Number of q-ary strings of length m which do not contain k consecutive zeros

Question

A finite q-ary-alphabet is given $$A_q = {0,1,2,...,q-1}.$$ Now I am considering the set of all finite strings over the alphabet $A_q$.

I am interested on the number $$N(m,k)_{A_q}$$ of strings of length $m$ which do not contain $k$ consecutive zeros. Is there a general formulary for this number? Or how can I determine $N(m,k)_{A_q}$?

Answer 1

Omitting the suffix $A_q$.

If $X$ is a letter that belongs to our alphabet that is not $0$ then any such string length $m\ge k$ may begin in $k$ mutually exclusive and exhaustive ways

$$X\: [\text{sequence length $m-1$ with no $k$ consecutive $0$s}]$$ $$0X\: [\text{sequence length $m-2$ with no $k$ consecutive $0$s}]$$ $$00X\: [\text{sequence length $m-3$ with no $k$ consecutive $0$s}]$$ $$\vdots$$ $$\underbrace{000\cdots 0}_{k-1 \text{ times}}X\: [\text{sequence length $m-k$ with no $k$ consecutive $0$s}]$$

Then, since $X$ can take $q-1$ different letters, there are $(q-1)N(m-1,k)$ sequences of the first type, $(q-1)N(m-2,k)$ of the second type and so on we have the recurrence

$$N(m,k)=(q-1)\sum_{i=1}^{k}N(m-i,k)$$

with initial conditions

$$N(m,k)= q^m \qquad\text{for}\quad 0\le m \le k-1$$

It is also possible to derive the generating function $f(x)$ for this in several different ways

$$f(x)=\frac{1+x+x^2+\cdots +x^{k-1}}{1-(q-1)(x+x^2+\cdots +x^{k})}=\frac{1-x^k}{1-qx-(q-1)x^{k+1}}$$

One of the nicest is to see that this function "builds" such a sequence from the irreducible "blocks", where the enumerator $x$ has its index representing word length

$$\{1,\, 2,\ldots ,\, q-1,\, 01,\, 02,\ldots ,\, 0(q-1),\, 001,\, 002,\ldots ,\, 00(q-1),\ldots ,\, \underbrace{00\cdots 0}_{\text{$k-1$ times}}1,\, \underbrace{00\cdots 0}_{\text{$k-1$ times}}2,\ldots ,\, \underbrace{00\cdots 0}_{\text{$k-1$ times}}(q-1)\}$$

There $q-1$ blocks of each length $1$ to $k$, this accounts for the denominator term, but then any such sequence built from these blocks may terminate with $1,2,3\ldots ,k-1$ concurrent $0$s or the empty word $\epsilon$, this accounts for the numerator term.

Answer 2

Here we are given a q-ary alphabet. We are asking for the number of words of length $m$ with character $0$ having runs at most length $k-1$.

Words with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.

A generating function for the number of Smirnov words over a q-ary alphabet is given by \begin{align*} \left(1-\frac{qz}{1+z}\right)^{-1} \end{align*}

Replacing occurrences of $0$ in a Smirnov word by one up to $k-1$ zeros generates words having runs of $0$ with length less than $k$. \begin{align*}\ z\longrightarrow z+z^2+\cdots+z^{k-1}=\frac{z\left(1-z^{k-1}\right)}{1-z} \end{align*}

Since there are no restrictions to other characters we can replace each occurrence of these characters by one or more occurrences of it. \begin{align*} z\longrightarrow z+z^2+\cdots=\frac{z}{1-z} \end{align*}

The resulting generating function is \begin{align*} \left(1-\frac{(q-1)\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{\frac{z\left(1-z^{k-1}\right)}{1-z}}{1+\frac{z\left(1-z^{k-1}\right)}{1-z}}\right)^{-1} &=\frac{1-z^k}{1-qz+(q-1)z^{k+1}} \end{align*}

Denoting with $[z^m]$ the coefficient of $z^m$ in a series we obtain the number of wanted words of length $m$ as \begin{align*} [z^m]\frac{1-z^k}{1-qz+(q-1)z^{k+1}} \end{align*}

Example: Let's look at an example. We take $A_q=\{0,1,2\}$ and $k=3$. We obtain with $q=3$ \begin{align*} \frac{1-z^3}{1-3z+2z^4}=1+3z+9z^2+26z^3+\color{blue}{76}z^4+222z^5+\cdots \end{align*}

The blue colored coefficient of $z^4$ shows there are $\color{blue}{76}$ words of length $4$ built from characters $\{0,1,2\}$ and runs of $0$ with length less than $k=3$. These words are listed lexicographically sorted below.

\begin{array}{cccccccccc} 0010&0110&0210&1011&1111&1211&2012&2112&2212\\ 0011&0111&0211&1012&1112&1212&2020&2120&2220\\ 0012&0112&0212&1020&1120&1220&2021&2121&2221\\ 0020&0120&0220&1021&1121&1221&2022&2122&2222\\ 0021&0121&0221&1022&1122&1222&2100&2200\\ 0022&0122&0222&1100&1200&2001&2101&2201\\ 0100&0200&1001&1101&1201&2002&2102&2202\\ 0101&0201&1001&1102&1202&2010&2110&2210\\ 0102&0202&1010&1110&1210&2011&2111&2211\\ \end{array}

[2017-10-22] Coefficients: We calculate the coefficients of the generating function in order to check OP's answer. We obtain

\begin{align*} \color{blue}{[z^m]}&\color{blue}{\frac{1-z^k}{1-qz+(q-1)z^{k+1}}}\\ &=[z^m](1-z^k)\sum_{j=0}^\infty\left(qz+(1-q)z^{k+1}\right)^j\tag{1}\\ &=[z^m]\sum_{j=0}^\infty z^j\left(q+(1-q)z^k\right)^j(1-z^k)\\ &=\sum_{j=0}^m [z^{m-j}]\left(q+(1-q)z^k\right)^j(1-z^k)\tag{2}\\ &=\sum_{j=0}^m [z^{j}]\left(q+(1-q)z^k\right)^{m-j}(1-z^k)\tag{3}\\ &=\sum_{j=0}^{\left\lfloor\frac{m}{k}\right\rfloor} [z^{kj}]\left(q+(1-q)z^k\right)^{m-kj}(1-z^k)\tag{4}\\ &=\sum_{j=0}^{\left\lfloor\frac{m}{k}\right\rfloor} [z^{kj}]\left(q+(1-q)z^k\right)^{m-kj} -\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor} [z^{k(j-1)}]\left(q+(1-q)z^k\right)^{m-kj}\tag{5}\\ &=\sum_{j=0}^{\left\lfloor\frac{m}{k}\right\rfloor} q^{m-kj-j}(1-q)^j\binom{m-kj}{j} -\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}q^{m-kj-(j-1)}(1-q)^{j-1}\binom{m-kj}{j-1}\tag{6}\\ &\color{blue}{=q^m+\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor} q^{m-(k+1)j} (1-q)^j\left[\binom{m-kj}{j}-\frac{q}{1-q}\binom{m-kj}{j-1}\right]}\tag{7} \end{align*} and OPs result follows easily from the representation (6).

Comment:

• In (1) we use the geometric series expansion.

• In (2) we apply the coefficient of operator rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and restrict the upper limit of the series with $m$ since other summands do not contribute.

• In (3) we change the order of summation $j\rightarrow m-j$.

• In (4) we respect that only multiples of $k$ of the exponent of $z$ do contribute.

• In (5) we factor out and note that the right hand sum starts with index $j=1$.

• In (6) we select the coefficient of $z^{kj}$ resp. $z^{k(j-1)}$ accordingly.

• In (7) we extract the summand with $j=0$ from the left-hand sum and collect both sums.

Answer 3

Let me present another approach to this interesting problem, which will allow to get a closed expression for $N(m,k,q)$.

Let's consider a word of length $m$ from the binary alphabet $\{0,X\}$, having a total of $s$ zero's.

$$ \begin{array}{*{20}c} X &| & {0,} & {0,} & {0,} &| & X &| & 0 &| & {X,} & X &| & {0,} & 0 &| & X \\ 0 &| & {1,} & {2,} & {3,} &| & 0 &| & 1 &| & {0,} & {0,} &| & {1,} & 2 &| & 0 \\ \end{array} $$

Imagine to sequentially scan the word and count the number of consecutive zeros, resetting the counter when the character is different from $0$, as in the exaple above.
Then we can bi-ject each word with an hystogram which has :
- $j$ bars;
- sum of the bars equal $s$;
- $m-s$ Xs.

Now, the number of ways to compose $j$ bars, summing to $s$, and with a heigth going from $1$ to max $r$ is given by $$ N_b (s - j,r - 1,j) = \text{No}\text{.}\,\text{of}\,\text{solutions}\,\text{to}\;\left\{ \begin{gathered} \text{0} \leqslant \text{integer}\;x_{\,j} \leqslant r - 1 \hfill \\ x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,j} = s - j \hfill \\ \end{gathered} \right. $$ which is expressed by $$ \bbox[lightyellow] { \begin{gathered} N_b (s - j,r - 1,j)\quad \left| \begin{gathered} \;1 \leqslant \text{integer }r \hfill \\ \;0 \leqslant \text{integer}\;j \leqslant \text{integer }s \hfill \\ \end{gathered} \right.\quad = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{{s - j}} {{r - 1}}\, \leqslant \,j} \right)} {\left( { - 1} \right)^k \left( \begin{gathered} j \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} s - 1 - k\,r \\ s - j - k\,r \\ \end{gathered} \right)} \hfill \\ \end{gathered} \tag{1} }$$ as explained in this post.

Then we can dispose the $m-s$ (undistinguishable) X place-holders and the $j$ (distinguishable) bars by
- reserving $j-1$ X's as separators between consecutive $0$ blocks
- putting the remaining X's in any of the $j+1$ interstices, thus in ${m-s+1} \choose{j}$ ways.

Finally we can compose the Xs in $(q-1)^{m-s}$ ways. We shall pay attention to that $j \leqslant s \leqslant \left\lceil {m/2} \right\rceil $

Thus the Number sought for, understood as the number of words which contains max $k-1$ consecutive zeros (in one or more runs) will be $$ \eqalign{ & N(m,k,q)\quad \left| \matrix{ \;1 \le {\rm integer }k,q \hfill \cr \;0 \le {\rm integer}\;m \hfill \cr} \right.\quad = \cr & = \sum\limits_{0\, \le \,\,s\,\, \le \,m} {\left( {q - 1} \right)^{\,m - s} \sum\limits_{0\, \le \,\,j\,\, \le \,s} {\left( \matrix{ m - s + 1 \cr j \cr} \right)\sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( {\, \le \,j} \right)} {\left( { - 1} \right)^i \left( \matrix{ j \cr i \cr} \right)\left( \matrix{ s - 1 - i\,\left( {k - 1} \right) \cr s - j - i\,\left( {k - 1} \right) \cr} \right)} } } = \cr & = \sum\limits_{0\, \le \,\,s\,\, \le \,m} {\left( \matrix{ \left( {q - 1} \right)^{\,m - s} \quad \cdot \hfill \cr \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( {\, \le \,m - s + 1} \right)} {\left( { - 1} \right)^i \left( \matrix{ m - s + 1 \cr i \cr} \right)\sum\limits_{\left( {0\, \le } \right)\,\,j - i\,\,\left( { \le \,s - i} \right)} {\left( \matrix{ m - s + 1 - i \cr j - i \cr} \right)\left( \matrix{ s - 1 - i\,\left( {k - 1} \right) \cr s - i\,k - \left( {j - i} \right) \cr} \right)} } \hfill \cr} \right)} = \cr & = \sum\limits_{0\, \le \,\,s\,\, \le \,m} {\left( {q - 1} \right)^{\,m - s} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( {\, \le \,m - s + 1} \right)} {\left( { - 1} \right)^i \left( \matrix{ m - s + 1 \cr i \cr} \right)\left( \matrix{ m - i\,k \cr s - i\,k \cr} \right)} } = \cr & = \sum\limits_{0\, \le \,\,s\,\, \le \,m} {\left( {q - 1} \right)^{\,m - s} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\, \le \,m/k} {\left( { - 1} \right)^i \left( \matrix{ m - s + 1 \cr i \cr} \right)\left( \matrix{ m - i\,k \cr m - s \cr} \right)} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\, \le \,m/k} {\left( { - 1} \right)^{\,i} \sum\limits_{0\, \le \,\,m - s\,\, \le \,m} {\left( \matrix{ m - i\,k \cr m - s \cr} \right)\left( {\left( \matrix{ m - s \cr i \cr} \right) + \left( \matrix{ m - s \cr i - 1 \cr} \right)} \right)\left( {q - 1} \right)^{\,m - s} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\, \le \,m/k} {\left( { - 1} \right)^{\,i} \left( {\left( \matrix{ m - i\,k \cr i \cr} \right)q^{\,m - i\,k - i} \left( {q - 1} \right)^{\,i} + \left( \matrix{ m - i\,k \cr i - 1 \cr} \right)q^{\,m - i\,k - i + 1} \left( {q - 1} \right)^{\,i - 1} } \right)} = \cr & = q^{\,m} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\, \le \,m/k} {\left( {\left( \matrix{ m - i\,k \cr i \cr} \right)\left( {{{1 - q} \over {q^{\,k + 1} }}} \right)^{\,i} } \right)} - q^{\,m - k} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\, \le \,\left( {m - k} \right)/k} {\left( {\left( \matrix{ m - k - i\,k \cr i \cr} \right)\left( {{{1 - q} \over {q^{\,k + 1} }}} \right)^{\,i} } \right)} = \cr & = q^{\,m} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( {\, \le \,m/k} \right)} {\left( {\left( \matrix{ m - i\,k \cr m - i\,\left( {k + 1} \right) \cr} \right)\left( {{{1 - q} \over {q^{\,r + 1} }}} \right)^{\,i} } \right)} - q^{\,\left( {m - k} \right)} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\,\left( { \le \,\left( {m - k} \right)/k} \right)} {\left( {\left( \matrix{ \left( {m - k} \right) - i\,k \cr \left( {m - k} \right) - i\,\left( {k + 1} \right) \cr} \right)\left( {{{1 - q} \over {q^{\,k + 1} }}} \right)^{\,i} } \right)} = \cr & = M(m,\;k,\;q) - M(m - k,\;k,\;q) \cr} $$ where:
- in the first passage we use the trinomial revision $\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ j \cr n \cr} \right) = \left( \matrix{ s \cr n \cr} \right)\left( \matrix{ s - n \cr j - n \cr} \right)$
- in the sum in $s$ we use $\sum\limits_k {\left( \matrix{ n \cr k \cr} \right)\left( \matrix{ k \cr m \cr} \right)y^{\,k} } = \left( \matrix{ n \cr m \cr} \right)\left( {1 + y} \right)^{\,n - m} y^{\,m} $ obtainable from $$ \left( {1 + y + y} \right)^{\,n} = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( {\, \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)\left( {1 + y} \right)^{\,k} y^{\,n - k} } = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( {\, \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( {\, \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)\left( \matrix{ k \cr j \cr} \right)y^{\,n - j} } } $$

In conclusion $$ \bbox[lightyellow] { \left\{ \matrix{ M(m,k,q) = q^{\,m} \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( {\, \le \,m/k} \right)} {\left( {\left( \matrix{ m - i\,k \cr m - i\,\left( {k + 1} \right) \cr} \right)\left( {{{1 - q} \over {q^{\,k + 1} }}} \right)^{\,i} } \right)} \hfill \cr N(m,k,q) = M(m,\;k,\;q) - M(m - k,\;k,\;q) \hfill \cr} \right.\quad \left| \matrix{ \;1 \le {\rm integer }k,q \hfill \cr \;0 \le {\rm integer}\;m \hfill \cr} \right. \tag{2} }$$

and it can be verified that the above expression

1. ${\bbox[#dfd,5px]{\text{satisfies the recursion provided in *N. Shales*'s answer}}}$
2. ${\bbox[#dfd,5px]{\text{gives the terms of the z-transform provided in *Markus Scheuer*'s answer}}}$ .

Answer 4

Based on Markus Scheuer's answer, I tried to derive a general formula for the asked number. The number of $q$-ary strings of length $m$ with no $k$ consecutive zeros is $$ N(m,k)_q = \sum_{j=0}^{\lfloor \frac{m}{k} \rfloor} \binom{m-kj}{j}(1-q)^j q^{m-(k+1)j} - \sum_{j=0}^{\lfloor \frac{m-k}{k} \rfloor} \binom{m-k-kj}{j}(1-q)^j q^{m-k-(k+1)j}, $$ where $1 \le k \le m$ is the constraint.

It would be nice, if some can confirm this formula. Is there a possibility to remove the constraint? So I would like to vary $m$ and $k$, even that the formula is valid for $ m \le k $. Or is it already general and the second sum is a "empty" sum if the upper limit is negative?