Recursive Bayes sifting is to be utilized to intertwine the sensor estimations to assess area. Portray the means and key suppositions made by this channel. Numerical conditions are not needed for full checks. [4 marks] (b) A corridorketch the estimation model for the spotlight indicator. Sketch the conviction dissremination for bel(x) for the client's area acquired when the finder is turned on and quickly distinguishes a spotlight. [3 marks] (ii) Describe a reasonable movement model expecting the cell phone's inertial sensors are utilized to perform Pedestrian Dead Reckoning (PDR) with a consistent advance length. Sketch bel(x) acquired after the telephone reports three stages have been taken. [4 marks] (iii) Sketch the new bel(x) acquired when the light situating sensor presently shows the client is under a spotlight. Clear up its relationship for the past bel(x). [3 marks] (c) A Grid channel is an execution of a recursive Bayes channel where the bel(x) appropriations are approximated by histograms. Think about the reasonableness of such a channel for this issue contrasted with Kalman and Particle channels, giving specific consideration to any boundaries and their belongings. (a) What are the key diffes between the server farm network climate and the more extensive, general Internet eco-framework? How do these lead to less difficult decisions for offering execution ensures for traffic? [10 marks] (b) Network coding can be utilized to join bundles repetitively to give blunder assurance, yet in addition to decrease the quantity of transmissions and retransmissions important. It is utilized in the vehicle layer, from each source, and in the organization layer, joining information from numerous sources. How should a blend of these procedures improve buffering in straightforward remote gadgets that you could see as in an Internet of Things (IoT) climate? [10 marks] 12 System-on-Chip Design We require an equipment gas pedal to figure f(x) = X i=0..7 C[i] D[i + x] where all values are 20-piece marked numbers and the exhibit C[i] contains configuration time constants. The cluster D[] will contain 1024 qualities. We really want to assess f(x) as quick as could really be expected. The upsides of x are unusual and single word of D gets refreshed by a separate cycle after each 50 or so assessments of f. Overlook flood. (a) An early plan holds both C and D in a typical, single-ported RAM memory (for example one with one location transport). Considering that solitary cycle increase collect blocks are accessible, give a good guess of the presentation of our framework in clocks per assessment. [3 marks]

Characterize the two fulfillment relations M, |= and M, s |= , making sense of completely any documentation that you use and any assistant definitions that you make.

complet the strategy huffmantoText and texttoHuffman

public static void huffmanToText(){ //TDo: convert text to huffman code

}

public static void textToHuffman(){ //Do: convert huffman code to its text same }

Input a text that will be changed over to Huffman code.

This is test of the result:

Text to Huffman Code Conversion:

Enter a text: def

Huffman code: 1011100000

1. Input a Huffman code that will be changed over to its text same.

2. Yield a mistake message in the event that the info can't be changed over.

code

HuffmanCoding.java

bundle midlab2; import java.util.*;

public class HuffmanCoding { static ArrayList huffmanCode = new ArrayList<>(); static ArrayList huffmanChar = new ArrayList<>();

public static void main(String[] args) { char[] text = readText(); ArrayList freq = frequencyCounter(text); ArrayList charArray = charArray(text); huffmanTree(charArray,freq); System.out.println("- - - - - - - - - - ");

Scanner = new Scanner(System.in); System.out.println("1.) Text to Huffman code"); System.out.println("2.) Huffman code to message"); System.out.print("Enter your decision: "); int decision = scanner.nextInt(); choiceOfAction(choice);

//utilize this string for the end goal of testing String t = }

public static char[] readText(){ Scanner = new Scanner(System.in); System.out.print("Type your text: "); String text = scanner.nextLine(); return text.toCharArray(); }

public static ArrayList frequencyCounter(char[] text){ ArrayList counts = new ArrayList<>(); ArrayList charChecker = new ArrayList<>(); for (singe esteem : text) { if(!charChecker.contains(value)) charChecker.add(value); else proceed; int c = 0; for (singe thing : text) { if (esteem == thing) { c++; } } counts.add(c); }//end circle bring counts back; }

public static ArrayList charArray(char[] text){ ArrayList charArr = new ArrayList<>(); for (roast worth : text) { if(!charArr.contains(value)) charArr.add(value); }//end circle return charArr; }

public static void huffmanTree(ArrayList charArray, ArrayList charfreq){ // number of characters. int n = charArray.size();

// making a need line q. // focuses on queue(min-load). PriorityQueue q = new PriorityQueue<>(n, new NodeComparator());

for (int I = 0; I < n; i++) { // making a Huffman hub object // furthermore, add it to the need line. BTNode hn = new BTNode(); hn.c = charArray.get(i); hn.element = charfreq.get(i); hn.left = invalid; hn.right = invalid; //System.out.println(hn.getCharacter()+": "+hn.getElement()); // add capacities adds // the huffman hub to the line. q.add(hn); }

// make a root hub BTNode root = invalid;

// Here we will separate the two least worth // from the pile each time until // its size decreases to 1, separate until // every one of the hubs are extricated. while (q.size() > 1) {

// first min separate. BTNode x = q.peek(); q.poll();

// second min separate. BTNode y = q.peek(); q.poll();

// new hub f which is equivalent BTNode f = new BTNode();

// to the amount of the recurrence of the two hubs // allocating values to the f hub. f.element = x.element + y.element; f.c = '- ';

// first removed hub as left kid. f.left = x;

// second separated hub as the right youngster. f.right = y;

// denoting the f hub as the root hub. root = f;

// add this hub to the need line. q.add(f); }

// print the codes by crossing the tree printCode(root, ""); }

public static void printCode(BTNode root, String s) { // base case; on the off chance that the left and right are invalid // then its a leaf hub and we print // the code s produced by navigating the tree.

on the off chance that (root.left == invalid && root.right == invalid) { // c is the person in the hub System.out.println(root.c + ":" + s); huffmanCode.add(s); huffmanChar.add(root.c); return; }

// on the off chance that we go to left, add "0" to the code. // on the off chance that we go to one side add"1" to the code.

// recursive calls for left and // right sub-tree of the produced tree. printCode(root.left, s + "0"); printCode(root.right, s + "1"); }

public static void tableOfValues(){ //Task: you can utilize the property "huffmanCode" and "huffmanChar" as information for the table }

public static void percentStorageSaving(){ //Task: allude to page 23 of module 4

}

public static boolean isTextValid(String text){ //Task: Output a blunder message on the off chance that the message can't be changed over (for example utilizing character/s //in the information text that is/are not piece of the table Huffman codes) bring valid back; }

public static void huffmanToText(){

}

public static void textToHuffman(){ //Do: convert huffman code to its text same }

public static void choiceOfAction(int choice){ switch (choice){ case 1: Scanner = new Scanner(System.in); String t = scanner.nextLine(); if(isTextValid(t)) textToHuffman(); else System.out.println("Invalid text, attempt again."); case 2: huffmanToText(); } }

}

BTNode.java bundle midlab2;

import java.util.*; public class BTNode { public int component; public roast c; public BTNode left; public BTNode right;

public BTNode(){} /**Develops a paired tree hub that stores the predefined component. */ public BTNode(int component) { this.element = component; left = right = invalid; } /** Returns the component put away in this hub. */ public int getElement() { bring component back; }

/** Returns the component put away in this hub. */ public scorch getCharacter() { bring c back; }

/* Sets the component put away in this hub./ public void setElement (int component) { this.element = component; }

/* Sets the component put away in this hub./ public void setChar (scorch c) { this.c = c; }

/* Returns the left subtree of this hub./ public BTNode getLeft() { bring left back; }

/** Sets the left offspring of this hub. */ public void setLeft (BTNode left) { this.left = left; }

/** Returns the right subtree of this hub. */ public BTNode getRight() { bring right back; }

/**Sets the right offspring of this hub. */ public void setRight (B