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CCC 2009 S3 - Degrees of Separation

CCC 2009 S3 - Degrees of Separation: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
201 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2009 S3 - Degrees of Separation, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 201 lines of Java from the credited upstream file ccc09s3.java.
  • The implementation visibly relies on sequence storage.
  • 14 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2009 S3 - Degrees of Separation · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2009//// J5/S3: Degrees of Separation//// this is relatively stright forward, except for the "s" command// which requires Warshall's Algorithm for transitive closure.//// a square 2D array holds the relationships. I think a triangular array// would have worked, but then I'd have to always order the two// inputs. So keeping it square and always doing 2 things (to represent// the dual nature of the connection) does simplify some things.////  import java.awt.*;import hsa.*; public class CCC2009S3DegreesofSeparation{     public static void main (String[] args)    {	Console c;	int[] [] g; 	c = new Console ();	TextInputFile f = new TextInputFile ("s3.4.in");  	// initialize the grid	g = new int [50] [50];	for (int i = 0 ; i < 50 ; i++)	    for (int j = 0 ; j < 50 ; j++)		g [i] [j] = 0;	g [1] [6] = 1;	g [6] [1] = 1;	g [2] [6] = 1;	g [6] [2] = 1;	g [3] [6] = 1;	g [6] [3] = 1;	g [4] [6] = 1;	g [6] [4] = 1;	g [5] [6] = 1;	g [6] [5] = 1;	g [7] [6] = 1;	g [6] [7] = 1;	g [3] [4] = 1;	g [4] [3] = 1;	g [4] [5] = 1;	g [5] [4] = 1;	g [3] [5] = 1;	g [5] [3] = 1;	g [3] [15] = 1;	g [15] [3] = 1;	g [13] [15] = 1;	g [15] [13] = 1;	g [14] [13] = 1;	g [13] [14] = 1;	g [12] [13] = 1;	g [13] [12] = 1;	g [7] [8] = 1;	g [8] [7] = 1;	g [8] [9] = 1;	g [9] [8] = 1;	g [9] [10] = 1;	g [10] [9] = 1;	g [9] [12] = 1;	g [12] [9] = 1;	g [10] [11] = 1;	g [11] [10] = 1;	g [11] [12] = 1;	g [12] [11] = 1;	g [16] [17] = 1;	g [17] [16] = 1;	g [16] [18] = 1;	g [18] [16] = 1;	g [18] [17] = 1;	g [17] [18] = 1;  	// main processing loop	char command;	int x, y;	command = f.readChar ();	while (command != 'q')	{	    if (command == 'i')	    {		x = f.readInt ();		y = f.readInt ();		g [x] [y] = 1;		g [y] [x] = 1;	    }	    else if (command == 'd')	    {		x = f.readInt ();		y = f.readInt ();		g [x] [y] = 0;		g [y] [x] = 0;	    }	    else if (command == 'n')	    {		x = f.readInt ();		int count = 0;		for (int i = 0 ; i < 50 ; i++)		    if (g [x] [i] == 1)			count++;		c.println (count);	    }	    else if (command == 'f')	    {		x = f.readInt ();		int count = 0;		count = friendofFriends (g, x);		c.println (count);	    }	    else if (command == 's')	    {		x = f.readInt ();		y = f.readInt ();		int count = 0;		count = shortestPath (g, x, y);		if (count == 999)		    c.println ("Not connected");		else		    c.println (count);	    }	    command = f.readChar ();	}    }      // Friends of friends    // If you can get from x to i and from i to j    // then j is a friend of a friend of x.    // (assuming you couldn't get directly from x to j,  and x not = j)    // The original array is copied to a temp array q,    // so as not to mess it up for furture commands.    public static int friendofFriends (int[] [] g, int x)    {	int[] [] q = new int [50] [50];	int count = 0;	for (int i = 0 ; i < 50 ; i++)	    for (int j = 0 ; j < 50 ; j++)		q [i] [j] = g [i] [j]; 	for (int i = 0 ; i < 50 ; i++)	    if (q [x] [i] == 1)		for (int j = 0 ; j < 50 ; j++)		    if (q [i] [j] == 1 && j != x && q [x] [j] == 0)			q [x] [j] = 2; 	for (int i = 0 ; i < 50 ; i++)	    if (q [x] [i] == 2)		count++;	return count;    }      // Degree of Separation: i.e. shortest path!    // This is Warshall's algorithm.     // (Prehaps a bit of overkill, but it works quickly)    // The temp array is filled with 999 to indicate no connection,    // if there is no direct connection.    // The main loop works on the idea:    // if you can get from i to j and from j to k    // then you can get from i to k.    // We are interested in the shortest path, so if    // i to j takes 3 moves and j to k takes 5 moves    // then i to k takes 8 moves, assuming it already had a value    // greater than that. (Don't forget to set both directions!)    public static int shortestPath (int[] [] g, int x, int y)    {	int[] [] q = new int [50] [50];	int count = 0;	for (int i = 0 ; i < 50 ; i++)	    for (int j = 0 ; j < 50 ; j++)		if (g [i] [j] == 1)		    q [i] [j] = g [i] [j];		else		    q [i] [j] = 999; 	for (int i = 0 ; i < 50 ; i++)	    for (int j = 0 ; j < 50 ; j++)		if (q [i] [j] > 0)		    for (int k = 0 ; k < 50 ; k++)			if ((q [j] [k] > 0) && (q [i] [j] + q [j] [k] < q [i] [k]))			{			    q [i] [k] = q [i] [j] + q [j] [k];			    q [k] [i] = q [i] [j] + q [j] [k];			} 	return q [x] [y];    }   }

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