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98 combinatorics/full.p


This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

98 combinatorics/full.p

Consider a string that contains all substrings of length n. For example,
for binary strings with n=2, a shortest string is 00110 -- it contains 00,
01, 10 and 11 as substrings. Find the shortest such strings for all n.


Knuth, Volume 2 Seminumerical Algorithms, section 3.2.2 discusses this problem.
He cites the following results:
Shortest length: m^n + n - 1, where m = number of symbols in the language.
[Exercise 7, W. Mantel, 1897]
The binary sequence is the LSB of X computed by the MIX program:

    LDA X
    JANZ *+2
    LDA A
    ADD X
    JNOV *+3
    JAZ *+2
    XOR A
    STA X

[Exercise 10, M. H. Martin, 1934]
Set x[1] = x[2] = ... = x[n] = 0. Set x[i+1] = largest value < n such that
substring of n digits ending at x[i+1] does not occur earlier in string.
Terminate when this is not possible.

If we instead consider the strings as circular, we have a well known
problem whose solution is given by any hamiltonian cycle in the de
Bruijn (or Good) graph of dimension K. (Or equivalently an eulerian
circuit in the de Bruijn graph of dimension K-1) As a string of length
2^K is produced, it must be optimal, and any shortest sequence must be
an eulerian circuit in a dB graph.

The de Bruijn graph Tn has as its vertex set the binary n-strings.
Directed edges join n-strings that may be derived by deleting the left
most digit and appending a 0 or 1 to the right end. de Bruijn + van
Ardenne-Ehrenfest (in 1951) counted the number of eulerian circuits in
Tn. There are 2^(2^(n-1)-n) of them.

Some examples:
K=2 1100
K=3 11101000
K=4 1111001011010000

The solution to the above problem (non-circular strings) can be found
by duplicating the first K-1 digits of the solution string at the end
of the string. These are not the only solutions, but they
are of minimum length: 2^K + K-1.

We can obtain a lower bound for the optimal sequence for the general case as

Consider first the simpler case of breaking into an answer machine which
accepts d+1 digits, values 0 to n-1. We wish to find the minimal universal
code that will allow us access to any such answering machine.

Let us construct a digraph G = (V,E), where the n^d vertices are labelled
with a d sequence of digits. Notation: let [v_{i,1},v_{i,2},...,v_{i,d}]
denote the labelling on node v_i. An edge e = (v_i, v_j) is in E iff for k
in 1, ..., d-1: v_{i,k+1} = v_{j,k}, i.e., the last d-1 digits in the
labelling of the initial vertex of e is identical with the first d-1 digits
in the labelling of the terminal vertex of e. We associate with each edge a
value, t(e) = v_{j,d}, the last digit in the labelling of the terminal

The intuition goes as follows: we are going to perform a Euler circuit of
the digraph, where the label on the current vertex gives the last d digits
in the output sequence so far. If we make a transition on edge e, we output
the tone/digit t(e) as the next output value, thus preserving the invariant
on the labelling.

How do we know that a Euler circuit exists? Simple: a connected digraph
has an Euler circuit iff for all vertices v: indegree(v) = outdegree(v).
This property is trivially true for this digraph.

So, in order to generate a universal code for the AM, we simply output 0^d
(to satisfy the precondition for being in vertex [0,...,0]), and perform an
Euler circuit starting at node [0,...,0].

Now, the total length of the universal sequence is just the number of edges
traversed in the Euler circuit plus the initial precondition sequence, or
n^d * n + d (number of vertices times the out-degree) or n^{d+1} + d. That
this is a minimal sequence is obvious.

Next, let us consider the machine AM' where the security code is of the form
[0...n-1]^d [0...m-1], i.e., d digits ranging from 0 to n-1, followed by a
terminal digit ranging from 0 to m-1, m < n.

We build a digraph G = (V, E) similar to the construction above, except for
the following: an edge e is in E iff t(e) in 0 to m-1. This digraph is
clearly non-Eulerian. In particular, there are two classes of vertices:

(1) v is of the form [0...n-1]^{d-1} [0...m-1] (``fat'' vertices)


(2) v is of the form [0...n-1]^{d-1} [m...n-1] (``thin'' vertices)

Observations: there are (n^{d-1} * m) fat vertices, and (n^{d-1} * (n-m))
thin vertices. All vertices have out-degree of m. Fat vertices have
in-degrees of n, and thin vertices have in-degrees of 0. Color all the
edges blue.

The question now becomes: can we put a bound on how many new (red) edges
must we add to G in order to make a blue edge covering path possible?
(Instead of thinking of edges being traversed multiple times in the blue
edge covering path, we allow multiple edges between vertices and allow each
edge to be traversed once.) Note that, in this procedure, we add edges only
if it is allowed (the vertex labelling constraint). We will first obtain a
lower bound on the length of a blue covering circuit, and then transform it
into a bound for arbitrary blue covering paths.

Clearly, we must add at least (n-m)*(n^{d-1}*m) edges incident from the fat
vertices. [ We need (n-m) new out-going edges for each of (n^{d-1}*m)
vertices to bring the out-degree up to the in-degree. ]

Let us partition our vertices into sets. Denote the range [0..m-1] by S,
the range [m..n-1] by L, and the range [0..n-1] by X.

Let S_0 = { v: v = [X^{d-1}S] }. S_0 is just the set of fat vertices.
Define in(S_0) = number of edges from vertices not in S to vertices in S.
Define out(S_0) in the corresponding fashion, and let excess(S_0) =
in(S_0)-out(S_0). Clearly, excess(S_0) = n^{d-1}m(n-m) from the argument
above. Generalizing the requirement for Eulerian digraphs, we see that we
must add excess(S_0) edges from S_0 if the blue edges connected to/within
S_0 are to be covered by some circuit (edges may not be travered multiple
times -- we add parallel edges to handle that case). In particular, edges
from S_0 will be incident on vertices of the form [X^{d-2}SX]. Furthermore,
they can not be [X^{d-2}SS] since that is a subset of S_0 and adding those
edges will not help excess(S_0). [Now, these edges may be needed if we are
to have a circuit, but we do not consider them since they do not help
excess(S_0).] So, we are forced to add excess(S_0) edges from S_0 to S_1 = {
v: v = [X^{d-2}SL] }. Color these newly added edges red.

Let us define in(S_1), out(S_1) and excess(S_1) as above for the modified
digraph, i.e., including the red excess(S_0) edges that we just added.
Clearly, in(S_1) = out(S_0) = n^{d-1}m(n-m), and out(S_1) = m*|S_1| =
m*n^{d-2}m(n-m), so excess(S_1) = n^{d-2}m(n-m)^2. Consider S_0 union S_1.
We must add excess(S_1) edges to S_0 union S_1 to make it possible for the
digraph to be covered by a circuit, and these edges must go from {S_0 union
S_1} to S_2 = { v: v = [X^{d-3}SL^2] } by a similar argument as before.
Repeating this partitioning process, eventually we get to S_{d-1} = { v: v =
[SL^{d-1}] }, where union of S_0 to S_{d-1} will need edges to S_d = { v: v
= [L^d] }, where this process terminates. Note that at this time,
excess(union of S_0 to S_{d-1}) = m(n-m)^d, but in(S_d) = 0 and out(S_d) =
m(n-m)^d, and the process terminates.

What have we shown? Adding up blue edges and the red edges gives us a lower
bound on the total number of edges in a blue-edges covering circuit (not
necessarily Eulerian) in the complete digraph. This comes out to be
n^{d+1}-(n-m)^{d+1} edges.

Next, we note that if we had an optimal path covering all the blue edges, we
can transform it into a circuit by adding d edges. So, a minimal path can
be no more than d edges shorter than the minimal circuit covering all blue
edges. [Otherwise, we add d extra edges to make it into a shorter circuit.]

So the shortest blue covering path through the digraph is at least
n^{d+1}-{n-m}^{d+1}-d. With an initial pre-condition sequence of length d
(to establish the transition invariant), the shortest universal answering
machine sequence is of length at least n^{d+1}-(n-m)^{d+1}.

While this has not been that constructive, it is easy to see that we can
achieve this bound. If we looked at the vertices in each of the S_i's, we
just add exactly the edges to S_{i+1} and no more. The resultant digraph
would be Eulerian, and to find the minimal path we need only start at the
vertex labelled [{n-1}^d], find the Euler circuit, and omit the last d edges
from the tour.


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