Renormalize files
This commit is contained in:
@ -1,49 +1,49 @@
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Strings and languages:
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- An alphabet is a finite set of symbols, such as {1,2,3} or {0,1}
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- 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}
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- A languge is a set of strings
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- L1 = {00, 01, 10, 11}
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- L2 = {a^n | n is an integer non-negative}
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Notation: a to the power of n just means a repeated n times
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Examples
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$L_8 = \{a^m b^n | m,n \in \mathbb{z}^{nonneg}\}$
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So valid strings would be
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aaabbb, aaabbbb, ab, abb
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$L_9 = \{a^n b^n | n \in \mathbb{z}^{nonneg}\}$
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So valid strings would be
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aabb, ab, aaaabbbb, aaaaaabbbbbb
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$L_{11} = \{w \in \{a, b\}^* | \text{The number of "a"s is equal to the number of "b"s}\}$
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This would include everything that L9 includes, but allows for the alphabet can be out of order
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You can also written this as
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$L_{11} = \{w \in \{a, b\}^* | n_a(w) = n_b(w)\}$
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You can extend the repetition notation (exponent) to include multiple symbols by wrapping it in parentheses
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Example:
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$L_{12} = \{(ab)^n | n \in \mathbb{z}^{nonneg}\}$
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This means we need to be careful as
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$L_{13} = \{ab^n | n \in \mathbb{z}^{nonneg}\}$
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is different, as L12 is asking for repetition of ab, while L13 is assigning a prefix of a and repetition of b
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$L_{14} = \{a^i b^j c^k | i,j,k \in \mathbb{z}^{nonneg}, j = 2i + 3k\}$
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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
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Intro to functions and relations
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Informally, a function (f) descrives an input output situation (A -> B)
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For every input element a in A, there is exactly one ouput element b in B
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f maps a to b
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A is the domain of f
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Strings and languages:
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- An alphabet is a finite set of symbols, such as {1,2,3} or {0,1}
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- 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}
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- A languge is a set of strings
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- L1 = {00, 01, 10, 11}
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- L2 = {a^n | n is an integer non-negative}
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Notation: a to the power of n just means a repeated n times
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Examples
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$L_8 = \{a^m b^n | m,n \in \mathbb{z}^{nonneg}\}$
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So valid strings would be
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aaabbb, aaabbbb, ab, abb
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$L_9 = \{a^n b^n | n \in \mathbb{z}^{nonneg}\}$
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So valid strings would be
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aabb, ab, aaaabbbb, aaaaaabbbbbb
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$L_{11} = \{w \in \{a, b\}^* | \text{The number of "a"s is equal to the number of "b"s}\}$
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This would include everything that L9 includes, but allows for the alphabet can be out of order
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You can also written this as
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$L_{11} = \{w \in \{a, b\}^* | n_a(w) = n_b(w)\}$
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You can extend the repetition notation (exponent) to include multiple symbols by wrapping it in parentheses
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||||
|
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Example:
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$L_{12} = \{(ab)^n | n \in \mathbb{z}^{nonneg}\}$
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|
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This means we need to be careful as
|
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$L_{13} = \{ab^n | n \in \mathbb{z}^{nonneg}\}$
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is different, as L12 is asking for repetition of ab, while L13 is assigning a prefix of a and repetition of b
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$L_{14} = \{a^i b^j c^k | i,j,k \in \mathbb{z}^{nonneg}, j = 2i + 3k\}$
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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
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Intro to functions and relations
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Informally, a function (f) descrives an input output situation (A -> B)
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For every input element a in A, there is exactly one ouput element b in B
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f maps a to b
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A is the domain of f
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B is the co-domain of f
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@ -1,52 +1,52 @@
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Lecture Topic: Functions
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For every input element $a \in A$ there is exactly one output element $b \in B$
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Jargon:
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- f maps a to b
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- the image of a under f is b
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functions are one-to-one (or injective):
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$\forall x,y \in A, x \neq y \rightarrow f(x) \neq f(y)$
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or in other terms
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$\forall x,y \in A, x = y \rightarrow f(x) = f(y)$
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Example:
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$f(x) = x^2$ is not one to one
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Proof by counter example:
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$x = -1$ 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
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To prove a function is one to one, it is a little more tricky, and needs to be general, for example:
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Let x and y be any arbitrary elements of the domain where f(x) = f(y)
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So it follows that x and y are equal (x = y)
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Example for a real function: $g(x) = 3x - 11$
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Real Numbers
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Let x and y be any real number such that $g(x) = g(y)$. We will show that $x = y$
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Since we know that g(x) = g(y), it means that $3x - 11 = 3y - 11$
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Add 11 to both sides: $\therefore 3x = 3y$
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Divide both sides by 3: $\therefore x = y$
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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
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To prove that a function A -> B is onto:
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- Let be represent and arbitrary element in the co-domain B
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- We want to find an element a in the domain and show that f(a) = b
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Example:
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Real Numbers, $f(x) = 5x + 2$ is onto
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Proof: Let b be any arbitrary real number (we want to find a real number a and show that f(a) = b)
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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 $a = \frac{b - 2}{5}$
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So, let $a = \frac{b - 2}{5}$ , which is a real number. Then plug this back into the function
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$f(a) = f(\frac{b - 2}{5}$) (Substitution)
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$= 5\frac{b-2}{5}+2$ (Definition of f)
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$= (b-2) +2)$ (Algebra)
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$=b$ (Algebra)
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Beginning of proving a function is not onto:
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Lecture Topic: Functions
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For every input element $a \in A$ there is exactly one output element $b \in B$
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Jargon:
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- f maps a to b
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- the image of a under f is b
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functions are one-to-one (or injective):
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$\forall x,y \in A, x \neq y \rightarrow f(x) \neq f(y)$
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or in other terms
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$\forall x,y \in A, x = y \rightarrow f(x) = f(y)$
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Example:
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$f(x) = x^2$ is not one to one
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Proof by counter example:
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$x = -1$ 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
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To prove a function is one to one, it is a little more tricky, and needs to be general, for example:
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Let x and y be any arbitrary elements of the domain where f(x) = f(y)
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So it follows that x and y are equal (x = y)
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Example for a real function: $g(x) = 3x - 11$
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Real Numbers
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Let x and y be any real number such that $g(x) = g(y)$. We will show that $x = y$
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Since we know that g(x) = g(y), it means that $3x - 11 = 3y - 11$
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Add 11 to both sides: $\therefore 3x = 3y$
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Divide both sides by 3: $\therefore x = y$
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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
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To prove that a function A -> B is onto:
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- Let be represent and arbitrary element in the co-domain B
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- We want to find an element a in the domain and show that f(a) = b
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Example:
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Real Numbers, $f(x) = 5x + 2$ is onto
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Proof: Let b be any arbitrary real number (we want to find a real number a and show that f(a) = b)
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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 $a = \frac{b - 2}{5}$
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So, let $a = \frac{b - 2}{5}$ , which is a real number. Then plug this back into the function
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$f(a) = f(\frac{b - 2}{5}$) (Substitution)
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$= 5\frac{b-2}{5}+2$ (Definition of f)
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$= (b-2) +2)$ (Algebra)
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$=b$ (Algebra)
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Beginning of proving a function is not onto:
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Find some specific co-domain element b and show that there is no domain element that could map to b
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@ -1,61 +1,61 @@
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Lecture Topic: Functions & Relations
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When a function is onto, it means that the entire co-domain of a function is the range of the function
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Proving a function is not onto:
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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.
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Example: $g: \mathbb{R} \rightarrow \mathbb{R}, g(x) = x^2 + 10$ is not onto
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Proof:
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For example, choose the element $-1$ from the co-domain $\mathbb{R}$
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Consider any domain element $a$
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The function will square $a$, and $a^2 \geq 0$ (because no squares are negative reals)
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Then $a^2 + 10 \geq 10$, so this means that $-1$ is not in the co-domain
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$\therefore$ The function is not onto
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You can also use a proof by contradiction by saying that:
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$g(a) = -1$
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$\therefore a^2 + 10 = -1$
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$\therefore a^2 = -11$
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Squares of a real number cannot be negative, so the function is not onto
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A function is not always numbers:
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Example: Favourite UNB Course : S -> C
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Where S is all students and C is all courses
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What would have to be true for this to be one to one or onto?:
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- One to one:
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- Onto:
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# Relations:
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A binary relation R on two sets A and B is a subset of $A \times B$
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- e.g. S = all students, C = all students, s is any given student, c is any given course
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- Relation: HasTaken: for (s, c) s has taken c
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- Relation Aplus: for (s,c) s got A+ in c
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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 $A \times A$
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- e.g. (John, Jane)
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A binary relation $R \subseteq A \times A$ is also called an equivalence relation if the following conditions hold:
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- reflexive: $\forall a \in A, (a,a) \in R$
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- symmetric: $\forall a,b \in A, (a,b) \in R \rightarrow (b,a) \in R$
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- transitive: $\forall a,b,c \in A, [(a,b) \in R \wedge (b,c) \in R] \rightarrow (a,c) \in R$
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Example:
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$G = \{(x,y) | x \in \mathbb{R}, y \in \mathbb{R}, x > y\}$
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Is G reflexive? (Is is true that every real number is relation to itself)
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No, because we can show that $(5,5) \notin \mathbb{R}$ because $5 \ngtr 5$
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Is G symmetric? (Is it the case that whenever $(a,b) \in G, (b,a)$ will also be in G)
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No, for example $(10,3) \in G$ because $10 > 3$ but $(3, 10) \notin G$ because $3 \ngtr 10$
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Is G transitive? (Is it true that whenever (a,b) and (b,c) are in G, (a,c) will also be in G)
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Yes, let a,b,c be any real numbers where $(a,b) \in G$ and $(b,c) \in G$
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Since $(a,b) \in G$, we know that $a > b$
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Since $(b,c) \in G$, we know that $b > c$
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Lecture Topic: Functions & Relations
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When a function is onto, it means that the entire co-domain of a function is the range of the function
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Proving a function is not onto:
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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.
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Example: $g: \mathbb{R} \rightarrow \mathbb{R}, g(x) = x^2 + 10$ is not onto
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Proof:
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For example, choose the element $-1$ from the co-domain $\mathbb{R}$
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Consider any domain element $a$
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The function will square $a$, and $a^2 \geq 0$ (because no squares are negative reals)
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Then $a^2 + 10 \geq 10$, so this means that $-1$ is not in the co-domain
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$\therefore$ The function is not onto
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You can also use a proof by contradiction by saying that:
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$g(a) = -1$
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$\therefore a^2 + 10 = -1$
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$\therefore a^2 = -11$
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Squares of a real number cannot be negative, so the function is not onto
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A function is not always numbers:
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Example: Favourite UNB Course : S -> C
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Where S is all students and C is all courses
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What would have to be true for this to be one to one or onto?:
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- One to one:
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- Onto:
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# Relations:
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A binary relation R on two sets A and B is a subset of $A \times B$
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- e.g. S = all students, C = all students, s is any given student, c is any given course
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- Relation: HasTaken: for (s, c) s has taken c
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- Relation Aplus: for (s,c) s got A+ in c
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|
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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 $A \times A$
|
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- e.g. (John, Jane)
|
||||
|
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A binary relation $R \subseteq A \times A$ is also called an equivalence relation if the following conditions hold:
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- reflexive: $\forall a \in A, (a,a) \in R$
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- symmetric: $\forall a,b \in A, (a,b) \in R \rightarrow (b,a) \in R$
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- transitive: $\forall a,b,c \in A, [(a,b) \in R \wedge (b,c) \in R] \rightarrow (a,c) \in R$
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Example:
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$G = \{(x,y) | x \in \mathbb{R}, y \in \mathbb{R}, x > y\}$
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|
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Is G reflexive? (Is is true that every real number is relation to itself)
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No, because we can show that $(5,5) \notin \mathbb{R}$ because $5 \ngtr 5$
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|
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Is G symmetric? (Is it the case that whenever $(a,b) \in G, (b,a)$ will also be in G)
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No, for example $(10,3) \in G$ because $10 > 3$ but $(3, 10) \notin G$ because $3 \ngtr 10$
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|
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Is G transitive? (Is it true that whenever (a,b) and (b,c) are in G, (a,c) will also be in G)
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Yes, let a,b,c be any real numbers where $(a,b) \in G$ and $(b,c) \in G$
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Since $(a,b) \in G$, we know that $a > b$
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Since $(b,c) \in G$, we know that $b > c$
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Then $a > b > c$, so that means that $a > c$, so also $(a,c) \in G$
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@ -1,76 +1,76 @@
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Lecture Topic: Relation examples
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# Relation Example 1
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$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$
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e.g $(22,7) \in R$ because $22-7 = 5(3)$
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e.g $(7,22) \in R$ because $7-22 = -15 = 5(-3)$
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e.g $(22,9) \notin R$ because $22-9 = 13$, which is not a multiple of 5
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Reflexive? Yes
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For every integer a, $a-a = 0 = 5(0)$
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Therefor $(a,a) \in R$ for every integer a
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Symmetric? Yes
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Let a, b be any integer, such that (a,b) is in the relation (can we prove that (b,a) is in relation as well?)
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Since (a,b) is in the relation, $a-b = 5n$, so, $b-a = -5(n) = 5(-n)$
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Note: $(b-a) = -(a-b)$
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Transitive? Yes
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Let a,b,c be any integers such that $(a,b) \in R$ and $(b,c) \in R$
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Can we prove that $(a,c) \in R$ ?
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Since $(a,b) \in R$, we know $a-b=5n$ for some $n \in \mathbb{Z}$
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Since $(b,c) \in R$, we know $b-c=5p$ for some $n \in \mathbb{Z}$
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Now, $(a-c) = (a-b) + (b-c) = 5n + 5p = 5(np)$
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Therefore, $(a-c) \in R$
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Note: This is an integer because it is the sum of integers
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# Relation Example 2
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$R = \{(i,j) | i \in \mathbb{R}, j \in \mathbb{R}, i + 2 > j\}$
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Examples:
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- $(5,2) \in R$ because $(5+2) > 2$
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- $(3,4) \in R$ because $(3+2) > 4$
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- $(0,7) \notin R$ because $(0+2) \ngtr 7$
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Reflexive? Yes
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Symmetric? No
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Transitive? No
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Proof: We need to find real numbers where $a,b,c$ where $(a,b) \in R$ and $(b,c) \in R$ but $(a,c) \notin R$
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For example, a = 1, b = 2, c = 3
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$(1,2) \in R$ because $1+2 > 2$
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$(2,3) \in R$ because $2+2 > 3$
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But $(1,3) \notin R$ because $(1+2) \ngtr 3$
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$\therefore$ R is not transitive
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# Equivalence relations
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Equivalence relations have all three of these properties
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Any equivalence relation R on a set A induces a partition of A
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- Splits A into equivalence classes
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- Each class contains elements that are related to themselves and to each other, but to nothing outside the class
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So, for the above relation:
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$$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$$
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This forms an equivalence relation
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|
||||
So for the set of all integers, this forms 5 equivalence classes:
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||||
- $..., 0, 5, 10, ...$
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||||
- $..., 1, 6, 11, ...$
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||||
- $..., 2, 7, 12, ...$
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||||
- $..., 3, 8, 13, ...$
|
||||
- $..., 4, 9, 14, ...$
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||||
|
||||
## Example
|
||||
$A = \{-3, -2, -1, 0, 1, 2, 3\}$
|
||||
$R = \{(i,j) | i \in \mathbb{A}, j \in \mathbb{A}, i^2 = j^2\}$
|
||||
|
||||
So the equivalence classes induced by this relation:
|
||||
- -3, 3
|
||||
- -2, 2
|
||||
- -1, 1
|
||||
Lecture Topic: Relation examples
|
||||
|
||||
# Relation Example 1
|
||||
$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$
|
||||
e.g $(22,7) \in R$ because $22-7 = 5(3)$
|
||||
e.g $(7,22) \in R$ because $7-22 = -15 = 5(-3)$
|
||||
e.g $(22,9) \notin R$ because $22-9 = 13$, which is not a multiple of 5
|
||||
|
||||
Reflexive? Yes
|
||||
For every integer a, $a-a = 0 = 5(0)$
|
||||
Therefor $(a,a) \in R$ for every integer a
|
||||
|
||||
Symmetric? Yes
|
||||
Let a, b be any integer, such that (a,b) is in the relation (can we prove that (b,a) is in relation as well?)
|
||||
|
||||
Since (a,b) is in the relation, $a-b = 5n$, so, $b-a = -5(n) = 5(-n)$
|
||||
Note: $(b-a) = -(a-b)$
|
||||
|
||||
Transitive? Yes
|
||||
Let a,b,c be any integers such that $(a,b) \in R$ and $(b,c) \in R$
|
||||
Can we prove that $(a,c) \in R$ ?
|
||||
|
||||
Since $(a,b) \in R$, we know $a-b=5n$ for some $n \in \mathbb{Z}$
|
||||
Since $(b,c) \in R$, we know $b-c=5p$ for some $n \in \mathbb{Z}$
|
||||
|
||||
Now, $(a-c) = (a-b) + (b-c) = 5n + 5p = 5(np)$
|
||||
Therefore, $(a-c) \in R$
|
||||
Note: This is an integer because it is the sum of integers
|
||||
|
||||
# Relation Example 2
|
||||
$R = \{(i,j) | i \in \mathbb{R}, j \in \mathbb{R}, i + 2 > j\}$
|
||||
|
||||
Examples:
|
||||
- $(5,2) \in R$ because $(5+2) > 2$
|
||||
- $(3,4) \in R$ because $(3+2) > 4$
|
||||
- $(0,7) \notin R$ because $(0+2) \ngtr 7$
|
||||
|
||||
Reflexive? Yes
|
||||
Symmetric? No
|
||||
Transitive? No
|
||||
|
||||
Proof: We need to find real numbers where $a,b,c$ where $(a,b) \in R$ and $(b,c) \in R$ but $(a,c) \notin R$
|
||||
|
||||
For example, a = 1, b = 2, c = 3
|
||||
$(1,2) \in R$ because $1+2 > 2$
|
||||
$(2,3) \in R$ because $2+2 > 3$
|
||||
|
||||
But $(1,3) \notin R$ because $(1+2) \ngtr 3$
|
||||
$\therefore$ R is not transitive
|
||||
|
||||
# Equivalence relations
|
||||
Equivalence relations have all three of these properties
|
||||
Any equivalence relation R on a set A induces a partition of A
|
||||
- Splits A into equivalence classes
|
||||
- Each class contains elements that are related to themselves and to each other, but to nothing outside the class
|
||||
|
||||
So, for the above relation:
|
||||
$$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$$
|
||||
This forms an equivalence relation
|
||||
|
||||
So for the set of all integers, this forms 5 equivalence classes:
|
||||
- $..., 0, 5, 10, ...$
|
||||
- $..., 1, 6, 11, ...$
|
||||
- $..., 2, 7, 12, ...$
|
||||
- $..., 3, 8, 13, ...$
|
||||
- $..., 4, 9, 14, ...$
|
||||
|
||||
## Example
|
||||
$A = \{-3, -2, -1, 0, 1, 2, 3\}$
|
||||
$R = \{(i,j) | i \in \mathbb{A}, j \in \mathbb{A}, i^2 = j^2\}$
|
||||
|
||||
So the equivalence classes induced by this relation:
|
||||
- -3, 3
|
||||
- -2, 2
|
||||
- -1, 1
|
||||
- 0
|
||||
@ -1,52 +1,52 @@
|
||||
Lecture Topic:
|
||||
|
||||
# Equivalence Classes
|
||||
You can represent equivalence classes by writing each one as a set:
|
||||
$$C_0 = \{..., -10, -5, 0, 5, 10, 15, ...\} \text{ or } \{5p | p \in \mathbb{Z}\}$$
|
||||
$$C_0 = \{..., -9, -4, 1, 6, 11, 16, ...\} \text{ or } \{5p + 1| p \in \mathbb{Z}\}$$
|
||||
$$\text{and so on...}$$
|
||||
# Graphs
|
||||
Definitions:
|
||||
- For each *vertex* v, the **degree** of v is the number of edges incident on v
|
||||
- A *path* is a **sequence** of vertices connected by edges
|
||||
- A *cycle* is a path that **starts and ends** at the same vertex
|
||||
- A *simple path* is a path that has **no repeated** vertices
|
||||
- A *connected graph* contains a path between **every pair** of vertices
|
||||
- A *disconnected graph* is the opposite (look at slides for real def)
|
||||
|
||||
# Regular Languages
|
||||
A class of languages that can be processed by simple computers called **finite automata**
|
||||
|
||||
## Simple Example of a finite automaton
|
||||
Vending machine/toll booth gate
|
||||
Requires 25 cents or more to be deposited (gives no change)
|
||||
|
||||
Takes as input, 5 cent, 10 cent, 25 cent coins
|
||||
Uses **states** to remember how much has been deposited so far
|
||||
|
||||
States:
|
||||
- q_0 : 0 cents deposited (START STATE)
|
||||
- q_5 : 5 cents
|
||||
- q_10 : 10 cents
|
||||
- q_15 : 15 cents
|
||||
- q_20 : 20 cents
|
||||
- q_25 : 25 cents (ACCEPT STATE)
|
||||
|
||||
We represent the input as a **string** of symbols.
|
||||
- n = nickel
|
||||
- d = dime
|
||||
- q = quarter
|
||||
Examples:
|
||||
- ddn
|
||||
- ndd
|
||||
- dnd
|
||||
- q
|
||||
Not accepted examples:
|
||||
- nd
|
||||
- d
|
||||
- $\epsilon$
|
||||
## State Diagram
|
||||
(not really possible without an image in obsidian)
|
||||
|
||||
# Finite Automata
|
||||
Lecture Topic:
|
||||
|
||||
# Equivalence Classes
|
||||
You can represent equivalence classes by writing each one as a set:
|
||||
$$C_0 = \{..., -10, -5, 0, 5, 10, 15, ...\} \text{ or } \{5p | p \in \mathbb{Z}\}$$
|
||||
$$C_0 = \{..., -9, -4, 1, 6, 11, 16, ...\} \text{ or } \{5p + 1| p \in \mathbb{Z}\}$$
|
||||
$$\text{and so on...}$$
|
||||
# Graphs
|
||||
Definitions:
|
||||
- For each *vertex* v, the **degree** of v is the number of edges incident on v
|
||||
- A *path* is a **sequence** of vertices connected by edges
|
||||
- A *cycle* is a path that **starts and ends** at the same vertex
|
||||
- A *simple path* is a path that has **no repeated** vertices
|
||||
- A *connected graph* contains a path between **every pair** of vertices
|
||||
- A *disconnected graph* is the opposite (look at slides for real def)
|
||||
|
||||
# Regular Languages
|
||||
A class of languages that can be processed by simple computers called **finite automata**
|
||||
|
||||
## Simple Example of a finite automaton
|
||||
Vending machine/toll booth gate
|
||||
Requires 25 cents or more to be deposited (gives no change)
|
||||
|
||||
Takes as input, 5 cent, 10 cent, 25 cent coins
|
||||
Uses **states** to remember how much has been deposited so far
|
||||
|
||||
States:
|
||||
- q_0 : 0 cents deposited (START STATE)
|
||||
- q_5 : 5 cents
|
||||
- q_10 : 10 cents
|
||||
- q_15 : 15 cents
|
||||
- q_20 : 20 cents
|
||||
- q_25 : 25 cents (ACCEPT STATE)
|
||||
|
||||
We represent the input as a **string** of symbols.
|
||||
- n = nickel
|
||||
- d = dime
|
||||
- q = quarter
|
||||
Examples:
|
||||
- ddn
|
||||
- ndd
|
||||
- dnd
|
||||
- q
|
||||
Not accepted examples:
|
||||
- nd
|
||||
- d
|
||||
- $\epsilon$
|
||||
## State Diagram
|
||||
(not really possible without an image in obsidian)
|
||||
|
||||
# Finite Automata
|
||||
(look at slides, didn't catch it)
|
||||
@ -1,85 +1,85 @@
|
||||
# Finite Automata
|
||||
Simple machines that take in strings (sequences of symbols) as input and recognize whether or note each input string satisfies some condition(s)
|
||||
|
||||
Finite automata use states to keep track of important information about symbols
|
||||
|
||||
## Mathematical definition of a finite automaton
|
||||
$M = (Q, \sum, \delta, q, F)$
|
||||
- $Q$ is a finite set of states
|
||||
- $\sum$ is a finite set of input symbols called the input alphabet
|
||||
- $\delta$ is the transition function
|
||||
- Takes every pair consisting of a state and an input smbol and returns the next state, this tells us everything we need to know about what the machine does in one computation step
|
||||
- $q \in Q$ is the start state
|
||||
- $F \subseteq Q$ is the set of accept states
|
||||
|
||||
## Acceptance by a finite automaton
|
||||
Let M be a finite automaton
|
||||
Let $w = w_1, w_2, ..., w_n$ be n input string over \sum
|
||||
As input string w is process by $M$, define the sequence of visited states $r_0, r_1, ..., r_n$ as follows:
|
||||
- $r_0 = q$
|
||||
- $\forall i = 0,1, ... n-1, r_{i+1} = \delta(r_i, w_{i+1})$
|
||||
If $r_n \in F$ then $M$ accepts $w$, otherwise $M$ rejects $w$
|
||||
|
||||
Note: The empty string $\epsilon$ has length $0$, it is accepted by $M$ if any only if the start state is an accept state
|
||||
|
||||
For a given finite automaton $M$, the set of stings accepted by $M$ is the language of $M$ and is denoted $L(M)$
|
||||
|
||||
A language $A$ is called regular if there exists a finite automaton $M$ such that $A = L(M)$
|
||||
|
||||
## Example of finite automata
|
||||
A finite automation $M_1$ that accepts
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) \geq 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start sate and C is the accept state
|
||||
A -> B (On 1)
|
||||
B -> C (On 1)
|
||||
A -> A (On 0)
|
||||
B -> B (On 0)
|
||||
C -> C (On 0, 1)
|
||||
|
||||
A: we have seen no ones yet
|
||||
B: we have seen exactly one 1
|
||||
C: we have seen two ore more 1's
|
||||
|
||||
## Example 2 of finite automata
|
||||
A finite automation $M_2$ that accepts
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) = 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and C is the accept state
|
||||
A -> B (On 1)
|
||||
B -> C (On 1)
|
||||
C -> D (On 1)
|
||||
A -> A (On 0)
|
||||
B -> B (On 0)
|
||||
C -> C (On 0)
|
||||
D -> D (On 0, 1)
|
||||
|
||||
Note: D is an example of a dead state, in which you cannot escape after
|
||||
|
||||
## Example 3 of finite automata
|
||||
A finite automation $M_3$ that accepts
|
||||
$$\{w \in \{a,b\}^* \ | \ \text{w starts with abb}\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and D is the accept state
|
||||
A -> B (On a)
|
||||
B -> C (On b)
|
||||
C -> D (On b)
|
||||
D -> D (On a, b)
|
||||
A -> X (On b)
|
||||
B -> X (On a)
|
||||
C -> X (On a)
|
||||
X -> X (On a ,b)
|
||||
|
||||
## Example 4 of finite automata
|
||||
A finite automation $M_4$ that accepts
|
||||
(Didn't catch the definition)
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) = 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and D is the accept state
|
||||
|
||||
(state transitions would go here)
|
||||
|
||||
A: the most recent symbol was 1 or we have seen no symbols
|
||||
B: the last symbol was 0, but we have noon seen 010 yet
|
||||
C: the last two symbols were 01, but we have not seen 010
|
||||
# Finite Automata
|
||||
Simple machines that take in strings (sequences of symbols) as input and recognize whether or note each input string satisfies some condition(s)
|
||||
|
||||
Finite automata use states to keep track of important information about symbols
|
||||
|
||||
## Mathematical definition of a finite automaton
|
||||
$M = (Q, \sum, \delta, q, F)$
|
||||
- $Q$ is a finite set of states
|
||||
- $\sum$ is a finite set of input symbols called the input alphabet
|
||||
- $\delta$ is the transition function
|
||||
- Takes every pair consisting of a state and an input smbol and returns the next state, this tells us everything we need to know about what the machine does in one computation step
|
||||
- $q \in Q$ is the start state
|
||||
- $F \subseteq Q$ is the set of accept states
|
||||
|
||||
## Acceptance by a finite automaton
|
||||
Let M be a finite automaton
|
||||
Let $w = w_1, w_2, ..., w_n$ be n input string over \sum
|
||||
As input string w is process by $M$, define the sequence of visited states $r_0, r_1, ..., r_n$ as follows:
|
||||
- $r_0 = q$
|
||||
- $\forall i = 0,1, ... n-1, r_{i+1} = \delta(r_i, w_{i+1})$
|
||||
If $r_n \in F$ then $M$ accepts $w$, otherwise $M$ rejects $w$
|
||||
|
||||
Note: The empty string $\epsilon$ has length $0$, it is accepted by $M$ if any only if the start state is an accept state
|
||||
|
||||
For a given finite automaton $M$, the set of stings accepted by $M$ is the language of $M$ and is denoted $L(M)$
|
||||
|
||||
A language $A$ is called regular if there exists a finite automaton $M$ such that $A = L(M)$
|
||||
|
||||
## Example of finite automata
|
||||
A finite automation $M_1$ that accepts
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) \geq 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start sate and C is the accept state
|
||||
A -> B (On 1)
|
||||
B -> C (On 1)
|
||||
A -> A (On 0)
|
||||
B -> B (On 0)
|
||||
C -> C (On 0, 1)
|
||||
|
||||
A: we have seen no ones yet
|
||||
B: we have seen exactly one 1
|
||||
C: we have seen two ore more 1's
|
||||
|
||||
## Example 2 of finite automata
|
||||
A finite automation $M_2$ that accepts
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) = 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and C is the accept state
|
||||
A -> B (On 1)
|
||||
B -> C (On 1)
|
||||
C -> D (On 1)
|
||||
A -> A (On 0)
|
||||
B -> B (On 0)
|
||||
C -> C (On 0)
|
||||
D -> D (On 0, 1)
|
||||
|
||||
Note: D is an example of a dead state, in which you cannot escape after
|
||||
|
||||
## Example 3 of finite automata
|
||||
A finite automation $M_3$ that accepts
|
||||
$$\{w \in \{a,b\}^* \ | \ \text{w starts with abb}\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and D is the accept state
|
||||
A -> B (On a)
|
||||
B -> C (On b)
|
||||
C -> D (On b)
|
||||
D -> D (On a, b)
|
||||
A -> X (On b)
|
||||
B -> X (On a)
|
||||
C -> X (On a)
|
||||
X -> X (On a ,b)
|
||||
|
||||
## Example 4 of finite automata
|
||||
A finite automation $M_4$ that accepts
|
||||
(Didn't catch the definition)
|
||||
$$\{w \in \{0,1\}^* \ | \ n_1(w) = 2\}$$
|
||||
(State diagram on board)
|
||||
A is the start state and D is the accept state
|
||||
|
||||
(state transitions would go here)
|
||||
|
||||
A: the most recent symbol was 1 or we have seen no symbols
|
||||
B: the last symbol was 0, but we have noon seen 010 yet
|
||||
C: the last two symbols were 01, but we have not seen 010
|
||||
D: we have seen the pattern 010
|
||||
@ -1,3 +1,3 @@
|
||||
Lecture Topic: More Finite Automata
|
||||
# Examples
|
||||
Lots of example of finite automata, easier to look at posted slides
|
||||
Lecture Topic: More Finite Automata
|
||||
# Examples
|
||||
Lots of example of finite automata, easier to look at posted slides
|
||||
|
||||
@ -1,3 +1,3 @@
|
||||
Lecture Topic: More FAs
|
||||
|
||||
Lecture Topic: More FAs
|
||||
|
||||
This lecture we also tried to determine the language from a visual example of a finite automata
|
||||
@ -1,15 +1,15 @@
|
||||
Lecture Topic: Non-determinism
|
||||
|
||||
# So Far
|
||||
All of the finite automata have been deterministic. This means that for any combination of state and input symbol there is exactly one transition defined
|
||||
|
||||
There are advantages to this
|
||||
- Easier to trace a string through the finite automata
|
||||
Also some disadvantages
|
||||
- Can make it harder to design
|
||||
|
||||
Example on whiteboard
|
||||
|
||||
In a non-deterministic finite automata (NFA), there are different paths, or choices you can make, so there can be more than one, or no transition defined for a given state.
|
||||
|
||||
Lecture Topic: Non-determinism
|
||||
|
||||
# So Far
|
||||
All of the finite automata have been deterministic. This means that for any combination of state and input symbol there is exactly one transition defined
|
||||
|
||||
There are advantages to this
|
||||
- Easier to trace a string through the finite automata
|
||||
Also some disadvantages
|
||||
- Can make it harder to design
|
||||
|
||||
Example on whiteboard
|
||||
|
||||
In a non-deterministic finite automata (NFA), there are different paths, or choices you can make, so there can be more than one, or no transition defined for a given state.
|
||||
|
||||
To verify a string in a NFA, you need to follow each "choice" until a string completes, (ie, for two transitions defined, fork and check both paths)
|
||||
@ -1,11 +1,11 @@
|
||||
Lecture Topic: More NFA
|
||||
|
||||
Design a state diagram for an NFA that accepts
|
||||
$$L = \{w \in \{a,b,c\}^* | \text{w starts with abc, ends with aab, and contains the substring bb}\}$$
|
||||
So, the first three states must be connecting a, b and c
|
||||
|
||||
(D) -a, b, c> (D)
|
||||
(F) -a, b, c> (F)
|
||||
(A) -a> (B) -b> (C) -c> (D) -b> (E) -b> (F) -a> (G) -a> (H) -b> ((I))
|
||||
|
||||
Lecture Topic: More NFA
|
||||
|
||||
Design a state diagram for an NFA that accepts
|
||||
$$L = \{w \in \{a,b,c\}^* | \text{w starts with abc, ends with aab, and contains the substring bb}\}$$
|
||||
So, the first three states must be connecting a, b and c
|
||||
|
||||
(D) -a, b, c> (D)
|
||||
(F) -a, b, c> (F)
|
||||
(A) -a> (B) -b> (C) -c> (D) -b> (E) -b> (F) -a> (G) -a> (H) -b> ((I))
|
||||
|
||||
It's important that on D and F, that b is allowed to loop, as if it did not, some valid strings would not be accepted, as it would only allow for two bs in the middle
|
||||
@ -1,2 +1,2 @@
|
||||
Lecture Topics: NFA examples
|
||||
|
||||
Lecture Topics: NFA examples
|
||||
|
||||
|
||||
@ -1,17 +1,17 @@
|
||||
Lecture Topic: Proofs
|
||||
|
||||
For any DFA, we already have an NFA, it just not happen to use any characteristics of NFAs like epsilon transitions or more/less than 1 transition per symbol per state
|
||||
|
||||
Suppose we have an NFA N that accepts language L
|
||||
We can construct a DFA D that accepts the same language
|
||||
* The DFA keeps track of all the possible states the NFA could be in after seeing any sequence of input symbols
|
||||
Example in slides
|
||||
|
||||
Reminder: Any language that can be accepted by a FA is called a regular language
|
||||
|
||||
Let A and B languages, we define the regular operations, union, concatenation and star
|
||||
Union: Is all the strings that are in either of the languages A or B
|
||||
Concatenation: All the strings that can be formed by the concatenation of A and B
|
||||
Star: Any sequence of strings formed from any combination strings in a language A
|
||||
|
||||
Lecture Topic: Proofs
|
||||
|
||||
For any DFA, we already have an NFA, it just not happen to use any characteristics of NFAs like epsilon transitions or more/less than 1 transition per symbol per state
|
||||
|
||||
Suppose we have an NFA N that accepts language L
|
||||
We can construct a DFA D that accepts the same language
|
||||
* The DFA keeps track of all the possible states the NFA could be in after seeing any sequence of input symbols
|
||||
Example in slides
|
||||
|
||||
Reminder: Any language that can be accepted by a FA is called a regular language
|
||||
|
||||
Let A and B languages, we define the regular operations, union, concatenation and star
|
||||
Union: Is all the strings that are in either of the languages A or B
|
||||
Concatenation: All the strings that can be formed by the concatenation of A and B
|
||||
Star: Any sequence of strings formed from any combination strings in a language A
|
||||
|
||||
Examples in slides
|
||||
@ -1,3 +1,3 @@
|
||||
Lecture Topic: NFA to DFA proof, Regex Intro
|
||||
|
||||
Lecture Topic: NFA to DFA proof, Regex Intro
|
||||
|
||||
Look at posted notes
|
||||
Reference in New Issue
Block a user