I understand this question has been answered before, but I'm looking for feedback on why my approach is incorrect. Consider the following:

How many bit strings of length n can be generated that contain 01? Write a recursive definition.

I started with $a_2 = 1$ because a length 2 bit string only has 1 string that contains 01... that string is simply 01. Then I noticed you could either tag a 1/0 on the beginning of it or a 1/0 on the end of it to get strings of length 3 that contain 01. This leaves you with 4 new bit strings of length 3. Then, each of these 4 you can either append a 1/0 to the beginning or 1/0 to the end. Resulting in 4 more for each of the 4 previously generated. This led me to believe the recurrence relation was:

$a_n = 4_{a_n-1}$. Because for each new string you generate that contains 01, you can generate 4 more. However the solution is:

$a_n = a_{n-1} + 2^{n-1} - 1$

Could someone please help me see where my reasoning is flawed?

  • $\begingroup$ You are counting some twice, e.g., 0101. $\endgroup$ – ryagami Nov 8 '16 at 22:23
  • $\begingroup$ @ryagami ok, thanks. Is there a way for me to approach this problem with the same strategy and not double count? $\endgroup$ – foobar512 Nov 8 '16 at 22:29

Your basic idea can be made to work if you add bits only at one end instead of adding them at both ends. Say that a bit string is good if it contains the substring $01$ and bad otherwise. Suppose that $\sigma$ is a string of length $n-1$.

  • If $\sigma$ is good, we can append either a $0$ or a $1$ to get a good string of length $n$; this accounts for $2a_{n-1}$ good strings of length $n$.

  • If $\sigma$ is bad, the only way to get a good string by appending one more bit is for $\sigma$ to end in $0$ and to append a $1$; this gives us one good string for every bad string of length $n-1$ that ends in $0$. There are $2^{n-1}$ strings of length $n-1$, and $a_{n-1}$ of them are good, so $2^{n-1}-a_{n-1}$ of them are bad. Only one bad string of length $n-1$ ends in $1$, the string $$\underbrace{111\ldots111}_{n-1\text{ bits}}\;,$$ so there are $2^{n-1}-a_{n-1}-1$ bad strings of length $n-1$ that can be extended to a good string of length $n$.

The total number of good strings of length $n$ is therefore


exactly the recurrence in the solution.

You can actually check this rather easily by working out a closed form for $a_n$. The bad strings of length $n$ are precisely the ones in which all of the zeroes, if any, are after all of the ones, if any. If there are $0$ ones, we get the string of $n$ zeroes. If there is $1$ one, we get a $1$ followed by $n-1$ zeroes. And so on: if there are $k$ ones, we get a block of $k$ ones followed by a block of $n-k$ zeroes. There is just one such string for each $k$ from $0$ through $n$, so there are $n+1$ bad strings of length $n$. It follows that


and it’s not hard to show by induction that $(2)$ satisfies the recurrence $(1)$:



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