Abstract | ||
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Let A be a finite alphabet, and H be a finite dictionary of words formed from the letters of this alphabet. In a conventional keyboard typically each key of the keyboard is assigned a unique symbol chosen from the alphabet. We consider the problem of assigning more than one symbol of the alphabet A to the same key on the keyboard. Let Ci be a subset of A. Every time a character in Ci is to be represented, the keyboard represents it using the digit (i.e. the index) 'i'. In a keyboard of this form, since multiple symbols of the alphabet A have the same representation, the representations of all the (distinct) words in the dictionary H need not be unique. A useful measure of the effectiveness of a particular keyboard is the number of words of H which are mapped to the same encoded form (i.e. they collide). This metric will be termed the ambiguity associated with the keyboard. The problem we study is one of optimally assigning the symbols of the alphabet to the keys of a given keyboard with a view to minimize the resulting ambiguity.The problem as stated above is proven to be NP-hard. This naturally leads us to seek fast and approximate solutions to the optimization problem on hand. After presenting the only reported solution to the problem, we report a fast learning automaton-based solution to this problem. Experimental results demonstrating the power of this solution have also been presented. |
Year | DOI | Venue |
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1990 | 10.1145/98894.99108 | IEA/AIE (Vol. 1) |
Keywords | Field | DocType |
finite dictionary,approximate solution,automaton solution,dictionary h,alphabet a,encoded form,optimization problem,particular keyboard,automaton-based solution,finite alphabet,keyboard optimization problem,conventional keyboard,indexation | Symbol,Computer science,Automaton,Artificial intelligence,Ambiguity,Optimization problem,Machine learning,Alphabet | Conference |
ISBN | Citations | PageRank |
0-89791-372-8 | 1 | 0.49 |
References | Authors | |
6 | 3 |
Name | Order | Citations | PageRank |
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B. John Oommen | 1 | 1255 | 222.20 |
R. S. Valiveti | 2 | 30 | 6.19 |
J. Zgierski | 3 | 1 | 0.49 |