details

property	value
status	complete
benchmark	thiemann18.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.800318003 seconds
cpu usage	311.193142319
max memory	5.319213056E9

stage attributes

key	value
output-size	6369
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) min(cons(x, nil)) -> x min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) empty(nil) -> true empty(cons(n, x)) -> false head(cons(n, x)) -> n tail(nil) -> nil tail(cons(n, x)) -> x sort(x) -> sortIter(x, nil) sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil))) if(true, x, y, z) -> y if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z) 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: eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) min(cons(x, nil)) -> x min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) empty(nil) -> true empty(cons(n, x)) -> false head(cons(n, x)) -> n tail(nil) -> nil tail(cons(n, x)) -> x sort(x) -> sortIter(x, nil) sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil))) if(true, x, y, z) -> y if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z) popout

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

job log

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actions

all output return to Runti Compl Full Rewri 10127