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Runtime_Complexity_Innermost_Rewriting 2019-04-01 06.40 pair #433313432
details
property
value
status
complete
benchmark
ExIntrod_GM04_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n112.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
2.59915 seconds
cpu usage
7.30165
user time
6.89267
system time
0.408982
max virtual memory
1.8279172E7
max residence set size
1198228.0
stage attributes
key
value
starexec-result
WORST_CASE(NON_POLY, ?)
output
6.91/2.56 WORST_CASE(NON_POLY, ?) 7.27/2.57 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 7.27/2.57 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 7.27/2.57 7.27/2.57 7.27/2.57 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 7.27/2.57 7.27/2.57 (0) CpxTRS 7.27/2.57 (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 7.27/2.57 (2) TRS for Loop Detection 7.27/2.57 (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 7.27/2.57 (4) BEST 7.27/2.57 (5) proven lower bound 7.27/2.57 (6) LowerBoundPropagationProof [FINISHED, 0 ms] 7.27/2.57 (7) BOUNDS(n^1, INF) 7.27/2.57 (8) TRS for Loop Detection 7.27/2.57 (9) InfiniteLowerBoundProof [FINISHED, 652 ms] 7.27/2.57 (10) BOUNDS(INF, INF) 7.27/2.57 7.27/2.57 7.27/2.57 ---------------------------------------- 7.27/2.57 7.27/2.57 (0) 7.27/2.57 Obligation: 7.27/2.57 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 7.27/2.57 7.27/2.57 7.27/2.57 The TRS R consists of the following rules: 7.27/2.57 7.27/2.57 nats -> adx(zeros) 7.27/2.57 zeros -> cons(n__0, n__zeros) 7.27/2.57 incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) 7.27/2.57 adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 7.27/2.57 hd(cons(X, Y)) -> activate(X) 7.27/2.57 tl(cons(X, Y)) -> activate(Y) 7.27/2.57 0 -> n__0 7.27/2.57 zeros -> n__zeros 7.27/2.57 s(X) -> n__s(X) 7.27/2.57 incr(X) -> n__incr(X) 7.27/2.57 adx(X) -> n__adx(X) 7.27/2.57 activate(n__0) -> 0 7.27/2.57 activate(n__zeros) -> zeros 7.27/2.57 activate(n__s(X)) -> s(X) 7.27/2.57 activate(n__incr(X)) -> incr(X) 7.27/2.57 activate(n__adx(X)) -> adx(X) 7.27/2.57 activate(X) -> X 7.27/2.57 7.27/2.57 S is empty. 7.27/2.57 Rewrite Strategy: INNERMOST 7.27/2.57 ---------------------------------------- 7.27/2.57 7.27/2.57 (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) 7.27/2.57 Transformed a relative TRS into a decreasing-loop problem. 7.27/2.57 ---------------------------------------- 7.27/2.57 7.27/2.57 (2) 7.27/2.57 Obligation: 7.27/2.57 Analyzing the following TRS for decreasing loops: 7.27/2.57 7.27/2.57 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 7.27/2.57 7.27/2.57 7.27/2.57 The TRS R consists of the following rules: 7.27/2.57 7.27/2.57 nats -> adx(zeros) 7.27/2.57 zeros -> cons(n__0, n__zeros) 7.27/2.57 incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) 7.27/2.57 adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 7.27/2.57 hd(cons(X, Y)) -> activate(X) 7.27/2.57 tl(cons(X, Y)) -> activate(Y) 7.27/2.57 0 -> n__0 7.27/2.57 zeros -> n__zeros 7.27/2.57 s(X) -> n__s(X) 7.27/2.57 incr(X) -> n__incr(X) 7.27/2.57 adx(X) -> n__adx(X) 7.27/2.57 activate(n__0) -> 0 7.27/2.57 activate(n__zeros) -> zeros 7.27/2.57 activate(n__s(X)) -> s(X) 7.27/2.57 activate(n__incr(X)) -> incr(X) 7.27/2.57 activate(n__adx(X)) -> adx(X) 7.27/2.57 activate(X) -> X 7.27/2.57 7.27/2.57 S is empty. 7.27/2.57 Rewrite Strategy: INNERMOST 7.27/2.57 ---------------------------------------- 7.27/2.57 7.27/2.57 (3) DecreasingLoopProof (LOWER BOUND(ID)) 7.27/2.57 The following loop(s) give(s) rise to the lower bound Omega(n^1): 7.27/2.57 7.27/2.57 The rewrite sequence 7.27/2.57 7.27/2.57 activate(n__incr(cons(X1_0, Y2_0))) ->^+ cons(n__s(activate(X1_0)), n__incr(activate(Y2_0))) 7.27/2.57 7.27/2.57 gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0]. 7.27/2.57 7.27/2.57 The pumping substitution is [X1_0 / n__incr(cons(X1_0, Y2_0))]. 7.27/2.57 7.27/2.57 The result substitution is [ ]. 7.27/2.57 7.27/2.57
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