Solving this Arbitage problem of UVA http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=40 but I am stuck with finding the negative cycle of shortest length(length here is number of vertices).Here is my code that successfully detects the negative cycle
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class _104 {
public static void main(String[] args) throws NumberFormatException,
IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(
System.in));
String input;
while ((input = reader.readLine()) != null) {
int n = Integer.parseInt(input);
double[][] cost = new double[n + 1][n + 1];
double[] spEstimate = new double[n + 1];
int parent[] = new int[n + 1];
for (int i = 0; i < n + 1; i++) {
spEstimate[i] = Double.MAX_VALUE;
cost[0][i] = 0;
cost[i][0] = Double.MAX_VALUE;
parent[i] = Integer.MAX_VALUE;
}
spEstimate[0] = 0.0;
parent[0] = 0;
for (int i = 1; i < n + 1; i++) {
String[] line = reader.readLine().split("\\s+");
for (int j = 1; j < n + 1; j++) {
if (i == j) {
cost[i][j] = 0;
} else if (i < j) {
cost[i][j] = -(Math
.log(Double.parseDouble(line[j - 2])) / Math
.log(2));
} else {
cost[i][j] = -(Math
.log(Double.parseDouble(line[j - 1])) / Math
.log(2));
}
}
}
int save = 1, s = 1;
boolean flag = BellmanFord(n, cost, spEstimate, parent);
display(cost);
// Relax all edges once more
boolean brk = true;
for (int i = 0; i < cost.length && brk; i++) {
for (int j = 0; j < cost.length && brk; j++) {
//relax(i, j, spEstimate, cost[i][j], parent);
}
}
ArrayList<Integer> path = new ArrayList<Integer>();
while (parent[save] != s) {
path.add(save);
save = parent[save];
}
if (flag) {
System.out.println("no arbitrage sequence exists");
} else {
path.add(0, path.get(path.size() - 1));
for (int i = path.size() - 1; i >= 0; --i) {
System.out.println(path.get(i));
}
}
}
reader.close();
}
public static boolean BellmanFord(int n, double[][] cost, double[] sp,
int[] parent) {
for (int k = 0; k < n - 1; k++) {
for (int i = 0; i < cost.length; i++) {
for (int j = 0; j < cost.length; j++) {
relax(i, j, sp, cost[i][j], parent);
}
}
}
// Relax all edges once more to detect cycle
for (int i = 0; i < cost.length; i++) {
for (int j = 0; j < cost.length; j++) {
if (sp[j] > (sp[i] + cost[i][j])) {
return false;
}
}
}
return true;
}
static void relax(int i, int j, double[] sp, double cij, int[] parent) {
if (sp[j] > (sp[i] + cij)) {
sp[j] = sp[i] + cij;
System.out.println("relaxed " + i + " " + j + " " + cij + " "
+ sp[i] + " " + sp[j]);
parent[j] = i;
}
}
static void display(double[][] cost) {
System.out.println("Display Cost");
for (int i = 0; i < cost.length; i++) {
for (int j = 0; j < cost.length; j++) {
System.out.print(cost[i][j] + "\t");
}
System.out.println();
}
}
static void display(double[] sp) {
for (int i = 0; i < sp.length; i++) {
System.out.println(sp[i]);
}
}
}
You can do it like that:
Fix the start vertex of the cycle(let's call it v).
Run Ford-Bellman algorithm assuming that dist[i] = 0 if i = v and INF otherwise.
If there is a negative cycle that contains v, after k iterations of the outer loop in Ford-Bellman algorithm dist[v] will become negative. So you can easily find such smallest k by simply checking if dist[v] is still non-negative or not after each iteration.
The smallest k among all v is the answer.
It is possible to solve this problem by considering cycles of increasing length as opposed to finding negative cycles as described by kraskevich. The worst case complexity for both approaches is O(n^4). This approach resembles Floyd-Warshall where you consider increasing lengths instead of intermediate vertices.
You can find a detailed explanation that includes diagrams and code here.
Related
I'm making a Java program where it counts the number of paths. I want it to only go up (north) and right (east). I also want it to output the paths taken. Is there something I can do?
import java.io.*;
public class PathsRecursion {
public static int countPaths(int n, int e) {
int dp[][] = new int[n + 1][e + 1];
for (int i = 0; i <= n; i++) {
dp[i][0] = 1;
}
for (int i = 0; i <= e; i++) {
dp[0][i] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= e; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
return dp[n][e];
}
public static void main(String[] args) {
int n = 1, e = 3;
System.out.println("Number of Paths: " + countPaths(n, e));
}
}
You added an extra row and column to you dp array. It needs to be a n by e array, not n + 1 by e + 1 array.
import java.io.*;
public class PathsRecursion {
public static int countPaths(int n, int e) {
int dp[][] = new int[n][e];
for (int i = 0; i <= n; i++) {
dp[i][0] = 1;
}
for (int i = 0; i <= e; i++) {
dp[0][i] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= e; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
return dp[n - 1][e - 1];
}
public static void main(String[] args) {
int n = 1, e = 3;
System.out.println("Number of Paths: " + countPaths(n, e));
}
}
For my university assignment in java I have been asked to provide "extra analytics functions" I decided to use Levenshtein distance but I have an issue where the number outputted to the console is one less than the actual answer. So the distance between "cat" and "hat" should be 1 but it's displaying as 0
public class Levenshtein {
public Levenshtein(String first, String second) {
char [] s = first.toCharArray();
char [] t = second .toCharArray();
int Subcost = 0;
int[][] array = new int[first.length()][second.length()];
for (int i = 0; i < array[0].length; i++)
{
array[0][i] = i;
}
for (int j = 0; j < array.length; j++)
{
array [j][0]= j;
}
for (int i = 1; i < second.length(); i++)
{
for (int j = 1; j < first.length(); j++)
{
if (s[j] == t [i])
{
Subcost = 0;
}
else
{
Subcost = 1;
}
array [j][i] = Math.min(array [j-1][i] +1,
Math.min(array [j][i-1] +1,
array [j-1][i-1] + Subcost) );
}
}
UI.output("The Levenshtein distance is -> " + array[first.length()-1][second.length()-1]);
}
}
Apparently you're using the following algorithm:
https://en.wikipedia.org/wiki/Levenshtein_distance#Iterative_with_full_matrix
I think you were not too accurate with indices. I'm not sure where exactly the problem is, but here is a working version:
public int calculateLevenshteinDistance(String first, String second) {
char[] s = first.toCharArray();
char[] t = second.toCharArray();
int substitutionCost = 0;
int m = first.length();
int n = second.length();
int[][] array = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
array[i][0] = i;
}
for (int j = 1; j <= n; j++) {
array[0][j] = j;
}
for (int j = 1; j <= n; j++) {
for (int i = 1; i <= m; i++) {
if (s[i - 1] == t[j - 1]) {
substitutionCost = 0;
} else {
substitutionCost = 1;
}
int deletion = array[i - 1][j] + 1;
int insertion = array[i][j - 1] + 1;
int substitution = array[i - 1][j - 1] + substitutionCost;
int cost = Math.min(
deletion,
Math.min(
insertion,
substitution));
array[i][j] = cost;
}
}
return array[m][n];
}
I have to find the minimum distance between two vertices and no. of edges traversed in finding that minimum distance in a adirected weighted graph. I did write a code on using Dijkstras algo. But my test cases are failing. What am i missing here.?
Ideal input: given any two vertices A,B
Output: min distance between two nodes, no. of edges traversed
If the node is unreachable then output should be -1, -1.
//This is a java program to find the shortest path between source vertex and destination vertex
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
public class Dijkstras_Shortest_Path
{
private int distances[];
private int Numnodes[];
private int numnodes;
private Set<Integer> settled;
private Set<Integer> unsettled;
private int number_of_nodes;
private int adjacencyMatrix[][];
public Dijkstras_Shortest_Path(int number_of_nodes)
{
this.number_of_nodes = number_of_nodes;
distances = new int[number_of_nodes + 1];
Numnodes=new int[number_of_nodes+1];
settled = new HashSet<Integer>();
unsettled = new HashSet<Integer>();
adjacencyMatrix = new int[number_of_nodes + 1][number_of_nodes + 1];
}
public void dijkstra_algorithm(int adjacency_matrix[][], int source)
{
int evaluationNode;
for (int i = 1; i <= number_of_nodes; i++)
for (int j = 1; j <= number_of_nodes; j++)
adjacencyMatrix[i][j] = adjacency_matrix[i][j];
for (int i = 1; i <= number_of_nodes; i++)
{
distances[i] = Integer.MAX_VALUE;
}
unsettled.add(source);
distances[source] = 0;
while (!unsettled.isEmpty())
{
evaluationNode = getNodeWithMinimumDistanceFromUnsettled();
unsettled.remove(evaluationNode);
settled.add(evaluationNode);
evaluateNeighbours(evaluationNode);
}
}
private int getNodeWithMinimumDistanceFromUnsettled()
{
int min;
int node = 0;
Iterator<Integer> iterator = unsettled.iterator();
node = iterator.next();
min = distances[node];
for (int i = 1; i <= distances.length; i++)
{
if (unsettled.contains(i))
{
if (distances[i] <= min)
{
min = distances[i];
node = i;
}
}
}
return node;
}
private void evaluateNeighbours(int evaluationNode)
{
int edgeDistance = -1;
int newDistance = -1;
int newNodes=1;
for (int destinationNode = 1; destinationNode <= number_of_nodes;
destinationNode++)
{
if (!settled.contains(destinationNode))
{
if (adjacencyMatrix[evaluationNode][destinationNode] !=
Integer.MAX_VALUE)
{
edgeDistance = adjacencyMatrix[evaluationNode]
[destinationNode];
newDistance = distances[evaluationNode] + edgeDistance;
newNodes=Numnodes[evaluationNode]+1;
if (newDistance < distances[destinationNode])
{
distances[destinationNode] = newDistance;
Numnodes[destinationNode]=newNodes;
}
unsettled.add(destinationNode);
}
}
}
}
public static void main(String... arg)
{
int adjacency_matrix[][];
int number_of_vertices;
int number_of_edges;
int source = 0, destination = 0;
Scanner scan = new Scanner(System.in);
try
{
number_of_vertices = scan.nextInt();
number_of_edges=scan.nextInt();
if (number_of_vertices<1||number_of_vertices>10000)
{
System.out.println("Number of vertices out of boundary");
}
if (number_of_edges<1||number_of_edges>10000)
{
System.out.println("Number of edges out of boundary");
}
adjacency_matrix = new int[number_of_vertices + 1]
[number_of_vertices + 1];
for (int i = 1; i <= number_of_vertices; i++)
{
for (int j = 1; j <= number_of_vertices; j++)
{
adjacency_matrix[i][j] = 0;
}
}
for(int i=1;i<=number_of_edges;i++)
{
int node1=scan.nextInt();
int node2=scan.nextInt();
adjacency_matrix[node1][node2]=scan.nextInt();
}
for (int i = 1; i <= number_of_vertices; i++)
{
for (int j = 1; j <= number_of_vertices; j++)
{
if (adjacency_matrix[i][j] == 0)
{
adjacency_matrix[i][j] = Integer.MAX_VALUE;
}
}
}
source = scan.nextInt();
destination = scan.nextInt();
Dijkstras_Shortest_Path dijkstrasAlgorithm = new
Dijkstras_Shortest_Path(
number_of_vertices);
dijkstrasAlgorithm.dijkstra_algorithm(adjacency_matrix, source);
for (int i = 1; i <= dijkstrasAlgorithm.distances.length - 1; i++)
{
if (i == destination)
System.out.println(dijkstrasAlgorithm.distances[i] +" "+
dijkstrasAlgorithm.Numnodes[i]);
}
} catch (InputMismatchException inputMismatch)
{
System.out.println("Wrong Input Format");
}
scan.close();
}
}
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I have created the following Simplex Algorithm that maximises on the objective function. I want the opposite to happen. In this example there are two variables and the algorithm must figure out what to multiply these two variables here (13.0 and 23.0) by in order to get the maximum possible result within the constraints set. I want the algorithm to figure out the lowest possible result instead.
My Code:
import java.util.*;
public class Simplex
{
private static final double EPSILON = 1.0E-10;
private double[][] tableaux;
private int numOfConstraints;
private int numOfVariables;
private int[] basis;
/**
* Constructor for objects of class Simplex
*/
public Simplex()
{
double[][] thisTableaux = {
{ 5.0, 15.0 },
{ 4.0, 4.0 },
{ 35.0, 20.0 },
};
double[] constraints = { 480.0, 160.0, 1190.0 };
double[] variables = { 13.0, 23.0 };
numOfConstraints = constraints.length;
numOfVariables = variables.length;
tableaux = new double[numOfConstraints+1][numOfVariables+numOfConstraints+1];
//adds all elements from thisTableaux to tableaux
for(int i=0; i < numOfConstraints; i++)
{
for(int j=0; j < numOfVariables; j++)
{
tableaux[i][j] = thisTableaux[i][j];
}
}
//adds a slack variable for each variable there is and sets it to 1.0
for(int i=0; i < numOfConstraints; i++)
{
tableaux[i][numOfVariables+i] = 1.0;
}
//adds variables into the second [] of tableux
for(int j=0; j < numOfVariables; j++)
{
tableaux[numOfConstraints][j] = variables[j];
}
//adds constraints to first [] of tableaux
for(int k=0; k < numOfConstraints; k++)
{
tableaux[k][numOfConstraints+numOfVariables] = constraints[k];
}
basis = new int[numOfConstraints];
for(int i=0; i < numOfConstraints; i++)
{
basis[i] = numOfVariables + i;
}
//show();
//optimise();
//assert check(thisTableaux, constraints, variables);
}
public void optimise() {
while(true) {
int q = findLowestNonBasicCol();
if(q == -1) {
break;
}
int p = getPivotRow(q);
if(p == -1) throw new ArithmeticException("Linear Program Unbounded");
pivot(p, q);
basis[p] = q;
}
}
public int findLowestNonBasicCol() {
for(int i=0; i < numOfConstraints + numOfVariables; i++)
{
if(tableaux[numOfConstraints][i] > 0) {
return i;
}
}
return -1;
}
public int findIndexOfLowestNonBasicCol() {
int q = 0;
for(int i=1; i < numOfConstraints + numOfVariables; i++)
{
if(tableaux[numOfConstraints][i] > tableaux[numOfConstraints][q]) {
q = i;
}
}
if(tableaux[numOfConstraints][q] <= 0) {
return -1;
}
else {
return q;
}
}
/**
* Finds row p which will be the pivot row using the minimum ratio rule.
* -1 if there is no pivot row
*/
public int getPivotRow(int q) {
int p = -1;
for(int i=0; i < numOfConstraints; i++) {
if (tableaux[i][q] <=0) {
continue;
}
else if (p == -1) {
p = i;
}
else if((tableaux[i][numOfConstraints+numOfVariables] / tableaux[i][q] < tableaux[p][numOfConstraints+numOfVariables] / tableaux[p][q])) {
p = i;
}
}
return p;
}
public void pivot(int p, int q) {
for(int i=0; i <= numOfConstraints; i++) {
for (int j=0; j <= numOfConstraints + numOfVariables; j++) {
if(i != p && j != q) {
tableaux[i][j] -= tableaux[p][j] * tableaux[i][q] / tableaux[p][q];
}
}
}
for(int i=0; i <= numOfConstraints; i++) {
if(i != p) {
tableaux[i][q] = 0.0;
}
}
for(int j=0; j <= numOfConstraints + numOfVariables; j++) {
if(j != q) {
tableaux[p][j] /= tableaux[p][q];
}
}
tableaux[p][q] = 1.0;
show();
}
public double result() {
return -tableaux[numOfConstraints][numOfConstraints+numOfVariables];
}
public double[] primal() {
double[] x = new double[numOfVariables];
for(int i=0; i < numOfConstraints; i++) {
if(basis[i] < numOfVariables) {
x[basis[i]] = tableaux[i][numOfConstraints+numOfVariables];
}
}
return x;
}
public double[] dual() {
double[] y = new double[numOfConstraints];
for(int i=0; i < numOfConstraints; i++) {
y[i] = -tableaux[numOfConstraints][numOfVariables];
}
return y;
}
public boolean isPrimalFeasible(double[][] thisTableaux, double[] constraints) {
double[] x = primal();
for(int j=0; j < x.length; j++) {
if(x[j] < 0.0) {
StdOut.println("x[" + j + "] = " + x[j] + " is negative");
return false;
}
}
for(int i=0; i < numOfConstraints; i++) {
double sum = 0.0;
for(int j=0; j < numOfVariables; j++) {
sum += thisTableaux[i][j] * x[j];
}
if(sum > constraints[i] + EPSILON) {
StdOut.println("not primal feasible");
StdOut.println("constraints[" + i + "] = " + constraints[i] + ", sum = " + sum);
return false;
}
}
return true;
}
private boolean isDualFeasible(double[][] thisTableaux, double[] variables) {
double[] y = dual();
for(int i=0; i < y.length; i++) {
if(y[i] < 0.0) {
StdOut.println("y[" + i + "] = " + y[i] + " is negative");
return false;
}
}
for(int j=0; j < numOfVariables; j++) {
double sum = 0.0;
for(int i=0; i < numOfConstraints; i++) {
sum += thisTableaux[i][j] * y[i];
}
if(sum < variables[j] - EPSILON) {
StdOut.println("not dual feasible");
StdOut.println("variables[" + j + "] = " + variables[j] + ", sum = " + sum);
return false;
}
}
return true;
}
private boolean isOptimal(double[] constraints, double[] variables) {
double[] x = primal();
double[] y = dual();
double value = result();
double value1 = 0.0;
for(int j=0; j < x.length; j++) {
value1 += variables[j] * x[j];
}
double value2 = 0.0;
for(int i=0; i < y.length; i++) {
value2 += y[i] * constraints[i];
}
if(Math.abs(value - value1) > EPSILON || Math.abs(value - value2) > EPSILON) {
StdOut.println("value = " + value + ", cx = " + value1 + ", yb = " + value2);
return true;
}
return true;
}
private boolean check(double[][] thisTableaux, double[] constraints, double [] variables) {
return isPrimalFeasible(thisTableaux, constraints) && isDualFeasible(thisTableaux, variables) && isOptimal(constraints, variables);
}
}
If you need any more info just ask. Any help appreciated thanks.
If you want to minimize f(x), this is equivalent to maximizing -f(x), so if your posted code solves maximization problems correctly, you can use it to minimize any objective function f(x) simply by maximizing its additive inverse -f(x).
Note that you do not change the constraints, only the objective function.
For example, minimizing f(x) = 3x + 5, x >= 1 is equivalent to maximizing -f(x) = -3x -5, x >= 1.
min[f(x), x>=1] = f(1) = 8 = -(-8) = -[-f(1)] = -max[-f(x), x>=1].
In general, min[f(x)] = f(Xmin) = -[-f(Xmax)] = -max[-f(x)] and Xmin = Xmax.
In the above example, min[f(x)] = -max[-f(x)] = 8 and Xmin = Xmax = 1.
In the particular example you give, you would simply need to change the line
double[] variables = { 13.0, 23.0 };
to
double[] variables = { -13.0, -23.0 };
The values of the variables returned should then be the same as for the minimum of the case where
double[] variables = { 13.0, 23.0 };
and multiplying the value of the objective function by -1 will give the minimum of the objective for the case where
double[] variables = { 13.0, 23.0 };
I have this problem, I need to generate from a given permutation not all combinations, but just those obtained after permuting 2 positions and without repetition. It's called the region of the a given permutation, for example given 1234 I want to generate :
2134
3214
4231
1324
1432
1243
the size of the region of any given permutation is , n(n-1)/2 , in this case it's 6 combinations .
Now, I have this programme , he does a little too much then what I want, he generates all 24 possible combinations :
public class PossibleCombinations {
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
System.out.println("Entrer a mumber");
int n=s.nextInt();
int[] currentab = new int[n];
// fill in the table 1 TO N
for (int i = 1; i <= n; i++) {
currentab[i - 1] = i;
}
int total = 0;
for (;;) {
total++;
boolean[] used = new boolean[n + 1];
Arrays.fill(used, true);
for (int i = 0; i < n; i++) {
System.out.print(currentab[i] + " ");
}
System.out.println();
used[currentab[n - 1]] = false;
int pos = -1;
for (int i = n - 2; i >= 0; i--) {
used[currentab[i]] = false;
if (currentab[i] < currentab[i + 1]) {
pos = i;
break;
}
}
if (pos == -1) {
break;
}
for (int i = currentab[pos] + 1; i <= n; i++) {
if (!used[i]) {
currentab[pos] = i;
used[i] = true;
break;
}
}
for (int i = 1; i <= n; i++) {
if (!used[i]) {
currentab[++pos] = i;
}
}
}
System.out.println(total);
}
}
the Question is how can I fix this programme to turn it into a programme that generates only the combinations wanted .
How about something simple like
public static void printSwapTwo(int n) {
int count = 0;
StringBuilder sb = new StringBuilder();
for(int i = 0; i < n - 1;i++)
for(int j = i + 1; j < n; j++) {
// gives all the pairs of i and j without repeats
sb.setLength(0);
for(int k = 1; k <= n; k++) sb.append(k);
char tmp = sb.charAt(i);
sb.setCharAt(i, sb.charAt(j));
sb.setCharAt(j, tmp);
System.out.println(sb);
count++;
}
System.out.println("total=" + count+" and should be " + n * (n - 1) / 2);
}