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Complexity_ITS 2019-03-21 04.46 pair #429990932
details
property
value
status
complete
benchmark
sect5-len.koat
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n068.star.cs.uiowa.edu
space
KoAT-2013
run statistics
property
value
solver
AProVE
configuration
complexity
runtime (wallclock)
1.83924 seconds
cpu usage
3.91777
user time
3.69016
system time
0.227613
max virtual memory
1.85207E7
max residence set size
221528.0
stage attributes
key
value
starexec-result
WORST_CASE(Omega(n^1), O(n^1))
output
3.74/1.81 WORST_CASE(Omega(n^1), O(n^1)) 3.88/1.81 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.88/1.81 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.88/1.81 3.88/1.81 3.88/1.81 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). 3.88/1.81 3.88/1.81 (0) CpxIntTrs 3.88/1.81 (1) Koat Proof [FINISHED, 78 ms] 3.88/1.81 (2) BOUNDS(1, n^1) 3.88/1.81 (3) Loat Proof [FINISHED, 144 ms] 3.88/1.81 (4) BOUNDS(n^1, INF) 3.88/1.81 3.88/1.81 3.88/1.81 ---------------------------------------- 3.88/1.81 3.88/1.81 (0) 3.88/1.81 Obligation: 3.88/1.81 Complexity Int TRS consisting of the following rules: 3.88/1.81 l0(A, B) -> Com_1(l1(0, B)) :|: TRUE 3.88/1.81 l1(A, B) -> Com_1(l1(A + 1, B - 1)) :|: B >= 1 3.88/1.81 l1(A, B) -> Com_1(l2(A, B)) :|: 0 >= B 3.88/1.81 3.88/1.81 The start-symbols are:[l0_2] 3.88/1.81 3.88/1.81 3.88/1.81 ---------------------------------------- 3.88/1.81 3.88/1.81 (1) Koat Proof (FINISHED) 3.88/1.81 YES(?, ar_1 + 2) 3.88/1.81 3.88/1.81 3.88/1.81 3.88/1.81 Initial complexity problem: 3.88/1.81 3.88/1.81 1: T: 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l0(ar_0, ar_1) -> Com_1(l1(0, ar_1)) 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l1(ar_0, ar_1) -> Com_1(l1(ar_0 + 1, ar_1 - 1)) [ ar_1 >= 1 ] 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l1(ar_0, ar_1) -> Com_1(l2(ar_0, ar_1)) [ 0 >= ar_1 ] 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(l0(ar_0, ar_1)) [ 0 <= 0 ] 3.88/1.81 3.88/1.81 start location: koat_start 3.88/1.81 3.88/1.81 leaf cost: 0 3.88/1.81 3.88/1.81 3.88/1.81 3.88/1.81 Repeatedly propagating knowledge in problem 1 produces the following problem: 3.88/1.81 3.88/1.81 2: T: 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 1) l0(ar_0, ar_1) -> Com_1(l1(0, ar_1)) 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l1(ar_0, ar_1) -> Com_1(l1(ar_0 + 1, ar_1 - 1)) [ ar_1 >= 1 ] 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l1(ar_0, ar_1) -> Com_1(l2(ar_0, ar_1)) [ 0 >= ar_1 ] 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(l0(ar_0, ar_1)) [ 0 <= 0 ] 3.88/1.81 3.88/1.81 start location: koat_start 3.88/1.81 3.88/1.81 leaf cost: 0 3.88/1.81 3.88/1.81 3.88/1.81 3.88/1.81 A polynomial rank function with 3.88/1.81 3.88/1.81 Pol(l0) = 1 3.88/1.81 3.88/1.81 Pol(l1) = 1 3.88/1.81 3.88/1.81 Pol(l2) = 0 3.88/1.81 3.88/1.81 Pol(koat_start) = 1 3.88/1.81 3.88/1.81 orients all transitions weakly and the transition 3.88/1.81 3.88/1.81 l1(ar_0, ar_1) -> Com_1(l2(ar_0, ar_1)) [ 0 >= ar_1 ] 3.88/1.81 3.88/1.81 strictly and produces the following problem: 3.88/1.81 3.88/1.81 3: T: 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 1) l0(ar_0, ar_1) -> Com_1(l1(0, ar_1)) 3.88/1.81 3.88/1.81 (Comp: ?, Cost: 1) l1(ar_0, ar_1) -> Com_1(l1(ar_0 + 1, ar_1 - 1)) [ ar_1 >= 1 ] 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 1) l1(ar_0, ar_1) -> Com_1(l2(ar_0, ar_1)) [ 0 >= ar_1 ] 3.88/1.81 3.88/1.81 (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(l0(ar_0, ar_1)) [ 0 <= 0 ] 3.88/1.82 3.88/1.82 start location: koat_start 3.88/1.82 3.88/1.82 leaf cost: 0 3.88/1.82 3.88/1.82
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