Question

Recursively define the set of bit strings that have more zeros than ones.

Short answer

It has been derived.

Step-by-step solution

Introduction

A string is just a finite sequence where S is a set of characters. Strings denoted by putting the characters together, for the example 12345. Besides that, strings are the fundamental object of computer science. Everything discrete can be described as a string of characters. Decimal numbers: 1010230824879.

Step 2: Define the set of bit strings

The objective is to recursively define the set of bit string that have more zeros than ones.

Consider, S as bit string that have more zeros than ones.

The recursive to create such bit strings is as follows;

Basic step:

$0\in S$.

Recursive step: if $w\mathit{\in }S$such that w = x.y for some strings x,y $\in${0,1} , then the string x0y, x01y and x10y are also in S.

Or,

Recursive step: Suppose $\mathrm{x},\mathrm{y}\in \mathrm{S}$then $\mathrm{x}1\mathrm{y}\in \mathrm{S}$and $\mathrm{x},\mathrm{y}1\in \mathrm{S}$ .

This is true because both x and y contain at least one zeros more than ones.

Thus, the concatenation of x and y, that is xy contains at least two more zeros than ones.

Hence, it has been derived.