191014K02 | Day 3 Lecture 1
191014K02: Day 3 Lecture 1
Work in progress.
Closest String
For two strings s_1, s_2 of the same length \ell, we define the Hamming Distance between S_1 & S_2 to be:
d\left(S_1, S_2\right):=\left|\left\{i: S_1[i] \neq S_2[i]\right\}\right|
so, for example, d(horse, force) =2.
For a string c and set S of stings of length \ell:
d(S, c):=\max _{t \in S} d(t,c)
Input: n strings of length \ell over an alphabet \Sigma1
Task: Find center sting c lot length \ell such that d(S, c) is minimized
GOAL: (1+\varepsilon) Approximation Algorithm
An Integer Linear Program for Closest String
Have an indicator variable for every position p and every letter \alpha \in \Sigma, introduce a binary variable x_{p,\alpha} with the following semantics:
\begin{equation*} x_{p,\alpha} = \begin{cases} 1 & \text{if } c[p] = \alpha\\ 0 & \text{otherwise.} \end{cases} \end{equation*}
The constraints:
- At every position we have one letter:
\forall p \in [\ell]: \sum_{\alpha \in \Sigma} x_{p,\alpha} = 1
We control the distance:
\forall t \in S: \quad \sum_{p=1}^\ell \left({\color{indianred}1-x_{p, t[p]}} \right) \leqslant d
…and ask the ILP to minimize d.
Note that:
\begin{equation*} 1 - x_{i,s[i]} = \begin{cases} 0 & \text{if the solution matches with } s[i] \text{at location } p,\\ 1 & \text{otherwise.} \end{cases} \end{equation*}
Now, as usual, relax the ILP and solve the associated LP.
Rounding. Think of the OPTLP variable values as “voting” for characters at each position. For example, if \ell = 5, \Sigma = \{A,C,G,T\}, and the OPTLP values turn out to be:
| A | C | G | T | |
|---|---|---|---|---|
| 1 | 0.9 | 0 | 0.05 | 0.05 |
| 2 | 0.1 | 0.2 | 0.4 | 0.3 |
| 3 | 0.5 | 0.1 | 0.1 | 0.3 |
| 4 | 0.1 | 0.7 | 0.1 | 0.1 |
| 5 | 0.2 | 0.2 | 0 | 0.6 |
Then you might be tempted to “round” the solution to AGACT because for each position p \in [5], the letters A, G, A, C, and T dominate the vote for that position. However: it turns out that this nautral rounding strategy can be arbitrarily bad!
Instead of picking the top choice, what we do instead is the following: for every position p \in [\ell], treat the x_{p,\alpha}’s as a probability distribution2 over \Sigma. Now the randomized rounding step involves sampling from this distribution to obtain the solution:
Set c[p] = \alpha with probability x_{p,\alpha}.
Define the indicator random variables q_{p,\alpha} as indicating for us when c is different from \alpha at position p, as a stepping stone to capturing distance eventually:
\begin{equation*} q_{p,\alpha} = \begin{cases} 1 & \text{if } {\color{indianred}c[p] \neq \alpha},\\ 0 & \text{otherwise}. \end{cases} \end{equation*}
Then we have the natural notion of a random variable to capture the distance between the string output by our randomized algorithm and any fixed string t \in S:
d[c,t] := \sum_{p \in \ell} q_{p,t[p]}
It’s time for our first cool claim.
Fix a t \in S. What is E[d(c,t)]?
It turns out: at most d, and therefore, at most OPTLP, and in turn, at most OPTILP = OPT
where the last inequality follows from the second LP constraint.
Note that d(c,t) is a sum of independent 0/1 random variables whose expectation is upper bounded by OPT. So our useful Chernoff variation applies here,
and we get the following:
\operatorname{Pr}[d(c,t)-\text{OPT} > \varepsilon \cdot \text{OPT}] \leqslant e^{-\varepsilon^2 \text{OPT} / 3}.
Now, applying union bound over all n choices of t \in S, we get:
\operatorname{Pr}[{\color{darkseagreen}d(c,S)} > (1+\varepsilon) \cdot \text{OPT}] \leqslant \frac{{\color{darkseagreen}n}}{e^{\varepsilon^2 \text{OPT} / 3}}.
So, if, for example:
\frac{n}{e^{\varepsilon^2 \text{OPT} / 3}} \leqslant \frac{1}{2},
then it’s a win!
We don’t have a win when OPT is really really small, in particular, if:
\text{OPT} \leqslant \frac{3\ln(2n)}{\varepsilon^2}
We handle this case with a local search algorithm.
Coming Soon.
