# Number of bit strings with 3 consecutive zeros or 4 consecutive 1s

I am trying to count the number of bit-strings of length 8 with 3 consecutive zeros or 4 consecutive ones. I was able to calculate it, but I am overcounting. The correct answer is $147$, I got $148$.

I calculated it as follows:

Number of strings with 3 consecutive zeros = $2^5+5\times2^4 = 112$, because the 3 zeros can start at bit number 1, 2, 3, .., 6

Number of strings with 4 consecutive ones = $2^4+4\times2^3 = 48$, I used the same reasoning.

Now I am trying to count the number of bit-strings that contain both 3 consecutive zeros and 4 consecutive 1s. I reasoned as follows:

the strings can be of the following forms: 0001111x, 000x1111, x0001111..thus there are $2+2+2 = 6$ possibilities for bit-strings where the 3 consecutive zeros come first. Symmetrically there are $6$ bit-strings where the 4 consecutive ones come first.

Thus the answer should be = $112+48-12 = 148$.

clearly there's something wrong with my reasoning, if someone could point it out, that would be awesome. Thanks

• Why can't the three consecutive zeros start at 7, or at 8? – Gerry Myerson Aug 4 '12 at 0:04
• because the string can have the following forms 000xxxxx, 1000xxxx, x1000xxx, xx1000xx, xxx1000x, xxxx1000 – turingcomplete Aug 4 '12 at 0:12
• You seem to want bit-strings of length 8, then. You've double counted the strings 00011111 and 00001111, to start, but that won't get us going in the right direction unfortunately. – Kevin Carlson Aug 4 '12 at 0:16
• ah yes, so sorry. I will edit the question now. Thanks for pointing that out. – turingcomplete Aug 4 '12 at 0:18
• I'm assuming you want the length to be $8$. The number of bitstrings with three consecutive zeroes is $107$; the number with four consecutive ones is $48$; the number with both is $8$; and $107 + 48 - 8 = 147$. I can't understand where your expression for the number with $3$ consecutive zeroes ($2^5 + 5 \times 2^4$) comes from at all, but I think your basic problem is overcounting strings with multiple copies of the target substring. – mjqxxxx Aug 4 '12 at 0:26

You've left out accounting for strings that have two triple zeroes. So $00010000,00010001,00001000,00011000,10001000$ were added to your total twice. This didn't cause any problems in your count of strings with four $1$s, however, since we can't put four $1$s in two separated places in an $8$-bit string. So the union now has $155$ elements, and cutting out the two duplicates from each symmetry of your intersection calculation turns that to $8$, for a total $107+48-8=147$.

• Aha, I see. I missed those duplicates. Thanks, now I understand it. – turingcomplete Aug 4 '12 at 1:09

Here's one way to get the 107 and the 48 in the comment by mjqxxxx.

Let $a_n$ be the number of bit-strings of length $n$ with 3 consecutive zeros. Then $$a_n=2a_{n-1}+2^{n-4}-a_{n-4}$$ because you can get such a string from a such a string of length $n-1$ by putting a zero or a one at the end, or from a string of length $n-4$ with no 3 consecutive zeros by tacking 1000 on at the end. Clearly $a_0=a_1=a_2=0$ and $a_3=1$, and then you can use the recursion to find $a_4=3$, $a_5=8$, $a_6=20$, $a_7=47$, $a_8=107$.

For bit-strings with 4 consecutive ones, the same logic gets you to $$b_n=2b_{n-1}+2^{n-5}-b_{n-5}$$ and then you proceed as before.

• This is another nice way to think about it, I did not think of that. Thank you. – turingcomplete Aug 4 '12 at 1:12
• @Garry Myseron can you please explain how you got above recurrences? How to further modify them based on change in question? – mcjoshi Dec 15 '16 at 15:13
• @thor, the sentence after the first recurrence explains how I got that recurrence. The second one follows the same logic, but with 4 everywhere instead of three. How to modify them depends on exactly what the change is in the question. Why not post a new question, and probably someone will answer, instead of tacking this on as a comment on a question from four years ago? – Gerry Myerson Dec 16 '16 at 4:01

Let $n$ be a nonnegative integer. Let $a_n$ be the number of bit strings of length $n$ with at least 3 consecutive zeros. Clearly $a_0 = 0$, $a_1 = 0$, $a_2 = 0.$ Let $n \geq 3.$ Denote by $S$ the set of all bit strings of length $n$ with at least 3 consecutive zeros. The set $S$ is a disjoint union of the following four sets:

• $S_3$ the set of all bit strings of length $n$ that end with $000$,
• $S_2$ the set of all bit strings in $S$ that end with $100$,
• $S_1$ the set of all bit strings in $S$ that end with $10$,
• $S_0$ the set of all bit strings in $S$ that end with $1$.

These sets have the following cardinalities: $|S_3| = 2^{n-3}$, $|S_2| = a_{n-3}$, $|S_1| = a_{n-2}$, $|S_0| = a_{n-1}$. Since $|S| =a_n$, by the sum rule $$a_n = 2^{n-3} + a_{n-3} + a_{n-2} + a_{n-1}.$$ Since $a_0 = 0$, $a_1 = 0$, $a_2 = 0$, we have \begin{align*} a_3 & = 2^0 + 0 + 0 + 0 = 1\\ a_4 & = 2^1 + 0 + 0 + 1 = 3\\ a_5 & = 2^2 + 0 + 1 + 3 = 8\\ a_6 & = 2^3 + 1 + 3 + 8 = 20\\ a_7 & = 2^4 + 3 + 8 + 20 = 47\\ a_8 & = 2^5 + 8 + 20 + 47 = 107. \end{align*}

For bit strings with at least 4 consecutive ones, the same reasoning leads to the recursion $$b_0 = b_1 = b_2 = b_3 = 0, \quad b_n=2^{n-4}+b_{n-4}+ b_{n-3}+ b_{n-2}+ b_{n-1}, \ n\geq 4.$$

$$\begin{array}{|l|r|l|} \hline format & N & exceptions \\ \hline 000***** & 32 & \\ 1000**** & 16 & \\ *1000*** & 16 & \\ **1000** & 16 & \\ ***1000* & 14 & 0001000* \\ ****1000 & 13 & 000*1000 , 10001000 \\ 1111**** & 13 & 1111000* , 11111000 \\ 01111*** & 7 & 01111000 \\ *01111** & 8 & \\ **01111* & 6 & 0001111* \\ ***01111 & 6 & *0001111 \\ \hline \end{array}$$

Total: $$147$$