1、Algorithms,國立清華大學資訊工程學系,CS1356 資訊工程導論,2018/9/29,2,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
2、 a circle and start killing one by skipping every two others. By luck, or maybe by the hand of God, Josephus and another man remained the last and gave up to the Romans.,3,Can You Find the Safe Place?,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,
3、37,38,39,40,41,Safe place,Can you find the safe place FASTER?,4,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
4、be achievable (by computer) The more effective, the better algorithm is. How to measure the “efficiency”?,5,Outline,The first algorithm Model, simulation, algorithm primitive The second algorithm Algorithm discovery, recursion The third algorithm Solving recursion, mathematical induction The central
5、 role of computer science Why study math?,6,The First Algorithm of the Josephus Problem,7,A Simpler Version,Lets 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? T
6、he input is n, the output f(n) is a number between 1 and n. Ex: f(8) = ?,1,2,3,4,5,6,7,8,8,1st 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 nee
7、ds (1) a model to represents “n people in a circle” (2) a way to simulate “kill every 2nd person” (3) knowing when to stop,9,Simulation Using Pebbles,Put n pebbles numbered 1,n in a circle clockwise sequentially From the pebble numbered 1, count pebbles, and remove every 2nd pebble, until there is o
8、nly 1 pebble left The number of the remaining pebble is f(n),1,2,3,4,5,6,7,8,10,Abstraction and Model,“put pebbles numbered 1,n in a circle” is a model to represent n people in a circle “remove every 2nd pebble” is a simulation for killing the every 2nd person. A model is an abstraction of objects,
9、systems, or concepts that capture important characters of the problem. The definition of important characters is varied for different problems.,11,Computer Simulation,How to perform simulations on computers? We need to use what computer can do to represent the model and perform actions.Algorithm pri
10、mitives A set of well-defined building blocks that computers can perform and human can understand easily.,12,Data Abstraction,There is nothing “circular” inside computer, and nothing can be “removed” physically. But we can arrange memory cells to “model” and “simulate” them data structure Some commo
11、n data structures,13,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”,14,Kill Every 2nd Person,Remove every 2nd node in the circular linked list. You need to maintain the circ
12、ular linked structure after removing node 2 The process can continue until ,1,2,3,5,6,7,8,4,15,Knowing When to Stop,Stop when there is only one node left How to know that? When the next is pointing to itself Its ID is f(n) f(8) = 1,1,3,5,6,7,8,4,16,Efficiency of an Algorithm,The efficiency of an alg
13、orithm is usually measured by the number of operations assignment, comparison, arithmetic operation Eventually the number of instructions in chap 2 Expressed in a function of problem size n. For example, 5n3+2n2+10nlogn (1) Only interested in the case of large n decided by the order of the dominant
14、term. n3 is the order of the dominant term of Eq. (1),17,How Fast to 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) i
15、s (n-1) We can just represent it as O(n).,18,The Second Algorithm of the Josephus Problem,19,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 Repea
16、t 1-3 until success,20,Observing a Case,Lets check out 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 Lets first consider the case when n is even,1,2,3,4,5,6,7,8,21,n Is Even,For n=
17、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, 11, 32, 53, 74, it becomes the n=4 problem. So, if f(4)=x, f(8) =2x 1,1,2,3,4,5,6,7,8,22,n Is Odd,Lets checkout the case n=9 The starting point becomes 3
18、If we can solve f(4), can we use it to solve f(9)? If we renumber the remaining people, 31, 52, 73, 94, it becomes the n=4 problem. So, if f(4)=x, f(9) =2x+1,1,2,3,4,5,6,7,8,9,23,Recursive Relation,Hypothesis: f(2n)=2f(n)1. f(2n+1)=2f(n)+1. How to prove or disprove it? This is called a recursive rel
19、ation. 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?,24,Recursive Relation Algorithm,With recursive relations, you can design a recursive algorithm Create a subroutine Inside the subroutine, include The base case. For the J
20、osephus problem, the base case is n=1 The recurrence part Make procedure calls to the same subroutine with smaller problem size Compose the result based on the recursive relation,25,Recursive Algorithm,Procedure Josephus (n, fn) /* n is the input, fn is the output (= f(n) */if (n = 1) then (fn 1) el
21、se( if (n is even) then( call Josephus (n/2 , fn)fn 2*fn 1) else ( call Josephus (n1)/2 , fn)fn 2*fn+1 ) ),Header,Base case,Recursive part,n is even f(n)=2f(n/2)-1,n is odd f(n)=2f(n-1)/2)+1,26,How Fast to Compute f(n)?,n reduces half at each step. Need log2(n) steps to reach n=1. Each step needs a
22、constant number of operations . The total number of operations is log2(n),27,The Third Algorithm of the Josephus Problem,28,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,29,n = 6, f(6) = ? n = 7, f(
23、7) = ? n = 8, f(8) = ? n = 9, f(9) = ?,Lets Make More Observations,n = 2, f(2) = ? n = 3, f(3) = ? n = 4, f(4) = ? n = 5, f(5) = ?,1,3,1,3,5,7,1,3,Have you observed the pattern of f(n)?,30,What Are the Patterns?,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
24、,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 ns group What is the pattern of k?,f(n)=2k+1,They are power of 2, 2m.,31,Guess the Solution,f(n) = 2k+1 k = n the first number in ns group The first number in ns group is 2m. 2m n 2m+1
25、 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.),32,Verify the Solution,If f(n)=2k+1, k=n2m is the solution, it must sati
26、sfy the recursionHow to prove it? Hint: using mathematical induction,33,Comparison of 3 Algorithms,Running time,n,34,The Central Role of Computer Science,Why study math?,35,Algorithms Studied so Far,In chap 1, binary and decimal conversion, 2s complement calculation, data compression, error correcti
27、on, 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),36,In chap 3, we saw examples of using algorithms to help manage resources Virtual me
28、mory mapping, scheduling, concurrent processes execution, etc. In chap 4, we learned protocols (distributed, 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 th
29、em,37,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,38,Why Study Math?,Train the logical thinking
30、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
31、,39,Try to Solve These 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 Kirchhoffs current law (線性代數) In CSMA/CD, what is the probability of collisions?
32、 Network analysis (機率),40,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,41,離散數學,Discrete structure, graph, integer, logic, abstract algebra, combinatorics
33、 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,42,線性代數,Vectors, matrices, tensors, vector spaces, linear transformation, system of e
34、quations Used in almost everywhere when dealing with more than one number 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,43,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 ex
35、amples: randomized algorithm, queue theory, computational finance, performance analysis,44,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 co
36、ntinuous 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,45,Reference,The Josephus problem: Graham, Knuth, and Patashnik, “Concrete Mathematics”, section 1.3 http:/en.wikipedia.org/wiki/Josephus_problem http:/ 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,
