What makes a good/bad lecture?

December 30, 2019

In the opinion of 45 Royal Holloway 2nd year students the answers, taking only those mentioned at least three times, are:

What makes a good lecture?

  • Engaging/enthusiastic lecturer (22)
  • Interactive (8)
  • Clear voice (7)
  • Eye contact with audience (5)
  • Checking for understanding (4)
  • Clear (4)
  • Clear handwriting (4)
  • Jokes/humour (4)
  • Seems interested in what they’re saying (4)
  • Well prepared (4)
  • Examples (3)
  • Sound excited/animated (3)
  • High quality notes (3)

What makes a bad lecture?

  • Too quiet (18)
  • Reading off board/book/page/slide (17)
  • Facing the board (9)
  • Not engaging with audience (8)
  • Monotone (7)
  • Poor handwriting (7)
  • Boring (5)
  • Not enough explanation (5)
  • Slow (3)
  • Too much talking (3)

Putnam 2018: determinant of an adjacency matrix

October 20, 2019

I took in the October issue of the American Mathematical Monthly to entertain me during the lulls while manning the mathematics desk at our open day on Saturday. As traditional, the first article was the questions in this year’s Putnam. Question A2 asked for the determinant of the adjacency matrix of the graph on the non-empty subsets of \{1,2,\ldots, n\} in which two subsets are connected if and only if they intersect.

Three hours later, hardly overwhelmed by student visits, although I did get to wheel out infinitely many primes and 0.\dot{9} = 1 to the evident interest of several potential students — and their parents, I had at most the outline of a fiddly solution that piled row operations on half-remembered determinantal identities.

Here is the far simpler solution I found when it came to write it up. The real purpose of this post is to make the connection with the Schur algebra and permutation representations of the symmetric group, and to record some related matrices that also have surprisingly simple determinants or spectral behaviour. I also include an explicit formula for the inverse of the adjacency matrix; this can be expressed using the Möbius function of the poset of subsets of \{1,2,\ldots, n\}, for reasons I do not yet fully understand.

A2. Let S_1, S_2, \ldots, S_{2^n-1} be the nonempty subsets of \{1,2,\ldots, n\} in some order and let M be the (2^n-1) \times (2^n-1) matrix whose (i,j) entry is

\displaystyle m_{ij} = \begin{cases} 0 & S_i \cap S_j = \varnothing \\ 1 & S_i \cap S_j \neq \varnothing \end{cases}

Calculate the determinant of M.

Solution. As is implicit in the question, we can order the sets as we like, since swapping two rows and then swapping the two corresponding columns does not change the determinant of M. We put the 2^{n-1} - 1 sets not having 1 first, then the 2^{n-1} sets having 1. With this order, continued recursively, the matrix M_n for n-subsets has the block form

\left( \begin{matrix} M_{n-1} & R_{n-1} \\ R_{n-1}^t & K_{n-1} \end{matrix} \right)

where R_k is the (2^k-1) \times 2^k matrix with first column all zero and then M_{k} to its right, and K_k is the k \times k all-ones matrix. For example,

M_3 = \left( \begin{matrix} 1 & 0 & 1 & {\bf 0} & 1 & 0 & 1 \\ 0 & 1 & 1 & {\bf 0} & 0 & 1 & 1 \\ 1 & 1 & 1 & {\bf 0} & 1 & 1 & 1 \\ {\bf 0} & {\bf 0} & {\bf 0} & 1 & 1 & 1 & 1 \\ 1& 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 &1 &1 \\ 1 & 1 & 1 & 1 & 1 &1 &1 \end{matrix} \right)

where the exceptional zero column in R_{2} and zero row in R_{2}^t are shown in bold.

Clearly \det M_1 = 1. We shall show, by induction on n, that \det M_n = -1 if n \ge 2. The 3 \times 3 matrix M_2 can be seen as a submatrix of M_3 above; computing the determinant as a sum over the 6 permutations of \{1,2,3\} gives 1 - 1 -1 = -1, as required.

For the inductive step, observe that row 2^{n-1} of M_n has 2^{n-1}-1 zeros (from the top row of R_{n-1}^t) followed by 2^{n-1} ones (from the top row of K_{n-1}). By row operations subtracting this row from each of the rows below, we obtain

\left( \begin{matrix} M_{n-1} & 0_{(2^{n-1}-1) \times 1} & M_{n-1} \\ 0_{1 \times (2^{n-1}-1)} & 1 & 1_{1 \times (2^{n-1}-1)} \\ M_{n-1} & 0_{(2^{n-1}-1) \times 1)} & 0_{(2^{n-1}-1) \times (2^{n-1}-1)} \end{matrix} \right).

The unique non-zero entry in row 2^{n-1} and column 2^{n-1} is the one in position (2^{n-1},2^{n-1}). Therefore deleting this row and column, we find that

\det M_n = \det \left( \begin{matrix} M_{n-1} & M_{n-1} \\ M_{n-1} & 0_{(2^{n-1}-1) \times (2^{n-1}-1)} \end{matrix} \right).

Given the zero matrix in the bottom right, it is impossible to obtain a non-zero contribution in the determinant by any permutation that picks an entry in the top-left copy of M_{n-1}. Therefore

\det M_n = -(\det M_{n-1})^2 = -1

as claimed.


The inverse matrix M_n^{-1} can be computed explicitly: labelling rows and columns by subsets of \{1,2,\ldots, n\} we have

\displaystyle -(M_n^{-1})_{XY} = \begin{cases} (-1)^{|Y \backslash X'|} & Y \supseteq X' \\ 0 & Y \not\supseteq X' \end{cases}

where X' is the complement of X in \{1,2,\ldots, n\}. Can the appearance of the Möbius function \mu(Z,Y) = (-1)^{Y \backslash Z} of the poset of subsets of \{1,2,\ldots, n\} be explained as an instance of Mobius inversion?

Symmetric group

Let \Omega be the set of all subsets of \{1,2,\ldots, n\} and let \mathbb{F} be an infinite field. The vector space \mathbb{F}\Omega with basis \Omega is a permutation module for the symmetric group S_n on \{1,2,\ldots, n\}. The Putnam matrix, extended by an extra zero row and column corresponding to the empty set, is the endomorphism of this permutation module that sends a subset to the sum of all the subsets that it intersects. This interpretation of the Putnam matrix suggests a rich family of variations on the theme.

Variations on the Putnam matrix

Diagonal blocks

Fix k \in \{1,\ldots, n\} with k \le n/2. Let \Omega^{(k)} be the set of k-subsets of \{1,2,\ldots, n\} and let M^{(k)}_n be the block of M_n corresponding to k-subsets of \{1,2,\ldots, n\}. Thus M^{(k)}_n is the endomorphism of the permutation module \mathbb{F}\Omega^{(k)} sending a k-subset to all the sum of all the k-subsets that it intersects. The permutation module \mathbb{F}\Omega^{(k)} has a multiplicity-free direct sum decomposition

U = U_0 \oplus U_1 \oplus \cdots U_k

where U_j has irreducible character canonically labelled by the partition (n-j,j). (Since k \le n/2 this is a partition.) It therefore follows from Schur’s Lemma that M_n^{(k)} has k+1 integer eigenvalues whose multiplicities are the dimensions of the U_i. Varying j and n, these numbers form a triangle whose diagonal entries, i.e. those for which j = \lfloor n/2 \rfloor, are the Catalan Numbers.

For example, by working with the filtration of \mathbb{F}\Omega^{(2)} by the Specht modules S^{(n-2,2)} \cong U_2, S^{(n-1,1)} \cong U_1 and S^{(n)} \cong U_0 one can show that the eigenvalues are 2n+1, n-3 and -1 with multiplicities 1, n-1 and n(n-3)/2, respectively. Computational data suggests that in general the eigenvalue associated to the partition (n-j,j) for j \ge 1 is (-1)^{j-1} \binom{n-k-j}{k-j}; probably this can be proved by a generalization of the filtration argument. (But maybe there is a better approach just using character theory.) Since the unique trivial submodule is spanned by the sum of all k-subsets, the eigenvalue for the partition (n) is simply the number of k-subsets meeting \{1,2,\ldots, k\}, namely \binom{n}{k} - \binom{n-k}{k}.

Weighting additions and removals by indeterminates

Let P_n(\alpha,\beta, \theta) be the generalization of the Putnam matrix where the entry remains zero for disjoint subsets, and if X meets Y then the entry is \alpha^{|X \backslash Y|}\beta^{|Y\backslash X|}\theta^{2|X\cap Y|}. For example, P_2(\alpha,\beta,\theta) is

\left( \begin{matrix} \theta^2 & 0 & \alpha\theta^2 \\ 0 & \theta^2 & \alpha \theta^2 \\ \beta\theta^2 & \beta\theta^2 & \theta^4 \end{matrix} \right)

where rows and columns are labelled \{1\}, \{2\}, \{1,2\}, and P_3(\alpha,\beta,1) is shown below, with rows and columns labelled \{1\}, \{2\}, \{1,2\}, \{3\}, \{1,3\}, \{2,3\}, \{1,2,3\}

\left( \begin{matrix} 1 & 0 & \alpha & 0 & \alpha & 0 & \alpha^2 \\ 0 & 1 & \alpha & 0 & 0 & \alpha & \alpha^2 \\ \beta & \beta & 1 & 0 & \alpha\beta & \alpha\beta & \alpha \\ 0 & 0 & 0 & 1 & \alpha & \alpha & \alpha^2 \\ \beta & 0 & \alpha\beta & \beta & 1 & \alpha\beta & \alpha \\ 0 & \beta & \alpha\beta & \beta & \alpha\beta & 1 & \alpha \\ \beta^2 & \beta^2 & \beta & \beta^2 & \beta & \beta & 1 \end{matrix} \right).

Specializing \alpha, \beta and \theta to 1 we get the Putnam matrix for n = 3 shown above. When P_n(\alpha,\beta,\theta) is written as a sum over the (2^n-1)! permutations of the collection of non-empty subsets of \{1,\ldots, n\}, each summand factorizes as a product over the disjoint cycles in the corresponding permutation. If (X_0, \ldots, X_{\ell-1}) is such a cycle then the power of \alpha is \sum_{i=1}^{\ell}|X_i \backslash X_{i-1}| (taking X_{\ell} = X_0), and similarly the power of \beta is \sum_{i=1}^{\ell} |X_{i-1}\backslash X_i|. Clearly these are the same. Moreover, by the identity

|X\backslash Y| + |Y\backslash X| + 2|X\cap Y| = |X| + |Y| = 2k,

after taking the product over all disjoint cycles, the total degree is \sum_{k=1}^n k \binom{n}{k} = n2^{n-1}. Therefore the determinant of the matrix a homogeneous polynomial in \alpha, \beta, \theta of degree n2^{n-1}. Since swapping \alpha and \beta corresponds to transposing the matrix, the determinant is a function of \alpha + \beta, \alpha\beta and \theta. We have seen that each summand of the determinant the same power of \alpha as \beta, and so the determinant depends only on \gamma = \alpha\beta. Therefore the determinant factorizes into homogeneous polynomials in \alpha\beta (degree 2) and \theta^2 (also degree 2). Hence we may specialize so that \alpha\beta = 1 and replace \theta^2 with \tau without losing any information. (This turns out to be the most convenient form for computation because, using MAGMA at least, it is fast to compute determinants of matrices whose coefficients lie in the single variable polynomial ring \mathbb{Q}[\tau] but much slower if the coefficients lie in \mathbb{Q}[\alpha,\beta,\tau].) The determinants for small n are shown below.

\begin{array}{ll} 1 & \tau \\ 2 & \tau^3(\tau-2) \\ 3 & \tau^7(\tau-2)^3(\tau^2-3\tau+3)\\ 4 & \tau^{15}(\tau-2)^7 (\tau^2-3\tau+3)^4 (\tau^2-2\tau+2)\end{array}

To continue, it is useful to define f_2 = \tau - 2 and for odd primes p, a polynomial f_p \in \mathbb{Z}[\tau] of degree p-1 by \tau f_p = (\tau-1)^p + 1. (The definition becomes uniform on changing 1 to (-1)^{p-1}.) For instance f_3 = \tau^2 - 3\tau + 3 and

f_5 = \tau^4 - 5\tau^3 + 10 \tau^2 - 10 \tau + 5.

We also define the following further polynomials:

\begin{aligned} f_4 &= \tau^2 - 2\tau + 2 \\ f_6 &= \tau^2 - \tau + 1 \\ f_8 &= \tau^4 - 4\tau^3 + 6\tau^2 -4\tau + 2 \\ f_{9} &= \tau^6 - 6\tau^5 + 15\tau^4 - 21\tau^3 + 18\tau^2 - 9\tau + 3 \\ f_{10} &= \tau^4 - 3\tau^3 + 4\tau^2 - 2\tau + 1 \\ f_{12} &= \tau^4 -4\tau^3 + 5\tau^2 - 2\tau + 1 \end{aligned}

Using this notation, the determinants for n \le 13 are

\begin{array}{ll} 1 & \tau \\ 2 & \tau^3 f_2 \\ 3 & \tau^7 f_2^3 f_3 \\ 4 & \tau^{15} f_2^7 f_3^4 f_4 \\ 5 & \tau^{31} f_2^{15} f_3^{10} f_4^{5} f_5 \\ 6 & \tau^{63} f_2^{31}f_3^{21}f_4^{15}f_5^6 f_6 \\ 7 & \tau^{127} f_2^{63} f_3^{42} f_4^{35} f_5^{21} f_6^{7} f_7 \\ 8 & \tau^{255} f_2^{127} f_3^{84} f_4^{71} f_5^{56}f_6^{28} f_7^8 f_8 \\ 9 & \tau^{511} f_2^{255} f_3^{169} f_4^{135} f_5^{126} f_6^{84} f_7^{36} f_8^{9} f_9 \\ 10 & \tau^{1023} f_2^{511} f_3^{340} f_4^{255} f_5^{253}f_6^{210}f_7^{120}f_8^{45} f_9^{10}f_{10}\\ 11 &\tau^{2047}f_2^{1023}f_3^{682}f_4^{495}f_5^{473}f_6^{462}f_7^{330}f_8^{165}f_9^{55}f_{10}^{11}f_{11} \\ 12 & \tau^{4095} f_2^{2047} f_3^{1365}f_4^{991}f_5^{858} f_6^{925}f_7^{792} f_8^{495} f_9^{220}f_{10}^{66} f_{11}^{12}f_{12} \\ 13 & \tau^{8191}f_2^{4095}f_3^{2730}f_4^{2015}f_5^{1573} f_6^{1729}f_7^{1716}f_8^{1287}f_9^{715}f_{10}^{286}f_{11}^{78}f_{12}^{13}f_{13} \end{array}

Note that f_n has degree \phi(n) in every case where it is defined, and that f_n(1) = 1 for all n except n =2, for which f_2(1) = -1. Assuming that this behaviour continues, and that f_2 always has exponent 2^{n-1}-1, as suggested by the data above, this gives another solution to the Putnam problem.

Weighting by integers

Given (w_0,w_1,\ldots, w_n) \in \mathbb{Z}^{n+1}, let Q_n(w) be the generalization of the Putnam matrix where if X meets Y then the entry is w_{|X\cap Y|}. When w_0 = 0 and w_i = (-1)^{i-1} for 1 \le i \le n, the determinants are given by specializing the matrix above by \alpha = 1, \beta = 1 and \theta = \sqrt{-1}, and so they are obtained from the factorizations in \mathbb{Q}[\tau] presented above by specializing \tau to -1.

The evaluations at -1 of the polynomials defined above are as follows: f_2(-1) = -3, f_p(-1) = 2^p - 1 for p an odd prime, and f_4(-1) = 5, f_6(-1) = 3, f_8(-1) = 18, f_9(-1) = 73, f_{10}(-1) = 11, f_{12}(-1) = 13.

Now suppose that w_i = (-1)^{i} for each i, so now w_0 = 1. For instance the matrix for n =3 is

\left( \begin{matrix} - & + & - & + & - & + & - \\ + & - & - & + & + & - & - \\ - & - & + & + & - & - & + \\ + & + & + & - & - & - & - \\ - & + & - & - & + & - & + \\ + & - & - & - & - & + & + \\ - & - & + & - & + & + & - \end{matrix}\right)

where + denotes 1 and - denotes -1. The determinants for n \in \{1,\ldots, 10\} are

1, 2^2, 2^9, 2^{28}, 2^{75}, 2^{186}, 2^{441}, 2^{1016}, 2^{2295}, 2^{5110}.

The sequence of exponents is in OEIS A05887, as the number of labelled acyclic digraphs with n vertices containing exactly n-1 points of in-degree zero. The determinant itself enumerates unions of directed cycles, with sign and a weighting, in the complete graph on n vertices; does this give the connection, or is it a new interpretation of sequence A05887?

Schur algebra

Let me finish by making the connection with an apparently completely unrelated object. Consider representations of the general linear group \mathrm{GL}_2(\mathbb{F}) in which the entries of the representing matrices are polynomials of a fixed degree in the four entries of each \mathrm{GL}_2(\mathbb{F}) matrix. For example, let E = \langle e_1, e_2 \rangle be the natural representation of \mathrm{GL}_2(\mathbb{F}). Then E itself is a polynomial representation of degree 1, and its symmetric square \mathrm{Sym}^2 \!E with basis e_1^2, e_1e_2, e_2^2 is a polynomial representation of degree 2, in which

\left( \begin{matrix} \alpha & \beta \\ \gamma & \delta \end{matrix} \right) \mapsto \left( \begin{matrix} \alpha^2 & 2\alpha\beta & \beta^2 \\ \alpha\gamma & \alpha\delta + \beta\gamma & \beta\delta \\ \gamma^2 & 2\gamma\delta & \delta^2 \end{matrix} \right).

By definition, \mathrm{Sym}^2 \! E is a quotient of E \otimes E. More generally, any polynomial representation of degree n is a subquotient of a direct sum of copies of E^{\otimes n}. Observe that there is a linear isomorphism E^{\otimes n} \rightarrow \mathbb{F}\Omega defined by

e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_n} \mapsto \bigl\{j \in \{1,2,\ldots, n\} : i_j = 2 \bigr\}.

The action of S_r on \mathbb{F}\Omega induces the place permutation action on E^{\otimes n}, defined (using right actions) by

e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_n} \sigma = e_{i_{1\sigma^{-1}}} \otimes e_{i_{2\sigma^{-1}}} \otimes \cdots \otimes e_{i_{n\sigma^{-1}}}.

For example, the 3-cycle (1,2,3) sends e_{a} \otimes e_{b} \otimes e_{c} to e_{c} \otimes e_{a} \otimes e_{b}, moving each factor one to the right (with wrap around). This gives some motivation for introducing the Schur algebra S(n,2), defined to be the algebra of linear endomorphisms of the polynomial representation E^{\otimes n} of \mathrm{GL}_2(\mathbb{F}) that commute with the place permutation action of S_n. In symbols,

S(n,2) = \mathrm{End}_{\mathbb{F}S_n}(E^{\otimes n}).

Somewhat remarkably one can show that the category of polynomial representations of \mathrm{GL}_2(\mathbb{F}) of degree n is equivalent to the module category of the Schur algebra. (See for instance the introduction to Green’s lecture notes.) The extended Putnam matrix corresponds to the element of the Schur algebra sending the pure tensor e_{i_1} \otimes \cdots \otimes e_{i_n} to the sum of all the pure tensors e_{j_1} \otimes \cdots \otimes e_{j_n} such that i_k = j_k = 1 for at least one k. For example, if n=2 then

\left( \begin{matrix} \alpha & \beta \\ \gamma & \delta \end{matrix} \right) \mapsto \left( \begin{matrix} \alpha^2 & \alpha\beta & \alpha\beta & \beta^2 \\ \alpha\gamma & \alpha\delta & \beta\gamma & \beta\delta \\ \beta^2 & \beta\gamma & \alpha\delta & \beta\delta \\ \gamma^2 & \gamma\delta & \gamma\delta & \delta^2 \end{matrix} \right), \  M_2 \mapsto \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \end{matrix} \right)

We see that S(2,n) is 10-dimensional, and, since the image of M_2 has equal coefficients in the required positions, it lies in S(2,n) as expected. For another example,

\left( \begin{matrix} 1 & 0 \\ \delta & 1 \end{matrix} \right) \mapsto \left( \begin{matrix} 1 & 0 & 0 & 0 \\ \delta & 1 & 0 & 0 \\ \delta & 1 & 1 & 0 \\ \delta^2 & \delta & \delta & 1 \end{matrix} \right)

corresponds to the endomorphism of \mathbb{F}\Omega sending a subset of \{1,2\} to the sum of all its subsets, weighted by a power of \delta recording how many elements were removed. This case is unusual in that the Schur algebra element comes from a single matrix in \mathrm{GL}_2(\mathbb{F}), rather than a linear combination of matrices lying in the group algebra \mathbb{F}\mathrm{GL}_2(\mathbb{F}), as would be required for the Putnam matrix. Passing to the Schur algebra replaces the infinite dimensional algebra \mathbb{F}\mathrm{GL}_2(\mathbb{F}) with a finite dimensional quotient.

Two proofs of Shannon’s Source Coding Theorem and an extension to lossy coding

September 29, 2019

The first purpose of this post is to record two approaches to Shannon’s Source Coding Theorem for memoryless sources, first by explicit codes and secondly using statistical properties of the source. After that I try, with mixed success, to give a motivated proof of a more general result on lossy coding.

Setup and statement.

Suppose that a memoryless source produces the symbols 1, \ldots, m with probabilities p(1), \ldots, p(m), respectively. Let

h = H(p(1),\ldots, p(m))

be the entropy of this probability dis tribution. Since the source is memoryless, the entropy of an r-tuple of symbols is then rh. Shannon’s Source Coding Theorem (or First Main Theorem, or Noiseless Coding Theorem) states that, given \epsilon > 0, provided r is sufficiently large, there is a variable length binary code of average length less than r(h + \epsilon) that can be used to encode all r-ruples of symbols.

Proof by Kraft–McMillan inequality

Choose r sufficiently large so that r > 1/\epsilon. Let q(1), \ldots, q(m^r) be the probabilities of the r-tuples and let \ell(j) \in \mathbb{N} be least such that \ell(j) \ge \log_2 1/q(j). Equivalently, \ell(j) is least such that

2^{-\ell(j)} \le q(j).

This equivalent characterisation shows that

\sum_{j=1}^{m^r} 2^{-\ell(j)} \le \sum_{j=1}^{m^r} q(j) =1

so by the Kraft–McMillan inequality, there exists a prefix free binary code with these codeword lengths. Moreover, since

\ell(j) = \lceil \log_2 1/q(j) \rceil \le 1 + \log_2 \frac{1}{q(j)}

we have

\begin{aligned}\sum_{j=1}^{m^r} q(j)\ell(j) &\le \sum_{j=1}^{m^r} q(j) \bigl( 1 + \log_2 \frac{1}{q(j)} \bigr) \\ &= \sum_{j=1}^{m^r} q(j) + \sum_{j=1}^{m^r} q(j) \log_2 \frac{1}{q(j)} &= 1 + hr \end{aligned}

as required.

The required prefix-free code is the Shannon code for the probabilities p(1), \ldots, p(m^r). Another algorithmic construction is given by the proof of the sufficiency of the Kraft–McMillan inequality on page 17 of Welsh Codes and cryptography. (The same proof is indicated, rather less clearly in my view, on the Wikipedia page.)

Proof using Asymptotic Equipartition Property (AEP).

Despite its forbidding name, for memoryless sources the AEP is little more than the Weak Law of Large Numbers, albeit applied in a slightly subtle way to random variables whose values are certain (functions of) probabilities. Let \mathcal{M} = \{1,\ldots, m\} be the probability space of symbols produced by the source. Let X \in \mathcal{M} be a random symbol produced by the source. Then \log p(X) is a random variable. Summing over the m outcomes in the probability space, we find that

\mathbb{E}[\log p(X)] = \sum_{i=1}^m \mathbb{P}[X=i] \log p(i) =  \sum_{i=1}^m p(i) \log p(i) = -h

where h = H(p(1), \ldots, p(m)) is the entropy already seen. While straightforward, and giving some motivation for considering \log p(X), the calculation hides some subtleties. For example, is it true that

\mathbb{P}[\log p(X) = \log p(i)] = p(i)?

In general the answer is ‘no’: in fact equality holds if and only if i is the unique symbol with probability p(i). (One further thing, that I certainly will not even dare to mention in my course, is that, by the formal definition of a random variable, X is the identity function on \mathcal{M}, and so p(X) is a composition of two functions; in Example 2.3.1 of Goldie and Pinch Communication Theory the authors remark that forming such compositions is a ‘characteristic feature of information theory, not found elsewhere in probability theory’.)

Anyway, returning to the AEP, let (X_1, \ldots, X_r) \in \mathcal{M}^r be a random r-tuple. For brevity we shall now call these r-tuples messages. Let P_r : \mathcal{M}^r \rightarrow [0,1] be defined by

P_r\bigl( (x_1, \ldots, x_r) \bigr) = p(x_1) \ldots p(x_r)

Since the source is memoryless, P_r\bigl( (x_1,\ldots, x_r)\bigr) is the probability that the first r symbols produced by the source are x_1, \ldots, x_r. Consider the random variable \log P_r\bigl( (X_1, \ldots, X_r) \bigr). By the displayed equation above,

\log P_r\bigl( (X_1,\ldots, X_r) \bigr) = \sum_{i=1}^r \log p(X_i).

This is a sum of r independent identically distributed random variables. We saw above that each has expected value -h. Hence, by the weak law, given any \eta > 0,

\displaystyle \mathbb{P}\Bigl[ -\frac{\log P_r\bigl( (X_1, \ldots, X_r) \bigr)}{r} \not\in (h-\eta, h+\eta) \Bigr] \rightarrow 0

as r \rightarrow \infty. By taking r sufficiently large so that the probability is less than \eta, we obtain a subset \mathcal{T} of \mathcal{M}^r such that (a)

\mathbb{P}[(X_1,\ldots, X_r) \in \mathcal{T}] > 1 - \eta

and (b) if (x_1,\ldots, x_r) \in \mathcal{T} then

P_r\bigl( (x_1, \ldots, x_r) \bigr) \in (2^{-r(H+\eta)}, 2^{-r(H-\eta)}).

The properties (a) and (b) say that memoryless sources satisfy the AEP. It is usual to call the elements of \mathcal{T} typical messages.

For example, if p(1) \ge \ldots \ge p(m) then the most likely message is (1,\ldots, 1)^r and the least likely is (m,\ldots, m); unless the distribution is uniform, both are atypical when \eta is small. A typical message will instead have each symbol i with frequency about rp(i). Indeed, the proof of the AEP for memoryless sources on page 80 of Welsh Codes and cryptography takes this as the definition as typical, using the Central Limit Theorem to replace the weak law so that ‘about’ can be defined precisely.

We can now prove the Source Coding Theorem in another way: if (x_1, \ldots, x_r) is typical then

P_r\bigl( (x_1,\ldots, x_r)\bigr) \ge 2^{-r(H+\eta)}

and so

\begin{aligned} 1 &\ge \mathbb{P}[(X_1,\ldots, X_r) \in \mathcal{T}] \\ &= \sum_{(x_1,\ldots, x_r) \in \mathcal{T}} P_r\bigl( (x_1, \ldots, x_r) \bigr) \\ &\ge |\mathcal{T}| 2^{-r(H+ \eta)}.\end{aligned}

Hence |\mathcal{T}| \;\text{\textless}\; r(h+\eta). For the atypical messages we make no attempt at efficiency, and instead encode them using binary words of length N for some N > r \log_2 m. Thus the code has (at most) two distinct lengths of codewords, and the expected length of a codeword is at most

L + \eta M.

By taking r sufficiently large and L = \lfloor r(h+ \eta)\rfloor + 1, we may assume that L < r(h + 2\eta). Therefore the expected length is at most

r(h + 2\eta) + \eta r \log_2 m;

when \eta is sufficiently small this is less than r(h+ \epsilon).

In Part C of my course I plan to give this proof before defining the AEP, and then take (a) and (b) as its definition for general sources. As a preliminary, I plan to prove the Weak Law of Large Numbers from Chebyshev’s Inequality.

AEP for general sources

Going beyond this, things quickly get difficult. Recall that a source producing symbols X_1, X_2, \ldots is stationary if (X_{i_1},\ldots,X_{i_t}) and (X_{{i_1}+s},\ldots X_{i_t+s}) have the same distribution for all i_1, \ldots, i_t and s, and ergodic if with probability 1 the observed frequencies of every t-tuple of symbols converges to its expected value. There is a nice proof using Fekete’s Lemma in Welsh Codes and cryptography (see page 78) that any stationary source has an entropy, defined by

\displaystyle \lim_{r \rightarrow \infty} \frac{H(X_1, X_2,\ldots, X_r)}{r}.

However the proof that a stationary ergodic source satisfies the AEP is too hard for my course. (For instance the sketch on Wikipedia blithely uses Lévy’s Martingale Convergence Theory.) And while fairly intuitive, it is hard to prove that a memoryless source is ergodic — in fact this is essentially the Strong Law of Large Numbers — so there are difficulties even in connecting the memoryless motivation to the general result. All in all this seems to be a place to give examples and state results without proof.

Lossy source coding

In §10 of Cover and Thomas, Elements of information theory, Wiley (2006) there is a very general but highly technical result on source encoding sources satisfying the AEP when a non-zero error probability is permitted. After much staring at the proof in §10.5 and disentangling of the definitions from rate-distortion theory, I was able to present a very simple version for the unbiased memoryless binary source in my course. As a rare question, someone asked about the biased memoryless binary source. Here is (one version of) the relevant result. Throughout let h(p) = H(p,1-p).

Theorem. Let 0 \;\textless\; D \;\textless\; p. Provided r is sufficiently large, there exists a code C \subseteq \{0,1\}^r such that

|C| = \lceil 2^{(h(p)-h(D)+\epsilon)r} \rceil

and an encoder f : \{0,1\}^r \rightarrow C such that

\mathbb{P}[d(f(S_1\ldots S_r), u) > Dr] < \epsilon.

Here d denotes Hamming distance. So the conclusion is that with probability \ge 1- \epsilon, a random word emitted by the source can be encoded by a codeword in C obtained by flipping at most Dr bits. Therefore each bit is correct with probability at least 1-D. By sending the number of a codeword we may therefore compress the source by a factor of h(p)-h(D)+\epsilon, for every \epsilon \;\textgreater\; 0. Note that we assume D < p, since if a bit error probability exceeding p is acceptable then there is no need for any encoding: the receiver simply guesses that every bit is 0.

In the lecture, I could not see any good way to motivate the appearance of h(p) - h(D), so instead I claimed that one could compress by a factor of h(p)\bigl(1-h(D)\bigr). The argument I had in mind was that by the AEP (or Shannon's Noiseless Coding Theorem) one can represent the typical words emitted by the source using all binary words of length h(p)r. Then compress to h(p)(1-h(D))r bits by thinking of these binary words as emitted by an unbiased source, using the special case of the theorem above when p = \frac{1}{2}. The problem is that decoding gives a bitwise error probability of D on the binary words of length h(p)r used to encode the source, and this does not give good control of the bitwise error probability on the original binary words of length r emitted by the source.

In an attempt to motivate the outline proof below, we take one key idea in the proof of the more general result in Cover and Thomas: the code C should be obtained by taking a random word X emitted by the source and flipping bits to obtain a codeword Y such that I(X;Y) is minimized, subject to the constraint that the expectation of d(X,Y) is \le Dr.

In the special case I presented in the lecture all words are equally likely to be emitted by the source, and so flipping bits at random gives codewords distributed uniformly at random on \{0,1\}^r. As a warm up, we use such codewords to prove the theorem in this special case.

Proof when p = \frac{1}{2}. Fix w \in \{0,1\}^r and let P be the probability that a codeword U, chosen uniformly at random from \{0,1\}^r, is within distance Dr of w. Since U is uniformly distributed, P is the size of the Hamming ball of radius Dr about U divided by 2^r. Standard bounds show that

\displaystyle \frac{1}{(r+1)} \frac{2^{h(D)r}}{2^r} \le P \le \frac{2^{h(D)r}}{2^r}.

By the lower bound, if we pick slightly more than 2^{(1-h(D))r} codewords uniformly at random, then the expected number of codewords within distance Dr of w is at least 1. More precisely, let M = \lceil 2^{(1-h(D)+\epsilon)r} \rceil be as in the statement of the theorem above. Let C be a random code constructed by chosing M codewords independently and uniformly at random from \{0,1\}^r. The probability that no codeword in C is within distance Dr of w is at most

\displaystyle \bigl( 1 - \frac{1}{2^{-(1-h(d))r}} \bigr)^{2^{(1-h(d)+\epsilon)r}} \le \exp \bigl( -\frac{2^{(1-h(d)+\epsilon)r}}{2^{(1-h(d))r}} \bigr) = \exp (-2^{\epsilon r})

which tends to 0 as r \rightarrow \infty. Since w was arbitrary, the above is also the probability that no codeword is within distance Dr of a word in \{0,1\}^r chosen uniformly at random. Therefore this average probability, \overline{P} say, can also be made less than \epsilon. Hence there exists a particular code with codewords u(1), \ldots, u(M) such that, for this code, \overline{P} \le \epsilon. Thus at least 1-\epsilon of all words in \{0,1\}^n are within distance Dr of some codeword in C. \Box

It seemed highly intuitive to me that, since by the AEP, we need only consider those w of weight about pr, the right construction in general would be to take such words and construct random codewords by randomly flipping Dr of their bits. But some thought shows this cannot work: for example, take p = \frac{1}{3} and D = \frac{1}{6}. Then flipping bits at random gives codewords of average weight

\frac{1}{3}(1-\frac{1}{6})r + \frac{2}{3}\frac{1}{6}r = \frac{7}{18}r.

Let w = 1\ldots 10\ldots 0 be of weight \frac{1}{3}r. Then w is at distance

\displaystyle \binom{\frac{1}{3}r}{\frac{1}{18}r} \binom{\frac{2}{3}r}{\frac{1}{9}r}  \approx 2^{\frac{1}{3} h(\frac{1}{6}r) + \frac{2}{3} h(\frac{1}{6})r} = 2^{h(\frac{1}{6})r}

words of weight \frac{7}{18}r, and to pick a random code from words of this weight so that the expectated number of codewords close to w is at least 1, we must choose M codewords where

\displaystyle M \times \frac{2^{h(\frac{1}{6})r} }{2^{h(\frac{7}{18})r}} \ge 1

If the denominator was 2^{h(\frac{1}{3}r)}, we could take M = 2^{(h(\frac{1}{3})-h(\frac{1}{6}))r} as required. But because we are stuck in the bigger space of words of weight (\frac{1}{3} + \frac{1}{18})r, we require a larger M.

Some of my confusion disappeared after doing the minimization of I(X;Y) more carefully. Since I(X;Y) = H(X) - H(X|Y) and H(X) = H(p) is constant, a good — although maybe not very intuitive — way to do this is to choose the probabilities \mathbb{P}[X=\alpha|Y=\beta] to maximize H(X|Y), consistent with the requirements that if \mathbb{P}[Y=1] = q then

(1-q)\mathbb{P}[X=0|Y=0] + q\mathbb{P}[X=0|Y=1] = p


(1-q)\mathbb{P}[X=1|Y=0] + q\mathbb{P}[X=0|Y=1] \le D.

Replacing the inequality with equality (it is a safe assumption that the expected distortion is maximized when the mutual information is minimized), and solving the linear equations, one finds that the flip probabilities are

\mathbb{P}[X=1|Y=0] = \frac{D+p-q}{2(1-q)},\ \mathbb{P}[X=0|Y=1] = \frac{D+q-p}{2q}.

Thus one may pick any q such that p-D \le q \le p+D. Since H(X|Y) = (1-q) h(\mathbb{P}[X=1|Y=0]) + qh(\mathbb{P}[X=0|Y=1]), substituting from the equations above gives a formula for H(X|Y) in terms of p, D and q. Differentiating with respect to q (this is fiddly enough to warrant computer algebra), one finds that

\begin{aligned} \displaystyle 2\frac{\mathrm{d}H(X|Y)}{\mathrm{d}q} = \log_2& \frac{(D+p-q)(2-D-p-q)}{(1-q)^2} \\ &- \log_2 \frac{(D-p+q)(-D+p+q)}{q^2}\end{aligned}

and so the extrema occur when

\displaystyle \frac{(D+p-q)(2-D-p-q)}{(1-q)^2} = \frac{(D-p+q)(-D+p+q)}{q^2}.

For no reason that I can see, on multiplying out, all powers of q^3 and q^4 cancel, and one is left with the factoring quadratic


Hence H(X|Y) is extremized only when q = (p-D)/(1-2D) and both flip probabilities are D and when q = (p-D)/(1-2p), when \mathbb{P}[X=1|Y=0]=(D-Dp-p^2)/(1+D-3p) and \mathbb{P}[X=0|Y=1]=p. In the first solution the second derivative satisfies

\displaystyle \frac{\mathrm{d}^2H(X|Y)}{\mathrm{d}q^2} = -\frac{(1-2D)^2}{2(1-D)D(p-D)(1-D-p)}

so it is a maximum. Consistent with the failed attempt above, we have q < p, so the space of codewords is smaller than the space of source words. The extreme value is H(X|Y) = h(D). Again this is somewhat intuitive: the minimum mutual information consistent with a distortion of D should be the uncertainty in the (D,1-D) probability distribution. But note that the distortion is applied to codewords, rather than words emitted by the source. (A related remark is given at the end of this post.)

To emphasise the difference, the left diagrams below show the probabilities going from Y to X, relevant to H(X|Y). The right diagram shows the probabilities going from X to Y, relevant to H(Y|X) and need in the proof below.

Flip probabilities

Note that the flip probabilities going from X to Y are not, in general equal. In the example above with p=\frac{1}{3} and D = \frac{1}{6} we have q = \frac{1}{4} and the flip probabilities from X to Y are \gamma = \frac{1}{16} and \delta = \frac{3}{8}. The extreme value is H(X|Y) = h(\frac{1}{6}) \approx 0.650.

The second solution appears to be spurious, i.e. not corresponding to a zero of the derivative. But since two maxima are separated by another zero of the derivative, it cannot be a maximum. This solution exists only when D-Dp-p^2 \ge 0, a condition equivalent to D \ge p \frac{p}{1-p}. In the example above q = \frac{1}{2} and (by complete chance) \mathbb{P}[X=1|Y=0] = 0 and \mathbb{P}[X=0|Y=1] = \frac{1}{3}. The extreme value is H(X|Y) = \frac{1}{2}\log_2 3 - \frac{1}{3} \approx 0.459.

Outline Proof. Let q = (p-D)/(1-2p). Let M = \lceil 2^{(h(p)-h(D)+\epsilon)r} \rceil as in the statement of the theorem. Choose codewords U(1), \ldots, U(M) independently and uniformly at random from the binary words of length r of weight qr. Let w be a fixed word of weight pr. When we flip \gamma r 0s and \delta r 1s in w we obtain a binary word of weight

\begin{aligned} pr + &(1-p) \delta r - p \epsilon r \\ &= \bigl( p + \frac{D(p-D)}{1-2D} - \frac{D(1-p-D)}{1-2D}\bigr)r \\ &= \bigl( \frac{p(1-2D) + D(p-D) - D(1-p-D)}{1-2D}\bigr)r \\ &= \bigl( \frac{p-2Dp + Dp - D^2 - D - Dp +D^2}{1-2D} \\ &= \frac{p-D}{1-2D}\bigr)r \\ &= qr \end{aligned}

as expected. Moreover, the number of binary words of weight qr we obtain in this way is \binom{pr}{\epsilon pr} \binom{(1-p)r}{\delta pr}. By the bounds above, the probability that a codeword of weight qr, chosen uniformly at random, is within distance Dr of w is about

2^{pr h(\delta) + (1-p)r h(\gamma)-h(q)}.

Considering all M codewords, the probability that no codeword is within distance Dr of w is, by the same exponential approximation used earlier, about

\exp (- 2^{(h(p)-h(D) + \epsilon + ph(\delta) - (1-p)h(\gamma) - h(q))r}).

Now, as one would hope, and as can be shown using Mathematica, with FullSimplify and the assumptions that 0 < p < 1 and 0 < d < p, we have h(p)-h(D) + ph(\delta) - (1-p)h(\gamma) - h(q) = 0. The end of the proof is therefore as before. \Box

Remark. It is very tempting to start on the other side, and argue that a codeword U chosen uniformly at random from the set of binary words of weight qr is within distance Dr of 2^{qrh(D) + (1-q)r h(D)} = 2^{h(D)r} source words of weight p by flipping exactly qrD 0s and (1-q)rD 1s. This avoids the horrendous algebraic simplification required in the proof above. But now it is the source word that is varying, not the codeword. One can argue this way by varying both, which is permitted by the AEP since typically source words have about pn 1s. This combines both sources of randomness: the random code and the random source word in a slightly subtle way, typical of the way the AEP is used in a practice.

Linearly recurrent sequences and LFSRs

September 17, 2019

Part B of my Cipher Systems course is on stream ciphers. Modern stream ciphers, such as Trivium, are typically constructed by judiciously introducing non-linearity into a long-period linear stream cipher. The linear case is therefore still important.

By one mathematical definition, a Linear Feedback Shift Register (LFSR) is a function F : \mathbb{F}_2^\ell \rightarrow \mathbb{F}_2^\ell of the form

(x_0,\ldots,  x_{\ell-2}, x_{\ell-1}) \mapsto \bigl(x_1,\ldots,x_{\ell-1}, f(x_0,,\ldots, x_{\ell-1})\bigr)

where the feedback function f is linear. If

f(x_0,x_1,\ldots, x_{\ell-1}) = \sum_{t \in T} x_{\ell -t}

then we say that f has taps T and that \ell is the width of F. The keystream generated by F for a key (k_0,\ldots,k_{\ell-1}) \in \mathbb{F}_2^\ell is the sequence k_0,\ldots, k_{\ell-1},k_{\ell}, k_{\ell+1}, \ldots , where each k_s for s \ge \ell is defined recursively by

k_s = f(k_{s-\ell}, \ldots, k_{s-1}) = \sum_{t \in T} k_{s-t}.

Equivalently, for s \ge \ell,

F\bigl( k_{s-\ell}, \ldots, k_{s-2}, k_{s-1} \bigr) = (k_{s-\ell+1}, \ldots, k_{s-1}, k_s) \quad(\star).

The equation (\star) is close to the hardware implementation of an LFSR: fill \ell registers with the key k_0k_1\ldots k_{\ell-1} and then repeatedly step by putting the sum of the bits k_{\ell-t} for each t \in T into the rightmost register, shifting the other registers to the left. (In hardware, these bits would be ‘tapped’ by physical wires.) The keystream can be read off from the rightmost register. This interpretation makes it clear that an LFSR of width \ell is invertible if and only if it taps the bit about to leave the register for good: that is \ell must be a tap.

The period of an invertible LFSR F is the least P \in \mathbb{N} such that F^P = F. One way to find P uses that in the canonical basis of \mathbb{F}_2^\ell, the matrix (acting on column vectors) representing F is the companion matrix of the polynomial X^\ell + \sum_{t \in T}X^{\ell - t}. Thus F^m = F if and only if X^\ell + \sum_{t \in T}X^{\ell -t} divides X^m + 1. This may seem concise written here, but in practice it takes well over a lecture to remind students of matrices, find the matrix representing F and determine its minimal polynomial. All this adds up to a substantial diversion from cryptography; I fear it lost most students in my Cipher Systems course.

Instead, this year I’m going to give up on linear algebra and use the theory of linearly recurrent sequences and infinite power series. My plan is to subsume all of the ring-theoretic background into the following definition.

Definition. A polynomial f(X) annihilates a power series K(X) if f(X)K(X) is a polynomial.

For example, the LFSR of width 4 with taps \{3,4\} generates the keystream {\bf 0001}00110101111{\bf 0001} \ldots of period 15. The corresponding power series is

\begin{aligned}(X^3 + X^6 + &X^7 + X^9 + X^{11} + X^{12} + X^{13} + X^{14})(1+X^{15}  + \cdots ) \\ &= \displaystyle \frac{X^3 + X^6 + X^7 + X^9 + X^{11} + X^{12} + X^{13} + X^{14}}{1+X^{15}}.\end{aligned}

Clearly the power series is annihilated by 1+X^{15}. But since k_s = k_{s-3} + k_{s-4} for s \ge 4, it is also annihilated by 1+X^3+X^4. Correspondingly,

X^3 + X^6 + X^7 + X^9 + X^{11} + X^{12} + X^{13} + X^{14} + \cdots = \frac{X^3}{1+X^3+X^4}.

More generally, the keystream generated by an LFSR with taps T is annihilated by the feedback polynomial f_T(X) = 1 + \sum_{t \in T} X^t. Indeed, the coefficient of X^s in the product is k_s + \sum_{t \in T} k_{s-t} = 0, and so f_T(X)K(X) is a polynomial of degree at most \ell-1. Hence the period of F is the minimum m \in \mathbb{N} such that X^m + 1 is divisible by 1 + \sum_{t \in T} X^t. (This is equivalent to the condition found before by taking reciprocal polynomials.)

One quickly gets further non-trivial results.

Keystream of maximum period

Given h(X) = h_0 + h_1X + \cdots + h_{\ell-1}X^{\ell-1}, we claim that the equation K(X) = h(X)/f_T(X) defines a corresponding keystream. To avoid the need to define inverses in the ring of power series, let’s agree to interpret this equation as equivalent to

(1+\sum_{t \in T}X^t)(k_0 + k_1X + k_2X^2 + \cdots + k_sX^s + \cdots ) = h(X).

In turn, this is equivalent to the linear equations

h_s = k_s + \sum_{t \in T : t \le s} k_{s-t}

that must be satisfied by k_0, k_1, k_2, \ldots; clearly these equations have a unique solution. In particular, there is a keystream such that K(x) = 1/f_T(X). (This holds even if the LFSR is not invertible.) It’s obvious that K(x) is annihilated only by the multiples of f_T(X). In the invertible case we have seen that f_T(X) divides X^m + 1 if and only if m is a multiple of the period P of the LFSR. Therefore there is a keystream of period P.

(For comparison, using the linear algebra method, one argues that the matrix representing an LFSR is cyclic and that, as is clear from the left-shift in the definition, the corresponding \mathbb{F}_2[x]-module is generated by (0,\ldots,0,1). The period of the keystream for (0,\ldots, 0,1) is therefore the period of F. Again this is an argument that becomes far more long-winded when enough detail is included to make it accessible to students.)

Sum of two keystreams

Suppose, in any attempt to increase the linear complexity, we add the keystream {\bf 001}0111{\bf 001} \ldots of the LFSR of width 3 with taps \{2,3\} to the keystream with period 15 in the example above. The new keystream has period 105 and is annihilated by

(1+X^3+X^4)(1+X^2+X^3) = 1+X^2+X^4+X^5+X^7.

It is therefore a keystream of the LFSR with taps \{2,4,5,7\}. I plan to get students to find these taps by the slow method of solving the linear equations (\star), and then, I hope, impress them with how much faster things are with a bit more algebra.

(Incidentally one can have endless fun with the two competing conventions for taps. While less standard, this example shows that our choice in which k_s = \sum_{t \in T}k_{s-t} is a natural fit for the annihilator method. It is also clearly best for the Berlekamp—Massey algorithm.)

Length of all keystreams

One computer demonstration that I think worked quite well last year used a Mathematica notebook to compute the lengths of all keystreams for all invertible LFSRs of width \ell \le 5. I then showed that the keystreams of maximum possible period 2^{\ell}-1 came from the LFSRs whose minimal polynomial is primitive. Of course most, quite reasonably, didn’t know what ‘primitive’ meant, but they could see Mathematica did, and that there was some underlying pattern related to the factorization of the minimal polynomial.

Since the matrix representing an LFSR with taps T in the canonical basis of \mathbb{F}_2^\ell is the companion matrix of X^\ell + \sum_{t \in T}X^{\ell -t}, the corresponding \mathbb{F}_2[X]-module is

\mathbb{F}_2[X]/\langle X^\ell + \sum_{t \in T} X_{\ell - t}\rangle.

Switching to the reciprocal polynomial, the lengths of the keystreams are the orbit sizes of X in its action on \mathbb{F}_2[X]/\langle f_T(X) \rangle. When f_T(X) is irreducible and primitive, we get a single non-zero orbit. The general case is quite intricate, and stated in Theorem 8.63 in Lidl and Niederreiter Finite fields.

For this retelling, we need the following function. Given r \in \mathbb{N} with 2^{m-1} < r \le 2^m, let b(r) = m. Thus the sequence b(r) starts

0, 1, 2, 2, 3, 3, 3, 3, 4, \ldots

and b(r) is the number of bits in the binary form of r-1.

In the next two results, let g(X) \in \mathbb{F}_2[X] be an irreducible polynomial of degree d and let t be the order of X in the finite field \mathbb{F}_2[X]/g(X).

Proposition. Let r \ge 2. The orbits of X on \mathbb{F}_2[X]/g(X)^r not contained in the unique maximal ideal g(X)\mathbb{F}_2[X]/g(X)^r have size 2^{b(r)}t and there are \bigl(2^{rd} - 2^{(r-1)d}\bigr)/ 2^{b(r)} t of them.

Proof. Let J be the complement of the maximal ideal. By hypothesis, X^t + 1 is divisible by g(X). Since t is odd,

X^{2^m t} + 1 = (X^t+1)^{2^r}

is divisible by g(X)^r if and only if 2^m \ge r. Therefore X has order 2^{b(r)} t in its action on J. Since r \ge 2, the elements of J are invertible. Hence if p(X) \in J then p(X)X^m = p(X) mod g(X)^r if and only if X^m = 1 mod g(X)^r. It follows that all the orbits of X on J have size 2^{b(r)}t. Since J has size 2^{dr} - 2^{d(r-1)}, the number of orbits is as claimed. \Box

Theorem. The non-zero orbits of X on \mathbb{F}_2[X]/g(X)^e have sizes 2^b t for b \in \mathbb{N}_0. If 2^b \le e then the number of orbits of size 2^b t is

\displaystyle \frac{2^{2^b d} - 2^{2^{b-1}d}}{2^b t}.

The number of orbits of size 2^{b(e)} t is

\displaystyle \frac{2^{e d} - 2^{2^{b(e)-1}}d}{2^{b(e)} t}.

Proof. Let b \in \mathbb{N}. Suppose that 2^b \le e. For each r such that 2^{b-1} < r \le 2^{b} we have b(r) = b and the count of orbits of size 2^b t in the previous proposition telescopes as required. A similar telescoping works to count the orbits of size 2^{b(e)} t. Finally we count the zero orbit \{0\} and (2^d-1)/t orbits of size t, which together form the minimal ideal g(X)^{d-1} \mathbb{F}_2[X]/g(X)^d isomorphic to the finite field \mathbb{F}_2[X]/g(X). \Box

For example, take g(X) = (1+X+X^2)^5. Then b(e) = 3, t = 3 and the orbits of X have sizes 1, 3; 6^2; 12^{20}; 24^{32}, where the semicolons indicate the split according to b \in \mathbb{N} used in the theorem, and the exponents indicate multiplicities. These orbit sizes form the multiset \{1,3,6^2, 12^{20}, 24^{32} \}.

Just for the next corollary, we define the product of two multisets to be the multiset obtained by multiplying the elements pairwise, respecting multiplicities. For example

\{1,3,6^2\} \times \{1,7\} = \{1,3,6^2,7,21,42^2\}.

Corollary. Let F be an invertible LFSR with taps T. Let f_T(X) = g_1^{e_1} \ldots g_c^{e_c} be the irreducible factorization of its feedback polynomial. Then the multiset of orbit sizes of F is the product of the multisets of orbit sizes for each factor g_i^{e_i}, as given by the previous theorem.

Proof. This is routine from the Chinese Remainder Theorem. \Box

A toy model for the proof of Shannon’s Noisy Coding Theorem for the BSC

September 6, 2019

In the Binary Symmetric Channel (BSC) with symbol error probability p, an input bit flips with probability p < 1/2. A binary code of length n can be used to communicate on the BSC by sending each bit of a codeword u, collecting the n received bits as a word v still of length n, and then decoding v as the nearest codeword u' to v with respect to Hamming distance. For instance, using the repetition code of length 2e+1, a decoding error occurs if and only if e+1 or more bits get flipped in the channel. This event has probability

\begin{aligned}\sum_{k=e+1}^n \binom{n}{k} p^k(1-p)^{n-k} &\le p^{e+1}(1-p)^e \sum_{k=e+1}^n \binom{n}{k} \\ &\le p^e (1-p)^e 2^{n-1}  \\ &\le \rho^e\end{aligned}

where \rho = 4p(1-p) < 1. Hence we can communicate with decoding error probability \log \epsilon / \log \rho. For example, if p = 1/4 then \rho = 3/4 and \rho^e < 5\% provided n \ge 23. With this solution we send 23 bits through the channel for each bit of the message. The information rate is therefore 1/23. Can we do better?

Capacity and mutual information

The capacity of the BSC is 1-h(p), where

h(p) = -p \log_2 p -(1-p) \log_2(1-p)

is Shannon’s entropy function. Capacity has a slightly technical definition in terms of mutual information. Recall that if X and Y are random variables then the conditional entropy H(X|Y) is the expected value

H(X|Y) = \sum_{y} \mathbb{P}[Y=y]H(X|Y=y),

where H(X|Y=y) is the entropy of the random variable X, conditioned on Y=y. Informally, H(X|Y) is the uncertainty in X that remains after learning Y. The mutual information I(X;Y) is then H(X) - H(X|Y); informally this is the amount of information about X we learn from Y. For example, if p =0 then Y determines X, so all uncertainty is removed, and I(X;Y) = H(X). By

I(X;Y) = H(X) - H(X|Y) = H(X) + H(Y) - H(X,Y) = H(Y) - H(Y|X).

mutual information is symmetric: the amount we learn about X from Y is also the amount we learn about Y from X. Is this intuitive?

Returning to the BSC, consider the most ‘informative’ source X^\star for which \mathbb{P}[X^\star=0] = \mathbb{P}[X^\star=1] = \frac{1}{2}. Let Y be the output bit. The capacity is then I(X^\star;Y). We saw a moment ago that, as expected, if p=0 then the capacity is H(X^\star) = 1. The most convenient way to compute the capacity for general p uses the right-hand side of the identity above. Since X^\star is uniform, so is Y, and hence H(Y) = 1. Moreover

\begin{aligned} H(Y|X^\star) &= \frac{1}{2} H(Y | X^\star = 0) + \frac{1}{2}H(Y|X^\star=1) \\ &= 2 \times \frac{1}{2} (-p \log_2 p -(1-p) \log_2(1-p)) \\ &= h(p)\end{aligned}.

Therefore the capacity is 1-h(p), as claimed.

Shannon’s Noisy Coding Theorem

By Shannon’s Noisy Coding Theorem capacity has the following interpretation: given any \epsilon > 0 and any \eta > 0, for all sufficiently large n there exists a binary code C of length n of size > 2^{n (1- h(p) - \eta)} such that the probability of a decoding error when C is used to communicate on the BSC, with each codeword equally likely to be sent, is < \epsilon. (To be precise, a decoding error means that the received word is not decoded as the sent codeword.) For example, if p = 1/4 then

1-h(p) \approx 0.189.

Therefore Shannon's Noisy Coding Theorem promises that, by taking n sufficiently large, we can communicate with negligible decoding error probability and information rate 0.188.

As far as I know, all known proofs of Shannon's Noisy Coding Theorem use the probabilistic method.

Outline proof

Let C = \{u(1),\ldots, u(N)\} be a binary code of length n and size about 2^{n (1- h(p) - \eta/2)}, obtained by choosing each codeword u(i) uniformly and independently at random from \mathbb{F}_2^n. Let p_{i}(C) be the probability (in the BSC model) that if u(i) is sent then the received word is wrongly decoded as some u(j) with j\not=i. An expectation argument shows that, provided n is sufficiently large,

\mathbb{E}_{\mathrm{code}}[p_{i}(C)] < \epsilon/2.

Note that here the expectation is taken with respect to the random choice of C and p_{i} is a random variable (that just happens to be a probability in the BSC model). We'll see shortly why \epsilon/2 is the right upper bound to take. It follows by averaging over all codewords that

\frac{1}{N}\sum_{i=1}^N\mathbb{E}_\mathrm{code}[p_i(C)] < \epsilon/2.

By the probabilistic method, there exists a particular code C such that

\frac{1}{N}\sum_{i=1}^N p_i(C) < \epsilon/2.

Does this mean that we can use C to communicate with decoding error probability < \epsilon/2? Unfortunately not, because the previous equation only says that on average, we meet this bound. There could be some codewords u(i), perhaps those unusually near to many other codewords, for which the decoding error probability is higher. But since the average is \epsilon/2, at most half the codewords have p_i(C) \ge \epsilon. Therefore by removing at most half the codewords, we get a code C' of size N' \ge N/2 with decoding error probability < \epsilon. It might seem bad that, at worst, |C'| = |C|/2 = N/2, but as N = 2^{(1-h(p)-\eta/2)n}, we have

\frac{N}{2} = 2^{(1-h(p) -\eta/2 - 1/n))n}

and provided n is sufficiently large, \eta + 1/n < \eta. This randomly constructed C' proves Shannon's Noisy Coding Theorem for the parameters \epsilon and \eta.


  1. The final expurgation step, and the need to work with \epsilon/2 rather than \epsilon can be avoided by choosing a random linear code. Then the multiset of Hamming distances \{ d(u(i),u(j)) : j\not=i \} which determines the probability p_i(C) of a decoding error on sending u(i) is the same for all i. Therefore all the p_i(C) are equal. Some linear codes will be uniformly terrible, but by the probabilistic method, some linear codes are uniformly good.
  2. Care is needed to turn the outline above into a complete proof. For example, in the first step bounding the p_i(C), one needs Chernoff’s bound or the Central Limit Theorem to justify that when a codeword u is sent, the received word v is, with probability tending to 1 with n, of distance between (p-\delta)n and (p+\delta)n of u, for any fixed \delta > 0.
  3. One subtle feature of the proof is that there are two different probability models in play. I tried to emphasise this above, for example by using \epsilon for the error bounds in the channel probability space, and \eta for the error bounds in the code probability space. A related feature is that we have to work with expectations, rather than probabilities, in the code’s probability space. For instance, it would be nice to have

    \mathbb{P}_\mathrm{code}\bigl[\mathbb{P}_\mathrm{channel}[A_i] > \epsilon\bigr] < \eta

    for sufficiently large n, where A_i is the event that a sent u(i) is wrongly decoded as some u(j) with j\not= i. But these probabilities seem hard to reason about directly.

A toy model

Here is a toy model that has the same split into two probability spaces, but is, I think, much easier to understand. To play Match! using a supplied deck of c cards each with some number between 1 and n, the two players each draw a card from the deck. If they find they have matching numbers, they both eat some cake. Let M be this ‘in game’ event. Otherwise they go to bed hungry.

Fix a deck and let f_r be the frequency of card r, for each r \in \{1,\ldots, n\}. We have

\mathbb{P}_\mathrm{in game}[M] = \sum_{r=1}^n \frac{f_r(f_r-1)}{c(c-1)}.

In the random process used by the manufacturer to construct decks, the cards are picked independently and uniformly. Hence (f_1,\ldots,f_r) has the multinomial distribution B(c,1/n,\ldots, 1/n). Therefore \mathbb{E}_\mathrm{deck}[f_r] = c/n and

\mathbb{E}_\mathrm{deck}[f_r^2] = \mathrm{Var}_\mathrm{deck}[f_r] + \mathbb{E}_\mathrm{deck}[f_r]^2 = \frac{c(n-1)}{n^2} + \frac{c^2}{n^2}.

It follows that

\begin{aligned}\mathbb{E}_\mathrm{deck}\bigl[\mathbb{P}_\mathrm{in game}[M]\bigr] &= \frac{1}{c(c-1)}\sum_{i=1}^r \bigl( \frac{c(n-1)}{n^2} + \frac{c^2}{n^2} - \frac{c}{n} \bigr) \\ &= \frac{n}{c(c-1)n^2} \bigl( cn- c + c^2 - cn \bigr) \\ &= \frac{1}{n}. \end{aligned}

This can also be seen without calculation: mixing up the two probability models, we can imagine that the manufacturer builds a two card deck, and issues the cards to the players. The probability they are equal is clearly 1/n.

We can also reason about probabilities in the deck model. For example \mathbb{P}_\mathrm{deck}\bigl[ \mathbb{P}_\mathrm{in game}[M=0] \bigr] is the probability that all cards are different, namely

\frac{n(n-1) \ldots (n-c+1)}{n^c} = \frac{c!}{n^c} \binom{n}{c}.

Appropriate: Donmar Warehouse

August 28, 2019

Appropriate, directed at the Donmar Warehouse by Ola Ince is a new production of a 2014 play by Branden Jacobs-Jenkins. The three central characters Toni, Bo and Franz meet in their deceased father’s plantation house to sort out his estate. Their tangled personal and family history, revealed in a long series of explosive dialogues, spans an impressive range of themes: most prominently, parent/child relationships and racism.

Here I’ll concentrate on Toni. Pathologically unable to accept help from anyone, but appointed as sole executor of the Lafayette estate, we meet her in the first act surrounded by the accumulated detritus of 20 years of her father’s hoarding. Toni is exhausted by caring for her father through a long illness, unaided by her younger brothers. Her engrained negativity is excusable. Monica Dolan’s superb performance is nuanced and brings out Toni’s capacity for love, and her need for it. She is however hard to like: hard for the audience, and impossible for her family members. The moment when she snaps is the climax of the play.

The reviews I read beforehand suggested that the album of appalling photographs of lynchings found part-way through the first act would dominate the play. While important, and central to the impressively developed racism theme, it is only one of many elements. When it turns out the photos are highly valuable (unlike the house, which, blighted by a graveyard, is worthless), we quickly see that none of the Lafayette children would hesitate to sell them. The moment when they reflexively reach for the polite formulations of the auction house and ‘private dealer’ is most revealing. Of them all, Toni is the most eager to excuse her father’s racism, even as the evidence for it becomes overwhelming. She is obliged to confront her own attitudes in several clever set pieces. The most impressive, however, concerns her youngest brother Franz, a diligent 12-stepper, and his fiancée River, who is firmly committed to helping him on his journey with whatever resources her relentless spirituality (neatly shading into meaningless psychobabble) can provide. When Bo’s wife Rachel wrongly assumes she is a native American, rather than a New Yorker née Tracy, there is a brilliant moment where we see both characters expose their darkest and most instinctive reactions.

The first act of the play, while highly entertaining (the witty touches begin even with the title of the play), did at times reduce to a sequence of monologues-as-dialogues, all delivered at a surprising high volume for normal conversation. The febrile atmosphere was convincing but less oppressive than I think the director intended. Still, it was effective as a scene-setting prologue to the longer second act. I’m not sure if it was the actors settling down, or a deliberate directorial decision, but anyway, the vocal and dramatic range vastly improved between the acts. The second act mixed comedy and tragic irony in as good a piece of theatre as I’ve seen in recent years.

The play deliberately avoids tying up its loose ends. I suspect the lack of any obvious character development, except perhaps in two of the teenage children, is also deliberate. At the end, we are left with the characters, all ruthlessly exposed, heading back to their normal lives. And instead of any resolution, we are left to think about them, their problems, and the many themes of the play. Overall, an excellent production of a remarkable play, that leaves one with much to chew over.

Why is Adobe Acrobat so innumerate?

July 12, 2019

The image below shows Adobe Acrobat’s conversion to Microsoft Powerpoint of a typical slide produced by LaTeX and Beamer. The original is shown below.

For the £70 annual fee Adobe charge for turning on this misfeature, I think one might expect something better. The execrable kerning is just one final kick in the teeth. Fortunately for my credit card, I was able to brandish the Consumer Contracts Regulations in an online chat and have been promised a full refund. I’ve seen similar glitches in journal papers: the culprit now seems clear. As an alternative on the Mac, I suggest using pdf-to-keynote and then exporting from Keynote. Or better still, have nothing to do with either Adobe or Microsoft — if only that were feasible. In fairness, I should note that the output from Adobes’s conversion is editable in Microsoft Powerpoint; unfortunately the only reasonable edit one could make it to delete the lot and start over.