details

property	value
status	complete
benchmark	qsortmiddle.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	AProVE_09_Inductive

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.513 seconds
cpu usage	313.308
user time	311.17
system time	2.13805
max virtual memory	1.8279408E7
max residence set size	5380788.0

stage attributes

key	value
starexec-result	WORST_CASE(Omega(n^1), ?)

output

WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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(xs) -> qs(half(length(xs)), xs) qs(n, nil) -> nil qs(n, cons(x, xs)) -> append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), 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)) length(nil) -> 0 length(cons(x, xs)) -> s(length(xs)) half(0) -> 0 half(s(0)) -> 0 half(s(s(x))) -> s(half(x)) get(n, nil) -> 0 get(n, cons(x, nil)) -> x get(0, cons(x, cons(y, xs))) -> x get(s(n), cons(x, cons(y, xs))) -> get(n, cons(y, xs)) 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(xs) -> qs(half(length(xs)), xs) qs(n, nil) -> nil qs(n, cons(x, xs)) -> append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), 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)) length(nil) -> 0 length(cons(x, xs)) -> s(length(xs)) half(0) -> 0 half(s(0)) -> 0 half(s(s(x))) -> s(half(x)) get(n, nil) -> 0 get(n, cons(x, nil)) -> x get(0, cons(x, cons(y, xs))) -> x get(s(n), cons(x, cons(y, xs))) -> get(n, cons(y, xs)) S is empty. Rewrite Strategy: FULL ---------------------------------------- popout

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

job log

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actions

all output return to Runtime Complexity: TRS