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Runti Compl Full Rewri 10127 pair #381902467
details
property
value
status
complete
benchmark
toList.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
292.047512054 seconds
cpu usage
305.354504679
max memory
5.28738304E9
stage attributes
key
value
output-size
5150
starexec-result
WORST_CASE(Omega(n^1), ?)
output
/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: isEmpty(empty) -> true isEmpty(node(l, x, r)) -> false left(empty) -> empty left(node(l, x, r)) -> l right(empty) -> empty right(node(l, x, r)) -> r elem(node(l, x, r)) -> x append(nil, x) -> cons(x, nil) append(cons(y, ys), x) -> cons(y, append(ys, x)) listify(n, xs) -> if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n)) if(true, b, n, m, xs, ys) -> xs if(false, false, n, m, xs, ys) -> listify(m, xs) if(false, true, n, m, xs, ys) -> listify(n, ys) toList(n) -> listify(n, nil) 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: isEmpty(empty) -> true isEmpty(node(l, x, r)) -> false left(empty) -> empty left(node(l, x, r)) -> l right(empty) -> empty right(node(l, x, r)) -> r elem(node(l, x, r)) -> x append(nil, x) -> cons(x, nil) append(cons(y, ys), x) -> cons(y, append(ys, x)) listify(n, xs) -> if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n)) if(true, b, n, m, xs, ys) -> xs if(false, false, n, m, xs, ys) -> listify(m, xs) if(false, true, n, m, xs, ys) -> listify(n, ys) toList(n) -> listify(n, nil) 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 append(cons(y, ys), x) ->^+ cons(y, append(ys, x)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [ys / cons(y, ys)]. The result substitution is [ ].
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