Narrow down chromatic number to range











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I am required to write a program that takes in an input file (which represents a graph) and outputs the chromatic number along with the vertices colors. I wrote a program that does this successfully, however I am given a large input file and the program runs indefinitely. I need to make the program narrow down the chromatic number to a range of ten values. Can anyone give me some advice as to how I can accomplish this. I will post my code below for reference.



Any help would be appreciated. Thanks!



import java.util.*;
import java.io.*;

public class chromatic_num {
public static int n;
public static int q;
public static int matrix;
public static List<Integer> list;

public static void main(String args) throws IOException {
//reads input from file
BufferedReader br;
try {
br = new BufferedReader(new InputStreamReader(new FileInputStream("file1.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}
String line;
boolean readFirstLine = false;
// new array list to store input
list = new ArrayList<Integer>();
// read file line by line
while ((line = br.readLine()) != null) {
// prints content
String toks = line.split(" ");
if (readFirstLine) {
list.add(Integer.parseInt(toks[0]));
list.add(Integer.parseInt(toks[1]));
}
else {
readFirstLine = true;
n = Integer.parseInt(toks[0]);
}
}
// finds max edges
n = Collections.max(list);
// for adjacency matrix
matrix = new int[n][n];
// constructs adj. matrix
int count = 0;
Integer v1 = 0, v2 = 0;
for (Integer v : list) {
if (count % 2 == 0) {
v1 = v - 1;
}
else {
v2 = v - 1;
matrix[v1][v2] = 1;
matrix[v2][v1] = 1;
}
count++;
}
// new array to store colors
q = new int[n];
chromatic_num chromatic = new chromatic_num();
//print solution to screen
System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
//print solution to file
try {
System.setOut(new PrintStream(new FileOutputStream("chromatic_num.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}

System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
br.close();
}
//color G using minimum number of colors, returns chromatic number
public int color()
{
//for (int t = 0
for (int i = 1; i <= n; i++)
{
//if G can be colored using i colors starting at vertex 0
//System.out.println("colored using " + i + " colors starting at vertex 1");
if (color(0,i)) {
//chromatic number is i, return it
return i;
}
else {
for (int j = 0; j < n; j++) {
q[j] = 0;
}
}
}
return 0;
}

//colors G using m colors starting at vertex v
public boolean color(int v, int m)
{
if (v > n - 1) {
//all vertices have been colored, success
return true;
}
else
{
for (int i = 1; i <= m; i++)
{
//assign color i to vertex v
q[v] = i;
//System.out.println("color " + i + " to vertex " + (v + 1));

//check whether colors of v and its neighbors conflict
boolean hasConflicted = hasConflicted(v);
//System.out.println("hasConflicted = " + hasConflicted);
if (hasConflicted == false)
{
//color the rest of G using m colors starting at vertex v+1, done by recursive call color(v+1, m)
boolean success = color(v + 1, m);

//if (the rest of G can be colored)
// all vertices have been colored, success
if (success) {
return true;
}
}
}
//assign color 0 to vertex v and fail, this happens when none of the m colors can be assigned to vertex v
q[v] = 0;
return false;
}
}
public boolean hasConflicted(int v) {
for (int i = 0; i < n; i++)
{
if (i != v && matrix[i][v] == 1 && q[i] == q[v]) {
return true;
}
}
return false;
}
}









share|improve this question






















  • How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
    – PhaseRush
    Nov 10 at 21:27










  • The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
    – wissam hammoud
    Nov 10 at 21:55

















up vote
0
down vote

favorite












I am required to write a program that takes in an input file (which represents a graph) and outputs the chromatic number along with the vertices colors. I wrote a program that does this successfully, however I am given a large input file and the program runs indefinitely. I need to make the program narrow down the chromatic number to a range of ten values. Can anyone give me some advice as to how I can accomplish this. I will post my code below for reference.



Any help would be appreciated. Thanks!



import java.util.*;
import java.io.*;

public class chromatic_num {
public static int n;
public static int q;
public static int matrix;
public static List<Integer> list;

public static void main(String args) throws IOException {
//reads input from file
BufferedReader br;
try {
br = new BufferedReader(new InputStreamReader(new FileInputStream("file1.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}
String line;
boolean readFirstLine = false;
// new array list to store input
list = new ArrayList<Integer>();
// read file line by line
while ((line = br.readLine()) != null) {
// prints content
String toks = line.split(" ");
if (readFirstLine) {
list.add(Integer.parseInt(toks[0]));
list.add(Integer.parseInt(toks[1]));
}
else {
readFirstLine = true;
n = Integer.parseInt(toks[0]);
}
}
// finds max edges
n = Collections.max(list);
// for adjacency matrix
matrix = new int[n][n];
// constructs adj. matrix
int count = 0;
Integer v1 = 0, v2 = 0;
for (Integer v : list) {
if (count % 2 == 0) {
v1 = v - 1;
}
else {
v2 = v - 1;
matrix[v1][v2] = 1;
matrix[v2][v1] = 1;
}
count++;
}
// new array to store colors
q = new int[n];
chromatic_num chromatic = new chromatic_num();
//print solution to screen
System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
//print solution to file
try {
System.setOut(new PrintStream(new FileOutputStream("chromatic_num.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}

System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
br.close();
}
//color G using minimum number of colors, returns chromatic number
public int color()
{
//for (int t = 0
for (int i = 1; i <= n; i++)
{
//if G can be colored using i colors starting at vertex 0
//System.out.println("colored using " + i + " colors starting at vertex 1");
if (color(0,i)) {
//chromatic number is i, return it
return i;
}
else {
for (int j = 0; j < n; j++) {
q[j] = 0;
}
}
}
return 0;
}

//colors G using m colors starting at vertex v
public boolean color(int v, int m)
{
if (v > n - 1) {
//all vertices have been colored, success
return true;
}
else
{
for (int i = 1; i <= m; i++)
{
//assign color i to vertex v
q[v] = i;
//System.out.println("color " + i + " to vertex " + (v + 1));

//check whether colors of v and its neighbors conflict
boolean hasConflicted = hasConflicted(v);
//System.out.println("hasConflicted = " + hasConflicted);
if (hasConflicted == false)
{
//color the rest of G using m colors starting at vertex v+1, done by recursive call color(v+1, m)
boolean success = color(v + 1, m);

//if (the rest of G can be colored)
// all vertices have been colored, success
if (success) {
return true;
}
}
}
//assign color 0 to vertex v and fail, this happens when none of the m colors can be assigned to vertex v
q[v] = 0;
return false;
}
}
public boolean hasConflicted(int v) {
for (int i = 0; i < n; i++)
{
if (i != v && matrix[i][v] == 1 && q[i] == q[v]) {
return true;
}
}
return false;
}
}









share|improve this question






















  • How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
    – PhaseRush
    Nov 10 at 21:27










  • The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
    – wissam hammoud
    Nov 10 at 21:55















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am required to write a program that takes in an input file (which represents a graph) and outputs the chromatic number along with the vertices colors. I wrote a program that does this successfully, however I am given a large input file and the program runs indefinitely. I need to make the program narrow down the chromatic number to a range of ten values. Can anyone give me some advice as to how I can accomplish this. I will post my code below for reference.



Any help would be appreciated. Thanks!



import java.util.*;
import java.io.*;

public class chromatic_num {
public static int n;
public static int q;
public static int matrix;
public static List<Integer> list;

public static void main(String args) throws IOException {
//reads input from file
BufferedReader br;
try {
br = new BufferedReader(new InputStreamReader(new FileInputStream("file1.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}
String line;
boolean readFirstLine = false;
// new array list to store input
list = new ArrayList<Integer>();
// read file line by line
while ((line = br.readLine()) != null) {
// prints content
String toks = line.split(" ");
if (readFirstLine) {
list.add(Integer.parseInt(toks[0]));
list.add(Integer.parseInt(toks[1]));
}
else {
readFirstLine = true;
n = Integer.parseInt(toks[0]);
}
}
// finds max edges
n = Collections.max(list);
// for adjacency matrix
matrix = new int[n][n];
// constructs adj. matrix
int count = 0;
Integer v1 = 0, v2 = 0;
for (Integer v : list) {
if (count % 2 == 0) {
v1 = v - 1;
}
else {
v2 = v - 1;
matrix[v1][v2] = 1;
matrix[v2][v1] = 1;
}
count++;
}
// new array to store colors
q = new int[n];
chromatic_num chromatic = new chromatic_num();
//print solution to screen
System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
//print solution to file
try {
System.setOut(new PrintStream(new FileOutputStream("chromatic_num.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}

System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
br.close();
}
//color G using minimum number of colors, returns chromatic number
public int color()
{
//for (int t = 0
for (int i = 1; i <= n; i++)
{
//if G can be colored using i colors starting at vertex 0
//System.out.println("colored using " + i + " colors starting at vertex 1");
if (color(0,i)) {
//chromatic number is i, return it
return i;
}
else {
for (int j = 0; j < n; j++) {
q[j] = 0;
}
}
}
return 0;
}

//colors G using m colors starting at vertex v
public boolean color(int v, int m)
{
if (v > n - 1) {
//all vertices have been colored, success
return true;
}
else
{
for (int i = 1; i <= m; i++)
{
//assign color i to vertex v
q[v] = i;
//System.out.println("color " + i + " to vertex " + (v + 1));

//check whether colors of v and its neighbors conflict
boolean hasConflicted = hasConflicted(v);
//System.out.println("hasConflicted = " + hasConflicted);
if (hasConflicted == false)
{
//color the rest of G using m colors starting at vertex v+1, done by recursive call color(v+1, m)
boolean success = color(v + 1, m);

//if (the rest of G can be colored)
// all vertices have been colored, success
if (success) {
return true;
}
}
}
//assign color 0 to vertex v and fail, this happens when none of the m colors can be assigned to vertex v
q[v] = 0;
return false;
}
}
public boolean hasConflicted(int v) {
for (int i = 0; i < n; i++)
{
if (i != v && matrix[i][v] == 1 && q[i] == q[v]) {
return true;
}
}
return false;
}
}









share|improve this question













I am required to write a program that takes in an input file (which represents a graph) and outputs the chromatic number along with the vertices colors. I wrote a program that does this successfully, however I am given a large input file and the program runs indefinitely. I need to make the program narrow down the chromatic number to a range of ten values. Can anyone give me some advice as to how I can accomplish this. I will post my code below for reference.



Any help would be appreciated. Thanks!



import java.util.*;
import java.io.*;

public class chromatic_num {
public static int n;
public static int q;
public static int matrix;
public static List<Integer> list;

public static void main(String args) throws IOException {
//reads input from file
BufferedReader br;
try {
br = new BufferedReader(new InputStreamReader(new FileInputStream("file1.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}
String line;
boolean readFirstLine = false;
// new array list to store input
list = new ArrayList<Integer>();
// read file line by line
while ((line = br.readLine()) != null) {
// prints content
String toks = line.split(" ");
if (readFirstLine) {
list.add(Integer.parseInt(toks[0]));
list.add(Integer.parseInt(toks[1]));
}
else {
readFirstLine = true;
n = Integer.parseInt(toks[0]);
}
}
// finds max edges
n = Collections.max(list);
// for adjacency matrix
matrix = new int[n][n];
// constructs adj. matrix
int count = 0;
Integer v1 = 0, v2 = 0;
for (Integer v : list) {
if (count % 2 == 0) {
v1 = v - 1;
}
else {
v2 = v - 1;
matrix[v1][v2] = 1;
matrix[v2][v1] = 1;
}
count++;
}
// new array to store colors
q = new int[n];
chromatic_num chromatic = new chromatic_num();
//print solution to screen
System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
//print solution to file
try {
System.setOut(new PrintStream(new FileOutputStream("chromatic_num.txt")));
} catch (FileNotFoundException e) {
throw new RuntimeException();
}

System.out.println(chromatic.color());
for (int i = 0; i < n; i++) {
System.out.println("" + (i + 1) + " " + q[i]);
}
br.close();
}
//color G using minimum number of colors, returns chromatic number
public int color()
{
//for (int t = 0
for (int i = 1; i <= n; i++)
{
//if G can be colored using i colors starting at vertex 0
//System.out.println("colored using " + i + " colors starting at vertex 1");
if (color(0,i)) {
//chromatic number is i, return it
return i;
}
else {
for (int j = 0; j < n; j++) {
q[j] = 0;
}
}
}
return 0;
}

//colors G using m colors starting at vertex v
public boolean color(int v, int m)
{
if (v > n - 1) {
//all vertices have been colored, success
return true;
}
else
{
for (int i = 1; i <= m; i++)
{
//assign color i to vertex v
q[v] = i;
//System.out.println("color " + i + " to vertex " + (v + 1));

//check whether colors of v and its neighbors conflict
boolean hasConflicted = hasConflicted(v);
//System.out.println("hasConflicted = " + hasConflicted);
if (hasConflicted == false)
{
//color the rest of G using m colors starting at vertex v+1, done by recursive call color(v+1, m)
boolean success = color(v + 1, m);

//if (the rest of G can be colored)
// all vertices have been colored, success
if (success) {
return true;
}
}
}
//assign color 0 to vertex v and fail, this happens when none of the m colors can be assigned to vertex v
q[v] = 0;
return false;
}
}
public boolean hasConflicted(int v) {
for (int i = 0; i < n; i++)
{
if (i != v && matrix[i][v] == 1 && q[i] == q[v]) {
return true;
}
}
return false;
}
}






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asked Nov 10 at 20:09









wissam hammoud

63




63












  • How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
    – PhaseRush
    Nov 10 at 21:27










  • The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
    – wissam hammoud
    Nov 10 at 21:55




















  • How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
    – PhaseRush
    Nov 10 at 21:27










  • The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
    – wissam hammoud
    Nov 10 at 21:55


















How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
– PhaseRush
Nov 10 at 21:27




How is the graph represented? CSV? Are you sure its indefinite, or is it just slow?
– PhaseRush
Nov 10 at 21:27












The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
– wissam hammoud
Nov 10 at 21:55






The input file contains the verticies in the top left, the edges in the top right, and the list of edges in the subsequent lines. I will post it down below. The assignment description tells us to narrown down the chromatic number to a range of ten values if the chromatic number is not found in one hour. pastebin.com/1Phq6PZJ
– wissam hammoud
Nov 10 at 21:55



















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