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Runtime Complexity: TRS Innermost pair #487111834
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
n144.star.cs.uiowa.edu
space
AProVE_09_Inductive
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
291.599 seconds
cpu usage
1086.49
user time
1071.94
system time
14.549
max virtual memory
3.7511908E7
max residence set size
1.501484E7
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), ?)
output
WORST_CASE(Omega(n^1), ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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 (innermost) 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: INNERMOST ---------------------------------------- (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 (innermost) 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: INNERMOST ---------------------------------------- (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 [ ]. ---------------------------------------- (4) Complex Obligation (BEST)
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