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Aug 98 Prog Challenge

Volume Number: 14 (1998)
Issue Number: 8
Column Tag: Programmer's Challenge

August 1998 Programmers Challenge

by by Bob Boonstra, Westford, MA

Block Buster

One of the Vice Presidents at the Real Job (we'll call him Don) has a collection of unusual puzzles at his conference table. The kind that require you to move a large block through a small hole, untangle a set of intertwined metal loops, do the impossible with rope, or otherwise do something that requires mastery of the fourth spatial dimension or a greater aptitude toward right-brain thinking than I possess. Don retired this week, and, in his honor, this month's Challenge is devoted to a spatial reasoning exercise. Not a puzzle that would have been worthy of his conference table, perhaps, but something in that same spirit.

Our puzzle is a variation of the Soma Cube, reportedly conceived by a writer from Denmark named Peter Hein as he was listening to a lecture on quantum physics. (That's one class I don't remember being able to daydream through!) The object of the Soma Cube is to form a 3x3 cube using seven shapes formed from three or four of the smaller cubes:

Of course, the Soma can be assembled into other shapes besides a cube, which forms the basis for our Challenge. Your job will be to write code that will assemble a larger number of potentially larger shapes into an even larger designated shape.

The prototype for the code you should write is:

#if defined(__cplusplus)
extern "C" {

#define kUnknown 0xFFFF

typedef struct Cubie {
   SInt16   xCoord;  /* x coordinate of cubie */
   SInt16   yCoord;  /* y coordinate of cubie */
   SInt16   zCoord;  /* z coordinate of cubie */
   UInt16   value;   /* ordinal value assigned to the Piece this cubie is part of */
} Cubie;

typedef struct Piece {
   UInt32   numCubies;    /* number of individual cubies in this piece */
   Cubie   **theCubies;   /* pointer to array of cubies in this piece */
} Piece;

Boolean BlockBuster(   
   /* return TRUE if you were able to solve the puzzle */
   long numPieces,      /* number of pieces to assemble */
   Piece thePieces[],   /* the pieces to assemble */
   Piece theGoal        /* the structure you should assemble */
      /* set (*theGoal.theCubies)[i].value to ((*thePiece->theCubies)+j)->value
         if piece j occupies cubie i in the reassembled puzzle */

#if defined(__cplusplus)
extern "C" {

The building block for our puzzle is the Piece, which is formed from numCubies individual cubes, provided to you in an array pointed to by theCubies. Each Piece has a unique nonzero identifier, which is associated with the value field in the Cubie structure. Cubies also have x, y, and z coordinates that define their relative orientation toward one another. There are no restrictions on the size or shape of a Piece, except that the constituent Cubies will all be connected (i.e., adjacent to another Cubie in the same Piece in x, y, or z.

Your BlockBuster routine is given an array (thePieces) of numPieces Pieces to work with. It is also provided with theGoal, an assembly of Cubies that you are to create from thePieces. For convenience, theGoal is also described using the Piece data structure. On input, the occupied coordinates are assigned a value of kUnknown. On output you should replace that value with the value of the Piece that occupies that coordinate. You may rotate and translate thePieces as you desire, but you may not break them by changing the relative orientation of the Cubies. You may use each Piece only once in assembling theGoal shape. All puzzles will be solvable, but if you feel you cannot solve a puzzle, BlockBuster should return FALSE.

As an example, the Pieces in the standard Soma Cube might be described as follows (with each 4-tuple representing the x, y, z, and value components of a cubie:

   {0,0,1, 5}, {1,0,1, 5}, {0,1,1, 5}, {0,0,0, 5},
   {0,0,0, 1}, {1,0,0, 1}, {0,1,0, 1}, 
   {0,0,0, 3}, {1,0,0, 3}, {2,0,0, 3}, {0,1,0, 3},
   {0,0,0, 7}, {1,1,1, 7}, {0,1,0, 7}, {0,1,1, 7},
   {0,0,0, 6}, {1,0,0, 6}, {0,1,0, 6}, {0,1,1, 6},
   {0,0,0, 4}, {1,0,0, 4}, {2,1,0, 4}, {1,1,0, 4},
   {0,0,0, 2}, {1,0,0, 2}, {2,0,0, 2}, {1,1,0, 2},

The winner will be the solution that assembles theGoal shape in the minimum amount of time. There are no storage constraints for this Challenge, except that it must execute on my 96MB 8500/200.

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

Sometimes Challenge solutions are very close to one another in performance, and other times the margin of victory is overwhelming. This month the winner falls into the latter category. Congratulations to Ernst Munter (Kanata, Ontario) for submitting an entry that soundly defeated the other four Challengers in the May Boggle contest. The objective was to score the most points by rearranging a Boggle® puzzle to form high scoring words, while at the same time minimizing the time penalty imposed for execution time. While Ernst took an order of magnitude more time to generate a solution, he found nearly four times as many unique words as his closest competitor, resulting in by far the highest score.

As you might suspect by looking at relative execution times, Ernst's strategy was a comparatively sophisticated one that he calls "simulated annealing". He begins by eliminating from the dictionary all words that cannot be formed by the letters available in the given instance of the puzzle. He then computes the frequency with which possible letter pairs occur in the dictionary, and uses this computation to exhaustively rearrange the puzzle letters and maximize what he calls "energy", the sum of the weights of the letter pairs in the rearranged puzzle. Only then does Ernst identify which dictionary words are present in the puzzle and compute a score. Optimization is done by randomly exchanging letter pairs until the expected gain for continuing exceeds the estimated penalty for continuing, after which the solution terminates.

One of the other solutions started by finding the longest (and therefore highest scoring) word in the dictionary that could be formed by the given letters and rearranging the puzzle to form that word. Another solution divided the board squares into three categories: corners, edges, and middle squares. It then rearranged the board so that "preferred" letters, determined a priori, were moved to the middle, and unfavored letters were moved to the corners.

There were 24 test cases used to evaluate this Challenge, 6 cases for each puzzle size (4, 5, 6, and 7). The table below lists, for each of the five solutions submitted, the cumulative number of words found, the number of those words that were unique, the associated points earned for the unique words, the time penalty (in points), and the net score. It also includes code size, data size, and programming language used. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges to date prior to this one.

Name Lang Total Words Unique Words Points Time Penalty Score Code Data
Ernst Munter (364) C++ 11560 11560 39932 5191 34741 5840 8
JG Heithcock (10) C 5055 3037 3823 7 3816 1688 148
Tim Gogolin C 2368 3275 3275 47 3228 3456 412
Randy Boring (77) C 3154 1976 2828 4 2824 2836 944
Eric Hangstefer (9) C 1858 1858 2594 477 2117 3832 68

Top Contestants

Here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated more than 10 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.

  1. Munter, Ernst 210
  2. Boring, Randy 74
  3. Cooper, Greg 54
  4. Mallett, Jeff 50
  5. Rieken, Willeke 47
  6. Nicolle, Ludovic 34
  7. Lewis, Peter 31
  8. Maurer, Sebastian 30
  9. Gregg, Xan 24
  10. Murphy, ACC 24
  11. Hart, Alan 21
  12. Antoniewicz, Andy 20
  13. Day, Mark 20
  14. Heithcock, JG 20
  15. Hostetter, Mat 20
  16. Studer, Thomas 20

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 solution to the Boggle Challenge:


Copyright © 1998, Ernst Munter

The Problem

A boggle board (Boggle®, a Parker Brothers game) shows a 2-D array of letters. The task is to find as many words as possible which can be traced on the puzzle, by going from letter to adjacent letter without using the same letter twice. Actually, it is the point count that should be maximized, where longer words count significantly higher than short words.

As a variant of the original game, an extra degree of freedom is available: the board may be rearranged to maximize the point count.

Solution Strategy

Finding all words that can be made on the given puzzle is fairly straight forward, with a scan of the dictionary. The real challenge is to find a way to rearrange the board efficiently. This seems to be a very hard problem to do exactly, that is to find the absolutely best arrangement.

I settled for a Monte Carlo approach, and try random shuffles until a good score is reached. The result is likely far from the optimum.

My strategy has a number of components.

First, I extract from the dictionary all candidate words for which the letters in the puzzle suffice, regardless of position.

Then I try to pre-arrange the puzzle into a state where letters that are adjacent in the candidate words are as much as possible adjacent also in the puzzle. To this end, I use a notion of "energy": Each possible 2-letter combination is given an energy value, equal to the sum of all candidate word values which contain this letter pair.

We can now compute a total energy for the boggle board which is maximized by a greedy algorithm as follows:

  • The energy of each letter position is computed as the sum of the energies of all pairings with the adjacent letters.
  • At each step, all letter pair exchanges are evaluated to see which would result in the largest increase in puzzle energy.
  • If no increase is possible we are done.
  • The selected pair is exchanged, and the previous steps are repeated.

The next phase is to evaluate the puzzle by finding all (candidate) words that match the board in its current state.

The score is then hopefully increased by a random search:

  • One letter pair is exchanged at random,
  • The puzzle is re-evaluated,
  • The configuration with the higher score is kept.
  • Try again, until we run out of time.

Time Management

The point score will be diminished by 1 point for each 20msec of running time. Rather than run for an arbitrary amount of time, or an fixed number of random swaps, an adaptive algorithm is used:

The elapsed time is measured after each run of random swaps, to establish an average cost for each swap, which includes the time to evaluate its effect on points.

The first run is for a fixed number of 32 swaps, to establish a baseline. Subsequent runs are limited in length by the following consideration. Using the known average time per run, I can fairly accurately predict the number of points to be lost by a run of N swaps. And I might gain some speculative number of points due to one or more lucky swaps.

As long as the expected gain exceeds the expected risk, we may continue to try random swaps. But it might be wise to limit the potential loss in each run to a small number such as 5 or 10% of the total points. Larger boards result in larger variations, and I take a larger risk, hopefully for a larger gain.


Memory required is for the Puzzle Structure (about 7KB) plus 12 bytes per "candidate", that is words which potentially fit the current puzzle.

No assumptions are made of persistence of the private storage between calls. Each call is handled completely independently. One could remember the puzzle size, and if the same, avoid recomputing the puzzle Neighborhood structure, but this is hardly worth the trouble.

#include "boggle.h"
#include <string.h>
#include <timer.h>

enum {   kMinLen=3,               // no points for words < kMinLen

enum {   kPenaltyRate=20000,   // 20msec per point
       kDefRiskFactor=13};      // 12 or 13 seem to work best

struct Timer {                  // microseconds system timer
 int T0;
 void StartTimer() {
  UnsignedWide UWT;
 int ReadTimer(){
  UnsignedWide UWT;
  return UWT.lo-T0;

// A word profile is a 32-bit bit map of letters used in word
static unsigned long GetProfile(const char* word,int num) {
 unsigned long p=0;
 for (int i=0;i<num;i++) {
  p |= 1<<(*word++ & 31);
 return p;

static unsigned long GetProfile(const char* word) {
 int c;
 unsigned long p=0;
 while (0 != (c=*word++)) {
  p |= 1<<(c & 31);
 return p;

struct Candidate {   // A dictionary word that could fit
 const char*       word;
 unsigned long      profile;
 unsigned short   value;
 unsigned char      fit;
 unsigned char      tempFit;

 Candidate(int len,const char* wd){
case 0:
case 1:
case 2:   value=0;break;
case 3:   
case 4:   value=1;break;
case 5:   value=2;break;
case 6:   value=3;break;

struct Neighbors {   
 char* next[9];   // pointers to the 8 neighbors of
};                              // a puzzle field, plus a NULL

struct Neighborhood {   
 Neighbors nb[kMaxDimSquare]; // 1 set per puzzle field
 void Init(int dim,char* c) {
  int dimSq=dim*dim;
  int col,row,x=0;
  for (col=1;col<dim-1;col++) Setup(x++,c++,1,dim,EDGE);
  for (row=1;row<dim-1;row++) {
   for (col=1;col<dim-1;col++) Setup(x++,c++,1,dim,MIDDLE);
  for (col=1;col<dim-1;col++) Setup(x++,c++,1,-dim,EDGE);
 void Setup(int from,char* c,int A,int B,int type) {
  char** np=nb[from].next;
  switch (type) {
case MIDDLE:
case EDGE:
case CORNER:

struct Tour { // Sequence of puzzle locations
 char* tour[1+kMaxDimSquare];
 void Clear(){tour[0]=0;}
 void Add(char* loc) {
  char** tp=tour;
  while (*tp) tp++;
 void Replace(char* loc1,char* loc2) {
  char** tp=tour;
  while (*tp) {
   if (*tp==loc1) {

struct Stack {   // used to remember the track when
 char** CP;      // tracing a word through the puzzle
 char   val;
 void Push(char** c,char v){CP=c;val=v;}
 void Pop(char** & c,char & v){c=CP;v=val;}

struct Puzzle {
 int       dimension;            // 4 to 7
 int      dimSquare;            // 16 to 49
 int      points;               // computed points
 int      riskFactor;
 unsigned    long profile;   // of letters in puzzle      
 Stack*    SP;   
 char*    P;                        // ref to the 'puzzle'
 Timer      tmr;               
 char      letterCount[32];      
 int       energy[kMaxDimSquare];   // energy of puzzle fields
 Stack    stack[kMaxDimSquare];
 Tour      TOUR[32];                  // list of puzzle fields
 Neighborhood N;
 int      pairs[32*32];               // values of letter pairs

 void    Init(int dim,char* puzzle) {
  for (int i=0;i<32;i++) TOUR[i].Clear();
  for (int i=0;i<dimSquare;i++) {
   char c=P[i];
  riskFactor=kDefRiskFactor-dim;   // more risk for larger puzzles

 int Trace(Candidate* cd) { // trace 1 word
  const char* wd=cd->word;
  char** tour=TOUR[*wd & 31].tour;
  char* loc=*tour;
   char letter;
   // loop unrolled for the first and second letters of the word
   if (letter=='Q') ++wd;      //skip 'U' after 'Q'

   // deal with the first and second letters
   *loc=0;               //hide 1st letter
   tour=N.nb[loc-P].next;      //goto 2nd tour
   loc=*tour; ++wd;
   if (letter==*wd) {
      if (letter=='Q') ++wd;      //skip 'U' after 'Q'
      if (wd[1]==0) goto success;
      *loc=0;                        //hide 2nd letter
      loc=*tour;   ++wd;
      goto third;                     //goto 3rd tour
   } else {
    loc=*++tour;            // next 2nd loc
      if (loc) goto second;
       **tour=letter;               //restore first letter
       if (letter=='Q') --wd;
       if (loc) goto first;
   return 0;                  //fail

  do {
   if (letter==*wd) {
      if (letter=='Q') ++wd;      //skip 'U' after 'Q'
      if (wd[1]==0) break;         //success
      *loc=0;               //hide this letter
      loc=*tour;   ++wd;
   } else do {
      if (loc==0) {
      if (SP<=stack) {
       return 0;
        **tour=letter;            //restore last hidden letter
        if (letter=='Q') --wd;
      } else break;
   } while(1);
  } while(1);
  while (SP>stack) {         // restore all letters
  return 1;

 int IsPossible(const char* word) {// all letters present
  int c,len=0;
  char myDist[32];
  while (0 != (c=31 & *word++)) {   
   if (letterCount[c]>myDist[c]) {
   } else return 0;
   if ((c==(31 & 'Q')) && (*word))
    word++;                        // don't need U after Q
  return len;

 void Swap(const int a,const int b){
  char ca=P[a],cb=P[b];

// The next set of functions are for points maximization

// RandomShake just swaps two letters in the puzzle,
// making sure they are different, and returns the
// profile of the two letters
 unsigned long RandomShake(int & a,int & b) {
  char ca,cb;
  do {
  } while ((a==b) || ((ca=P[a])==(cb=P[b])));
  unsigned long shakeProfile=(1L<<(ca&31)) | (1L<<(cb&31));
  return shakeProfile;

// The function Evaluate checks all candidates for fit,
// and computes the total point score for the puzzle
 void Evaluate(Candidate* cd,int numCandidates) {
  int pts=0;
  for (int i=0;i<numCandidates;i++) {
   if (0 != (cd->fit=Trace(cd))) {   

// Improves() answers the question: would the proposed shakup
// increase the score. Computes candidates.tempFit.
// Uses shakeProfile to avoid recomputing candidates that
// are unaffected by the swap.
 bool Improves(
    unsigned long shakeProfile,
    Candidate* cd,int numCandidates) {
  int pts=points;
  for (int i=0;i<numCandidates;i++) {      
   if ((shakeProfile & cd->profile)) {   
   } else {
  if (pts>points) {
   return true;
  return false;

// UpdateFits commits tempFit to fit, after a swap is accepted
 void UpdateFits(Candidate* cd,int numCandidates) {
  for (int i=0;i<numCandidates;i++) {

// Rearrange shakes the puzzle a specified number of times
// and retains the configuration yielding the highest score
 void Rearrange(int howOften,Candidate* cd,int numCandidates) {
  for (int i=0;i<howOften;i++) {
   int s1,s2;
   unsigned long shakeProfile=RandomShake(s1,s2);
   if (Improves(shakeProfile,cd,numCandidates)) {
   } else {
// Optimize calls Rearrange a number of times, to strike a
// compromise between points gained by random shakes, and
// points lost due to runtime cost.
 void Optimize(Candidate* cd,int numCandidates) {
  int run=kFirstRun;
  int countRuns=0;
  int runningRate;
  int oldPoints=points;
  int basePoints=points;
  int runTime=tmr.ReadTimer();
  int lost=(runTime+kPenaltyRate/2)/kPenaltyRate;
  do {
   int pointsToRisk=points/riskFactor-lost;
   int run=kPenaltyRate*pointsToRisk/runningRate;
   if (run<1) break;
  } while (points > oldPoints);

// The next set of functions relate to energy maximization

// PrimePairs uses letter pairs in all candidate words
// to construct an array of basic energy values for each.
 void PrimePairs(Candidate* cd,int numCandidates) {
  for (int i=0;i<numCandidates;i++) {
   const char* wd=cd->word;
   int a=*wd & 31,b;
   do {
    b=*++wd & 31;
    if (b==0) break;
   } while(1);

// Computes the energy of each puzzle field
 void InitEnergy() {
  for (int i=0;i<dimSquare;i++)

// Get the energy for one field, using the Neighborhood
// struct to trace the field's neighbors.
 int GetEnergy(int k) {
  char** tour=N.nb[k].next;
  char* loc;
  int* pairP=pairs+32*(P[k]&31);
  int e=0;
  while (0 != (loc=*tour++)) {
  return e;

// This swap did some good, let's keep it
 void CommitSwap(int a,int b){

// ComputeDelta makes a tentative swap and returns
// the increase or decrease in energy.
 int ComputeDelta(int a,int b) {
  if ((a==b)||(P[a]==P[b])) return 0;
  int ea=GetEnergy(a);
  int eb=GetEnergy(b);
  Swap(a,b);                        // restore previous state
  return ea+eb-energy[a]-energy[b];

// ComputeBestSwap tries all possible swaps and returns
// the swap yielding the highest energy increase
 int ComputeBestSwap(int & aa,int & bb) {
  int bestDelta=-10000;
  for (int a=1;a<dimSquare;a++) {
   for (int b=0;b<a;b++) {
    if (P[a]^P[b]) {
     int delta=ComputeDelta(a,b);
     if (delta>bestDelta) {
  return bestDelta;

// MaximizeEnergy keeps swapping puzzle letters until no
// more increase in puzzle energy can be obtained.
 void MaximizeEnergy() {
  int a,b,delta=ComputeBestSwap(a,b);
  do {
   if (delta>0) CommitSwap(a,b);
   else break;
  } while (1);

struct PrivateData {   // just the puzzle and the candidate words
 Puzzle   pzl;
 int      numCandidates;
 Candidate   candidates[1];    // open ended array

 void Init(int dimension,char* puzzle) {

 void SelectCandidates(int dictSize,const char *dictionary[]) {
  for (int i=0;i<dictSize;i++) {
   const char* word=dictionary[i];
   int len=pzl.IsPossible(word);
   if (len>=kMinLen)

 void Solve() {

 int CopyResults(const char *wordsFound[]) {
  int n=0;
  Candidate* cd=candidates;
  for (int i=0;i<numCandidates;i++) {
  return n;

// the function Boggle is published in boggle.h
long Boggle(
   long dimension,   
   char puzzle[],   
   long dictSize,   
   const char *dictionary[],
   const char *wordsFound[],   
   void *privStorage   
   ) {   
 int nFound=0;
 srand(10001);   // just to make it repeatable

 if ((unsigned long)dimension <= kMaxDim) {
   PrivateData* PD=(PrivateData*)privStorage; // assumed to be enough

 return nFound;   

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B&H Photo has the new 2016 15″ Apple Touch Bar MacBook Pros in stock today and on sale for up to $150 off MSRP. Shipping is free, and B&H charges NY sales tax only: - 15″ 2.7GHz Touch Bar... Read more
Apple refurbished iPad Pros available for up...
Apple has Certified Refurbished 9″ and 12″ Apple iPad Pros available for up to $160 off the cost of new iPads. An Apple one-year warranty is included with each model, and shipping is free: - 32GB 9″... Read more
16GB iPad Air 2, Apple refurbished, available...
Apple has Certified Refurbished 16GB iPad Air 2s available for $319 including free shipping. A standard Apple one-year is included. Their price is $60 off original MSRP for this model. Read more
Apple iMacs on sale for up to $120 off MSRP
B&H Photo has 21″ and 27″ Apple iMacs on sale for up to $120 off MSRP, each including free shipping plus NY sales tax only: - 27″ 3.3GHz iMac 5K: $2199 $100 off MSRP - 27″ 3.2GHz/1TB Fusion iMac... Read more
Apple refurbished Apple TVs available for up...
Apple has Certified Refurbished 32GB and 64GB Apple TVs available for up to $30 off the cost of new models. Apple’s standard one-year warranty is included with each model, and shipping is free: -... Read more
Save up to $350 with Apple Certified Refurbis...
Apple has Certified Refurbished 2015 21″ & 27″ iMacs available for up to $350 off MSRP. Apple’s one-year warranty is standard, and shipping is free. The following models are available: - 21″ 3.... Read more
2015 12-inch Retina MacBooks, Apple refurbish...
Apple has Certified Refurbished 2015 12″ Retina MacBooks available for up to $410 off original MSRP. Apple will include a standard one-year warranty with each MacBook, and shipping is free. The... Read more

Jobs Board

*Apple* & PC Desktop Support Technician...
Apple & PC Desktop Support Technician job in Manhattan, NY Introduction: We have immediate job openings for several Desktop Support Technicians with one of our most Read more
*Apple* & PC Desktop Support Technician...
Apple & PC Desktop Support Technician job in Stamford, CT We have immediate job openings for several Desktop Support Technicians with one of our most well-known Read more
*Apple* Retail - Multiple Positions - Apple,...
Job Description: Sales Specialist - Retail Customer Service and Sales Transform Apple Store visitors into loyal Apple customers. When customers enter the store, Read more
*Apple* Site Security Manager - Apple (Unite...
# Apple Site Security Manager Job Number: 54692472 Culver City, California, United States Posted: Jan. 19, 2017 Weekly Hours: 40.00 **Job Summary** The Apple Read more
*Apple* macOS Systems Integration Administra...
…most exceptional support available in the industry. SCI is seeking an Junior Apple macOS systems integration administrator that will be responsible for providing Read more
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