Write a Verilog program which implements and tests a four element cache with four bit memory addresses in three configurations (required whether or not the bonus eight-element code is provided): 1. Direct mapped, 2. Two-way set associative, and 3. Fully associative. Up to fifteen additional bonus points may be acquired by extending the program to support an eight element cache. This option would require the submission of two separate source code files.: one for the standard requirements and a second for the extension. The report must include both versions and a discussion of the differents. Both programs must build, execute, and produce correct results in order to receive any of these points. The program reads a memory trace to test the configurations and report the number of of hits and misses. A LRU mechanism must be implemented for both the two-way and fully associative configurations.

This question involves a class named textfile that represents a text file. public class textfile { private string filename; private string filename; private arraylist words; // constructors not shown // postcondition: returns the number of bytes in this file public int filesize() { } // precondition: 0 < = index < words.size() // postcondition: removes numwords words from the words arraylist beginning at // index. public void deletewords(int index, int numwords) { } // precondition: 0 < = index < = words.size() // postcondition: adds elements from newwords array to words arraylist beginning // at index. pub lic voidaddwords(int index, string[] newwords) { } // other methods not shown } complete the filesize() method. the filesize() is computed in bytes. in a text file, each character in each word counts as one byte. in addition, there is a space in between each word in the words arraylist, and each of those spaces also counts as one byte. for example, suppose the words arraylist stores the following words: { mary had a little lamb; its fleece was white as snow. } the filesize() method would compute 4 + 3 + 1 + 6 + 5 + 4 + 6 + 3 + 5 + 2 + 5 as the sum of the lengths of each string in the arraylist. the value returned would be this sum plus 10, because there would also be 10 spaces in between the 11 words. complete the filesize() method below: // postcondition: returns the number of bytes in this file public int filesize() { }