CS135601 Introduction to Information Engineering Algorithms 12/9/2018 Che-Rung Lee 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Josephus problem Flavius Josephus is a Jewish historian living in the 1st century. According to his account, he and his 40 comrade soldiers were trapped in a cave, surrounded by Romans. They chose suicide over capture and decided that they would form a circle and start killing themselves using a step of three. As Josephus did not want to die, he was able to find the safe place, and stayed alive with his comrade, later joining the Romans who captured them. 12/9/2018 CS135601 Introduction to Information Engineering

Can you find the safe place? 40 41 1 2 39 3 38 37 4 5 36 6 35 7 34 8 33 9 32 Safe place 10 31 11 30 Can you find the safe place FASTER? 12 29 13 28 14 27 15 26 16 25 17 24 18 23 19 22 21 20 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Algorithm An effective method for solving a problem using a finite sequence of instructions. It need be able to solve the problem. (correctness) It can be represented by a finite number of (computer) instructions. Each instruction must be achievable (by computer) The more effective, the better algorithm is. How to measure the “efficiency”? 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Outline The first algorithm Model, simulation, algorithm primitive The second algorithm Algorithm discovery, recursion The third algorithm Solving recursion, mathematical induction The central role of computer science Why study math? 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering The first algorithm 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering A simpler version Let’s consider a similar problem There are n person in a circle, numbered from 1 to n sequentially. Starting from the number 1 person, every 2nd person will be killed. What is the safe place? The input is n, the output f(n) is a number between 1 and n. Ex: f(8) = ? 1 2 8 3 7 4 6 5 12/9/2018 CS135601 Introduction to Information Engineering

The first algorithm: simulation We can find f(n) using simulation. Simulation is a process to imitate the real objects, states of affairs, or process. We do not need to “kill” anyone to find f(n). The simulation needs (1) a model to represents “n people in a circle” (2) a way to simulate “kill every 2nd person” (3) knowing when to stop 12/9/2018 CS135601 Introduction to Information Engineering

Model n people in a circle We can use “data structure” to model it. This is called a “circular linked list”. Each node is of some “struct” data type Each link is a “pointer” 1 2 3 5 6 7 8 4 struct PIC { int ID; struct PIC *next; } 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Kill every 2nd person Remove every 2nd node in the circular liked list. You need to maintain the circular linked structure after removing node 2 The process can continue until … 1 8 2 7 3 6 4 5 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Knowing when to stop Stop when there is only one node left How to know that? When the *next is pointing to itself It’s ID is f(n) f(8) = 1 1 8 7 3 6 4 5 12/9/2018 CS135601 Introduction to Information Engineering

How fast can it compute f(n)? Since the first algorithm removes one node at a time, it needs n-1 steps to compute f(n) The cost of each step (removing one node) is a constant time  (independent of n) So the total number of operations to find f(n) is (n-1) We can just represent it as O(n). 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering The Second Algorithm 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Can we do better? Simulation is a brute-force method. Faster algorithms are usually expected. The scientific approach Observing some cases Making some hypotheses (generalization) Testing the hypotheses Repeat 1-3 until success 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Observing a case Let’s checkout the case n = 8. What have you observed? All even numbered people die This is true for all kinds of n. The starting point is back to 1 This is only true when n is even Let’s first consider the case when n is even 1 2 8 3 7 4 6 5 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering n is even For n=8, after the first “round”, there are only 4 people left. If we can solve f(4), can we use it to solve f(8)? If we renumber the remaining people, 11, 32, 53, 74, it becomes the n=4 problem. So, if f(4)=x, f(8) =2x – 1 1 2 8 3 7 4 6 5 1 2 3 4 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering n is odd Let’s checkout the case n=9 The starting point becomes 3 If we can solve f(4), can we use it to solve f(9)? If we renumber the remaining people, 31, 52, 73, 94, it becomes the n=4 problem. So, if f(4)=x, f(9) =2x+1 1 9 2 8 3 7 4 6 5 1 2 3 4 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Recursive relation Hypothesis: If f(n)=x, f(2n)=2x–1. If f(n)=x, f(2n+1)=2x+1. How to prove or disprove it? This is called a recursive relation. You can design a recursive algorithm Compute f(8) uses f(4); Compute f(4) uses f(2); Compute f(2) uses f(1); f(1) =1. Why? 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Recursive function call Josephus(9) int Josephus (n) { /* base case */ if (n == 1) return 1; if (n%2 == 0) /*n is even*/ return 2*Josephus (n/2)-1; else /*n is odd*/ return 2*Josephus((n–1)/2)+1; } output fn fn=3 call Josephus(4) fn  2*fn+1 fn=1 call Josephus(2) fn  2*fn–1 fn=1 call Josephus(1) fn  2*fn–1 fn=1 if (n = 1) return 1 12/9/2018 CS135601 Introduction to Information Engineering

How fast can it compute f(n)? n reduces at least half at each step. Need log2(n) steps to reach n=1. Each step needs a constant number of operations . The total number of operations is log2(n) 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering The Third Algorithm 12/9/2018 CS135601 Introduction to Information Engineering 21

CS135601 Introduction to Information Engineering Can we do better? Can we solve the recursion? If we can, we may have a better algorithm to find f(n) Remember: observation, hypotheses, verification 12/9/2018 CS135601 Introduction to Information Engineering

Let’s make more observations n = 2, f(2) = ? n = 3, f(3) = ? n = 4, f(4) = ? n = 5, f(5) = ? n = 6, f(6) = ? n = 7, f(7) = ? n = 8, f(8) = ? n = 9, f(9) = ? 5 1 3 7 1 1 3 3 5 10 15 20 25 30 35 Have you observed the pattern of f(n)? 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering What are the patterns? n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 f(n) k 1 2 3 4 5 6 7 f(1)=f(2)=f(4)=f(8)=f(16)=1. What are they in common? If we group the sequence [1], [2,3], [4,7],[8,15], f(n) in each group is a sequence of consecutive odd numbers starting from 1. Let k = n – the first number in n’s group What is the pattern of k? They are power of 2, 2m. f(n)=2k+1 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Guess the solution f(n) = 2k+1 k = n – the first number in n’s group The first number in n’s group is 2m. 2m  n < 2m+1 How fast can you find m? (1) use “binary search” on the bit pattern of n Since n has log2(n) bits, it takes log2(log2(n)) steps. (2) use “log2” function and take its integer part It takes constant time only. (independent of n.) 12/9/2018 CS135601 Introduction to Information Engineering

Running time of three algorithms 10 3 4 5 6 7 8 -8 -6 -4 -2 2 Algorithm 1 Algorithm 2 Algorithm 3 Running time n 12/9/2018 CS135601 Introduction to Information Engineering

The Central Role of Computer Science Why study math? 12/9/2018 CS135601 Introduction to Information Engineering 27

Algorithms we studied so far In chap 1, binary and decimal conversion, 2’s complement calculation, data compression, error correction, encryption… In chap 2, we studied how to use computer to implement algorithms In homework 4, we have problems for pipeline, prefix sum (parallel algorithm), and virtual memory page replacement (online algorithm) 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering In chap 3, we see some examples of using algorithms to help manage resources Virtual memory mapping, scheduling, concurrent processes execution, etc. In chap 4, we learned protocols(distributed algorithms, randomized algorithms) CSMA/CD, CSMA/CA, routing, handshaking, flow control, etc. In chap 5, we learned some basic algorithmic strategies, such as simulation, recursion, and how to analyze them 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Learning algorithms Every class in CS has some relations with algorithms But some classes are not obviously related to algorithms, such as math classes. 微積分,離散數學,線性代數,機率,工程數學,… Why should we study them? They are difficult, and boring, and difficult… 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Why study math? Train the logical thinking and reasoning Algorithm is a result of logical thinking: correctness, efficiency, … Good programming needs logical thinking and reasoning. For example, debugging. Learn some basic tricks Many beautiful properties of problems can be revealed through the study of math Standing on the shoulders of giants 12/9/2018 CS135601 Introduction to Information Engineering

Try to solve those problems Given a network of thousands nodes, find the shortest path from node A to node B The shortest path problem (離散數學) Given a circuit of million transistors, find out the current on each wire Kirchhoff's current law (線性代數) In CSMA/CD, what is the probability of collisions? Network analysis (機率) 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering 微積分 Limits, derivatives, and integrals of continuous functions. Used in almost every field 網路分析, 效能分析 訊號處理, 影像處理, 類比電路設計 科學計算, 人工智慧, 電腦視覺, 電腦圖學 … Also the foundation of many other math 機率, 工程數學 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering 離散數學 Discrete structure, graph, integer, logic, abstract algebra, combinatorics The foundation of computer science Every field in computer science needs it Particularly, 演算法, 數位邏輯設計, 密碼學, 編碼理論, 計算機網路, CAD, 計算理論 Extended courses Special topics on discrete structure, graph theory, concrete math 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering 線性代數 Vectors, matrices, vector spaces, linear transformation, system of equations Used in almost everywhere when dealing with more than one variables 網路分析, 效能分析 訊號處理, 影像處理 科學計算, 人工智慧, 電腦視覺, 電腦圖學 CAD, 電路設計 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering 機率 Probability models, random variables, probability functions, stochastic processes Every field uses it Particularly, 網路分析, 效能分析, 訊號處理, 影像處理, 科學計算, 人工智慧, 電腦視覺… Other examples: randomized algorithm, queue theory, computational finance, performance analysis 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering 工程數學 A condensed course containing essential math tools for most engineering disciplines Partial differential equations, Fourier analysis, Vector calculus and analysis Used in the fields that need to handle continuous functions 網路分析, 效能分析, 訊號處理, 影像處理, 科學計算, 人工智慧, 電腦視覺, CAD, 電路設計… 12/9/2018 CS135601 Introduction to Information Engineering

CS135601 Introduction to Information Engineering Reference The Josephus problem: Graham, Knuth, and Patashnik, “Concrete Mathematics”, section 1.3 http://en.wikipedia.org/wiki/Josephus_problem http://mathworld.wolfram.com/JosephusProblem.html Algorithm representation Textbook: 5.2 The related courses are those listed in the last part of slides, and of course, 演算法設計,高等程式設計實作 12/9/2018 CS135601 Introduction to Information Engineering