Sequence Comparison.pdf

6.3 Distance between Non-Overlapping Hits 101
and
By Theorem 6.3,
C 11 (x)=
|π|
∑
k=1
p |π|−k x |π|−k ,
w−1
A θ =[ p i ],
∑
i=0
[ ]
0 1
M θ =
−p w ∑i=0 w−1 .
pi
µ θ =
w
∑
i=1
(1/p) i .
Example 6.5. Continue Example 6.3. For the spaced seed π = 1 a ∗1 b , a ≥ b ≥ 1, we
have
⎡
A π = ⎣ ∑b−1 i=0 pa+i q + 1 ∑ a−1 ⎤
i=0 pb+i q
⎦.
∑ b−1
i=0 pa+1+i ∑ a+b
i=0 pi
Therefore,
µ π = ∑a+b i=0 pi + ∑ b i=0 ∑ b−1
j=0 pa+i+ j q
p a+b (1 + p(1 − p b .
))
6.3.2 An Upper Bound for µ π
A spaced seed is uniform if its matching positions form an arithmetic sequence.
For example, 1**1**1 is uniform with matching position set {0,3,6} in which the
difference between two successive positions is 3. The unique spaced seed of weight
2 and length m is 1 ∗ m−2 1. Therefore, all the spaced seeds of weight 2 are uniform.
In general, a uniform seed is of form (1∗ k ) l 1, l ≥ 1 and k ≥ 0. In Example 6.2, we
have showed that Π i ≤ Θ i for any uniformly spaced seed π and the consecutive seed
θ of the same weight. By (6.9), µ π ≥ µ θ .
Now we consider non-uniformly spaced seeds. We have proved that µ θ =
∑ w θ
i=1 (1/p)i for consecutive seed θ. For any spaced seed π, by definition, µ π ≥|π|.
Thus, for any fixed probability p and the consecutive seed θ of the same weight,
µ π can be larger than µ θ when the length |π| of π is large. In this subsection, we
shall show that when |π| is not too big, µ π is smaller than µ θ for any non-uniformly
spaced seed π.
For any 0 ≤ j ≤|π|−1, define
RP(π)+ j = {i 1 + j,i 2 + j,···,i wπ + j}
and let
o π ( j)=|RP(π) ∩ (RP(π)+ j)|.