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COS2601 Assignment 2 Year 2023
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Complete Solutions to COS2601 Assignment 2 for the year 2023. Please note this is programming-language fluid and may not be 100% as per your chosen programming language.
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By: willemcierenberg • 1 year ago
By: Learn2Progress • 1 year ago
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Question 1aUniversal Set:
The appropriate universal set for this language is the set of all possible strings over the alphabet Σ =
{a, b}. Let's denote this set as U.
Question 1b
The generators of ODDnotAB can be defined as follows:
The empty string ε is a generator of ODDnotAB.
For any string x in ODDnotAB, the strings xa and xb are also generators of ODDnotAB.
Question 1c
Function on the Universal Set:
We need to define a function on the universal set that determines whether a given string belongs to
ODDnotAB or not. Let's denote this function as isInODDnotAB(x), where x is a string from the
universal set U. The function isInODDnotAB(x) returns true if x belongs to ODDnotAB, and false
otherwise.
Question 1d
Recursive Definition of ODDnotAB:
Using the concepts described above, we can write the recursive definition for the language
ODDnotAB as follows:
1. Base Cases:
a) The empty string ε is in ODDnotAB.
isInODDnotAB(ε) = true
b) Strings of length 1:
For any single character string x, isInODDnotAB(x) = true.
2. Recursive Cases:
a) If x is a string in ODDnotAB, then the strings xa and xb are also in ODDnotAB.
isInODDnotAB(xa) = isInODDnotAB(x)
isInODDnotAB(xb) = isInODDnotAB(x)
b) If x is a string in ODDnotAB, then the strings xaa and xab are also in ODDnotAB.
isInODDnotAB(xaa) = isInODDnotAB(x)
isInODDnotAB(xab) = isInODDnotAB(x)
c) If x is a string in ODDnotAB, then the strings xba and xbb are also in ODDnotAB.
isInODDnotAB(xba) = isInODDnotAB(x)
isInODDnotAB(xbb) = isInODDnotAB(x)
By using these base cases and recursive cases, we can determine whether a given string belongs to the
language ODDnotAB or not.
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