) -> if(lt(x, length(orig)), x, l, accu, orig) if(true, x, l, accu, orig) -> rev(s(x), tail(l), cons(head(l), accu), orig) if(false, x, l, accu, orig) -> accu S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence lt(s(x), s(y)) ->^+ lt(x, y) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [x / s(x), y / s(y)]. The result substitution is [ ]. ----------------------------------------
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