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Problem solution · Java

CCC 2008 S4 - Twenty-four

CCC 2008 S4 - Twenty-four: 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
140 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 S4 - Twenty-four, 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

  • 140 lines of Java from the credited upstream file ccc08s4.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 S4 - Twenty-four · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2008//// S4: TwentyFour//// This is PURE BRUTE FORCE. I'm sure someone will quickly// show me a better algorithm.//// the idea here is://    Generate all possible answers.//    look for biggest <= 24. done. :-)//// to generate all answers I do all possible operations// for example if the hand is: 2 6 8 10// then I go thru ALL permutations of two numbers and do// the possible operations, creating new hands of 3 numbers.//// In the above example I'd get://  8 8 10   (added 2+6)//  -4 8 10  (subtracted 2-6)//  12 8 10  (multiplied 2*6)  (couldn't divide 2/60//  10 6 10  (added 2+8)//  -6 6 10  (subtracted 2-8)//  ...//// in this way a hand of 4 become many hands of 3// then those hands of 3 become hands of 2 and then 1.// finally it looks for the biggest number <= 24.//// I calculated a max possible number of hands as://   1 hand of 4    (given)//  48 hands of 3   ( = 12 perumtations * 4 operations)// 1152 hands of 2  ( = 48 hands * 6 permutations * 4 operations)// 9216 hands of 1  ( = 1152 * 2 numbers * 4 operations)//// It ain't pretty, but it works! :-)//  import java.awt.*;import hsa.*; public class CCC2008S4TwentyFour{    static Console c;      // this takes a group of handSize numbers and combines 2 of the numbers    // using all operations in all possible ways, creating a new series    // of hands, with one fewer numbers in each.    static int doOperations (int[] hand, int handSize, int[] [] newHand, int k)    {	for (int i = 0 ; i < handSize ; i++)	    for (int j = 0 ; j < handSize ; j++)		if (i != j)		{		    // do +		    newHand [k] [0] = hand [i] + hand [j];		    // fill the hand with the other numbers		    int q = 1;		    for (int p = 0 ; p < handSize ; p++)			if (!(p == i || p == j))			    newHand [k] [q++] = hand [p];		    k++; 		    // do -		    newHand [k] [0] = hand [i] - hand [j];		    // fill the hand with the other numbers		    q = 1;		    for (int p = 0 ; p < handSize ; p++)			if (!(p == i || p == j))			    newHand [k] [q++] = hand [p];		    k++; 		    // do *		    newHand [k] [0] = hand [i] * hand [j];		    // fill the hand with the other numbers		    q = 1;		    for (int p = 0 ; p < handSize ; p++)			if (!(p == i || p == j))			    newHand [k] [q++] = hand [p];		    k++; 		    // do / if you can		    if (hand [j] != 0 && hand [i] % hand [j] == 0)		    {			newHand [k] [0] = hand [i] / hand [j];			// fill the hand with the other numbers			q = 1;			for (int p = 0 ; p < handSize ; p++)			    if (!(p == i || p == j))				newHand [k] [q++] = hand [p];			k++;		    }		}	return k;    }      public static void main (String[] args)    {	int n, size, newSize;	int[] [] hand, newHand;	c = new Console ();	TextInputFile f = new TextInputFile ("s4.5.in");	n = f.readInt ();	for (int i = 0 ; i < n ; i++)	{	    newHand = new int [1] [4];	    newHand [0] [0] = f.readInt ();	    newHand [0] [1] = f.readInt ();	    newHand [0] [2] = f.readInt ();	    newHand [0] [3] = f.readInt ();	    newSize = 1; 	    // generate all the hands, size 4 to 3, 3 to 2 and 2 to 1.	    for (int k = 4 ; k > 1 ; k--)	    {		size = newSize;		hand = newHand;		newHand = new int [10000] [3];		newSize = 0;		for (int j = 0 ; j < size ; j++)		    newSize = doOperations (hand [j], k, newHand, newSize);	    } 	    // find the biggest <= 24 and print it.	    size = newSize;	    int biggest = 0;	    for (int j = 0 ; j < size && biggest < 24 ; j++)		if (newHand [j] [0] > biggest && newHand [j] [0] <= 24)		    biggest = newHand [j] [0];	    c.println (biggest);	}     }}

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