Typesetting math equations in Latex

While writing any document in $\LaTeX$ using the amsmath package many wishes need to fullfilled. The most cumbersome task that troubles me frequently is typing the right and left delimitors (say \lvert \rvert) seperately. Why can't there be a command that does that on its own. Turns  out that somebody did write a package to improve upon amsmath : mathtools .

The most useful feature I found is, allowing the user to define paired delimitors For example, instead of typing,

    \left\{ \frac{a}{b+c} \right\}

You can just define paired delimitors and use them as follows,

    \usepackage{mathtools}
    \DeclarePairedDelimiter{\set}{\lbrace}{\rbrace}
    \set*{ \frac{a}{b+c} }

The output is the same as before,
$$\left\{ \frac{a}{b+c} \right\}$$

Using the * after the command resizes the delimitors to fit the vertical length of the expression.

Another interesting tool in this package is allowing tags for equations to have user-defined labels. See section 3.2.2 in  mathtools .

Another great tool for publishing documents on the web is mathjax. Recently mathjax allowed users to use mathjax scripts through their plugins on popular web platforms. 

Subadditivity, Limits and Lim Inf

Consider a sub-additive sequence $a_n$ for $n \geq 1$, where sub-additivity is defined as follows :
$$a_{n+m} \leq a_n + a_m$$

Fekete's Lemma says that for any such sequence $\lim_{n \rightarrow \infty} a_n/n = \lim \inf_{n \rightarrow \infty} a_n/n $, which can be easily proved. The proof is as follows-

Let $\lim \inf_{n \rightarrow \infty} a_n/n  = l$. Therefore, $\exists K$ s.t.
$$\left\vert \frac{a_K}{K} - l \right\vert = \epsilon/2$$

Now consider a large enough $L$ such that $ \frac{a_r}{KL} < \epsilon/2, \forall  r < K$. Thus, for every $n \geq KL$ we can write $n$ as $n = Kq + r$, where $q \geq L$ and $r<K$. Therefore,

\begin{align}
\frac{a_n}{n} & \leq \frac{q a_K}{Kq+r} + \frac{a_r}{Kq+r} \\
                        & \leq \frac{q a_K}{Kq} + \frac{a_r}{Kq} \\
                        & \leq \frac{a_K}{K} + \frac{a_r}{Kq} \\
                        & \leq  l + \frac{\epsilon}{2} + \frac{\epsilon}{2}\\
\end{align}

Since this is true for all $n > KL$ the limit exists and is equal to $l$. An alternate but similar proof is given here.

Pairwise Independence, Mutual Independence and Coding Theory

Consider a set of variables $\{X_1, X_2, ..., X_n\}$. A pair of variables $X_i, X_j$ are pairwise independent  iff $P(X_i,X_j) = P(X_i)  P(X_j)$. And any subset $A_k =\{X_{i_1},X_{i_2},...X_{i_k}\}$ of random variables is mutually independent if

                            $P(X_{i_1},X_{i_2},...,X_{i_k}) = \prod_j P(X_{i_j}) $

Bernstein gave an example of a set of random variables, of which any two are pairwise independent, but the set as a whole is not. A general class of examples which can be easily constructed from this set is the following -

$X_1 = A_1$
$X_2 = A_2$
$X_3 = A_3$
$\vdots$
$X_k = A_k$
$X_{k+1} = A_1+A_2+...+A_k$

where $A_i$'s are a set of i.i.d. mutually independent random variables, uniformly distributed in $\mathbb{F}_2$. Now it can be easily seen that any size $k$ subset $S_k$ of  $\{X_1, X_2, ..., X_{k+1}\}$ is a mutually independent set, but the $k+1$  set  $\{X_1, X_2, ..., X_{k+1}\}$  has a degenerate probability distribution given $A_k$ i.e.
                               $P(X_1, X_2, ..., X_{k+1} | A_k) \in \{0,1\}$

Now one can observe from the example that this is an example of parity check codes (wiki).  Also it is easy to observe that parity check codes are MDS codes. (wiki)

Now, given a general set of $n$ random variables which is $k$ independent (any $k$ subset is mutually independent) is it possible to construct an MDS code from these?

I give below conditions in which MDS code are constructable :

• The random variables $X_i$ are uniformly distributed in a field $\mathbb{F}_q$.
• The set $A_n = \{X_1, X_2, ..., X_n\}$ is $k$ independent i.e. any $k$ subset $A_k$ is mutually independent
• The probability distribution of $A_n$ given $A_k$ is degenerate i.e.

                                                    $P(A_n | A_k) \in \{0,1\}$

Now, given these conditions it can be easily seen from the proof of the Singleton bound (for general non-linear codes) that $\{X_1, X_2,...,X_n\}$ is an MDS code [length, information bits, min. distance] = $[ n, k, n-k+1]$.

Pirates, treasure, and data security

Suppose you have a file that you want to divide between $n$ users. Now the division should be such that whenever any $k$ of the $n$ persons collaborate they know the complete file otherwise they must not have any information about the file.

Interestingly the solution to this problem is related to the solution to the following problem :

Thirteen pirates go on an extended voyage, pillaging and plundering from Africa to Asia. By the end they have quite a stash--too much to take back with them. They decide to lock it in a chest, leave the chest on an island, and come back for it a year later. Of course, not being terribly trusting, they want to ensure that none of them can come back early and claim the treasure for himself. They could just put 13 locks on it and each take a key, but a pirate's life is dangerous--they may not all be around in a year. What they want is for any majority of the original thirteen to be able to open the chest, while any fewer will be locked out. How many locks will it take for them to achieve this? The locks they are using are quite simple. Each key opens only one lock (no master keys), but keys can be duplicated, so multiple pirates can have keys to the same lock.

Solution1:

For the pirate problem you should choose ${n \choose {k-1}}$ locks and distribute the keys such that for every group of ${k-1}$ users there is a lock that they do not have the key for.

This solution easily extends to the first problem. Let the data file be $x \in \mathbb{F}_q$. Now generate ${n \choose {k-1}}$ random variables $R_i, i \in \{1, \ldots, {n\choose {k-1}}\}$ and give each user the information $x + \sum_{i=1}^{n \choose {k-1}} R_i$. Now, the values of the random variables can be distributed like the keys of the lock in the previous puzzle, and we are done.

Solution2:

This is the solution proposed by Adi Shamir (the guy behind Bitcoins), here.

Basic Idea : If you are given any $k$ points of a $k-1$ degree polynomial $f(x)$, you can determine the polynomial. So,

secret := polynomial coefficients
user data := $n$ points of the polynomial

Ofcourse, the polynomial is over a finite field. You need only $k$ random variables in this scheme to share one bit ($a_0 = [x^0] f(x)$) instead on ${n \choose (k-1)}$ in the previous scheme.