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Problem solution · Java

CCC 2006 S4 - Groups

CCC 2006 S4 - Groups: 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
101 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 2006 S4 - Groups, 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

  • 101 lines of Java from the credited upstream file ccc06s4.java.
  • The implementation visibly relies on sequence storage.
  • 11 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 2006 S4 - Groups · JavaJava
Use this to learn the idea, then write your own version.
// CCC2006s4Groups//// this is a rather straight forward 2D array problem//// read and test for closure// find the identity, if there is one// test for inverses// test for associativity//// quit as soon as any test fails// import java.awt.*;import hsa.*; public class CCC2006s4Groups{    static Console c;     public static void main (String[] args)    {	c = new Console ();	TextInputFile f = new TextInputFile ("s4.1.in");	int[] [] g;	int n;	int identity;	boolean group, found; 	n = f.readInt ();	group = false;	while (n > 0)	{	    group = true;	    g = new int [n + 1] [n + 1]; 	    // step 1: read the group and test for closure	    for (int i = 1 ; i <= n && group ; i++)		for (int j = 1 ; j <= n && group ; j++)		{		    g [i] [j] = f.readInt ();		    if (g [i] [j] < 1 || g [i] [j] > n)			group = false;		} 	    // step 2: find indentity, if there is one	    // (one row must equal 1,2,3...) and its column must also	    //  equal 1,2,3...) 	    identity = 0;	    found = false;	    for (int i = 1 ; i <= n && group && identity == 0 ; i++)	    {		found = true;		int j;		for (j = 1 ; j <= n && found ; j++)		{		    if (g [i] [j] != j)			found = false;		}		if (found)		{ 		    identity = i;		    for (int k = 1 ; k <= n && found ; k++)			if (g [identity] [k] != k)			    found = false;		}	    }	    group = group && found; 	    // step 3: test for inverses	    // every row MUST have an identity value	    // and if its at i,j then j,i must also be the identity	    for (int i = 1 ; i <= n && group ; i++)	    {		found = false;		for (int j = 1 ; j <= n && !found ; j++)		    if (g [i] [j] == identity && g [j] [i] == identity)			found = true;		group = group && found;	    } 	    // step 4: test for associativity	    // test every combo...	    for (int i = 1 ; i <= n && group ; i++)		for (int j = 1 ; j <= n && group ; j++)		    for (int k = 1 ; k <= n && group ; k++)			if (g [g [i] [j]] [k] != g [i] [g [j] [k]])			    group = false; 	    if (group)		c.println ("yes");	    else		c.println ("no");	    n = f.readInt ();	}    }}

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