details

property	value
status	complete
benchmark	Ex9_BLR02_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n042.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	291.777760029 seconds
cpu usage	310.94900534
max memory	6.216495104E9

stage attributes

key	value
output-size	5128
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).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), 46 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: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) sieve(X) -> n__sieve(X) nats(X) -> n__nats(X) activate(n__filter(X1, X2, X3)) -> filter(X1, X2, X3) activate(n__sieve(X)) -> sieve(X) activate(n__nats(X)) -> nats(X) activate(X) -> X 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: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) sieve(X) -> n__sieve(X) nats(X) -> n__nats(X) activate(n__filter(X1, X2, X3)) -> filter(X1, X2, X3) activate(n__sieve(X)) -> sieve(X) activate(n__nats(X)) -> nats(X) activate(X) -> X 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 sieve(cons(s(N), n__sieve(X1_0))) ->^+ cons(s(N), n__sieve(filter(sieve(X1_0), N, N))) gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0]. The pumping substitution is [X1_0 / cons(s(N), n__sieve(X1_0))]. The result substitution is [ ]. ---------------------------------------- popout

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all output return to Runti Compl Full Rewri 10127