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Runti Compl Full Rewri 10127 pair #381903427
details
property
value
status
complete
benchmark
Ex6_15_AEL02_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n040.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.88119101524 seconds
cpu usage
8.439295642
max memory
1.376247808E9
stage attributes
key
value
output-size
8433
starexec-result
WORST_CASE(NON_POLY, ?)
output
/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).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 (9) InfiniteLowerBoundProof [FINISHED, 818 ms] (10) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) sel(0, cons(X, Z)) -> X first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(s(X))) sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z)) sel1(0, cons(X, Z)) -> quote(X) first1(0, Z) -> nil1 first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z))) quote(n__0) -> 01 quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z))) quote1(n__nil) -> nil1 quote(n__s(X)) -> s1(quote(activate(X))) quote(n__sel(X, Z)) -> sel1(activate(X), activate(Z)) quote1(n__first(X, Z)) -> first1(activate(X), activate(Z)) unquote(01) -> 0 unquote(s1(X)) -> s(unquote(X)) unquote1(nil1) -> nil unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) fcons(X, Z) -> cons(X, Z) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) 0 -> n__0 cons(X1, X2) -> n__cons(X1, X2) nil -> n__nil s(X) -> n__s(X) sel(X1, X2) -> n__sel(X1, X2) activate(n__first(X1, X2)) -> first(X1, X2) activate(n__from(X)) -> from(X) activate(n__0) -> 0 activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__nil) -> nil activate(n__s(X)) -> s(X) activate(n__sel(X1, X2)) -> sel(X1, X2) 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(INF, INF). The TRS R consists of the following rules: sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) sel(0, cons(X, Z)) -> X first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(s(X))) sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z)) sel1(0, cons(X, Z)) -> quote(X) first1(0, Z) -> nil1 first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z))) quote(n__0) -> 01 quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z))) quote1(n__nil) -> nil1
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