Cost-Based Optimization of Integration Flows - Datenbanken ...

4 Vectorizing Integration Flows
In the following, we formally analyze the time complexity for exhaustively solving this
problem. For this purpose, we analyze the complexity of the best and worst case and
combine this to the general result.
Lemma 4.1. The cost-based plan vectorization problem exhibits an exponential time complexity
of O(2 m ) for the best-case plan of an operator sequence.
Proof. The distribution function D of the number of possible plans over k is a symmetric
function according to Pascal’s Triangle 9 , where the condition l bi = l bk−i+1 with i ≤ m/2
holds. Based on Definition 2.1, a plan contains m operators. Due to Definition 4.2, we
search for k execution buckets b i with l bi ≥ 1 ∧ l bi ≤ m and ∑ |b|
i=1 l b i
= m. Hence, different
numbers of buckets k ∈ [1, m] have to be evaluated. From now on, we fix m ′ as m ′ = m−1
and k ′ as k ′ = k − 1. In fact, there is only one possible plan for k = 1 (all operators in
one bucket) and k = m (each operator in a different bucket), respectively:
( ) m
′
|P | k ′ =0 =
0
= 1 and |P ′ | k ′ =m ′ = ( m
′
m ′ )
= 1 . (4.9)
Now, without loss of generality, we fix a specific m. The number of possible plans for a
given k is then computed with
|P ′′ | k =
( m
′
k ′ )
=
( m ′ ) (
− 1 m
k ′ +
′ )
− 1
− 1 k ′ =
k ′
∏
i=1
m ′ + 1 − i
. (4.10)
i
In order to compute the total number of possible plans, we have to sum up the possible
plans for each k, with 1 ≤ k ≤ m:
|P ′′ | =
m ′
∑
k ′ =0
( m
′
k ′ )
with k ′ = k − 1 and m ′ = m − 1. (4.11)
Finally, ∑ ( )
n n
k=0
is known to be equal to 2
k
n . Hence, by changing the index k from
k ′ = 0 to k = 1, we can write:
|P ′′ | =
m ′
∑
k ′ =0
( m
′
k ′ )
=
m∑
k=1
( ) m − 1
= 2 m−1 . (4.12)
k − 1
In conclusion, there are 2 m−1 possible plans that must be evaluated. Due to the linear complexity
of O(m) for determining the costs of a plan, the cost-based plan vectorization problem
exhibits an exponential best-case overall time complexity of O (2 m ) = O ( m · 2 m−1) .
Hence, Lemma 4.1 holds.
Lemma 4.2. The cost-based plan vectorization problem exhibits an exponential time complexity
of O(2 m ) for the worst-case plan of a set of operators.
9 As an alternative to Pascal’s Triangle, we could also consider the m − 1 virtual delimiters between
operators. Due to the binary decision for each delimiter to be set or not, we get 2 m−1 different plans.
However, we used Pascal’s Triangle in order to be able to determine the number of plans for a given
number of execution buckets k.
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