# day24 - COP 3503 Computer Science II Spring 2000 CLASS...

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COP 3503 – Computer Science II – Spring 2000 - CLASS NOTES - DAY #24 Hashed Tables Hash tables (files) rely on hashing to perform insertion, deletion, and retrieval in constant time. Hashing functions are a mapping between a key value and a location in the table. The location is typically produced as an offset from the first position in the table. The key value is the data field on which retrieval, insertion, and deletion will be based. To be retrieved from, inserted to, or deleted from the file, an item’s key value must be specified. If the hashing function is a 1:1 function from key value to location then any (retrieval) can be done in constant time. If the hashing function is not 1:1, but M:1, then it is possible for more than key value to map to the same location in the hash table. This is called a collision. Typically, a restriction on the possible key values that may be used will be expected. Without any restrictions the size of the set of possible key values is infinite. [Domain is infinite and the range will normally be finite]. Hashing Methods Method #1 Convert the key value (a string, let’s say) into an integer by adding the product of the character (its ASCII or Unicode value) and some number (say 128) raised to the position of the character. Note: 128 is used since the typical character set requires 7 bits in which to encode the character set (ASCII, not extended ASCII or Unicode). Since 2 7 = 128 this means that we can encode 128 different characters using the integer numbers 0 through 127. Example: CSII = (C * 128 3 ) + (S*128 2 ) + (I*128 1 ) + (I+128 0 ) This method has a serious potential for overflow! For example using ASCII code the value for the work “junk” is 224,229,227! A long string will generate a huge number. Also note that this technique is not a 1:1 mapping so collisions will be possible. Collision resolution will be discussed later. To prevent calculating a number such as 128 i directly for some applies to general polynomials can be used. A general polynomial: A 3 X 3 + A 2 X 2 + A 1 X 1 + A 0 X 0 Day 24 - 1

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can be evaluated as: (((A 3 ) X + A 2 ) X + A 1 )X + A 0 This has three distinct advantages over the earlier method. (1) A large intermediate result that will overflow is deferred until the end of the calculation, (2) only three multiplications and three additions are required to evaluate the polynomial, and (3) the entire calculation proceeds from left to right (exponentiation is from right to left). A better solution is : public int hash1 (string s, int tablesize) { int HashVal = 0; for (int i = 0; i < s.length; i++) HashVal = (HashVal * 128 + s.charAT(i)) % tablesize; return HashVal; }// end hash1 Note that the only improvement in this solution compared to the first is that modulo arithmetic has been applied. However, modulo operations are very expensive. An even better solution:
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## This note was uploaded on 02/22/2009 for the course COP 3503c taught by Professor Staff during the Spring '08 term at University of Central Florida.

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day24 - COP 3503 Computer Science II Spring 2000 CLASS...

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