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July 01 Challenge

Volume Number: 17 (2001)
Issue Number: 07
Column Tag: Programmer's Challenge

by Bob Boonstra, Westford, MA

Down-N-Out

George Warner earns two Challenge points for suggesting another interesting board game, this time the solitaire game known as Down-N-Out. The Down-N-Out board is a 10x30 rectangular array of cells, initially populated randomly with 100 cells of each of three colors. The object is to score as many points as possible by removing cells from the board. A cell can be removed if it is adjacent (horizontally or vertically, but not diagonally) with a cell of the same color. When a cell is removed, all cells connected to it by transitive adjacency are removed. That is, all adjacent cells are removed, and all cells adjacent to those cells, etc. The number of points earned for each move is equal to the square of the number of cells removed (e.g., 2 cells = 4 points, 3 cells = 9 points, etc.). There is an obvious advantage to planning moves that maximize the number of connected cells removed simultaneously. After each move, the board is compacted by sliding all cells downward to fill any empty cells, and then by sliding all columns to the center to fill any empty columns. The game continues as long as cells can be removed.

For those of you interested in trying the game, there is a shareware version available at
http://www.peciva.com/software/downout.shtml.

The prototype for the code you should write is:

typedef char CellColor;   /* 0==empty, 1..numColors are valid colors */

void InitDownNOut(
   short boardSizeRows,   /* number of rows in the game */
   short boardSizeCols,   /* number of columns in the game */
   short numColors,         /* number of colors in the game */
   WindowPtr wdw    /* window where results of your moves should be displayed */
);

void HandleUpdateEvent(EventRecord theEvent);

Boolean /* able to play       */ PlayOneDownNOutMove(
   CellColor board[],   /* board[row*boardSizeCols + col] is color of cell at [row][col] */
   long score,                  /* points earned prior to this move */
   short *moveRow,            /* return row of your next move */
   short *moveCol            /* return col of your next move */
);

void TermDownNOut(void);

Each game begins with a call to your InitDownNOut routine, where you are given the dimensions of the game board (boardSizeRows and boardSizeCols), the number of colors in the game (numColors), and a pointer (gameWindow) to a WindowRecord where you must display the game state as it progresses. Finally, you will be given the initial state of the game board, fully populated with equal numbers of each color cell, subject to rounding limitations. InitDownNOut should allocate any dynamic memory needed by your solution, and that memory should be returned at the end of the game when your TermDownNOut routine is called.

Your PlayOneDownNOutMove routine will be called repeatedly, once for each move you make. You will be given your current point score as calculated by the test code and the state of the game board. You should determine the most advantageous move and return it in moveRow and moveCol. You should update the game board, eliminating cells removed by your move and compacting the board vertically and then horizontally. You should calculate the number of cells removed and return it in numberOfCellsRemoved.

The last time we ran a Challenge that involved maintaining a display, contestants asked how the window would be redrawn in response to an update event. This time, I'm asking you to write a routine to do that. Your HandleUpdateEvent routine will be called by the test code whenever an update event is received for your gameWindow.

During the call to InitDownNOut, and after each of your moves, you should display the updated game state in the gameWindow. The details of the display are up to you, as long as the display correctly and completely represents the state of the board.

The winner will be the best scoring entry, as determined by the sum of the point score of each game, minus a penalty of 1% for each millisecond of execution time used for that game. The Challenge prize will be divided between the overall winner and the best scoring entry from a contestant that has not won the Challenge recently.

This will be a native PowerPC Challenge, using the CodeWarrior Pro 6 environment. Solutions may be coded in C or C++. I've deleted Pascal from the list of permissible languages, both because it isn't supported by CW6 (without heroics) and because no one has submitted a Pascal solution in a long time.

Three Months Ago Winner

Congratulations to Ernst Munter (Kanata, Ontario, Canada) for submitting the best scoring solution in the April Crossword II Challenge. This Challenge was inspired by a classroom exercise to construct a 20x20 crossword puzzle using the names of the elements in the periodic table, valuing each word according to the atomic number of the corresponding element, with the objective of maximizing the total value of the puzzle. We generalized the problem by making the word list, word values, and puzzle size parameters of the problem. And to incorporate the usual emphasis on efficiency, we penalized each test case by 1% for each minute of execution time required to generate the puzzle.

The winning solution starts by assigning a strength value to each word in the word list. The strength of a word is a scaled version of the value assigned by the problem input, divided by the length of a word. This heuristic favors shorter words of a given value over longer words of the same value. Then the Board::Solve routine tries to place words until a time limit (set to 15 seconds) expires or there are no more valid moves to explore. The moves are attempted in order of decreasing value, where the value of a move is the assigned value of word being placed, divided by the length of the word minus the number of letters that intersect other words. Again, this gives priority to placement of shorter high value words over longer ones, and to placements that efficiently use the board space by intersecting other words.

I evaluated the four entries received using a set of ten test cases ranging in size from 20x20 to 50x50. Ernst's solution packed 10% more word value into his puzzles than the second place entry by Ron Nepsund, taking significantly more execution time as well. For the original 20x20 problem based on the periodic table, Ernst's entry produced the following crossword, valued at 2470 points:

LAWRENCIUM__XENON_P_
_S__E______B______L_
_T___O_____AMERICIUM
_ACTINIUM__R______T_
_T______E__I___ARGON
SILVER_ERBIUM___H_N_
_N______C__M____E_I_
_E__BISMUTH_POLONIUM
________R__G____I_M_
_F_C____Y__O_R__U___
_E_A_C_____LEADM_U__
_RADIUM__T_D_D_B__R_
_M_M_R___H___O_E__A_
_IRIDIUM_O_TIN_R__N_
_U_U_U___R_____K__I_
_M_M_M_I_I__NOBELIUM
_______R_U_____L__M_
CERIUM_OSMIUM__ZINC_
_______N________U___
HAFNIUM__FRANCIUM___

Ernst would have won by an even wider margin were it not for an ambiguity in the problem statement. The problem specified that each word could only occur once in the puzzle, and that each sequence of letters in the puzzle had to form a word. What I meant to say, however, but didn't, was that each word in the puzzle needed to be distinct. Two of the contestants took advantage of this loophole, for example, to claim credit for the word "tin" embedded in the longer word "actinium". Fortunately for my sense of fairness if nothing else, when I ran the tests both allowing and not allowing the loophole, the scores were such that the ranking of the entries was unchanged. The results as presented reflect the actual wording of the puzzle, and allow a word to be embedded in another word.

As the best-placing entry from someone who has not won a Challenge in the past two years, Ron Nepsund wins a share of this month's Challenge prize. You don't need to defeat the Challenge points leaders to claim a part of the prize, so enter the Challenge and win Developer Depot credits!

The table below lists, for each of the solutions submitted, the number of points earned by each entry, and the total time in seconds. It also lists the code size, data size, and programming language used for each entry. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges prior to this one.

Name Points Time(secs) Code Size
Ernst Munter(731) 52378 151.4 3940
Ron Nepsund(47) 47489 6.7 44520
Jan Schotsman(7) 44888 32.9 12708
Ken Slezak(26) [LATE] 4927937.4 3160
Name Data Size Lang
Ernst Munter 174 C++
Ron Nepsund 6530 C++
Jan Schotsman 448 C++
Ken Slezak 47 C++

Top Contestants...

Listed here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated 20 or more points during the past two years. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants, the number of wins over the past 24 months, and the total number of career Challenge points.

Rank Name Points(24 mo)
1. Munter, Ernst 304
2. Rieken, Willeke 83
3. Saxton, Tom 76
4. Taylor, Jonathan 56
5. Shearer, Rob 55
6. Wihlborg, Claes 49
7. Maurer, Sebastian 48
Name Wins(24 mo) Total Points
Munter, Ernst 12 751
Rieken, Willeke 3 134
Saxton, Tom 2 185
Taylor, Jonathan 2 56
Shearer, Rob 1 62
Wihlborg, Claes 2 49
Maurer, Sebastian 1 108

...and the Top Contestants Looking for a Recent Win

In order to give some recognition to other participants in the Challenge, we also list the high scores for contestants who have accumulated points without taking first place in a Challenge during the past two years. Listed here are all of those contestants who have accumulated 6 or more points during the past two years.

Rank Name Points Points
(24 mo) Total
8. Boring, Randy 32 142
9. Schotsman, Jan 14 14
10. Sadetsky, Gregory 12 14
11. Nepsund, Ronald 10 57
12. Day, Mark 10 30
13. Jones, Dennis 10 22
14. Downs, Andrew 10 12
15. Duga, Brady 10 10
16. Fazekas, Miklos 10 10
17. Flowers, Sue 10 10
18. Strout, Joe 10 10
19. Nicolle, Ludovic 7 55
20. Hala, Ladislav 7 7
21. Miller, Mike 7 7
22. Widyatama, Yudhi 7 7
23. Heithcock, JG 6 43

There are three ways to earn points: (1) scoring in the top 5 of any Challenge, (2) being the first person to find a bug in a published winning solution or, (3) being the first person to suggest a Challenge that I use. The points you can win are:

1st place 20 points
2nd place 10 points
3rd place 7 points
4th place 4 points
5th place 2 points
finding bug 2 points
suggesting Challenge 2 points

Here is Ernst's winning CrosswordII solution:

CrosswordII.cp
Copyright © 2001
Ernst Munter, Kanata, ON, Canada


To avoid any significant point penalty (of 1% per minute), processing stops
after 15 seconds.

A private copy of the puzzle is built where each cell is an unsigned character,
with value of 0, c, or 2*c.  An empty cell is 0, an placing a word is done
by adding each of the word’s character into the corresponding cell.  Similarly,
removal of a word is done with subtraction.

*/
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <Events.h>
#include “CrosswordII.h”

typedef unsigned long ulong;
typedef unsigned short ushort;
typedef unsigned char uchar;

static int N=0;

enum {
   kDown   = 0,
   kAcross   = 1,
   kMaxMoves = 5,
   kTicksPerSecond = 60,
   kMaxSeconds   = 15
};

struct MyWord
struct MyWord
// Encapsulation of Words
{
   const Words* word;
   ulong   length;
   ulong    strength;// length-relative value
   bool   used;
   MyWord(){}
   MyWord(const Words* wp) :
      word(wp),
      length(strlen(wp->theWord)),
      strength((0x10000L*wp->value)/(1+length)),
      used(false)
   {}
   const Words* Word() const {return word;}
   const char* Chars() const {return word->theWord;}
   long Value() const {return word->value;}
   int Length() const {return length;}
   bool IsAvailable() const {return !used;} 
   void SetUsed() {used = true;}
   void ClearUsed() {used = false;}
   ulong Strength() const {return strength;}
}; 

static int CmpWord(const void* a,const void* b)
{
   MyWord* ap=(MyWord*)a;
   MyWord* bp=(MyWord*)b;
   return bp->strength - ap->strength;
}

struct MyMove
// A Move is a placement of a word
{
   MyWord* w;
   ulong   value;
   ushort   row;
   ushort   col;
   ushort   delta;
   ushort   size;
   ulong    Value() const {return value;}
   ulong   Points() const {return w->word->value;}
   void    Init(int numIntersects,MyWord* wx,
                              int r,int c,int d,int s)
   {
      w=wx;
      value=(0x10000 * w->word->value) / 
                              (1+w->Length()-numIntersects);
      row=r;
      col=c;
      delta=d;
      size=s;
   }
   void    Clear() {value=0;}
   ulong    IsValid() const {return value;}// != 0
   void Convert(const Words* words,WordPositions* p)
   // Converts this instance of “MyMove” to a “WordPosition” as defined in 
   // “CrosswordII.h”
   {
      p->whichWord=w->word-words;
      p->row=row;
      p->col=col;
      p->orientation=(delta==1)?kAcross:kDown;
   }
   void RemoveWord(char* puzzle)
   {
      char* p=puzzle+row*size+col;
      char* str=w->word->theWord;
      for (int i=0;i<w->Length();i++)
      {
         *p -= *str++;
         p+=delta;
      }
      w->ClearUsed();
   }
   void PlaceWord(char* puzzle)
   {
      char* p=puzzle+row*size+col;
      char* str=w->word->theWord;
      for (int i=0;i<w->Length();i++)
      {
         *p += *str++;
         p+=delta;
      }
      w->SetUsed();
   }
   int IntersectAcross(MyWord* w,int r,int c,
                     char* puzzle,int puzzleSize)
   {
// returns -1(no fit), 0 (fit, no intersects) or n>0 (n intersects with other words)
      
      // insertion point p
      char* p=puzzle+r*puzzleSize+c;
      int len=w->Length();
      
      // cell before the word must be a border or blank
      char* rowStart=puzzle+r*puzzleSize;
      char* cellBefore=p-1;
      if ((cellBefore >= rowStart) && (0 != *cellBefore)) 
         return -1;
         
      // cell after the word must be a border or blank
      char* rowEnd=rowStart+puzzleSize;
      char* cellAfter=p+len;
      if ((cellAfter < rowEnd) && (0 != *cellAfter)) 
         return -1;
         
      // all cells to the side of the word must be
      //      (a) either blank
      //      (b) or part of a crossing word   
      //   we know case b applies only if the cell the current word is
      //   to occupy is already occupied - with a letter equal to str[x]
      
      char* str=w->word->theWord;
      char* puzzleEnd=puzzle+puzzleSize*puzzleSize;
      int numIntersects=0;
      for (int i=0;i<len;i++,str++,p++)
      {
         if (*p == 0)// crossing a blank
         {
            // cell above must be outside border, or blank
            char* cellAbove=p-puzzleSize;
            if ((cellAbove >= puzzle) && (0 != *cellAbove)) 
               return -1;
            // cell below must be outside border, or blank
            char* cellBelow=p+puzzleSize;
            if ((cellBelow < puzzleEnd) && (0 != *cellBelow)) 
               return -1;
         } else if (*p == *str)// crossing a word, matching
         {
            numIntersects++;
         } else   // crossing, but no match
         {
            return -1;
         }
      }
      Init(numIntersects,w,r,c,1,puzzleSize);
      return numIntersects;
   }
   int IntersectDown(MyWord* w,int r,int c,
                        char* puzzle,int puzzleSize)
   {
// returns -1(no fit), 0 (fit, no intersects) or n>0 (n intersects with other words)
            
      // insertion point p
      char* p=puzzle+r*puzzleSize+c;
      int len=w->Length();
      
      // cell before the word must be a border or blank
      char* colStart=puzzle+c;
      char* cellBefore=p-puzzleSize;
      if ((cellBefore >= colStart) && (0 != *cellBefore)) 
         return -1;
         
      // cell after the word must be a border or blank
      char* bottomBorder=puzzle+puzzleSize*puzzleSize;
      char* cellAfter=p+len*puzzleSize;
      if ((cellAfter < bottomBorder) && (0 != *cellAfter)) 
         return -1;
         
      // all cells to the side of the word must be
      //      (a) either blank
      //      (b) or part of a crossing word   
      //   we know case b applies only if the cell the current str would
      //   occupy is already occupied - with a letter equal to str[x]
      
      char* str=w->word->theWord;
      char* puzzleEnd=puzzle+puzzleSize*puzzleSize;
      char* leftEdge=puzzle+r*puzzleSize;
      int numIntersects=0;
      for (int i=0; i<len; 
                  i++,str++,p+=puzzleSize,leftEdge+=puzzleSize)
      {
         if (*p == 0)// crossing a blank
         {
            // cell on the left must be the left border, or blank
            char* cellLeft=p-1;
            if ((cellLeft >= leftEdge) && (0 != *cellLeft)) 
               return -1;
            // cell on right must be on the right edge, or blank
            char* cellRight=p+1;
            char* rightEdge=leftEdge+puzzleSize;
            if ((cellRight < rightEdge) && (0 != *cellRight)) 
               return -1;
         } else if (*p == *str)// crossing a word, matching
         {
            numIntersects++;
         } else   // crossing, but no match
         {
            return -1;
         }
      }
      Init(numIntersects,w,r,c,puzzleSize,puzzleSize);
      return numIntersects;
   }
};

typedef MyMove* MyMovePtr;

inline bool operator > (const MyMove & a,const MyMove & b) 
{
   return a.Value() > b.Value();
}

struct MyMoveArray
struct MyMoveArray
{
   int numMoves;
   int maxMoves;
   MyMove   moves[kMaxMoves];
   MyMoveArray(int max) :
      numMoves(0),
      maxMoves(max)
   {}
   int NumMoves() const {return numMoves;}
   MyMove* Moves() {return moves;} 
   void Insert(MyMove & m)
   {
      if (numMoves==0)
      {
         numMoves=1;
         moves[0]=m;
      } else if (numMoves<maxMoves)
      {
         MyMove* mx=moves+numMoves;
         while ((mx>moves) && (*(mx-1) > m))
         {
            *mx=*(mx-1);
            mx=mx-1;
         }   
         *mx=m;
         numMoves++;
      } else if (m > moves[numMoves-1])
      {
         numMoves—;
         Insert(m);
      }
   }
};

struct Board
struct Board
{
   long   puzzleSize;
   char*    puzzle;
   long   numWords;
   MyWord* myWords;
   long   numPositions;
   WordPositions* bestPositions;
   
   MyMove*      movePool;   //   single pool allocated for movelists
   MyMove*     endMovePool;   
   MyMovePtr*   moveStack;   //   move stack tracks the history of executed moves
   MyMovePtr*   moveStackPointer;
   MyMovePtr*   lastMoveStack;
   
   Board(long pSize,const Words* words,long nWords) :
      puzzleSize(pSize),
      puzzle(new char[(pSize)*(pSize)]),
      numWords(nWords),
      myWords(new MyWord[nWords]),
      numPositions(0),
      bestPositions(new WordPositions[numWords]),
      
      
      movePool(new MyMove[numWords*kMaxMoves]),
      endMovePool(movePool+numWords*kMaxMoves),
      moveStack(new MyMovePtr[numWords]),moveStackPointer(moveStack),
      lastMoveStack(moveStack+numWords-1)
      
   {
      for (long i=0;i<numWords;i++)
         myWords[i]=MyWord(words+i);
      
// sort words by strength
      qsort(myWords,numWords,sizeof(MyWord),CmpWord);

// remove all 0-value words      
      long i=numWords;
      while ((i>0) && (myWords[i-1].Value()<=0))
         i=i-1;
         
      numWords=i; 
   }
   ~Board()
   {
      delete [] bestPositions;
      delete [] myWords;
      delete [] puzzle;
   }
   void Clear() 
   {
      memset(puzzle,0,sizeof(char)*(puzzleSize)*(puzzleSize));
   }
   int Solve(const Words* words,WordPositions* positions);
   
   void SetPosition(const Words* words,MyWord* w,WordPositions* pos,
      int row,int col,int o)
   {
      pos->whichWord=w->Word()-words;
      pos->row=row;
      pos->col=col;
      pos->orientation=o; 
   }
   
   void PushMove(MyMove* mp){
      *moveStackPointer++=mp;
   }
   
   MyMove* PopMove()
   {
      return *—moveStackPointer;
   } 
   
   MyMove* GenerateMoveList(MyMove* mp)
   {
//   Lists all legal moves in a list, starting with a null-move;
//   sorts the moves and returns the highest value move on the list 
//   Each move is given a “value” reflecting its relative merit. 
      if (mp+kMaxMoves >= endMovePool)             
         return 0; // no room for movelist, should not really happen
                 // but if it does, we just have to backtrack   
      MyMove m;
      int i,row,col,maxRow,maxCol,drow,dcol;
         
// create moves
      MyMoveArray ma(kMaxMoves);
      
      MyWord* w=myWords;
      ulong bestStrength=0;
      for (i=0;i<numWords;i++,w++)
      {
         if (!w->IsAvailable()) continue;
         ulong strength=w->Strength();
         if (strength < bestStrength) continue;
         maxCol=maxRow=puzzleSize-w->Length();
//   find every legal position
         drow=0;
         for (row=puzzleSize/2;(row>=0)&&(row<puzzleSize);row+=drow)
         {
            dcol=0;
            for (col=puzzleSize/2;(col>=0) && (col<puzzleSize);col+=dcol)
            {
               if ((col<=maxCol) &&
                  (m.IntersectAcross(w,row,col,puzzle,puzzleSize)>=0))
               { 
                  ma.Insert(m);
                  bestStrength=strength;
               }
               
               if ((row<=maxRow) &&
                  (m.IntersectDown(w,row,col,puzzle,puzzleSize)>=0))
               { 
                  ma.Insert(m); 
                  bestStrength=strength;
               }
               
               if (dcol>=0) dcol=-1-dcol; else dcol=1-dcol;
            }
            if (drow>=0) drow=-1-drow; else drow=1-drow;
         }
      }
      
// put a sentinel 0-move at the start of the movelist
      mp->Clear();
// copy moves from the moves array into the movelist space
      MyMove* mx=ma.Moves();
      for (int i=0;i<ma.NumMoves();i++)
         *(++mp) = *mx++;
      
      return mp;
   }
      
   long Execute(MyMove* mp)
   {
      mp->PlaceWord(puzzle);
      PushMove(mp);
      return mp->Points();   
   }
   
   MyMove* Undo(long & points)
// Undoes the last stacked move, returns this move, or 0 if no move found   
   {
      MyMove* mp=PopMove();
      if (mp==0) return mp;
      mp->RemoveWord(puzzle);
      points -= mp->Points();
      return mp;
   }
   
   long CopyMovesBack(const Words* words,WordPositions* positions)
//    Scans the movestack, converts MyMoves to positions.
//   Returns the number of positions   
   {
      int numMoves=0;
      for (MyMovePtr* index=moveStack+1;index<moveStackPointer;index++)
      {
         MyMove* mp=*index;
         mp->Convert(words,positions+numMoves);
         numMoves++;
      }
      return numMoves;
   }
};

Board::Solve
int Board::Solve(const Words* words,WordPositions* positions)
{
   WordPositions* pos=positions;
   long numPositions=0;
   long bestPoints=0;
   long start=TickCount();
   
   Clear();
   moveStackPointer=moveStack;   
   // Put a sentinel null move at start of move stack      
   PushMove(0);
   MyMove* moveList=movePool;
   long points=0;
         
   MyMove* nextMove=GenerateMoveList(moveList);
   // moveList to nextMove defines a movelist which always starts with a 0-move
   // and is processed in order nextMove, nextMove-1, ... until 0-move is found
   if (!nextMove)
      return 0;
      
   for (;;) 
   {
      while (nextMove && nextMove->IsValid())
      {
         points+=Execute(nextMove);
         if (points > bestPoints)
         {
            bestPoints=points;
            numPositions=CopyMovesBack(words,positions);
         } 
         moveList=1+nextMove;
         long numTicks=TickCount()-start;
         if (numTicks>kMaxSeconds*kTicksPerSecond)
            break;
         nextMove=GenerateMoveList(moveList);
                     
      } // end while
      
      do {
         MyMove* prevMove=Undo(points);
         if (!prevMove)  // stack is completely unwound, exhausted
            break;
               
      // try to use the last move:
         nextMove = prevMove-1;
      } while (!nextMove->IsValid());
            
      moveList=nextMove;
      while ((moveList>=movePool) && (moveList->IsValid()))
         moveList—;
         
      if (moveList<=movePool)
         break;
   }   
   return numPositions;   
}

CrosswordII
short /* numberOfWordPositions */ CrosswordII  (
   short puzzleSize,            /* puzzle has puzzleSize rows and columns */
   const Words words[],      /* words to be used to form the puzzle */
   short numWords,               /* number of words[] available */
   WordPositions positions[]   /* placement of words in puzzle */
) {
   if (numWords <= 0)
      return 0;
      
   Board B(puzzleSize,words,numWords);
   
   long numberOfWordPositions=B.Solve(words,positions);
      
   return numberOfWordPositions;
}


 
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Viber lets you send free messages and make free calls to other Viber users, on any device and network, in any country! Viber syncs your contacts, messages and call history with your mobile device,... Read more
Apple iOS 7.1.1 - The latest version of...
The latest version of iOS can be downloaded through iTunes. Apple iOS 7 brings an all-new design and all-new features. Simplicity Simplicity is often equated with minimalism. Yet true simplicity is... Read more
1Password 4.3 - Powerful password manage...
1Password is a password manager that uniquely brings you both security and convenience. It is the only program that provides anti-phishing protection and goes beyond password management by adding Web... Read more
Lens Blur 1.3.0 - True out-of-focus boke...
Let Lens Blur transform your existing photo into true SLR-quality out-of-focus bokeh effect! Everyone needs a gorgeous personalized background for a social profile, blog, Web/UI design, presentation... Read more
VMware Fusion 6.0.3 - Run Windows apps a...
VMware Fusion allows you to create a Virtual Machine on your Mac and run Windows (including Windows 8.1) and Windows software on your Mac. Run your favorite Windows applications alongside Mac... Read more
BitTorrent Sync 1.3.93 - Sync files secu...
BitTorrent Sync allows you to sync unlimited files between your own devices, or share a folder with friends and family to automatically sync anything. File transfers are encrypted. Your information... Read more

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Spin It Review
Spin It Review By Jordan Minor on April 24th, 2014 Our Rating: :: SPIN ME RIGHT ROUNDUniversal App - Designed for iPhone and iPad Spin It has a fine puzzle game model, but its execution lacks energy.   | Read more »
The Sandbox Gets Update, Receives New Ca...
The Sandbox Gets Update, Receives New Campaign and New Elements Posted by Tre Lawrence on April 24th, 2014 [ permalink ] Universal App - Designed for iPhone and iPad | Read more »
Football Management Simulator One For El...
Football Management Simulator One For Eleven Released Worldwide Today for iOS Posted by Simon Reed on April 24th, 2014 [ permalink ] Free-To-Play football management title One For E | Read more »
Leo’s Fortune Review
Leo’s Fortune Review By Jordan Minor on April 24th, 2014 Our Rating: :: FORTUNATE SONUniversal App - Designed for iPhone and iPad Leo’s Fortune delivers a platforming experience as creative and refined as any console game.   | Read more »
Suited Up (Games)
Suited Up 1.0 Device: iOS Universal Category: Games Price: $1.99, Version: 1.0 (iTunes) Description: Suited Up is a difficult, one-touch platformer that requires players to visualize each jump. The controls in Suited Up are simple,... | Read more »
MyTP One Mountain - Ski, Freeski and Sno...
MyTP One Mountain - Ski, Freeski and Snowboard 1.0.0 Device: iOS Universal Category: Games Price: $.99, Version: 1.0.0 (iTunes) Description: As real snow is melting away in the snow parks around the northern hemisphere, it's now time... | Read more »
Tank Battle: East Front 1943 (Games)
Tank Battle: East Front 1943 1.0 Device: iOS Universal Category: Games Price: $1.99, Version: 1.0 (iTunes) Description: Tank Battle: East Front 1943 is the third in the successful ‘Tank Battle: East Front’ series of games for iPhone... | Read more »
Third Eye Crime: Act 1 (Games)
Third Eye Crime: Act 1 1.0 Device: iOS Universal Category: Games Price: $2.99, Version: 1.0 (iTunes) Description: "You have to see Third Eye Crime" Touch Arcade "Third Eye Crime is unlike anything currently available for mobile... | Read more »
Wayward Souls (Games)
Wayward Souls 1.00 Device: iOS Universal Category: Games Price: $4.99, Version: 1.00 (iTunes) Description: **Buy the game now at an introductory sale price of $4.99 USD. Every time we do a big content update, we will raise the price... | Read more »
Leo's Fortune (Games)
Leo's Fortune 1.0.2 Device: iOS Universal Category: Games Price: $4.99, Version: 1.0.2 (iTunes) Description: Leo’s Fortune is a platform adventure game where you hunt down the cunning and mysterious thief that stole your gold.... | Read more »

Price Scanner via MacPrices.net

16GB 1st generation iPad mini available for $...
Radio Shack has a select number of refurbished 1st generation 16GB WiFi iPad minis available for $199.99 on their online store. Choose free shipping or free ship-to-store. We expect these to sell out... Read more
13-inch 2.5GHz MacBook Pro on sale for $100 o...
B&H Photo has the 13″ 2.5GHz MacBook Pro on sale for $1099 including free shipping plus NY sales tax only. Their price is $100 off MSRP. Read more
iPad Sales “Lull” A Reality Correction Of Unm...
I have lots of time for Jean-Louis Gassée, the former Apple Computer executive (1981 to 1990) who succeeded Steve Jobs as head of Macintosh development when the latter was dismissed in 1985. Mr.... Read more
Apple Makes OS X Betas Available To All – Wit...
Apple’s OS X Beta Seed Program, which lets you install the latest pre-release builds, try it out, and submit your feedback, is now open to anyone who wants to sign on rather than to developers and... Read more
Apple Releases iOS 7.1.1 Update
The latest iOS 7.1.1 update contains improvements, bug fixes and security updates, including: • Further improvements to Touch ID fingerprint recognition • Fixes a bug that could impact keyboard... Read more
Logitech Announces Thinner, Lighter, More Fle...
Logitech has announced an update to its Ultrathin for iPad Air, iPad mini and iPad mini with Retina display, improving the flexibility and design of its award-winning predecessor with an even thinner... Read more
Logitech Introduces Hinge, Big Bang and Turna...
Logitech has announced expansion of its tablet product line with three new cases – the Logitech Hinge, the Logitech Big Bang and the Logitech Turnaround – each for the iPad Air, iPad mini and iPad... Read more
WaterField’s Rough Rider Leather Messenger Ba...
WaterField Designs have announced the new 15-inch size of their popular Rough Rider leather messenger bag, a vintage-looking bag that combines Old West charm and ruggedness with distinctly modern... Read more
New Mac Pro on sale, save $100 on the 4-Core...
J&R has the new 4-Core Mac Pro in stock today and on sale for $2899 including free shipping plus NY sales tax only. Their price is $100 off MSRP, and it’s the lowest price available for this... Read more
Apple refurbished iMacs available for up to $...
The Apple Store has Apple Certified Refurbished 2013 iMacs available for up to $300 off the cost of new models. Apple’s one-year warranty is standard, and shipping is free. - 27″ 3.4GHz iMac – $1699... Read more

Jobs Board

*Apple* Solutions Consultant (ASC) - Apple (...
**Job Summary** The ASC is an Apple employee who serves as an Apple brand ambassador and influencer in a Reseller's store. The ASC's role is to grow Apple Read more
Position Opening at *Apple* - Apple (United...
…customers purchase our products, you're the one who helps them get more out of their new Apple technology. Your day in the Apple Store is filled with a range of Read more
*Apple* Solutions Consultant (ASC) - Apple (...
**Job Summary** The ASC is an Apple employee who serves as an Apple brand ambassador and influencer in a Reseller's store. The ASC's role is to grow Apple Read more
*Apple* Inc. Research Data Specialist - Appl...
…of Worldwide Market Research & Intelligence. The team is responsible for conducting Apple branded consumer market research. It is also responsible for analyzing data Read more
*Apple* Automotive Parts Department position...
Apple Automotive is one of the fastest growing dealer…and it shows. Consider making the switch to the Apple Automotive Group today! At Apple Automotive, we Read more
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