Data Structures and Algorithms. The material for this lecture is drawn, in part, from The Practice of Programming (Kernighan & Pike) Chapter 2

Transcription

1 Daa Srucures and Algorihms The maerial for his lecure is drawn, in par, from The Pracice of Programming (Kernighan & Pike) Chaper 2 1

2 Moivaing Quoaion Every program depends on algorihms and daa srucures, bu few programs depend on he invenion of brand new ones. -- Kernighan & Pike 2

3 Goals of his Lecure Help you learn abou: Common daa srucures and algorihms Why? Shallow moivaion: Provide examples of poiner-relaed C code Why? Deeper moivaion: Common daa srucures and algorihms serve as high level building blocks A power programmer: Rarely creaes programs from scrach Ofen creaes programs using high level building blocks

4 A Common Task Mainain a able of key/value pairs Each key is a sring; each value is an in Unknown number of key-value pairs For simpliciy, allow duplicae keys (clien responsibiliy) In Assignmen #, mus check for duplicae keys! Examples (suden name, grade) ( john smih, 8), ( jane doe, 9), ( myungbak lee, 81) (baseball player, number) ( Ruh, ), ( Gehrig, ), ( Manle, 7) (variable name, value) ( maxlengh, 2000), ( i, 7), ( j, -10)

5 Daa Srucures and Algorihms Daa srucures Linked lis of key/value pairs Hash able of key/value pairs Algorihms Creae: Creae he daa srucure Add: Add a key/value pair Search: Search for a key/value pair, by key Free: Free he daa srucure 5

6 Daa Srucure #1: Linked Lis Daa srucure: Nodes; each conains key/value pair and poiner o nex node "Manle" 7 Algorihms: Creae: Allocae Table srucure o poin o firs node Add: Inser new node a fron of lis Search: Linear search hrough he lis Free: Free nodes while raversing; free Table srucure 6

7 Linked Lis: Daa Srucure sruc Node { cons char *key; in value; sruc Node *nex; ; sruc Table { sruc Node *firs; ; sruc Table sruc Node sruc Node 7

8 Linked Lis: Creae (1) sruc Table *Table_creae(void) { = (sruc Table *) malloc(sizeof(sruc Table)); ->firs = ; reurn ; = Table_creae(); 8

9 Linked Lis: Creae (2) sruc Table *Table_creae(void) { = (sruc Table *) malloc(sizeof(sruc Table)); ->firs = ; reurn ; = Table_creae(); 9

10 Linked Lis: Add (1) void Table_add(sruc Table *,cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); /* we omi error checking here (e.g., p == ) */ p->key = key; p->value = value; p->nex = ->firs; ->firs = p; These are poiners o srings ha exis in he RODATA secion Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); 10

11 Linked Lis: Add (2) void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = key; p->value = value; p->nex = ->firs; ->firs = p; p Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); 11

12 Linked Lis: Add () void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = key; p->value = value; p->nex = ->firs; ->firs = p; p "Manle" 7 Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); 12

13 Linked Lis: Add () void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = key; p->value = value; p->nex = ->firs; ->firs = p; p "Manle" 7 Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); 1

14 Linked Lis: Add (5) void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = key; p->value = value; p->nex = ->firs; ->firs = p; p "Manle" 7 Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); 1

15 Linked Lis: Search (1) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); "Manle" 7 15

16 Linked Lis: Search (2) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 16

17 Linked Lis: Search () in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 17

18 Linked Lis: Search () in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 18

19 Linked Lis: Search (5) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 19

20 Linked Lis: Search (6) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; for (p = ->firs; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; 1 in value; in found; found = Table_search(,, &value); p "Manle" 7 20

21 Linked Lis: Free (1) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); "Manle" 7 21

22 Linked Lis: Free (2) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p "Manle" 7 22

23 Linked Lis: Free () void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 2

24 Linked Lis: Free () void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 2

25 Linked Lis: Free (5) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 25

26 Linked Lis: Free (6) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 26

27 Linked Lis: Free (7) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 27

28 Linked Lis: Free (8) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp "Manle" 7 28

29 Linked Lis: Free (9) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; for (p = ->firs; p!= ; p = nexp) { nexp = p->nex; free(p); free() Table_free(); p nexp "Manle" 7 29

30 Linked Lis Performance Creae: fas Add: fas Search: slow Free: slow Wha are he asympoic run imes (big-oh noaion)? Would i be beer o keep he nodes sored by key? 0

31 Daa Srucure #2: Hash Table Fixed-size array where each elemen poins o a linked lis 0 ARRAYSIZE-1 sruc Node *array[arraysize]; Funcion maps each key o an array index For example, for an ineger key h Hash funcion: i = h % ARRAYSIZE (mod funcion) Go o array elemen i, i.e., he linked lis hashab[i] Search for elemen, add elemen, remove elemen, ec. 1

32 Hash Table Example Ineger keys, array of size 5 wih hash funcion h mod % 5 is % 5 is % 5 is Revoluion 199 WW Civil 2

33 How Large an Array? Large enough ha average bucke size is 1 Shor buckes mean fas search Long buckes mean slow search Small enough o be memory efficien No an excessive number of elemens Forunaely, each array elemen is jus soring a poiner 0 This is OK: ARRAYSIZE-1

34 Wha Kind of Hash Funcion? Good a disribuing elemens across he array Disribue resuls over he range 0, 1,, ARRAYSIZE-1 Disribue resuls evenly o avoid very long buckes This is no so good: 0 ARRAYSIZE-1 Wha would be he wors possible hash funcion?

35 Hashing Sring Keys o Inegers Simple schemes don disribue he keys evenly enough Number of characers, mod ARRAYSIZE Sum he ASCII values of all characers, mod ARRAYSIZE Here s a reasonably good hash funcion Weighed sum of characers x i in he sring ( a i x i ) mod ARRAYSIZE Bes if a and ARRAYSIZE are relaively prime E.g., a = 65599, ARRAYSIZE = 102 5

36 Implemening Hash Funcion Poenially expensive o compue a i for each value of i Compuing a i for each value of I Insead, do (((x[0] * x[1]) * x[2]) * x[]) * unsigned in hash(cons char *x) { in i; unsigned in h = 0U; for (i=0; x[i]!='\0'; i++) h = h * (unsigned char)x[i]; reurn h % 102; Can be more clever han his for powers of wo! (Described in Appendix) 6

37 Hash Table Example Example: ARRAYSIZE = 7 Lookup (and ener, if no presen) hese srings: he, ca, in, he, ha Hash able iniially empy. Firs word: he. hash( he ) = % 7 = 1. Search he linked lis able[1] for he sring he ; no found

38 Hash Table Example (con.) Example: ARRAYSIZE = 7 Lookup (and ener, if no presen) hese srings: he, ca, in, he, ha Hash able iniially empy. Firs word: he. hash( he ) = % 7 = 1. Search he linked lis able[1] for he sring he ; no found Now: able[1] = makelink(key, value, able[1]) he 8

39 Hash Table Example (con.) Second word: ca. hash( ca ) = % 7 = 2. Search he linked lis able[2] for he sring ca ; no found Now: able[2] = makelink(key, value, able[2]) he 9

40 Hash Table Example (con.) Third word: in. hash( in ) = % 7 = 5. Search he linked lis able[5] for he sring in ; no found Now: able[5] = makelink(key, value, able[5]) he ca 0

41 Hash Table Example (con.) Fourh word: he. hash( he ) = % 7 = 1. Search he linked lis able[1] for he sring he ; found i! he in ca 1

42 Hash Table Example (con.) Fourh word: ha. hash( ha ) = % 7 = 2. Search he linked lis able[2] for he sring ha ; no found. Now, inser ha ino he linked lis able[2]. A beginning or end? Doesn maer he in ca 2

43 Hash Table Example (con.) Insering a he fron is easier, so add ha a he fron he in ha ca

44 Hash Table: Daa Srucure enum {BUCKET_COUNT = 102; sruc Node { cons char *key; in value; sruc Node *nex; ; sruc Table { sruc Node *array[bucket_count]; ; sruc Table sruc Node sruc Node

45 Hash Table: Creae sruc Table *Table_creae(void) { = (sruc Table*)calloc(1, sizeof(sruc Table)); reurn ; = Table_creae();

46 Hash Table: Add (1) void Table_add(sruc Table *,cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); in h = hash(key); p->key = key; p->value = value; p->nex = ->array[h]; ->array[h] = p; Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); These are poiners o srings ha exis in he RODATA secion Preend ha Ruh hashed o 2 and Gehrig o 72 6

47 Hash Table: Add (2) void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); in h = hash(key); p->key = key; p->value = value; p->nex = ->array[h]; ->array[h] = p; Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); p 7

48 Hash Table: Add () void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); in h = hash(key); p->key = key; p->value = value; p->nex = ->array[h]; ->array[h] = p; Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); Preend ha Manle hashed o 806, and so h = 806 p "Manle" 7 8

49 Hash Table: Add () void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); in h = hash(key); p->key = key; p->value = value; p->nex = ->array[h]; ->array[h] = p; Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); h = 806 p "Manle" 7 9

50 Hash Table: Add (5) void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); in h = hash(key); p->key = key; p->value = value; p->nex = ->array[h]; ->array[h] = p; Table_add(,, ); Table_add(,, ); Table_add(, "Manle", 7); h = 806 p "Manle" 7 50

51 Hash Table: Search (1) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); for (p = ->array[h]; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); "Manle" 7 51

52 Hash Table: Search (2) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); for (p = ->array[h]; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); Preend ha Gehrig hashed o 72, and so h = 72 "Manle" 7 52

53 Hash Table: Search () in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); for (p = ->array[h]; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 h = 72 5

54 Hash Table: Search () in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); for (p = ->array[h]; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p "Manle" 7 h = 72 5

55 Hash Table: Search (5) in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); for (p = ->array[h]; p!= ; p = p->nex) if (srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; in value; in found; found = Table_search(,, &value); p h = 72 "Manle" 7 55

56 Hash Table: Free (1) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); "Manle" 7 56

57 Hash Table: Free (2) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); b = 0 "Manle" 7 57

58 Hash Table: Free () void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p b = 0 "Manle" 7 58

59 Hash Table: Free () void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p b = 1,, 2 "Manle" 7 59

60 Hash Table: Free (5) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p b = 2 "Manle" 7 60

61 Hash Table: Free (6) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp b = 2 "Manle" 7 61

62 Hash Table: Free (7) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); p nexp b = 2 "Manle" 7 62

63 Hash Table: Free (8) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); b = 2,, 72 b = 72,, 806 b = 807,, 102 "Manle" 7 6

64 Hash Table: Free (9) void Table_free(sruc Table *) { sruc Node *p; sruc Node *nexp; in b; for (b = 0; b < BUCKET_COUNT; b++) for (p = ->array[b]; p!= ; p = nexp) { nexp = p->nex; free(p); free(); Table_free(); b = 102 "Manle" 7 6

65 Hash Table Performance Creae: fas Add: fas Search: fas Free: slow Wha are he asympoic run imes (big-oh noaion)? Is hash able search always fas? 65

66 Key Ownership Noe: Table_add() funcions conain his code: void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = key; Caller passes key, which is a poiner o memory where a sring resides Table_add() funcion sores wihin he able he address where he sring resides 66

67 Key Ownership (con.) Problem: Consider his calling code: sruc Table ; char k[100] = ; Table_add(, k, ); srcpy(k, ); Via Table_add(), able conains memory address k Clien changes sring a memory address k Thus clien changes key wihin able Wha happens if he clien searches for Ruh? Wha happens if he clien searches for Gehrig? 67

68 Key Ownership (con.) Soluion: Table_add() saves copy of given key void Table_add(sruc Table *, cons char *key, in value) { sruc Node *p = (sruc Node*)malloc(sizeof(sruc Node)); p->key = (cons char*)malloc(srlen(key) + 1); srcpy(p->key, key); Why add 1? If clien changes sring a memory address k, daa srucure is no affeced Then he daa srucure owns he copy, ha is: The daa srucure is responsible for freeing he memory in which he copy resides The Table_free() funcion mus free he copy 68

69 Summary Common daa srucures and associaed algorihms Linked lis fas inser, slow search Hash able Fas inser, (poenially) fas search Invaluable for soring key/value pairs Very common Relaed issues Hashing algorihms Memory ownership 69

70 Appendix Supid programmer ricks relaed o hash ables 70

71 Revisiing Hash Funcions Poenially expensive o compue mod c Involves division by c and keeping he remainder Easier when c is a power of 2 (e.g., 16 = 2 ) An alernaive (by example) 5 = % 16 is 5, he las four bis of he number Would like an easy way o isolae he las four bis

72 Recall: Biwise Operaors in C Biwise AND (&) 5 & 15 & Mod on he cheap! E.g., h = 5 & 15; Biwise OR ( ) One s complemen (~) Turns 0 o 1, and 1 o 0 E.g., se las hree bis o 0 x = x & ~7;

73 A Faser Hash Funcion unsigned in hash(cons char *x) { in i; unsigned in h = 0U; for (i=0; x[i]!='\0'; i++) h = h * (unsigned char)x[i]; reurn h % 102; Previous version unsigned in hash(cons char *x) { in i; unsigned in h = 0U; for (i=0; x[i]!='\0'; i++) h = h * (unsigned char)x[i]; reurn h & 102; Wha happens if you misakenly wrie h & 102? Faser 7

74 Speeding Up Key Comparisons Speeding up key comparisons For any non-rivial value comparison funcion Trick: sore full hash resul in srucure in Table_search(sruc Table *, cons char *key, in *value) { sruc Node *p; in h = hash(key); /* No % in hash funcion */ for (p = ->array[h%102]; p!= ; p = p->nex) if ((p->hash == h) && srcmp(p->key, key) == 0) { *value = p->value; reurn 1; reurn 0; Why is his so much faser? 7