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What we learn with pleasure we never forget. Alfred Mercier Smitha N Pai

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Identifiers are stored in fixed sized table called hash table. Arithmetic function f is used to determine the address or location of an identifier, x in the table. f(x) gives the home or hash address of x in the table The hash table ht is stored in sequential memory locations that are partitioned into b buckets ht[0] ht[1], ht[b-1]. Each bucket has s slots s=1 means each bucket holds one record Hash function is used to transform the identifier x into an address in the hash table f(x) maps the set of possible identifiers onto integers 0 through b-1

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Identifier density is the ratio of n/T where n is the number of identifiers in the table. T represents the set of all possible identifiers including the ones which may not exist in real life situation The loading density or loading factor of a hash table is n/sb Hash function may map several identifiers into the same bucket. Two identifiers i1 and i2 are synonym wth respect to f if f(i1)=f(i2) An overflow occurs when we hash a new identifier i into a full bucket A collision occurs when we hash two non identical identifiers into the same bucket When bucket size is 1, collision and overflows occur simultaneously

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If key range too large, use hash table with fewer buckets and a hash function which maps multiple keys to same bucket: h(k 1 ) = = h(k 2 ): k 1 and k 2 have collision at slot Popular hash functions: hashing by division h(k) = k%D, where D number of buckets in hash table Example: hash table with 11 buckets h(k) = k%11 80 3 (80%11= 3), 40 7, 65 10 58 3 collision!

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If we choose an identifier x form the identifier space the probability that f(x) = I is 1/b for all buckets i which means that a random x has equal chance of hashing into any of the b buckets- we call such a function having this property of uniform hashing. The time required to enter, delete or search for identifiers using hashing depends only on the time required to compute the hash function and to search one bucket.

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Midsquare- Square the identifier and using appropriate number of bits from the middle of the square to obtain the bucket address Since middle bits fo the square depend upon all the characters in an identifier there is a high probability that different identifiers will produce different hash addresses even when some characters are same Number of bits used depends on the table size If r bits are used then the range of values is 2 r Size of hash table should be power of 2

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Mid-Square Method - Concat, Square and Remove the Middle! E.G. 32 character identifiers being hashed - Table size [0..99] A..Z ---> 1,2,...26 0..9 ----> 27,...36 Identifier: CS1 --->3+19+28 (concat) = 31,928 (31,928) 2 = 1,019,397,184 - 10 digits extract middle 2 digits (5th and 6th) get 39, so: H(CS1) = 39

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Divide the identifier x by some number M and use the remainder as the hash address for x F(x)= x%M Bucket address range from 0 to M-1 where M is the table size Choice of M results in a biased use of hash table when several of the identifiers in use have the same result With M=2, odd keys are mapped to odd buckets and even keys to even Choose M as a prime number Value of M should be such that it has no prime divisors less than 20

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E.g. - Table size of 100 3 Digit numbers are the keys 999 possible items Indices 0..99 on the table 999 % 100 = 99 (100 is Table size) 524 % 100 = 24 199 % 100 = 99 (COLLISION)

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00 191 292 30 40 550 60 70 80 90 100 1141 12 1358 140 Basic Division Method- H(Key) = Key % 15, Values to be hashed all > 0 0 indicates null value - or nothing there Results after hashing 41, 58, 12, 92, 50 and 91: This is a nice distribution of values, no collisions!

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030 60 (collision) 10 20 30 40 520 50 (collision) 60 70 80 90 10 40, 70 (collisions ) 110 120 130 140 Conclusion: % 15 is a BAD HASH FUNCTION for this particular set of values! In general: Choose the nearest prime number 1.5 times greater than the size of the data set you are hashing! Now with values 10, 20, 30, 40, 50, 60, 70

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Partition the identifier x into several parts. All except for the last one have the same length Add the parts together to obtain the hash address Shift folding Shift all parts except for the last one so that the least significant bit of each part lines up with the corresponding bit of the last part. Add the parts together to obtain f(x) Folding at the boundaries Reverses every other partition before adding Add the partitions which gives the hash address

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Folding Method: Chopping the key k into two parts and adding yields the following hash addresses: H(3205)=32+05=37, H(7148)=71+48=19, H(2345)=23+45=68 Observe that the leading digit 1 in H(7148) is ignored. Alternatively, one may want to reverse the second part before adding, thus producing the following hash addresses: H(3205)=32+50=82, H(7148)=71+84=55, H(2345)=23+54=77

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Key: 3455677234 Partitioning: 345 |567 | 723 |4 Adding: 345 + 567 + 723 + 4 = 1639 Address: 639 (the carry is ignored)

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enter number:12345 enter folding combination: 12+34+5 12+34+5 =51/*get the last two digits*/ final answer=51

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Used with static files- in which all the identifiers are know in advance Transform the identifiers into numbers using some radix r Examine the digit of each identifier, deleting those digits that have the most skewed distributions Continue deleting the digits until the number of remaining digits is small enough to give an address in the range of the hash table

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Example: Keys: 2234, 3452, 2784 Distribution: Positions 1 and 2 : best distribution. Hence we have: Addresses: 23, 45, 78

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Handling Collisions - Techniques: Two Major Strategies: 1) Open Addressing - Find another spot in the "Table" (same contiguous address space) 2) Chaining - Find another spot outside the "Table" Open Addressing Techniques: a) Linear Probing - search sequentially (with wrap-around) until you find the first vacant slot b) Quadratic Probing -

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Quadratic Probing - Hashed value to index i - slot i is occupied! 1st try after i ---> try i+1 2nd try after i ---> try i + 2 2 3rd try after i ---> try i + 3 2 (Always % tablesize, of course) ETC. Rehashing: When see spot is occupied, hash original key over with a second hash function - this to find another spot in the table.

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Chaining Techniques: This technique "Chains" the item that collided to a location outside the "Table" - to another block of memory (It is done with Dynamic, Extendible Hashing Techniques) Problems with both Open Addressing and Chaining - can have very long searches for an item that collided a bunch with other items!

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0 1 2......... 38160H(160)=38 39204H(204)=38 40219H(219)=38 41119H(119)=39 42412H(412)=39 43390H(390)=39 44263H(263)=40............... size Open Addressing with Linear Probing Clustering can occur : Suppose keys 160, 204, 219, 119, 412, 390, 263 are loaded and H is biased for returning 38-40! Conclusion: Clustering can occur due to a biased hash function with linear probing as a collision resolution technique!

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046 1 213 33 4 5 624 790 8 9 10 0 1 2 3 X - Initial hash position, plus 4th, 8th, 12th.... probes 4 1st, 3rd, 5th, 7th, 9th.... probes after collision at position 3 5 6 7 2nd, 6th, 10th.... probes after collision at position 3 Quadratic Probing: H(Key) = Key %11, Hashing values 13, 3, 24, 46, 90: Note the wraparound calculations! Also, quadratic probing may never "get anywhere"- H(K) = K % 8......

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Compute f(x) for identifier x and then examine the hash table buckets ht[f(x)+j)%TABLE_SIZE] for 0<=j<=TABLE_SIZE Outcomes Bucket contains x. Already in the table, -Report duplicate identifier or update information in the other fields of the element Bucket contains the empty string- Bucket is empty and we may inset the new element into it Bucket contains a nonempty string other than x- Proceed ot the examine the next bucket Return to home bucket ht[f(x)](j=TABLE_SIZE) –Home bucket is being examined for the second time and all remaining buckets have been examined. –Table is full and we report an error condition and exit

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E.g. of Chaining - Index Hashtable Chains 0 400 ---->310 ----->20 ----->50 1 501 ---->211..... 9 89

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Designed to prevent primary clustering. It does this by increasing the step by increasingly large amounts as more probes are required to insert a record. This prevents clusters from building up. In quadratic probing the step is equal to the square of the probe number. With linear probing the step values for a sequence of probes would be {1, 2, 3, 4, etc}. For quadratic probing the step values would be {1, 2 2, 3 2, 4 2, etc}, i.e. {1, 4, 9, 16, etc}. Disadvantage of this method: After a number of probes the sequence of steps repeats itself (remember that the step will be probe number 2 mod the size of the hash table). This repetition occurs when the probe number is roughly half the size of the hash table.

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