Question

Imprecise Counting - Long Runs in Binary Strings

Let n=2^k for some positive integer k and consider the set Sn of all n-bit binary strings.

1. Let c be an integer in {0,…,n−k}. Consider any j∈{1,…,n−k−c+1}. How many strings b1,…,bn∈Sn have bj,bj+1,…,bj+k+c−1=00…0? In other words, how many strings in Sn have k+c consecutive zeros beginning at position j?
2. For each j∈{1,…,n−k+c+1}, let Xj be the subset of Sn consisting only of the strings counted in the previous question. Show that

   (n−k−c+1)∑(j=1) {|Xj|≤2^n / 2^c}

   (Hint: Remember that 2^k=n.)
3. This leads to the following conclusion: The number of n-bit binary strings containing a substring of k+c consecutive zeros is at most 2^n / 2^c. Explain how to arrive at this conclusion.

Answer 1

• We need to consider all strings of length n in which k+c positions are occupied by 0s. Each of the remaining n-(k+c) places of the string are either occupied by a 0 or by a 1. So, the total number of possible strings which satisfy the given condition is .
• Each

so,

which must be the same as and from it follows,

. Since we obtain the given inequality
.

• Any n-bit binary string which contains a sub-string of k+c consecutive 0s will be having the start of the sub-string at a certain where . So, the number of all possible binary strings will be the sum of all the strings with the sub-string starting at each of the s which is precisely . So, by the last inequality this must always be less than .