Sequence Comparison.pdf

6.2 Basic Formulas on Hit Probability 95
s[i]s[k + i]s[2k + i]···s[(l − 1)k + i], i = r + 1,r + 2,...,k − 1,
where l = ⌊ n k
⌋,r = n − kl − 1. Because π hits the first r + 1 sequences with probability
Π l+1 and the last k − 1 − r sequences with probability Π l , we have that
For any i, j ≥|π|, ( ∩ i−1
t=0Āt
subevent of ( ∩ i−1 ) (
0
Ā t=0 ∩
¯Π n ′ =(¯Π l+1 ) r+1 ( ¯Π l ) k−1−r . (6.4)
)
i+ j−1
are independent and ∩t=0
Ā t is a
)
∩ . Hence,
)
and
i+ j−1
(∩
(∩ ) (
i−1
¯Π i ¯Π j = Pr[
0
Ā t=0 ∩ ∩
t=i+|π|−1Āt
i+ j−1
t=i+|π|−1Āt
i+ j−1
t=i+|π|−1Āt
)] [
> Pr
∩
i+ j−1
t=0
Ā t
]
= ¯Π i+ j .
Hence, formula (6.4) implies that ¯Π ′ n > ¯Π n or equivalently Π ′ n < Π n for any n ≥|π ′ |.
6.2.1 A Recurrence System for Hit Probability
We have shown that the non-hit probability of a consecutive seed θ satisfies equation
(6.3). Given a consecutive seed θ and n > |θ|, it takes linear-time to compute
the hit probability Θ n . However, calculating the hit probability for an arbitrary seed
is rather complicated. In this section, we generalize the recurrence relation (6.3) to
a recurrence system in the general case.
For a spaced seed π,wesetm = 2 |π|−w π
.LetW π be the set of all m distinct strings
obtained from π by filling 0 or 1 in the ∗’s positions. For example, for π = 1∗11∗1,
W π = {101101,101111,111101,111111}.
The seed π hits the random sequence R at position n − 1 if and only if a unique
W j ∈ W π occurs at the position. For each j, letB ( n
j) denote the event that W j occurs
at the position n − 1. Because A n denotes the event that π hits the sequences R at
position n − 1, we have that A n = ∪ 1≤ j≤m B ( n j) and B ( n
j) ’s are disjoint. Setting
]
π n
( j) = Pr
[Ā0 Ā 1 ···Ā n−2 B ( j)
n−1
, j = 1,2,···,m.
We have
and hence formula (6.2) becomes
π n =
∑
1≤ j≤m
π ( j)
n
¯Π n = ¯Π n−1 − π n (1) − π n
(2) −···−π n (m) . (6.5)
Recall that, for any W j ∈ W π and a,b such that 0 ≤ a < b ≤|π|−1, W j [a,b] denotes
the substring of W j from position a to position b inclusively. For a string s, weuse