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How many address bits are required to represent 4K memory ?
Ans. 12 address bits are required for representing 4K memory, as4K = 22 x 210 = 212
Therefore 1K = 1024 = 210
compare 2g , 3g , and 4g also with advantage of 3g and 4g.
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Q. Example on Sorting using Combinational Circuit? Example: Think about a unsorted list having element values like given below {3,9,8,5,10,12,14,20,90,95,60,40,23,35,18,0}
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need tutorial on data paralellism.
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