Basics 1

1. Basics 1 Prof. Michael Tsai 2012/02/21

2. Variables • 變數是什麼? • 變數: Names(x和y)對應到某個data • 每個變數可能有不同的資料型態(data type)

3. Data Type • What is a data type? • A data type in a programming language is a set of data with values having pre-defined characteristics. • A data type is a collection of objects and a set of operations that act on those objects. • 每種data type 有所占的記憶體大小及可表示的資料數值範圍 • Data types in C • char, int, float, long, double(unsigned, signed, …) • Array • Structure(User-defined) intiarray[16]; struct { int a; int b; char str[16]; int * iptr; } blah;

4. Data Types’ Operations • Operations • +, -, *, /, %, == • =, +=, -= • ? : • sizeof, - (negative) • giligulu(int a, int b)

5. Data Type • Representation of the objects of the data type • Example: char • char blah=‘A’; (‘A’: ASCII code is 65(dec), or 0x41 (hex)) Q: The maximum number which can be represented with a char variable? A: 255. • How about char, int, long, float? 1 byte of memory: 01000001

6. Data type • Q: Do we need to know about the representation of a data type? • A: It depends. • Algorithms could be made more efficient. • But, If the representation of the data type is changed, the program needs to be verified, revised, or completely re-written. 囧 • Porting to a different platform (x86, ARM, embedded system, …) • Changing the specification of a program or a library(ex. 16-bit int 32-bit long)

7. Abstract Data Type • “Abstract Data Type” (ADT): • Separate the specifications from the representation and the implementation User Specification (Interface) Representation and Implementation

8. Abstract Data Type • Specifications: • Operations: • Name of the function and the description of what the function does • The type of the argument(s) • The type of the result(s) (return value) • Data (usually hidden) • Function categories: • Creator/constructor • Transformers • Observer/reporter

9. What is a Data Structure? • An organization of information, usually in memory, for better algorithm efficiency. (演算法跑起來比較快) • Or, a way to store and organize data in order to facilitate access and modifications. (比較好存取和更改) • 例子: 存多項式: • 又可分為: • Linear data structure: 必須循序地存取 (如linked list, stack, queue) • Non-linear data structure: 可以不循序的存取 (如tree, graph)

10. What is an algorithm? • “The step-by-step instructions to solve a given problem” • 解決某一個問題的詳細、一步一步地指令 • “A computable set of steps to achieve a desired result.” • All algorithm must satisfy the following criteria: • Input: 外部給的資訊(可以是零個或多個) • Output: 產生的結果(至少一個) • Definiteness: 每一個指令都是清楚而不模糊的 • Finiteness: 所有的狀況下(所有的input), 演算法會在有限步驟之後結束 • Effectiveness: 每一個指令都必須是簡單可以直接執行的 (必須可以執行)

11. 例子 • Statement 1:“Is n=2 the largest value of n for which there exist positive integers x, y, and z such that has a solution?” • Statement 2: “Store 5 divided by zero into x and go to statement ㄆ.” • Which criterion do they violate? • Input • Output • Definiteness • Finiteness • Effectiveness

12. How do we describe an algorithm? • Human language (English, Chinese, …) • Programming language • A mix of the above • 拿平底鍋 • 拿沙拉油 • 我們有油嗎? • 有的話, 倒一茶匙的沙拉油到鍋子裡 • 沒有的話, 我們想要買油嗎? • 是的話, 就去全聯買一罐沙拉油 • 如果不想的話, 只好先不煮了. • 打開火爐, …

13. Example: Selection Sort • Integers are stored in an array, list. The i-th integer is stored in list[i], 0<i<n. • Goal: Devise a program to sort a set of integers • Solution: From those integers that are currently unsorted, find the smallest and place it next in the sorted list. ㄅ ㄆ 1 1 ㄆ 2 ㄅ 2 1 ㄆ ㄅ

14. Example: Selection Sort • First attempt: for(i=0; i<n;++i){ Examine list[i] to list[n-1]and suppose that the smallest integer is at list[min]; Interchange list[i]and list[min]; } Task 1 Task 2

15. Task 2 Task 2 voidswap(int*x,int*y){ int temp =*x; *x=*y; *y=temp; } Or #define SWAP(x,y,t) ((t)=(x), (x)=(y), (y)=(t))

16. Task 1 min=i; for(j=i;j<n;++j) if(list[j]<list[min]) min=j; Task 1

17. How do we prove that it is correct? • (這一頁回家自己看) • [Theorem] Function sort(list,n) correctly sorts a set of n>=1 integers. The result remains in list[0], …, list[n-1] such that . • Proof: When the outer for loop completes its iteration for i=q, we have . Further, on subsequent iterations, i>q and list[0] through list[q] are unchanged. Hence following the last iteration of the outer for loop (i.e., i=n-2), we have .

18. Example: Binary Search • Input: • searchnum: the number to be found • list: sorted array, size n, and • Output: • -1 if searchnum is not found in list • the index of searchnum in list if searchnum is found

19. Example: searchnum=13; 0 1 2 3 4 5 6 7 8 9 10 11 1 3 4 4 7 11 13 13 13 18 19 6

20. Example: searchnum=13; 0 1 2 3 4 5 6 7 8 9 10 11 1 3 4 4 7 11 13 13 13 18 19 6 right left middle middle=(left+right)/2; left=middle+1;

21. Example: searchnum=5; 0 1 2 3 4 5 6 7 8 9 10 11 1 3 4 4 7 11 13 13 13 18 19 6 return -1;

22. intbinsearch(int list[], intsearchnum, int left, int right) { int middle; while(left<=right) { middle=(left+right)/2; switch(COMPARE(list[middle], searchnum)) { case -1: left=middle+1; break; case 0: return middle; case 1: right=middle-1; } } return -1; } list: 存sort好數字的array searchnum: 要找的數字 left, right: 正在找的範圍左邊和右邊邊界

23. 怎麼評估一個程式寫得好不好? • Does the program meet the original specifications of the task? • Does it work correctly? • Does the program contain documentation that shows how to use it and how it works? • Does the program effectively use functions to create logical units? • Is the program’s code readable?

24. 怎麼評估一個程式寫得好不好? • Does the program efficiently use primary and secondary storage? Primary storage: memory? Secondary storage: Hard drive, flash disk, etc. • Is the program’s running time acceptable for the task? Example: Network intrusion detection system (1) 99.8% detection rate, 50 minutes to finish analysis of a minute of traffic (2) 85% detection rate, 20 seconds to finish analysis of a minute of traffic

25. 怎麼評估一個程式寫得好不好? • 程式是否有效地使用主要及次要的儲存? • 程式的執行時間是否適合所需解決的工作內容? Space complexity Time complexity

26. 空間及時間複雜度 • 程式的空間複雜度: • 程式執行完畢所需使用的所有空間(記憶體) • 程式的時間複雜度: • 程式執行完畢所需使用的(執行)時間 • Goal: 找出執行時間/使用空間”如何”隨著input size變長 (成長的有多快) • 什麼是input size? • 問題給的input的”元素數量”, 如: • Array大小 • 多項式最高項的次方 • 矩陣的長寬 • 二進位數的位元數目

27. 空間複雜度 • 程式所需空間: • 固定的空間 • 和input/output的大小及內容無關 • 變動的空間 • 和待解問題P的某個input instance I(某一個input)有關 • 跟recursivefunction會使用到的額外空間有關

28. 時間複雜度 • 一個程式P所需使用的時間: • Compile所需時間 • 執行時間 (execution time or run time) • Compile時間: 固定的. (例外?) • C (and other compiled programming languages) One Compilation  Multiple Executions • Run time: • 和input instance的特性有關!

29. 如何得到? • 總執行時間 • 所執行的程式指令的數目 • 比較數學的方法(使用function來代表程式執行的時間)

30. 如何得到? • 方法1: • 數出整個程式每個operation各總共花了多少時間. Is it good to use?(方法1)

31. 如何得到? • 方法2: • 把程式分解為一個一個步驟 • 每個步驟的執行時間變成跟input的特性無關了 • 然後數數看總共有幾個步驟 • 但是… • 不同的程式步驟可能有不同的執行時間 • a=2; • a=2*b+3*c/d-e+f/g/a/b/c; • 要數出所有步驟的數目需要解某一個input instance

32. Example float sum(float list[], int n) { float tempsum = 0; count++; //assignment int i; for (i=0;i<n;i++) { count++; // for loop tempsum+=list[i]; count++; //assignment } count++; //last iteration of for count++; return tempsum; }

33. Example float sum(float list[], int n) { float tempsum = 0; int i; for (i=0;i<n;i++) { count+=2; } count+=3; return tempsum; } count = 2n+3 (steps)

34. Example 2 float rsum(float list[], int n) { count++; // if if (n) { count++; //return return rsum(list,n-1)+list[n-1]; } count++; //return return list[0]; } 2n+2 < 2n+3. Does this mean rsum is faster than sum ? No! count = 2n+2 (steps)

35. 比較數學的方法—Asymptotic analysis • “預測”當input的性質改變(通常是input size改變)時, 執行時間的成長速度(growth of run time) • 比較兩個做相同事情的程式的時間複雜度 • “程式步驟”不是那麼精確: • “3n+3” 比“3n+5” 快? • 通常 “3n+3”, “7n+2”, or “2n+15” 執行時間都相差不遠. • 我們將使用一個比較不準確的方式來描述執行時間…

36. Example Usually no (if the constants are close). • Program P and Q • N很大的時候, Q 會比P 快,不管的數值是什麼. • Example: • , then for . • , then for . • 需要知道的數值嗎? Break even point

37. Asymptotic Notation – Big OH • Definition [Big “oh”]: • “f of n is big oh of g of n” (是集合的成員) • “=“ is “is” not “equal” • 可以想成是Upper Bound(王先生打了太多全壘打, 所以是upper bound)

38. Example • ? • Yes, since for all . • ? • Yes, since for all . • ? • Yes, since for all . • ? • Yes, since for all . for all

39. Example • ? • Yes, since for all . • ? • Yes, since for all . • ? • Yes, since for all . • ? • Yes, since for all . • ? • No. Cannot find for all

40. The World of Big Oh • constant • linear • quadratic • cubic • exponential • , , ,, Faster Slower

41. Big Oh 是 Upper Bound • 但是沒有說它是多好的upper bound • 通常我們會把它取得越小(緊)越好.

42. AUseful Theorem • Theorem: If , then . • Proof: (自己閱讀) , for all . So,