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Week 11 - Friday
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What did we talk about last time? Graph isomorphism Trees
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You are the despotic ruler of an ancient empire Tomorrow is the 25th anniversary of your reign You have 1,000 bottles of wine you were planning to open for the celebration Your Grand Vizier uncovered a plot to murder you: 10 years ago, a rebel worked in the vineyard that makes your wine He poisoned one of the 1,000 bottles you are going to serve tonight and has been waiting for revenge The rebel died an accidental death, and his papers revealed his plan, but not which bottle was poisoned The poison exhibits no symptoms until death Death occurs within ten to twenty hours after consuming even the tiniest amount of poison You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned You have a few hundred prisoners, sentenced to death Can you use just the prisoners and risk no slaves? If so, what's the smallest number of prisoners you can risk and still be sure to find the bottle?
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A grammar for a formal language (such as we will discuss next week or the week after) is made up of rules that allow non- terminals to be turned into other non-terminals or terminals For example: 1. 2. | 3. 4. a | an | the 5. funky 6. DJ | beat 7. plays | spins Make a parse tree corresponding to the sentence, "The DJ plays a funky beat"
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Any tree with more than one vertex has at least one vertex of degree 1 If a vertex in a tree has degree 1 it is called a terminal vertex (or leaf) All vertices of degree greater than 1 in a tree are called internal vertices (or branch vertices)
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For any positive integer n, a tree with n vertices must have n – 1 edges Prove this by mathematical induction Hint: Any tree with 2 or more nodes has a vertex of degree 1. What happens when you remove that vertex?
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In a rooted tree, one vertex is distinguished and called the root The level of a vertex is the number of edges along the unique path between it and the root The height of a rooted tree is the maximum level of any vertex of the tree The children of any vertex v in a rooted tree are all those nodes that are adjacent to v and one level further away from the root than v Two distinct vertices that are children of the same parent are called siblings If w is a child of v, then v is the parent of w If v is on the unique path from w to the root, then v is an ancestor of w and w is a descendant of v
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Consider the following tree, rooted at 0 What is the level of 5? What is the level of 0? What is the height of this tree? What are the children of 3? What is the parent of 2? What are the siblings of 8? What are the descendants of 3? 3 3 0 0 5 5 2 2 1 1 7 7 9 9 6 6 4 4 8 8
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A binary tree is a rooted tree in which every parent has at most two children Each child is designated either the left child or the right child In a full binary tree, each parent has exactly two children Given a parent v in a binary tree, the left subtree of v is the binary tree whose root is the left child of v Ditto for right subtree
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As we all know from data structures, a binary tree can be used to make a data structure that is efficient for insertions, deletions, and searches But, it doesn't stop there! We can represent binary arithmetic with a binary tree Make a binary tree for the expression ((a – b)∙c) + (d/e) The root of each subtree is an operator Each subtree is either a single operand or another expression
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If k is a positive integer and T is a full binary tree with k internal vertices, then T has a total of 2k + 1 vertices and has k + 1 terminal vertices Prove it! Hint: Induction isn't needed. We just need to relate the number of non-terminal nodes to the number of terminal nodes
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If T is a perfect binary tree with height h, then it has 2 h+1 – 1 vertices Prove it using induction!
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If T is a binary tree with t terminal vertices and height h, then t 2 h Prove it using strong induction on the height of the tree Hint: Consider cases where the root of the tree has 1 child and 2 children separately
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I am assuming that you have been graphing functions since about 6 th or 7 th grade Still, the formal definition of a graph of a function is: Let f be a real-valued function of a real variable The graph of f is the set of all points (x, y) in the Cartesian coordinate plane such that x is in the domain of f and y = f(x)
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In computer science, we are often interested in power functions of x written p a (x) where p a (x) = x a where x and a are nonnegative Power functions are the building blocks of polynomial functions Graph the following on the same coordinate axes p0p0 p 1/2 p1p1 p2p2
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Recall the definition of the floor of x: x = the largest integer that is less than or equal to x Graph f(x) = x Defining functions on integers instead of real values affects their graphs a great deal Graph p 1 (x) = x, x R Graph f(n) = n, n N
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There is a strong visual (and of course mathematical) correlation a function that is the multiple of another function Examples: g(x) = x + 2 2g(x) = 2x + 4 Given f graphed below, sketch 2f -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 2121 -2
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Consider the absolute value function f(x) = |x| Left of the origin it is constantly decreasing Right of the origin it is constantly increasing -6 -5 -4 -3 -2 -11 2 3 4 5 6 2121 -2
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We say that f is decreasing on the set S iff for all real numbers x 1 and x 2 in S, if x 1 f(x 2 ) We say that f is increasing on the set S iff for all real numbers x 1 and x 2 in S, if x 1 < x 2, then f(x 1 ) < f(x 2 ) We say that f is an increasing (or decreasing) function iff f is increasing (or decreasing) on its entire domain Clearly, a positive multiple of an increasing function is increasing Virtually all running time functions are increasing functions
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Student Lecture
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Mathematicians worry about the growth of various functions They usually express such things in terms of limits, maybe with derivatives We are focused primarily on functions that bound running time spent and memory consumed We just need a rough guide We want to know the order of the growth
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Let f and g be real-valued functions defined on the same set of nonnegative real numbers f is of order at least g, written f(x) is (g(x)), iff there is a positive A R and a nonnegative a R such that A|g(x)| ≤ |f(x)| for all x > a f is of order at most g, written f(x) is O(g(x)), iff there is a positive B R and a nonnegative b R such that |f(x)| ≤ B|g(x)| for all x > b f is of order g, written f(x) is (g(x)), iff there are positive A, B R and a nonnegative k R such that A|g(x)| ≤ |f(x)| ≤ B|g(x)| for all x > k
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Express the following statements using appropriate notation: 10|x 6 | ≤ |17x 6 – 45x 3 + 2x + 8| ≤ 30|x 6 |, for x > 2 Justify the following: is ( x)
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Let f and g be real-valued functions defined on the same set of nonnegative real numbers 1. f(x) is (g(x)) and f(x) is O(g(x)) iff f(x) is (g(x)) 2. f(x) is (g(x)) iff g(x) is O(f(x)) 3. f(x) is (f(x)), f(x) is O(f(x)), and f(x) is (f(x)) 4. If f(x) is O(g(x)) and g(x) is O(h(x)) then f(x) is O(h(x)) 5. If f(x) is O(g(x)) and c is a positive real, then c f(x) is O(g(x)) 6. If f(x) is O(h(x)) and g(x) is O(k(x)) then f(x) + g(x) is O(G(x)) where G(x) = max(|h(x)|,|k(x)|) for all x 7. If f(x) is O(h(x)) and g(x) is O(k(x)) then f(x)g(x) is O(h(x)k(x))
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If 1 < x, then x < x 2 x 2 < x 3 …… So, for r, s R, where r 1, x r < x s By extension, x r is O(x s )
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Prove a bound for g(x) = (1/4)(x – 1)(x + 1) for x R Prove that x 2 is not O(x) Hint: Proof by contradiction
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Let f(x) be a polynomial with degree n f(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 … + a 1 x + a 0 By extension from the previous results, if a n is a positive real, then f(x) is O(x s ) for all integers s n f(x) is (x r ) for all integers r ≤ n f(x) is (x n ) Furthermore, let g(x) be a polynomial with degree m g(x) = b m x m + b m-1 x m-1 + b m-2 x m-2 … + b 1 x + b 0 If a n and b m are positive reals, then f(x)/g(x) is O(x c ) for real numbers c > n - m f(x)/g(x) is not O(x c ) for all integers c > n - m f(x)/g(x) is (x n - m )
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More on asymptotic notation
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Keep reading Chapter 11 Keep working on Assignment 9 Due next Friday
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