2024-01-22 10:12:47

UNB/Year 4/Semester 2/CS2333/2024-01-15.md

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+Strings and languages:
+- An alphabet is a finite set of symbols, such as {1,2,3} or {0,1}
+- A string is an alphabet $\sum$ is a finite sequence of symblols from $\sum$, for example 010010 and 10010101 are strings over the alphabet {0,1}
+- A languge is a set of strings
+	- L1 = {00, 01, 10, 11}
+	- L2 = {a^n | n is an integer non-negative}
+Notation: a to the power of n just means a repeated n times
+
+Examples
+$L_8 = \{a^m b^n | m,n \in \mathbb{z}^{nonneg}\}$
+
+So valid strings would be
+aaabbb, aaabbbb, ab, abb
+
+$L_9 = \{a^n b^n | n \in \mathbb{z}^{nonneg}\}$
+
+So valid strings would be
+aabb, ab, aaaabbbb, aaaaaabbbbbb
+
+
+$L_{11} = \{w \in \{a, b\}^* | \text{The number of "a"s is equal to the number of "b"s}\}$
+This would include everything that L9 includes, but allows for the alphabet can be out of order
+
+You can also written this as
+$L_{11} = \{w \in \{a, b\}^* | n_a(w) = n_b(w)\}$
+
+
+You can extend the repetition notation (exponent) to include multiple symbols by wrapping it in parentheses
+
+Example:
+$L_{12} = \{(ab)^n | n \in \mathbb{z}^{nonneg}\}$
+
+This means we need to be careful as 
+$L_{13} = \{ab^n | n \in \mathbb{z}^{nonneg}\}$
+is different, as L12 is asking for repetition of ab, while L13 is assigning a prefix of a and repetition of b 
+
+$L_{14} = \{a^i b^j c^k | i,j,k \in \mathbb{z}^{nonneg}, j = 2i + 3k\}$
+
+So this means that for any number of "a"s and "c"s, the number of "b"s is predetermined if you have selected your "a"s and "c"s
+
+Intro to functions and relations
+
+Informally, a function (f) descrives an input output situation (A -> B)
+For every input element a in A, there is exactly one ouput element b in B
+
+f maps a to b
+
+A is the domain of f
+B is the co-domain of f

UNB/Year 4/Semester 2/CS2333/2024-01-17.md

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+Lecture Topic: Functions
+
+For every input element $a \in A$ there is exactly one output element $b \in B$
+
+Jargon:
+- f maps a to b
+- the image of a under f is b
+
+functions are one-to-one (or injective):
+	$\forall x,y \in A, x \neq y \rightarrow f(x) \neq f(y)$
+	or in other terms
+	$\forall x,y \in A, x = y \rightarrow f(x) = f(y)$
+
+Example:
+$f(x) = x^2$ is not one to one
+
+Proof by counter example:
+$x = -1$ and $y = 1$ are two integers in the domain that $x=y$ but $f(-1)$ and $f(1)$  are both equal to 1, so it is not one to one
+
+To prove a function is one to one, it is a little more tricky, and needs to be general, for example:
+
+Let x and y be any arbitrary elements of the domain where f(x) = f(y)
+So it follows that x and y are equal (x = y)
+
+Example for a real function: $g(x) = 3x - 11$
+Real Numbers
+Let x and y be any real number such that $g(x) = g(y)$. We will show that $x = y$
+Since we know that g(x) = g(y), it means that $3x - 11 = 3y - 11$
+Add 11 to both sides: $\therefore 3x = 3y$
+Divide both sides by 3: $\therefore x = y$
+
+Important to note: The definition (domain and co-domain) can change if a function is one to one, for example the square function on all integers vs all positive integers
+
+To prove that a function A -> B is onto:
+- Let be represent and arbitrary element in the co-domain B
+- We want to find an element a in the domain and show that f(a) = b
+
+Example: 
+Real Numbers, $f(x) = 5x + 2$ is onto
+
+Proof: Let b be any arbitrary real number (we want to find a real number a and show that f(a) = b)
+
+A through process might follow that, I want a value a such that f(a) = b, I want $5a + 2 = b$, then $5a = b - 2$, then $a = \frac{b - 2}{5}$ 
+
+So, let $a = \frac{b - 2}{5}$ , which is a real number. Then plug this back into the function
+$f(a) = f(\frac{b - 2}{5}$) (Substitution)
+$= 5\frac{b-2}{5}+2$  (Definition of f)
+$= (b-2) +2)$ (Algebra)
+$=b$  (Algebra)
+
+Beginning of proving a function is not onto:
+Find some specific co-domain element b and show that there is no domain element that could map to b

UNB/Year 4/Semester 2/CS2333/2024-01-19.md

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+Lecture Topic: Functions & Relations
+
+When a function is onto, it means that the entire co-domain of a function is the range of the function
+
+Proving a function is not onto:
+
+Just prove for any element b in the co-domain B and show that b cannot be equal to f(a) for any a in the domain.
+
+Example: $g: \mathbb{R} \rightarrow \mathbb{R}, g(x) = x^2 + 10$ is not onto
+
+Proof:
+For example, choose the element $-1$ from the co-domain $\mathbb{R}$ 
+Consider any domain element $a$
+The function will square $a$, and $a^2 \geq 0$ (because no squares are negative reals)
+Then $a^2 + 10 \geq 10$, so this means that $-1$ is not in the co-domain
+$\therefore$ The function is not onto
+
+You can also use a proof by contradiction by saying that:
+$g(a) = -1$
+$\therefore a^2 + 10 = -1$
+$\therefore a^2 = -11$
+Squares of a real number cannot be negative, so the function is not onto
+
+A function is not always numbers:
+
+Example: Favourite UNB Course : S -> C
+Where S is all students and C is all courses
+
+What would have to be true for this to be one to one or onto?:
+- One to one:
+- Onto:
+
+# Relations:
+A binary relation R on two sets A and B is a subset of $A \times B$ 
+- e.g. S = all students, C = all students, s is any given student, c is any given course
+- Relation: HasTaken: for (s, c) s has taken c
+- Relation Aplus: for (s,c) s got A+ in c
+
+A relation can be between the same set, for example the set of A = {John, Jane, Bill, Sue, Mary, Betty}, we can define a relation that is the relationship between who loves who $A \times A$
+- e.g. (John, Jane)
+
+A binary relation $R \subseteq A \times A$ is also called an equivalence relation if the following conditions hold:
+- reflexive: $\forall a \in A, (a,a) \in R$
+- symmetric: $\forall a,b \in A, (a,b) \in R \rightarrow (b,a) \in R$
+- transitive: $\forall a,b,c \in A, [(a,b) \in R \wedge (b,c) \in R] \rightarrow (a,c) \in R$
+
+Example:
+
+$G = \{(x,y) | x \in \mathbb{R}, y \in \mathbb{R}, x > y\}$
+
+Is G reflexive? (Is is true that every real number is relation to itself)
+No, because we can show that $(5,5) \notin \mathbb{R}$ because $5 \ngtr 5$
+
+Is G symmetric? (Is it the case that whenever $(a,b) \in G, (b,a)$ will also be in G)
+No, for example $(10,3) \in G$ because $10 > 3$ but $(3, 10) \notin G$ because $3 \ngtr 10$
+
+Is G transitive? (Is it true that whenever (a,b) and (b,c) are in G, (a,c) will also be in G)
+Yes, let a,b,c be any real numbers where $(a,b) \in G$ and $(b,c) \in G$
+Since $(a,b) \in G$, we know that $a > b$
+Since $(b,c) \in G$, we know that $b > c$
+Then $a > b > c$, so that means that $a > c$, so also $(a,c) \in G$