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Chapter 8 1

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Symbol Table Symbol table is used widely in many applications. dictionary is a kind of symbol table data dictionary is database management In general, the following operations are performed on a symbol table determine if a particular name is in the table retrieve the attribute of that name modify the attributes of that name insert a new name and its attributes delete a name and its attributes 2

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Symbol Table Popular operations on a symbol table include search, insertion, and deletion A binary search tree could be used to represent a symbol table. The worse-case complexities for the operations are O(n). Hashing: insertions, deletions & finds in O(1). Static hashing Dynamic hashing 3

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Static Hashing Dictionary pairs are stored in a fixed-size table called hash table. The address of location of an dictionary pairs, x, is obtained by computing some arithmetic function h(x). The memory available to maintain the symbol table (hash table) is assumed to be sequential. The hash table consists of b buckets and each bucket contains s records. h(x) maps the set of possible dictionary pairs onto the integers 0 through b-1. 4

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Hash Tables The key density of a hash table is the ratio n/T, where n is the number of dictionary pairs in the table and T is the total number of possible keys. The loading density or loading factor of a hash table is α= n/(sb). 5

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Hash Tables Two keys, I 1, and I 2, are said to be synonyms with respect to h if h(I 1 ) = h(I 2 ). An overflow occurs when a new key i is mapped or hashed by h into a full bucket. A collision occurs when two non-identical keys are hashed into the same bucket. If the bucket size is 1, collisions and overflows occur at the same time. 6

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Example 8.1 b=26, s=2 Assume there are 10 distinct keys GA, A,G,L,A2,A1,A3,A4 and E If no overflow occur, the time required for hashing depends only on the time required to compute the hash function h. A1? 7

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Hash Functions Requirements Simple to compute Minimize the number of collisions Dependent upon all the characters in the keys Uniform hash function If there are b buckets, we hope to have h(x)=i with the probability being (1/b) Mid-square, division, folding, digit analysis 8

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Division Using the modulo (%) operator. A key k is divided by some number D and the remainder is used as the hash address for k. The bucket addresses are in the range of 0 through D-1. If D is a power of 2, then h(k) depends only on the least significant bits of k. 9

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Division If a division function h is used as the hash function, the table size should not be a power of two. Since programmers have a tendency to use many variables with the same suffix too many collisions If D is divisible by two, the odd keys are mapped to odd buckets and even keys are mapped to even buckets. Thus, the hash table is biased. 10

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Mid-Square Mid-Square function It is computed by squaring the key and then using an appropriate number of bits from the middle of the square to obtain the bucket address. Table size is a power of two 11

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Folding The key k is partitioned into several parts, all but the last being of the same length. All partitions are added together to obtain the hash address for k. Shift folding: different partitions are added together to get h(k). Folding at the boundaries: key is folded at the partition boundaries, and digits falling into the same position are added together to obtain h(k). This is similar to reversing every other partition and then adding. 12

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Example 8.2 13

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Secure Hash Functions Can be applied to any sized message M Produces fixed-length output h Is easy to compute h=H(M) for any message M Given h is infeasible to find x s.t. H(x)=h one-way property Given x is infeasible to find y s.t. H(y)=H(x) weak collision resistance Is infeasible to find any x,y s.t. H(y)=H(x) strong collision resistance 14

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Secure Hash Algorithm (SHA) 步驟 1: 對訊息做前處理使得它的長度成為 q*512 位元，其中 q 是 某個整數。前處理的程序可能得在訊息的尾端加上許多個 0 的字 串。 步驟 2: 初始化 160- 位元的輸出緩衝區 OB ，它包含五個 32- 位元的 暫存器 A, B, C, D, E ，其中 A = 67453401 ， B = efcdab89 ， C = 98badcfe ， D = 10325476 ， E = c3d2e1f0 。（所有的值為 16 進位）。 步驟 3: for (int i = 1; i <= q; i++) { 令 B i = 訊息中第 i 個 512 位元區塊 ; OB = F (OB, B i ); } 步驟 4: 輸出 OB 。 15

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Atomic SHA Operation 16 t= 步驟數； 0 ≤ t ≤ 79 f t (B, C, D)= 步驟 t 的基本（位元的）邏輯函數； 例如， (B^C) v (B^D) S k = 把 32 位元的暫存器左迴旋 k 個位元 W t = 從 Bi 推導出來的 32 位元值 K t = 常數

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Example 8.4 A = 67453401 ， B = efcdab89 ， C = 98badcfe ， D = 10325476 ， E = c3d2e1f0 f t (B,C,D)=(B C D), K 0 =5a82799, W 0 =0000000 B=1110 1111 1100 1101 1010 1011 1000 1001 S 30 (B)=0111 1011 1111 0011 0110 1010 1110 0010=7bf36ae2=C A=6e3edeb6 B=67453401, D=98badcfe, E=10325476 17

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Overflow Handling There are two ways to handle overflow: Open addressing Linear probing Quadratic probing Rehasing Chaining 18

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Open Addressing Assumes the hash table is an array The hash table is initialized so that each slot contains the null key. When a new key is hashed into a full bucket, find the closest unfilled bucket. linear probing or linear open addressing 19

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Linear Probing (Example 8.6) Assume 26-bucket table with one slot per bucket and the following keys: GA, D, A, G, L, A2, A1, A3, A4, Z, ZA, E. Let the hash function h(k) = first character of k. When entering G, G collides with GA and is entered at ht[7] instead of ht[6]. 0A 1A2 2A1 3D 4A3 5A4 6GA 7G 8ZA 9E 10 11L 12 13 ~… 24 25Z 20

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Linear Probing When linear open address is used to handle overflows, a hash table search for key k proceeds as follows: compute h(k) examine key at positions ht[h(k)], ht[(h(k)+1) % b], …, ht[(h(k)+j) % b], in this order until one of the following condition happens: ht[(h(k)+j) % b]=k; in this case k is found ht[h(k)+j] is empty; k is not in the table We return to the starting position ht[h(k)]; the table is full and k is not in the table 21

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Linear Probing When linear probing is used to resolve overflows, keys tend to cluster together. This increases search time. e.g., to find ZA, you need to examine ht[25], ht[0], …, ht[8] (total of 10 comparisons). The number of comparisons to look up a key is approximated (2-α)/(2-2α), where α is the load density 0A 1A2 2A1 3D 4A3 5A4 6GA 7G 8ZA 9E 10 11L 12 13 ~… 24 25Z 22

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Quadratic Probing One of the problems of linear open addressing is that it tends to create clusters of keys. These clusters tend to merge as more keys are entered, leading to bigger clusters. It is difficult to find an unused bucket. A quadratic probing scheme improves the growth of clusters. A quadratic function of i is used as the increment when searching through buckets. Perform search by examining bucket h(k), (h(k)+i 2 )%b, (h(k)-i 2 )%b for 1 ≤ i ≤ (b-1)/2. 23

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Example (h(k)+i 2 )%b, (h(k)-i 2 )%b for 1 ≤ i ≤ (b-1)/2. Example: b=10, insert: 89, 18, 49, 58, 69 89 % 10=9, 18 % 10=8 49 % 10=9, (9+1 2 ) % 10=0 58 % 10=8, (8+1 2 ) % 10=9, (8+2 2 ) % 10=2 69 % 10=9, (9+1 2 ) % 10=0, (9+2 2 ) % 10=3 24 4958691889 Index0123456789

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Rehashing Another way to control the growth of clusters is to use a series of hash functions h 1, h 2, …, h m. This is called rehashing. Buckets h i (k), 1 ≤ i ≤ m are examined in that order. 25

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Chaining Linear probing performs poorly because the search for a key involves comparisons with keys that have different hash values. Unnecessary comparisons can be avoided if all the synonyms are put in the same list, where one list per bucket. Each chain has a head node. Head nodes are stored sequentially. 26

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Example 27 Average search length is (6*1+3*2+1*3+1*4+1*5)/12 = 2

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Dynamic Hashing Retain the fast retrieval time Dynamically increasing and decreasing file size without penalty Dynamic hashing is to minimize access to pages 28

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An Example Hash Function kh(k) A0100 000 A1100 001 B0101 000 B1101 001 C1110 001 C2110 101 C3110 011 C5110 101 29 A: 100, B: 101, C: 110

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Hashing With Directory Accessing any page only requires two steps: First step: use the hash function to find the address of the directory entry. Second step: retrieve the page associated with the address 30

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Dynamic Hash Tables with Directories 31 Insert C5 into the hash table: h(C5,2)=01, d[01] A1, B1 overflow determine the least u such that h(k,u) is not the same for all keys in the overflow bucket suppose that u=3 h(A1,3)=001, h(B1,3)=001, h(C5,3)=101 Insert C1 into the hash table: h(C1,2)=01, d[01] A1, B1 overflow determine the least u such that h(k,u) is not the same for all keys in the overflow bucket suppose that u=4 h(A1,4)=0001, h(B1,4)=1001, h(C1,4)=0001

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Directoryless Dynamic Hashing Hashing with directory requires at least one level of indirection Directoryless hashing assume a continuous address space in the memory to hold all the records. Therefore, the directory is not needed. Thus, the hash function must produce the actual address of a page containing the key. Contrasting to the directory scheme in which a single page might be pointed at by several directory entries, in the directoryless scheme one unique page is assigned to every possible address. 32

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Inserting into a Directoryless Dynamic Hash Table 33 0~2 r +q, 0 ≤q<2 r Insert C5 (110101) into the hash table: h(C5,2)=01, d[01] A1, B5 overflow activated 2 r +q reallocating (b) 000, 100 indexed using h(k,3) 0~q-1 and 2 r ~2 r +q-1 are indexed using h(k,r+1); else, h(k,r)

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