details

property	value
status	complete
benchmark	logarithm.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n029.star.cs.uiowa.edu
space	AProVE_06

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.589321136 seconds
cpu usage	308.35209866
max memory	5.316022272E9

stage attributes

key	value
output-size	4541
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) inc(s(x)) -> s(inc(x)) inc(0) -> s(0) log(x) -> logIter(x, 0) logIter(x, y) -> if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y)) if(false, b, x, y) -> logZeroError if(true, false, x, s(y)) -> y if(true, true, x, y) -> logIter(x, y) 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: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) inc(s(x)) -> s(inc(x)) inc(0) -> s(0) log(x) -> logIter(x, 0) logIter(x, y) -> if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y)) if(false, b, x, y) -> logZeroError if(true, false, x, s(y)) -> y if(true, true, x, y) -> logIter(x, y) 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 le(s(x), s(y)) ->^+ le(x, y) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [x / s(x), y / s(y)]. The result substitution is [ ]. popout

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

job log

popout

actions

all output return to Runti Compl Full Rewri 10127