Counting Bit Strings Consisting Entirely of 1s

When dealing with bit strings, one common challenge is to count the number of strings of a specific form within a given length constraint. In this scenario, we are interested in counting the number of bit strings with a length not exceeding a positive integer n that consist entirely of 1s. This includes strings such as '1', '11', '111', and so forth, up to the maximum length of n. This article delves into the mathematical approach to solve this problem and provides insights into how these bit strings can be counted efficiently.

The Problem

Given a positive integer n, we need to determine the total number of bit strings that:

Consist entirely of the digit '1'. Have a length not exceeding n. Include the empty string.

The Solution

First, let's define the problem more formally. We are tasked with counting all the bit strings that consist entirely of 1s and have a length from 1 to n. This count includes all possible lengths from 1 to n, as well as the empty string, which is a special case since it has a length of 0.

To solve this problem, we can use a mathematical approach. Let's denote the number of such bit strings by C(n).

When n is 1, the only possible bit string consists of a single 1. Hence, C(1) 1.

For any positive integer n, the bit strings can be formed by adding a '1' to all bit strings with lengths from 1 to n-1. Therefore, we can express the count recursively as:

C(n) C(n-1) 1

The ' 1' accounts for the empty string, which has a length of 0 and is the only bit string of length 0 that consists entirely of 1s (in this context).

The Recursive Formula

To understand the recursive formula better, let's break it down:

C(1) 1 (only the empty string and '1') C(2) C(1) 1 2 (the empty string, '1', and '11') C(3) C(2) 1 3 (the empty string, '1', '11', and '111') C(4) C(3) 1 4 (the empty string, '1', '11', '111', and '1111')

From this, we can observe a clear pattern: C(n) n 1. This can be proven using induction:

Proof by Induction

Step 1: Basis

For n 1, C(1) 1, which is true.

Step 2: Inductive Hypothesis

Assume that C(k) k 1 for some positive integer k.

Step 3: Inductive Step

We need to show that C(k 1) (k 1) 1.

By the recursive formula:

C(k 1) C(k) 1

Using the inductive hypothesis:

C(k 1) (k 1) 1

This completes the inductive step, and thus we have:

C(n) n 1

Conclusion

The number of bit strings with a length not exceeding n that consist entirely of 1s is given by C(n) n 1. This solution is efficient and easy to implement, making it a valuable tool in various computational and algorithmic scenarios. By mastering this concept, one can better understand and solve similar problems involving bit strings and their properties.

References

Knuth, D. E. (1998). The Art of Computer Programming, Volume 4, Fascicle 0: Bitwise Operations on Words. Addison-Wesley Professional. Roberts, D. E., Stirzaker, D. (2003). Steps in Dirichlet Series. Cambridge University Press.