For problem 11.24a and b 11.6. PROBLEMS 11. GRAPHS of numbers in (0, 1,2,3] (0 i

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For problem 11.24a and b 11.6. PROBLEMS 11. GRAPHS of numbers in (0, 1,2,3] (0 is blank). We placed some of the H dominos in a ring so that touching dominos meet at the same number. The ring does not include all the 10 dominos. (a) Can you place all the dominos in a ring? (b) How many dominos are there for [0,...,n? (c) For which n can you place all the dominos in a ring? Hints: Create a graph in which every number is a vertex. Problem 11.20. Problem 11.23. We show the possible dominos using pairs DOEEDE Problem 11.24 (Ramsey Numbers). Recall that any social network with 6 people has a 3-person friend clique or a 3-person war. This is remarkable. No matter how random the network here is always some structure. Can we insist on more structure? Yes, by increasing the size (a) Show that any social network with 10 people has a 4-person friend clique or a 3-person war al network with 9 people has a 4-person friend clique or a 3-person war Hints: Assume no 3-person war and use contradiction to show that there is a 4-person end clique. To get a contradiction, show that every vertex has 3 enemies and 5 friends.] (c) Ramsey asked if you can insist on any amount of structure. For numbers k,s 0 there is a (smallest) number R(k, s) for which any social network with R(k, s) people has either a k-person friend clique or an 8-person war. R(3.3) 6 and R(4.3) 9 (Ö) Prove that R(k,s) R(s,k) and R(k, s) S R(k-1,s)+ R(k,s -1). Explain why this ult: there is as much structure as you wish in large enough graphs. proves Ramsey's res (üi) What is R(k, 1)? Prove by induction that R(k, s) compute Ramsey(5.5) or face extinction. We could marshal the world's best year we might have the value. If, instead, the aliens demanded Aliens give us a year to minds and fastest computers, and within a Ramsey(6.6), we should launch a preemptive attack. -Paul Erdós (paraphrased) Problem 11.25. Give the adjacency matrix A for the graph on the right te matrices D, whose (i,j) entry is the number of paths that DA for k 2 1. (The kth s if an only if they are not adjacent in G. Give the (a) For k = 1.2, 3, compu of length k from vertex i to vertex j (b) Compute Akfor k= 1, 2, 3 and compare with Dk (c) For a general graph, prove by induction t ' , power of A giv es the number of paths of length k between vertices lement Graph). For graph G, the complement G has .) in G are the complement of the edges in G: the c only if the same vertices, but the edges vertices u and v are adjacent in G complement of the graph on the right and the comple adjacent in Problem 11.27. Answer the following questions abou distince a graph and its com (a) If G is regular, prove that G is also regular nected graph G for which the complement

Explanation / Answer

#include <iostream>
#include <stdio.h>
#include
#include
#include

// This part reads data in.

main()
{

    FILE *myFile;
    myFile = fopen("datafile.txt", "r");

    //Find null
    if (myFile == NULL)
    {
        printf("File open failed");
return -1 ;
    }

    //read file into array

    string[] Line = sr.ReadLine().Split(',');
    if (Row == 0)
    {
        data = new string[Line.Length, Line.Length];
    }
    for (int column = 0; column < Line.Length; column++)
    {
        data[Row, column] = Line[column];
    }
    Row++;
    Console.WriteLine(Row);
}

//MOVING FUNCTION where x = h and w = y.
void up ()
{
setArray();
take ();
box [x][y+1];
result = box [x][y+1];
showupdatedpuzzle ();
}

void down ()
{
setArray();
take ();
box [x][y-1];
result = box [x][y-1];
}

void left ()
{
setArray();
take();
box [x-1][y];
result = box [x-1] [y];
}

void right ()
{
setArray();
take();
box [x+1][y];
result = box [x+1][y];
}

THIS is how I think I select the type of move.

char choice;
cout << "How would you like to move the blank? (select #)"<< endl;
cout << "u: Up" << endl;
cout << "d: Down" << endl;
cout << "l: Left" << endl;
cout << "r: Right" << endl;
cin>> choice; cout< switch (choice)
{
case 'u': up(); break;
case 'd': down(); break;
case 'l': left (); break;
case 'r': right (); break;

}

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