CPMagical CP Library

Problem solution · Java

CCC 2008 S5 - Nukit

CCC 2008 S5 - Nukit: 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
127 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 2008 S5 - Nukit, 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

  • 127 lines of Java from the credited upstream file ccc08s5.java.
  • The implementation visibly relies on sequence storage.
  • 10 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 2008 S5 - Nukit · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2008//// S5: Nukit//// This is Dynamic Programming approach//// Four people, independently, sent me this same algorithm// Konstantin Lopyrev, Amlesh Jayakumar, Matthew Lai and one other. // This is Konstantin's version.//// The idea is to have a 4D array representing all// possible positions, and determine if they are winning// positions or not. This is created first. Then when  // a starting position is read, you can just look up the// answer. //// The array is called winningPosition and is filled with // false to begin with. // SO, (and this is key.) //       We assume all positions are loosing ones.// (that's the first set of loops in main)//// All that remains to to find the winning ones. // and that's easy://       a winning Position is one which has a move//                        which leads to a loosingPosition.// (that's the second set of loops in the main)//// That's it, except for the method loosingPosition.// This method was tough to wrap my head around.// "A loosing position is one which is NOT a winning one." // (and by the way, if the given position is an impossible position// its not a loosing position.) //// The problem I have with this is: I'm trying to determine the value// of winningPosition, which calls the loosingPosition method// and it uses winningPosition to determine that...// That's circular reasoning!!! But its not.   // As in all DP methods, we work from the ground up so to speak. // // Look at it another way. The lowest positions like 0,0,0,0 are // loosingPositions, Because if you apply any rule to it, it gives// negatives and the loosingPosition method for that will return false and // the value of winningPosition[0][0][0][0] retains unchanged, at false.// So the "can't move" positions are essentially taken care of with// the if in loosingPosition() and the initial assumption// that all positions are loosing ones. //// Later on as you move up to a bigger position, that has moves to be made,// if you are looking at some move a,b,c,d then the call to loosingPosition// will involve SMALLER values, (a lower position), that is, one that// we already figured out, for sure, is a winning or loosing position,// so we CAN reliably return a value that we can use to establish what// the original a,b,c,d is. QED.//// if that still doesn't make sense, sorry.  I tried :-)// (This DOES work!)// import java.awt.*;import hsa.*; public class CCC2008S5NukitDP{    static Console cc;     static boolean[] [] [] [] winningPosition;    static int[] [] moves = {{2, 1, 0, 2}, {1, 1, 1, 1}, {0, 0, 2, 1}, {0, 3, 0, 0}, {1, 0, 0, 1}};      // its a loosing position if its NOT a winning position :-)    // The "if" is there to handle the "can't move" situations, the false it    // returns is not to be taken as "this is a winning position",    // but as "the position that lead to this, IS a loosing position."    static boolean loosingPosition (int a, int b, int c, int d)    {	if (a < 0 || b < 0 || c < 0 || d < 0)	    return false;	else	    return !winningPosition [a] [b] [c] [d];    }      public static void main (String[] args)    {	int n, a, b, c, d;	cc = new Console ();	TextInputFile f = new TextInputFile ("s5.4.in"); 	winningPosition = new boolean [31] [31] [31] [31];		// Step 1: Assume all positions are loosing	for (int i = 0 ; i < 31 ; i++)	    for (int j = 0 ; j < 31 ; j++)		for (int k = 0 ; k < 31 ; k++)		    for (int l = 0 ; l < 31 ; l++)			winningPosition [i] [j] [k] [l] = false; 	// Identify all winning positions. A winning position is	// one which has at least one move which leads to a	// loosing position.	for (int i = 0 ; i < 31 ; i++)	    for (int j = 0 ; j < 31 ; j++)		for (int k = 0 ; k < 31 ; k++)		    for (int l = 0 ; l < 31 ; l++)			for (int m = 0 ; m < 5 ; m++)			    if (loosingPosition (i - moves [m] [0], j - moves [m] [1], k - moves [m] [2], l - moves [m] [3]))				winningPosition [i] [j] [k] [l] = true; 	n = f.readInt ();	for (int i = 0 ; i < n ; i++)	{	    a = f.readInt ();	    b = f.readInt ();	    c = f.readInt ();	    d = f.readInt ();	    if (winningPosition [a] [b] [c] [d])		cc.println ("Patrick");	    else		cc.println ("Roland");	}    }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗