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May 99 Challenge

Volume Number: 15 (1999)
Issue Number: 5
Column Tag: Programmer's Challenge

May 99 Challenge

by Bob Boonstra, Westford, MA


Every once in a while, good fortune not only comes your way, but actually reaches out of your computer monitor, and grabs you by the throat. I felt a little like that while reading a recent issue of TidBITS. In it was a column by Rick Holzgrafe reflecting on the increasing speed of computers, in which Rick described a program he wrote for a PDP-11/60 to solve a word puzzle. The idea behind the puzzle was to take a phrase and map it onto a rectangular grid, with the objective being to map the phrase into a rectangle of the smallest possible area. The word puzzle looked like a good idea for a Challenge, and Rick and TidBITS agreed to let me use it.

In more detail, the puzzle works like this. To start, you are given a null-terminated string consisting of printable characters. You process the characters in order, ignoring any non-alphabetic characters and ignoring case. The first alphabetic character can be mapped to any square in the grid. The next letter can be mapped to any adjacent square, where adjacent is any of the eight neighboring squares in a horizontal, vertical, or diagonal direction. You may reuse a square if it is adjacent and already has the letter you are mapping. If the same letter occurs twice in a row in the input string, the letters must still be mapped to adjacent (but distinct) squares.

The prototype for the code you should write is:

#if defined(__cplusplus)
extern "C" {

typedef struct GridPoint {
   long x;
   long y;
} GridPoint;

void Piper (
   char *s,
   GridPoint pt[]

#if defined(__cplusplus)

For example, your Piper routine might be provided with the string:

            How much wood would a woodchuck chuck if a woodchuck 
            could chuck wood?

You might place the letters of that string into a 4x14 rectangle like:

                  ULD  ADLU
                K   WO

Or, you might compact them into an 4x4 rectangle:


You must return the GridPoint coordinates of where each character is mapped, with pt[i] containing the coordinates of input character s[i]. The origin of your coordinate system should be the cell where the first character is placed. The winner will be the solution that stores the input string in a rectangle of minimal area. Note that you are minimizing rectangle area, not the number of occupied squares. A time penalty of 1% for each second of execution time will be added

This will be a native PowerPC Challenge, using the latest CodeWarrior environment. Solutions may be coded in C, C++, or Pascal.

Three Months Ago Winner

The February Challenge invited readers to write a player for the game of Chinese Checkers. Played on a hexagonal board with six appended triangles, Chinese Checkers pits between 2 and 6 players against one another, with the objective being to move one's pieces from the home triangle to the opposite triangle. In the traditional game, the home triangles are usually 3 or 4 positions on a side; the Challenge extended the game to triangles of up to 64 positions. Pieces can either be moved to an immediately adjacent position or jumped over an adjacent piece. A single piece is permitted to make a sequence of jumps in a single move.

As simple as the game sounds, readers found it to be very difficult, so difficult that no solutions were submitted for the Chinese Checkers Challenge. Which left Yours Truly in something of a difficult spot. I could stop the column at this point, which wouldn't be very interesting for readers, not to mention not very satisfying for the magazine. Or I could write a solution for the Challenge myself, something I haven't done since I retired from Competition four plus years ago. Somewhat against my better judgement, I selected the latter option.

The first thing I noticed in solving the Challenge was that the board coordinate system specified in the problem wasn't very useful in generating a solution. I needed a coordinate system that could be easily rotated in 60-degree increments, enabling my solution to play any of the six possible player positions. After some thought, I came up with a more symmetric coordinate system, called CanonPosition in the code, along with conversion routines ConvertPositionToCanonPosition and ConvertCanonPositionToPosition. The commentary in the code illustrates the coordinate system for a board of size 4. Then I needed a way to evaluate board positions. I decided on a simple metric that summed the distances of all pieces from the apex of their goal triangle. That metric could be improved upon by taking possible jump moves into account. The solution starts by computing all possible moves for our player. It then tries each of those moves, and then recursively calls MakeNextMove to process the next player. It computes and tries all moves for that player, and recurses for the next player. Recursion terminates when kMaxPlys turns have been taken for all players. Positions are evaluated using a min-max technique, where each player selects the position that maximizes his position relative to the best position of the other players.

The code could be improved in many ways. Instead of trying all possible positions, it could prune some obviously bad moves in the backward direction. This is complicated by the fact that some good jump multi-moves can include individual jumps that appear to be moving backward. The code might also be improved by evaluating moves by progressive deepening, rather than the depth-first search currently used, and by ordering move evaluation based on the scores at the prior depth. This technique is used in chess programs to prune the move tree to a manageable size. These and other optimizations are left to the reader. :-)

Remember, you can't win if you don't play. To ensure that you have as much time as possible to solve the Challenge, subscribe to the Programmer's Challenge mailing list. To subscribe, see the Challenge web page at <>. The Challenge is sent to the list around the 12th of the month before the solutions are due, often in advance of when the physical magazine is delivered.

Here is our sample Chinese Checkers solution:

Chinese Checkers.c
Copyright © 1999 J. Robert Boonstra

*  Example solution for the February 1999 Programmer's Challenge
 *  This solution is provided because no solutions were submitted
 *  for the ChineseCheckers Challenge.  This solution leaves a 
 *  great deal to be desired: it is not optimized, it does not 
 *  prune prospective moves efficiently, and it does not employ
 *  any of the classic alpha-beta techniques for efficiently
 *  selecting a move.

#include <stdio.h>
#include <stdlib.h>
#include <Quickdraw.h>
#include "ChineseCheckers.h"

/* Position coordinates specified in problem (size==4)

  0:                           0
  1:                         0   1 
  2:                      -1   0   1 
  3:                    -1   0   1   2
  4:  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6 
  5:    -5  -4  -3  -2  -1   0   1   2   3   4   5   6
  6:      -5  -4  -3  -2  -1   0   1   2   3   4   5  
  7:        -4  -3  -2  -1   0   1   2   3   4   5
  8:          -4  -3  -2  -1   0   1   2   3   4  
  9:        -4  -3  -2  -1   0   1   2   3   4   5
 10:      -5  -4  -3  -2  -1   0   1   2   3   4   5  
 11:    -5  -4  -3  -2  -1   0   1   2   3   4   5   6
 12:  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6 
 13:                    -1   0   1   2
 14:                      -1   0   1 
 15:                         0   1 
 16:                           0

#define kMaxPlys 1
#define kEmpty -1

/* CanonPosition uses units with (0,0) at the middle of the board */
typedef struct CanonPosition {
   long row;
   long col;
} CanonPosition;

/* Canonical position coordinates (size==4)

 -8:                           0
 -7:                        -1   1 
 -6:                      -2   0   2 
 -5:                    -3  -1   1   3
 -4: -12 -10  -8  -6  -4  -2   0   2   4   6   8  10  12 
 -3:   -11  -9  -7  -5  -3  -1   1   3   5   7   9  11
 -2:     -10  -8  -6  -4  -2   0   2   4   6   8  10  
 -1:        -9  -7  -5  -3  -1   1   3   5   7   9
  0:          -8  -6  -4  -2   0   2   4   6   8  
  1:        -9  -7  -5  -3  -1   1   3   5   7   9
  2:     -10  -8  -6  -4  -2   0   2   4   6   8  10  
  3:   -11  -9  -7  -5  -3  -1   1   3   5   7   9  11
  4: -12 -10  -8  -6  -4  -2   0   2   4   6   8  10  12 
  5:                    -3  -1   1   3
  6:                      -2   0   2 
  7:                        -1   1 
  8:                           0

/* PlayerPos is used to store the location of each of a players pieces */
typedef struct PlayerPos {
   CanonPosition pos;
} PlayerPos;

typedef char *CanonBoard;

static long myNumPlayers,myNumPieces,myIndex,myGameSize,
static PlayerPos *myPositions;

/* global board */
static CanonBoard myBoard;

/* moveIncrement is added to a CanonPosition to find the adjacent square in the
 * six possible directions 0..6, with 0==horizontal right, 1==down right, ...
 * 5==up right
static CanonPosition moveIncrement[6] = 
   { { 0, 2}, { 1, 1}, { 1,-1}, { 0,-2}, {-1,-1}, {-1, 1} };

/* macros to access the board */
#define CanonRowSize(size) (6*(size)+1)
#define CanonBoardPos(row,col) (3*(myGameSize) +    \
         (2*(myGameSize)+(row))*(CanonRowSize(myGameSize)) + (col))
#define CanonBoardVal(board,row,col)          \
#define IsEmpty(board,row,col)             \
         (kEmpty == board[CanonBoardPos(row,col)])

/* Convert coordinates from problem statement to canonical coordinates */
static CanonPosition ConvertPositionToCanonPosition (
            Position *pos, long size) {
   CanonPosition canon;
   canon.row = pos->row - 2 * size;
   if (pos->row == 2 * (int)(pos->row/2)) {
      canon.col = 2 * pos->col;
   } else {
      canon.col = 2 * pos->col - 1;
   return canon;

/* Convert canonical coordinates to coordinates from problem statement */
static Position ConvertCanonPositionToPosition (CanonPosition *canon, long size) {
   Position pos;
   pos.row = canon->row + 2 * size;
   if (canon->row == 2 * (int)(canon->row/2)) {
      pos.col = canon->col / 2;
   } else {
      pos.col = (canon->col + 1) / 2;
   return pos;

/* rotate board by posNum increments of player positions (60 degrees) */
static CanonPosition RotateCanonPosition0ToN(
         CanonPosition oldPos, long posNum) {
   CanonPosition newPos;
   while (posNum<0) posNum+=6;   /* normalize to 0..5 */
   while (posNum>5) posNum-=6;
   switch (posNum) {
   case 0:
      newPos.row = oldPos.row;
      newPos.col = oldPos.col;
   case 1:
      newPos.row = (oldPos.row + oldPos.col)/2;
      newPos.col = (oldPos.col - 3*oldPos.row)/2;
   case 2:
      newPos.row =  (oldPos.col - oldPos.row)/2;
      newPos.col = -(oldPos.col + 3*oldPos.row)/2;
   case 3:
      newPos.row = -oldPos.row;
      newPos.col = -oldPos.col;
   case 4:
      newPos.row = -(oldPos.row + oldPos.col)/2;
      newPos.col = -(oldPos.col - 3*oldPos.row)/2;
   case 5:
      newPos.row = -(oldPos.col - oldPos.row)/2;
      newPos.col =  (oldPos.col + 3*oldPos.row)/2;
   return newPos;

/* return the max column number in a given row */
static long MaxColInRow(long row, long size) {
   long maxCol;
   if (row<-size) {
      maxCol = row+2*size;
   } else if (row<0) {
      maxCol = 2*size-row;
   } else if (row<=size) {
      maxCol = row+2*size;
   } else /* if (row<=2*size) */ {
      maxCol = 2*size-row;
   return maxCol;

/* determine if a row,col coordinate represents a legal position */
static Boolean IsLegalPosition(CanonPosition *pos) {
   long maxCol;
   if ((pos->row<-2*myGameSize) || (pos->row>2*myGameSize)) 
         return false;
   if ((pos->row + pos->col) != 
               2 * (int)((pos->row + pos->col)/2)) 
      return false;
   maxCol = MaxColInRow(pos->row,myGameSize);
   if ((pos->col<-maxCol) || (pos->col>maxCol)) 
      return false;
   return true;

/* move a piece between positions from and to, does not check legality of move */
static void MoveFromTo(CanonBoard b,CanonPosition *from,CanonPosition *to,long newValue) {
   PlayerPos *p = &myPositions[newValue*myNumPieces];
   long oldValue = CanonBoardVal(b,from->row,from->col);
   int i;
   if (oldValue != newValue) {
      DebugStr("\p check err");
   if ( IsLegalPosition(from) && IsLegalPosition(to) ) {
      CanonBoardVal(b,from->row,from->col) = kEmpty;
      CanonBoardVal(b,to->row,to->col) = (char)newValue;
      for (i=0; i<myNumPieces; i++) {
         if (    (p[i].pos.row == from->row) && 
                     (p[i].pos.col == from->col)) {
            p[i].pos.row = to->row;
            p[i].pos.col = to->col;

/* return the distance of a given position from a goal postion for player 0 */
static long PositionDistFromGoal (const CanonPosition *a, const CanonPosition *goal) {
   long rowDelta,colDelta;
   rowDelta = a->row - goal->row;
   if (rowDelta<0) rowDelta = -rowDelta;
   colDelta = a->col - goal->col;
   if (colDelta<0) colDelta = -colDelta;
   if (rowDelta>=colDelta) return rowDelta;
   else return rowDelta + (colDelta-rowDelta)/2;

/* return the cumulative distance of a player from his goal postion */
static long PlayerDistFromGoal(long player) {
   long cumDist;
   int i;
   CanonPosition goal;
   PlayerPos *p = &myPositions[player*myNumPieces];
   goal.row = -2*myGameSize;
   goal.col = 0;
   goal = 
   for (i=0, cumDist=0; i<myNumPieces; i++) {
      CanonPosition *cp = &p[i].pos;
      long dist = PositionDistFromGoal(cp,&goal);
      cumDist += dist;
   return cumDist;

/* initialize the positions for a player at a given position */
static void InitPlayer(char *b,PlayerPos *piecePositions, long player, long position,long size) {
   CanonPosition pos,newPos;
   int col,maxCol,pieceCount;
   PlayerPos *piecePos;
   pieceCount = 0;
   for (maxCol=0; maxCol<size; maxCol++) {
      pos.row = -2*size+maxCol;
      maxCol = maxCol;
      for (col=-maxCol; col<=maxCol; col+=2) {
         pos.col = col;
         newPos = RotateCanonPosition0ToN(pos,position);
         CanonBoardVal(b,newPos.row,newPos.col) = (char)player;
   piecePos = &piecePositions[myNumPieces*player + pieceCount];
         piecePos->pos = newPos;

/* some variables to record the history of multi-jump moves, to
   prevent them from repeating infinitely */
static CanonPosition gMoveHistoryPos[6*64];
static long gMoveHistoryDirection[6*64];
static long gMoveHistoryCtr = -1;

/* Calculate a move for a given piece for a given player in a given moveDir.
 * Return true there is a legal move.
 * Return doneWithThisDirection==true if there are no more moves in this direction.
static Boolean CalcMove(
         CanonBoard b, long player, long pieceNum, long moveDir, 
         CanonPosition *p1, CanonPosition *p2, 
         Boolean *doneWithThisDirection, long iterationLimit) {
   Boolean legalMove = true;
   if (gMoveHistoryCtr<0) {
      PlayerPos *p = &myPositions[player*myNumPieces];
      *p2 = *p1 = p[pieceNum].pos;
      p2->row += moveIncrement[moveDir].row;
      p2->col += moveIncrement[moveDir].col;
      if (!IsLegalPosition(p2)) legalMove = false;
      else if (IsEmpty(b,p2->row,p2->col)) {
      } else {
         long oldVal = CanonBoardVal(b,p2->row,p2->col);
         p2->row += moveIncrement[moveDir].row;
         p2->col += moveIncrement[moveDir].col;
         if (!IsLegalPosition(p2)) legalMove = false;
         else if (IsEmpty(b,p2->row,p2->col)) {
            if (gMoveHistoryCtr>iterationLimit) 
                        DebugStr("\p limit exceeded");
            gMoveHistoryPos[++gMoveHistoryCtr] = *p1;
            gMoveHistoryDirection[gMoveHistoryCtr] = 6;
            gMoveHistoryPos[++gMoveHistoryCtr] = *p2;
            gMoveHistoryDirection[gMoveHistoryCtr] = -1;
         } else {
            legalMove = false;
   } else {
      CanonPosition pStart,pTemp;
      long newDir;
      pStart = pTemp = gMoveHistoryPos[gMoveHistoryCtr];
      while (   (gMoveHistoryCtr>=0) && 
                     (gMoveHistoryDirection[gMoveHistoryCtr]>=6) ) {
      if (gMoveHistoryCtr>0) {
         newDir = gMoveHistoryDirection[gMoveHistoryCtr];
         pTemp.row += moveIncrement[newDir].row;
         pTemp.col += moveIncrement[newDir].col;
         if (!IsLegalPosition(&pTemp)) legalMove=false;
         else if (IsEmpty(b,pTemp.row,pTemp.col)) legalMove=false;
         else {
            pTemp.row += moveIncrement[newDir].row;
            pTemp.col += moveIncrement[newDir].col;
            if (!IsLegalPosition(&pTemp)) legalMove=false;
            else if (!IsEmpty(b,pTemp.row,pTemp.col)) legalMove=false;
            else {
               int i;
               for (i=0; i<=gMoveHistoryCtr; i++)
                  if (   (pTemp.row == gMoveHistoryPos[i].row) && 
                           (pTemp.col == gMoveHistoryPos[i].col) )
               if (legalMove) {
                  gMoveHistoryDirection[++gMoveHistoryCtr] = -1;
                  gMoveHistoryPos[gMoveHistoryCtr] = pTemp;
                  *p1 = gMoveHistoryPos[0];
                  *p2 = pTemp;
      } else {
   *doneWithThisDirection = (gMoveHistoryCtr<0);
   return legalMove;

/* multiplier to determine how much storage to reserves for moves for each piece */
#define kMemAllocFudge 12

static long EnumerateMoves(CanonBoard b, long player, CanonPosition moveFrom[], CanonPosition moveTo[]) {
   long numMoves = 0;
   int piece,moveDir;
   Boolean legalMove,doneWithThisDirection;
   for (piece=0; piece<myNumPieces; piece++) {
      int pieceCounter=0;
      int firstPieceMoveIndex = numMoves;
      moveDir = 0;
      do {
         legalMove = CalcMove(b,player,piece,moveDir,
         if (doneWithThisDirection)
         if (!legalMove) continue;
         if (numMoves>=kMemAllocFudge*myNumPieces-1) {
            DebugStr("\pnumMoves limit exceeded");
         } else {
            int i;
            for (i=firstPieceMoveIndex; i<numMoves; i++)
               if ( (moveTo[i].row==moveTo[numMoves].row) && 
                   (moveTo[i].col==moveTo[numMoves].col) )
                     legalMove = false;
            if (!legalMove) continue;
      } while (moveDir<6);
   return numMoves;

 * Recursive routine to explore move tree.
 * MakeNextMove iterates over all possible moves for a player.
 * If ply limit is not yet reached, it recurses by calling for the next player.
 * Ply limit is decreased by 1 when the "me" player is called.
 * Recursion terminates when ply limit is reached.
 * Score is assigned on return based on the perspective of the player making the move.
 * Simple-minded score algorithm is used: 
 *   difference between player score and the best other player score, where
 *   a player's score is the number of spaces he is away from the final state
 * No alpha-beta pruning is employed - search is exhaustive.

static long MakeNextMove(CanonBoard b, long me, long player, long playerDistances[6], long numPlys,
   Boolean firstTime, CanonPosition *from, CanonPosition *to) {
   long theMove,nextPlayer,numMoves,
               bestScore=0x7FFFFFFF, myBestDistance=0x7FFFFFFF;
   CanonPosition pFrom,pTo,bestFrom,bestTo;
   int newPlys;
   CanonPosition *moveFrom,*moveTo;

   /* allocate storage for possible moves */
   moveFrom = (CanonPosition *)
   if (0==moveFrom) 
         DebugStr("\pproblem allocation moveFrom memory");
   moveTo = (CanonPosition *)
   if (0==moveTo) 
         DebugStr("\pproblem allocation moveTo memory");
   /* prime best move with starting move */
   bestFrom = *from;
   bestTo = *to;
   /* enumerate all legal moves for this player */
   numMoves = EnumerateMoves(b,player,moveFrom,moveTo);
   /* examine all of the enumerated moves */
   for (theMove=0; theMove<numMoves; theMove++) {
      long opponent,scoreDifference,minOpponentDistance,
      int thePlayer;
      pFrom = moveFrom[theMove];
      pTo = moveTo[theMove];
      if (firstTime) {
         *from = pFrom;
         *to = pTo;
      nextPlayer = (player+1)%myNumPlayers;
      newPlys = (player==me) ? numPlys-1 : numPlys;
      /* record move in the simulated board */
      myDistance = PlayerDistFromGoal(player);
      /* recurse if ply limit not reached */
      if ( (newPlys>=0) && (myDistance>myMinDist) ){
         /* MakeNextMove returns each player's distance from the goal in 
            returnedDistances, and the score from nextPlayer's perspective in 
            returnScore is ignored except by nonrecursive callers to MakeNextMove 
         long returnScore;
         returnScore = 
               MakeNextMove(b, me, nextPlayer, returnedDistances, 
                           newPlys, false, from, to);
      } else /*if (player==me)*/ {
         /* terminating recursion, calculate position values for each player */
         /* compute distances for all players */
         for (thePlayer=0; thePlayer<myNumPlayers; thePlayer++)
            returnedDistances[thePlayer] = 
      /* compute best opponent score from this player perspective */
      for (thePlayer=0,minOpponentDistance=0x7fffffff; 
                     thePlayer<myNumPlayers; thePlayer++) {
         if (   (thePlayer != player) && 
            (returnedDistances[thePlayer]<minOpponentDistance) ) 
            minOpponentDistance = returnedDistances[thePlayer];
      scoreDifference = 
      /* Save score if it is the best for this player.
         This move is best if
         (1) our distance from goal minus best opponents distance is smallest, or
         (2) goal distance difference is equal, but our absolute distance is smallest, or
         (3) goal distance difference is equal, and coin flip says pick this move 
                                          (commented out) */
      if ( (scoreDifference < bestScore) ||
          ((scoreDifference==bestScore) && 
                  (returnedDistances[player]<myBestDistance)) /*||
((scoreDifference==bestScore) && ((rand()&0x0080)==1))*/ ) {
         bestScore = scoreDifference;
         myBestDistance = returnedDistances[player];
         for (opponent=0; opponent<myNumPlayers; opponent++)
      playerDistances[opponent] = returnedDistances[opponent];
         bestFrom = pFrom;
         bestTo = pTo;
      /* reverse move to clear board for mext move */
   /* free dynamically allocated move storage */
   /* return best move */
   *from = bestFrom;
   *to = bestTo;

   return bestScore;

/* find best move for player me from this position on the board */
static long FindBestMove(CanonBoard b, long me, long numPlys, CanonPosition *from, CanonPosition *to) {
   long playerDistances[6];

   return MakeNextMove(b,me,me,playerDistances,numPlys,true,from,to);

void InitChineseCheckers(
   long numPlayers,      /* 2..6  */
   long gameSize, /* base of home triangle, 3..63, you have size*(size+1)/2 pieces */
   long playerPosition[6],   /* 0..5, entries 0..numPlayers-1 are valid */
   long yourIndex /* 0..numPlayers-1, your position is playerPosition[yourIndex] */
) {
   int i,numPositions;
   /* allocate memory for board */
   numPositions = 6*(1+gameSize)*4*(1+gameSize);
   myBoard = (char *)malloc(numPositions*sizeof(char));
   if (myBoard==0) DebugStr("\p could not allocate board");
   myNumPieces = gameSize*(gameSize+1)/2;
   myPositions = (PlayerPos *)
   if (myPositions==0) 
               DebugStr("\p could not allocate myPositions");

   /* copy parameters */
   for (i=0;i<6; i++) myPlayerPosition[i] = playerPosition[i];
   myIndex = yourIndex;
   myNumPlayers = numPlayers;
   myGameSize = gameSize;
   /* initialize board */
   for (i=0; i<numPositions; i++) myBoard[i] = kEmpty;
   for (i=0; i<numPlayers; i++) {
   /* calculate distance metric at goal position */
   for (i=1, myMinDist=0; i<gameSize; i++) myMinDist += i*(i+1);

void YourMove(
   Position *fromPos,   /* originating position */
   Position *toPos   /* destination position */
) {
   CanonPosition from,to;
   long numPlys = kMaxPlys;
   *fromPos = ConvertCanonPositionToPosition(&from,myGameSize);
   *toPos = ConvertCanonPositionToPosition(&to,myGameSize);

void OpponentMove(
   long opponent,   /* index in playerPosition[] of the player making move */
   Position fromPos,   /* originating position */
   Position toPos      /* destination position */
) {
   CanonPosition from,to;
   from = ConvertPositionToCanonPosition(&fromPos,myGameSize);
   to =   ConvertPositionToCanonPosition(&toPos,myGameSize);

void TermChineseCheckers(void) {
   free (myPositions);
   free (myBoard);

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Square Rave is an awesome little music-oriented puzzle game that smacks of games like Lumines, but with its own unique sense of gameplay. To help wrap your head around the game, keep the following tips and tricks in mind. [Read more] | Read more »
Snowboard Party 2 (Games)
Snowboard Party 2 1.0 Device: iOS Universal Category: Games Price: $1.99, Version: 1.0 (iTunes) Description: Crowned the best snowboarding game available on the market, Snowboard Party is back to fulfill all your adrenaline needs in... | Read more »
One Button Travel (Games)
One Button Travel 1.0 Device: iOS Universal Category: Games Price: $2.99, Version: 1.0 (iTunes) Description: “To cut a long story short, If you like interactive fiction, just go buy this one.” - “Oozes the polish that... | Read more »

Price Scanner via

Holiday weekend Mac sales roundup: B&H Ph...
B&H Photo continues to have all new Macs on sale for up to $500 off MSRP as part of their Black Friday/Holiday weekend sale. Shipping is free, and B&H charges NY tax only: - 15″ 2.2GHz Retina... Read more
iMobie Releases its Ace iOS Cleaner PhoneClea...
iMobie Inc. has announced the new update of PhoneClean 4, its iOS cleaner designed to reclaim wasted space on iPhone/iPad for use and keep the device fast. Alongside, iMobie hosts a 3-day giveaway of... Read more
U.S. Cellular Offering iPad Pro
U.S. Cellular today announced that it is offering the new iPad Pro with Wi-Fi + Cellular, featuring a 12.9-inch Retina display with 5.6 million pixels — the most ever in an iOS device. U.S. Cellular... Read more
Newegg Canada Unveils Black Friday Deals for...
Newegg Canada is offering more than 1,000 deep discounts to Canadian customers this Black Friday, available now through Cyber Monday, with new deals posted throughout the week. “Black Friday is... Read more
Black Friday: Macs on sale for up to $500 off...
BLACK FRIDAY B&H Photo has all new Macs on sale for up to $500 off MSRP as part of their early Black Friday sale including free shipping plus NY sales tax only: - 15″ 2.2GHz Retina MacBook Pro: $... Read more
Black Friday: Up to $125 off iPad Air 2s at B...
BLACK FRIDAY Walmart has the 16GB iPad Air 2 WiFi on sale for $100 off MSRP on their online store. Choose free shipping or free local store pickup (if available): - 16GB iPad Air 2 WiFi: $399, save $... Read more
Black Friday: iPad mini 4s on sale for $100 o...
BLACK FRIDAY Best Buy has iPad mini 4s on sale for $100 off MSRP on their online store for Black Friday. Choose free shipping or free local store pickup (if available): - 16GB iPad mini 4 WiFi: $299.... Read more
Black Friday: Apple Watch for up to $100 off...
BLACK FRIDAY Apple resellers are offering discounts and bundles with the purchase of an Apple Watch this Black Friday. Below is a roundup of the deals being offered by authorized Watch resellers:... Read more
Black Friday: Target offers 6th Generation iP...
BLACK FRIDAY Save $40 to $60 on a 6th generation iPod touch at Target with free shipping or free local store pickup (if available). Sale prices for online orders only, in-store prices may vary: -... Read more
Black Friday: Walmart and Target offer iPod n...
BLACK FRIDAY Walmart has the 16GB iPod nano (various colors) on sale for $119.20 on their online store for a limited time. That’s $30 off MSRP. Choose free shipping or free local store pickup (if... Read more

Jobs Board

*Apple* Site Security Manager - Apple (Unite...
# Apple Site Security Manager Job Number: 42975010 Culver City, Califo ia, United States Posted: Oct. 2, 2015 Weekly Hours: 40.00 **Job Summary** The Apple Site Read more
WiSE *Apple* Pay Quality Engineer - Apple (...
# WiSE Apple Pay Quality Engineer Job Number: 44313381 Santa Clara Valley, Califo ia, United States Posted: Nov. 13, 2015 Weekly Hours: 40.00 **Job Summary** Join our Read more
Holiday Retail Associate with *Apple* Knowl...
…and assertive.Someone who can troubleshoot iOS devices (iPhone and iPad) and Apple Mail issues.Someone who can offer solutions.Someone who can work weekends.Someone with Read more
*Apple* Systems Engineer (Mclean, VA and NYC...
Summary:Assist in providing strategic direction and technical leadership within the Apple portfolio, including desktops, laptops, and printing environment. This person Read more
Simply Mac *Apple* Specialist- Service Repa...
Simply Mac is the largest premier retailer of Apple products in the nation. In order to support our growing customer base, we are currently looking for a driven Read more
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