details

property	value
status	complete
benchmark	qsort.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	AProVE_09_Inductive

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.534778118 seconds
cpu usage	305.616992813
max memory	5.390462976E9

stage attributes

key	value
output-size	5304
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: qsort(nil) -> nil qsort(cons(x, xs)) -> append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs))))) filterlow(n, nil) -> nil filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs) if1(true, n, x, xs) -> filterlow(n, xs) if1(false, n, x, xs) -> cons(x, filterlow(n, xs)) filterhigh(n, nil) -> nil filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs) if2(true, n, x, xs) -> filterhigh(n, xs) if2(false, n, x, xs) -> cons(x, filterhigh(n, xs)) ge(x, 0) -> true ge(0, s(x)) -> false ge(s(x), s(y)) -> ge(x, y) append(nil, ys) -> ys append(cons(x, xs), ys) -> cons(x, append(xs, ys)) 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: qsort(nil) -> nil qsort(cons(x, xs)) -> append(qsort(filterlow(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs))))) filterlow(n, nil) -> nil filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs) if1(true, n, x, xs) -> filterlow(n, xs) if1(false, n, x, xs) -> cons(x, filterlow(n, xs)) filterhigh(n, nil) -> nil filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs) if2(true, n, x, xs) -> filterhigh(n, xs) if2(false, n, x, xs) -> cons(x, filterhigh(n, xs)) ge(x, 0) -> true ge(0, s(x)) -> false ge(s(x), s(y)) -> ge(x, y) append(nil, ys) -> ys append(cons(x, xs), ys) -> cons(x, append(xs, ys)) 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(x, xs), ys) ->^+ cons(x, append(xs, ys)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [xs / cons(x, xs)]. The result substitution is [ ]. popout

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job log

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actions

all output return to Runti Compl Full Rewri 10127