TweetFollow Us on Twitter

Nov 92 Challenge
Volume Number:8
Issue Number:7
Column Tag:Programmers' Challenge

Programmers' Challenge

By Mike Scanlin, MacTutor Regular Contributing Author

November 92 Programming Challenge of the Month

Millions of Colors?

Ever wonder how many of the 16,777,216 possible colors are really used in a typical 24-bit color image? Do you really need to run in 24-bit mode to appreciate certain images? Could some images be accurately represented as indexed color images instead, without any loss of color? Hmmm... The first thing you’d need to know is how many unique RGB values there are in the image, which would tell you how big your color lookup table would need to be. Let’s try it.

This month’s challenge is to quickly determine how many unique RGB values there are in a given 24-bit color image. The input to your function will be the dimensions and base address of an interleaved ARGB image:

unsigned long UniqueRGBValues(baseAddress, numRows, numCols)
PtrbaseAddress;
short   numRows, numCols;

The byte pointed to by baseAddress is the alpha byte of the upper left pixel. Following that are bytes for red, green and blue, which are then followed by the next set of ARGB values. You can ignore the alpha bytes completely when calculating unique RGB values. If you feel the need to allocate an array of 16,777,216 bits then, yes, you can assume your routine will have at least 2.5MBs free memory when called to do so (but remember that the time to initialize such an array is non-zero; there may be faster methods...).

Let’s say the maximum value for numCols is 640 and for numRows it’s 480. Your routine should be very fast for the average case but also be able to deal with the worst case where you have 640x480 unique RGB values.

We goofed

No two ways about it. In our rush to get the October issue out the door we were a little too hasty in determining the winner of the August MacTutor Challenge. Not more than 48 hours after the issue had gone to press (but before the stated challenge deadline had passed) we received a solution that was better than the one declared to be the winner. Our apologies to Greg Landweber (Princeton, NJ) who was the actual winner (and will also be receiving the prize). In order to prevent this from happening again in the future, we have moved the deadline up (see below).

The “deadline has now passed” winner of the “How many ways can you spell ‘CAT’” challenge is Will Galway (Salt Lake City, UT) whose entry was the only non-recursive one received. Take note recursion fanatics: Although recursion is a Good Thing conceptually (and in many cases practically), these monthly challenges are primarily about speed. We don’t need to contribute to the large body of existing slow code; we need faster code. Take a couple of No-Doze and study Will’s non-recursive alternative.

Thanks to Bob Barnhart (San Diego, CA) for his entertaining animated solution. Too bad it wasn’t as fast as it was fun to watch. Bob’s entry brings up another point as well: Please use a column width of 79 characters or less in your code. AppleLink (or the internet-AppleLink gateway, I’m not sure which) breaks lines longer than 80 characters and it’s a pain to manually fix them up when I get them. Thanks.

Here are Greg’s winning solution to the August Challenge (the real winner) and Will’s winning solution to the September Challenge (some comments have been removed for space reasons. The complete sources are on the source code disk):

Banded Pegs

/* Solution to the August 1992 MacTutor
 * Programmers' Challenge
 *
 * by Greg Landweber
 */

/* The number of holes in a row or column. */
#define max 13

void BandedPegs (numPegs, pegsPtr, numEdgePegsPtr,
 edgePegsPtr, areaPtr)
short numPegs;
Point *pegsPtr;
short *numEdgePegsPtr;
Point *edgePegsPtr;
Fixed *areaPtr;
{
 /* leftmost and rightmost peg in each row */
    short   xLeft[max],xRight[max];
 /* top and bottom rows containing pegs */
    short   top,bottom;
 /* horizontal and vertical coords. of peg */
    short   x,y;
 /* used to compute twice the enclosed area */
    short   area;
 /* number of pegs on left and right side */
    short   numLeft,numRight;
 /* array of pegs on left and right */
    Point   leftPegs[max],rightPegs[max];
 /* general use array index */
    short   index;
 /* for stepping through arrays of Points */
    Point   *pegPtr1,*pegPtr2;
 
/* Fill xLeft[v] and xRight[v] with the h-coords
 * of the leftmost and rightmost pegs in row v.
 * If there are no pegs in row v, then set
 *  xLeft[v]  = max, and
 *      xRight[v] = -1.
 * Note that any pegs inbetween the leftmost and
 * rightmost pegs in a row will automatically be
 * in the interior of the rubber band polygon.
 * This reduces the maximum number of pegs to 26.
 */
 
    for ( index = 0; index < max; index++ ) {
        xLeft [index] = max;
        xRight[index] = -1;
    }
 
    pegPtr1 = pegsPtr;
    for ( index = numPegs; index > 0; index-- ) {
        y = pegPtr1->v;
        x = pegPtr1->h;
        if ( x < xLeft [y] )
            xLeft [y] = x;
        if ( x > xRight[y] )
            xRight[y] = x;
        pegPtr1++;
    }
 
/* Find the bottom (lowest v) and top
 * (highest v) rows containing pegs. */

    bottom = -1;
    while ( xLeft [++bottom] == max );
 
    top = max;
    while ( xLeft [--top] == max );
 
/* Fill leftPegs[] with a list of all the pegs
 * on the left side of the convex polygon from
 * the top (hi v) to the bottom (lo v), and put
 * the number of those pegs - 1 in numLeft. */

 /* leftPegs[0] is the topmost (highest v) */
    leftPegs[0].h = xLeft[top];
 /* point on the left side of the polygon. */
    leftPegs[0].v = top;
 /* Index of the last peg in leftPegs[]. */
    numLeft = 0;
 
 /* Add pegs from the top to the bottom. */
    for (y = top - 1; y >= bottom; y--)
    /* Check if there is a peg in row y. */
        if ( (x = xLeft[y]) != max ) {
        /* Note thatpegPtr2 is the current
        * peg in the list and pegPtr1 is the
        * next. */ 
            pegPtr1 = leftPegs;
            pegPtr2 = pegPtr1++;
            for ( index = 0; index < numLeft; index++ )
            /* Is the peg at {x,y} to the left of
             * the line from *pegPtr1 to *pegPtr2? */
                if ( ( (x - pegPtr1->h) 
                    (pegPtr2->v - pegPtr1->v) ) <
                    ( (pegPtr2->h - pegPtr1->h) *
                    (y  - pegPtr1->v) ) )
                /* If so, all the pegs from pegPtr1 on
                 * will be to the right of the line
                 * from {x,y} to *pegPtr2, and so we
                 * remove them from the left peg list. */
                    numLeft = index;
                else
                /* If not, we go on to the next peg. */
                    pegPtr2 = pegPtr1++;
            /* Tack {x,y} onto the end of the list. */
            numLeft++;
            pegPtr1->v = y;
            pegPtr1->h = x;
        }

/* Fill rightPegs[] with a list of all the pegs
 * on the right side of the convex polygon from
 * the top (hi v) to the bottom (lo v), and put
 * the number of those pegs - 1 in numRight.
 */
 
 /* rightPegs[0] is the topmost (highest v)
  * point on the right side of the polygon. */
    rightPegs[0].h = xRight[top];
    rightPegs[0].v = top;

 /* Index of the last peg in rightPegs[]. */
    numRight = 0;

 /* Add pegs from the top to the bottom. */
    for (y = top - 1; y >= bottom; y--)
    /* Check if there is a peg in row y. */
        if ( (x = xRight[y]) != max ) { 
        /* Note that pegPtr2is the current peg */
        /* in the list and pegPtr1 is the next. */
            pegPtr1 = rightPegs;        
            pegPtr2 = pegPtr1++;        
            for ( index = 0; index < numRight; index++ )
            /* Is the peg at {x,y} to the right of
             * the line from *pegPtr1 to *pegPtr2?*/
                if ( ( (x - pegPtr1->h) *
                    (pegPtr2->v - pegPtr1->v) ) >
                    ( (pegPtr2->h - pegPtr1->h) *
                    (y - pegPtr1->v) ) )
               /* If so, all the pegs from pegPtr1 on
                * will be to the left of the line
                * from {x,y} to *pegPtr2, and so we
                * remove them from the right peg list. */
                    numRight = index;   
                else
                /* If not, we go on to the next peg.*/
                    pegPtr2 = pegPtr1++;
            numRight++;                 
            /* Tack {x,y} onto the end of the list. */
            pegPtr1->v = y;
            pegPtr1->h = x;
        }
 
/* Copy the contents of numLeft[] and
 * numRight[] into edgePegsPtr. */
    pegPtr2 = edgePegsPtr;

    pegPtr1 = leftPegs + 1;
    for ( index = numLeft - 1; index > 0; index-- )
        *(pegPtr2++) = *(pegPtr1++);

/* Do the pegs all lie on the same line?
 * If so, the left and right are the same.  */
    if ( *( (long *)leftPegs + 1 ) !=
        *( (long *)rightPegs + 1 ) ) {
        pegPtr1 = rightPegs + 1;
        for ( index = numRight - 1; index > 0; index-- )
            *(pegPtr2++) = *(pegPtr1++);
    }
 
/* Put all the pegs in the top and bottom
 * rows into edgePegsPtr. */
    pegPtr1 = pegsPtr;
    for ( index = numPegs; index > 0; index-- ) {
        if ( (pegPtr1->v == top) || (pegPtr1->v == bottom) )
            *(pegPtr2++) = *pegPtr1;
        pegPtr1++;
    }
 
/* Figure out how many pegs there are touching
 * the edge of the polygon. */
    *numEdgePegsPtr = pegPtr2 - edgePegsPtr;
 
/* Compute twice the area to the left of the
 * right side of the polygon. */
    area = 0;
 
/* The area of a trapezoid with height h and\
 * parallel sides of length a and b is h*(a+b)/2.
 * Here we have h = pegPtr2->v - pegPtr1->v,
 * a = pegPtr2->h, and  b = pegPtr1->h. */
    pegPtr1 = rightPegs;

/* Loop through all of the line segments on
 * the right side of the convex polygon. */        
    for ( index = numRight; index > 0; index-- ) {
        pegPtr2 = pegPtr1++;        
        area += (pegPtr2->v - pegPtr1->v) *
            (pegPtr2->h + pegPtr1->h);
    }
 
/* Subtract twice the area to the left of the
 * left side of the polygon. */
    pegPtr1 = leftPegs;             

/* Loop through all of the line segments on
 * the left side of the convex polygon. */
    for ( index = numLeft; index > 0; index-- ) {
        pegPtr2 = pegPtr1++;        
        area -= (pegPtr2->v - pegPtr1->v) *
            (pegPtr2->h + pegPtr1->h);
    }
 
/* Finally, divide by two and convert the
 * result to type Fixed. */
    *areaPtr = FixRatio( area, 2 );
}

How Many ways can you spell ’CAT‘

/* count-paths.h:  Declarations for count-paths.c
 *
 * Copyright (C) 1992,  William F. Galway
 *
 * Anyone can do what they like with this code,
 * as long as they acknowledge its author,
 * and include this message in their code.
 */
 
typedef int BOOL;
 
#define TRUE 1
#define FALSE 0
 
/* Possible target systems/compilers...  */
#define ThinkC 0
#define GnUnix 1
 
#if !defined(TARGET)
#define TARGET ThinkC
#endif
 
#if !defined(DEBUG)
#define DEBUG FALSE
#endif
 
#if !defined(VERBOSE)
#define VERBOSE FALSE
#endif
 
/* Maximum dimensions of the "matrix". */
#define MAXORDER 10
 
#if (TARGET==GnUnix)
/* This is the "Mac" StringPtr type.  The first
 * byte gives the length, the rest of the bytes
 * make up the string. */
typedef unsigned char Str255[256], *StringPtr;
 
/* Native is the type most naturally addressed,
 * roughly speaking...  */
typedef void Native;
#endif

#if (TARGET==ThinkC)
/* Native is the type most naturally addressed,
 * roughly speaking...  */
typedef char Native;
#endif
 
typedef struct locnode {
  /* Next node in list for a given character. */
    struct locnode *next;
} LocNode;
 
typedef struct {
    /* Number of entries per row...  */
    long dy;

    /* Vector of LocNodes indexed by
     * character code, giving first location
     * of character. */
    LocNode char_index[256];

    /* "Matrix" of LocNodes giving further
     * locations of each character. */
    LocNode index_matrix[(2+MAXORDER)*(2+MAXORDER)];
} Index;
 
/* BuildIndex builds up index for matrix of
 * given order. */
void BuildIndex(long order, const char *matrix,
 Index *index);
 
/* count_paths counts paths using previously
 * built index. */
long count_paths(const Index *index,
 const StringPtr word);
 
/* CountPaths is the "top level" path counting
 * routine. */
long CountPaths(short order, char *matrix,
 const StringPtr inputWordPtr);
 
/*-----------------------------------------*/
 
/* count-paths.c
 *
 * Copyright (C) 1992,  William F. Galway
 *
 * Anyone can do what they like with this code,
 * as long as they acknowledge its author,
 * and include this message in their code.
 */
 
/* The algorithm used by this implementation
 * avoids "combinatorial blowup" by working
 * backwards through the input word, keeping a
 * "count table" showing the number of paths for
 * the substring at each node.  For example, for
 * the string "CAR" we would get the following
 * counts (count tables) at each stage:
 *  -for "r":
 *     0  0  0
 *     0  1  0
 *     0  0  0
 *  -for "ar":
 *     0  0  0
 *     1  0  1
 *     0  1  0
 *  -for "car":
 *     1  0  1
 *     0  0  0
 *     0  0  2
 * giving a total of 4 solutions found at the
 * final stage. (This non-recursive approach is
 * reminiscent of the iterative versus the
 * recursive method of computing Fibonacci
 * numbers.)
 *
 * We actually keep two count tables around, one
 * giving counts for the "previous stage" (the
 * "previous table"), and one being built up for
 * the "current stage" (the "current table"). We
 * build the current table by locating occurrences
 * of the leading character of the substring, and
 * then summing the counts from the four
 * neighboring locations in the previous table. 
 * To ease the problem of dealing with the edges
 * of the tables, we allocate "dummy" rows and
 * columns at the edges of our count tables. The
 * counts at the edges always remain zero, while
 * the interesting stuff goes on in the interior
 * of the tables.
 *
 * To simplify (and speed up) the task of locating
 * occurrences of characters in the matrix, we
 * first build an "index" for the matrix which is
 * basically a linked list of pointers and then
 * index into the index (!) by the character that
 * we need the location(s) of. The index needs
 * building only once for a given matrix, after
 * which the count_paths routine may be called
 * (see how CountPaths invokes count_paths below).
 *
 * Other points to note:
 *
 *  -- Use of "Native" pointers for less "pointer
 *     arithmetic".
 *  -- The result returned by CountPaths is more
 *     properly interpreted as an unsigned long
 *     rather than as a signed long.
 *  -- These routines are not robust when called
 *     with matrices of order outside the range
 *     1..MAXORDER.
 */
 
#include "count-paths.h"
#include <stdio.h>
 
/* Build up index for matrix of given order.  */
void BuildIndex(long order, const char *matrix,
 Index *index)
{
    register unsigned char *chrp;
    register LocNode *spot, *spot2;
    long i,j;
 
    /* Zero out the char_index (256 entries).  */
    spot = index->char_index;
    spot2 = spot+256;
    do {
        (spot++)->next = NULL;
    } while (spot < spot2);
 
    /* Build up the index... The c'th entry in
     * char_index points to a chain of pointers
     * residing in index_matrix...  Note that
     * "edge" rows and columns are allowed to
     * contain nonsense. */

    spot = index->index_matrix+order+3;
    chrp = (unsigned char *)matrix;
    i = order;
    do {
        j = order;
        do {
            /* char_index[char] points to head of
             * chain for char. Set spot pointed at
             * to point to "next" spot with ch in
             * it (as previously stored in
             * char_index). */
            spot2 = &index->char_index[*chrp++];
            spot->next = spot2->next;
            spot2->next = spot++;
        } while (--j);

        /* Skip last & first columns of row. */
        spot += 2;
    } while (--i);
  
    index->dy = order+2;
 
    return;
}
 
/* Count paths using previously built index. */
long count_paths(const Index *index, const StringPtr word)
{
    register unsigned char *chrp;
    register long dyoffset;

    /* tbl_offset gives offset from "current
     * counts" table to "previous counts" table. 
     * i.e., previous_counts =
     * current_counts+tbl_offset. */
    long tbl_offset;

    /* current_offset, previous_offset give
     * offset from index->index_matrix to
     * current/previous count tables. */
    register long current_offset;
    register long previous_offset;
    LocNode *spot;
    long *countp;
    register long total;
    long count_tables[2*(2+MAXORDER)*(2+MAXORDER)];
 
    /* Point chrp to last char of word. */
    chrp = word + *word;
 
    /* Initialize misc offsets, pointers. */
    dyoffset = index->dy*sizeof(long);

    /* (short) avoids subroutine call for
     * multiply for some systems. */
    tbl_offset = (short)(index->dy)*(short)dyoffset;
    current_offset = (Native *)count_tables-
        (Native *)(index->index_matrix);
    previous_offset = tbl_offset+current_offset;
 
    /* Zero out the count tables. */
    countp=count_tables;
    do {
        *countp++ = 0;
    } while (countp < (long *)((Native *)
        count_tables+2*tbl_offset));
  
    total = 0;
 
    /* Initialize counts for "previous table".
     * (It will soon be previous!) */
    for (spot=(index->char_index)[*chrp].next;
        spot!=NULL; spot=spot->next) {
        *(long *)((Native *)spot+previous_offset) = 1;
        total++;
    }
 
    if (total==0 || --chrp<=word)
        return total;
 
    while (TRUE) {
        total = 0;
        for (spot=(index->char_index)[*chrp].next;
            spot!=NULL; spot=spot->next) {
            countp = (long *)((Native *)spot +
                previous_offset);

            /* Hairy expression avoids variable,
             * may free up register... */
            total += *(long *)((Native *)spot +
                current_offset) = *(countp-1) +
                *(countp+1) + *(long *)((Native *)
                countp-dyoffset) + *(long *)
                ((Native*)countp+dyoffset);
        }

        if (total==0 || --chrp<=word)
            return total;
 
      /* Swap "current" and "previous" count
       * tables. */
        current_offset += tbl_offset;
        previous_offset -= tbl_offset;
        tbl_offset = - tbl_offset;
         /* Zero out current counts, only need
         * touch non-zero entries. */
        for (spot=(index->char_index)[*(chrp+2)].next;
            spot!=NULL; spot=spot->next) {
            *(long *)((Native *)spot + current_offset) = 0;
        }
    }
}


long CountPaths(short order, char *matrix,
 const StringPtr inputWordPtr)
{
    long ord=order;
    Index index;
 
    /* Problem statement restricts word length to
     * be >0, but be paranoid since
     * count_paths(...) is not robust for 0 length
     * words. Return 0 if empty (zero length)
     * word. */
    if (*inputWordPtr == 0) {
        return 0;
    } else if (*inputWordPtr == 1) {
        /* Avoid work of building index, etc. for
         * length one words. */
        register char ch=(char)inputWordPtr[1];
        char *chrp = matrix;
        long total=0;
 
        do {
            if (ch == *chrp++) {
                total++;
            }
        } while (chrp < matrix+order*order);
        return total;
    } else {
        /* Invoke count_paths after building the
         * index... */
        BuildIndex(ord, matrix, &index);
        return count_paths(&index, inputWordPtr);
    }
}
 

Community Search:
MacTech Search:

Software Updates via MacUpdate

Geekbench 4.4.0 - Measure processor and...
Geekbench provides a comprehensive set of benchmarks engineered to quickly and accurately measure processor and memory performance. Designed to make benchmarks easy to run and easy to understand,... Read more
CleanMyMac X 4.4.4 - Delete files that w...
CleanMyMac makes space for the things you love. Sporting a range of ingenious new features, CleanMyMac lets you safely and intelligently scan and clean your entire system, delete large, unused files... Read more
TeamViewer 14.4.2669 - Establish remote...
TeamViewer gives you remote control of any computer or Mac over the Internet within seconds or can be used for online meetings. Find out why more than 200 million users trust TeamViewer! Free for non... Read more
Paperless 3.0.6 - $69.95
Paperless is a digital documents manager. Remember when everyone talked about how we would soon be a paperless society? Now it seems like we use paper more than ever. Let's face it - we need and we... Read more
BetterTouchTool 3.141 - Customize multi-...
BetterTouchTool adds many new, fully customizable gestures to the Magic Mouse, Multi-Touch MacBook trackpad, and Magic Trackpad. These gestures are customizable: Magic Mouse: Pinch in / out (zoom)... Read more
TextMate 2.0.rc.29 - Code/markup editor...
TextMate is a versatile plain text editor with a unique and innovative feature set which caused it to win an Apple Design Award for Best Mac OS X Developer Tool in August 2006 A rapidly growing... Read more
Little Snitch 4.4.1 - Alerts you about o...
Little Snitch gives you control over your private outgoing data. Track background activity As soon as your computer connects to the Internet, applications often have permission to send any... Read more
Little Snitch 4.4 - Alerts you about out...
Little Snitch gives you control over your private outgoing data. Track background activity As soon as your computer connects to the Internet, applications often have permission to send any... Read more
MPlayer OSX Extended 16 - Multimedia pla...
MPlayer OSX Extended is the future of MPlayer OSX. Leveraging the power of the MPlayer and FFmpeg open source projects, MPlayer OSX Extended aims to deliver a powerful, functional and no frills video... Read more
Google Chrome 75.0.3770.142 - Modern and...
Google Chrome is a Web browser by Google, created to be a modern platform for Web pages and applications. It utilizes very fast loading of Web pages and has a V8 engine, which is a custom built... Read more

Latest Forum Discussions

See All

TEPPEN guide - Tips and tricks for new p...
TEPPEN is a wild game that nobody asked for, but I’m sure glad it exists. Who would’ve thought that a CCG featuring Capcom characters could be so cool and weird? In case you’re not completely sure what TEPPEN is, make sure to check out our review... | Read more »
Dr. Mario World guide - Other games that...
We now live in a post-Dr. Mario World world, and I gotta say, things don’t feel too different. Nintendo continues to squirt out bad games on phones, causing all but the most stalwart fans of mobile games to question why they even bother... | Read more »
Strategy RPG Brown Dust introduces its b...
Epic turn-based RPG Brown Dust is set to turn 500 days old next week, and to celebrate, Neowiz has just unveiled its biggest and most exciting update yet, offering a host of new rewards, increased gacha rates, and a brand new feature that will... | Read more »
Dr. Mario World is yet another disappoin...
As soon as I booted up Dr. Mario World, I knew I wasn’t going to have fun with it. Nintendo’s record on phones thus far has been pretty spotty, with things trending downward as of late. [Read more] | Read more »
Retro Space Shooter P.3 is now available...
Shoot-em-ups tend to be a dime a dozen on the App Store, but every so often you come across one gem that aims to shake up the genre in a unique way. Developer Devjgame’s P.3 is the latest game seeking to do so this, working as a love letter to the... | Read more »
Void Tyrant guide - Guildins guide
I’ve still been putting a lot of time into Void Tyrant since it officially released last week, and it’s surprising how much stuff there is to uncover in such a simple-looking game. Just toray, I finished spending my Guildins on all available... | Read more »
Tactical RPG Brown Dust celebrates the s...
Neowiz is set to celebrate the summer by launching a 2-month long festival in its smash-hit RPG Brown Dust. The event kicks off today, and it’s divided into 4 parts, each of which will last two weeks. Brown Dust is all about collecting, upgrading,... | Read more »
Flappy Royale is an incredibly clever ta...
I spent the better part of my weekend playing Flappy Royale. I didn’t necessarily want to. I just felt like I had to. It’s a hypnotic experience that’s way too easy to just keep playing. | Read more »
Void Tyrant guide - General tips and tri...
Void Tyrant is a card-based dungeon-crawler that doesn’t fit in the mold of other games in the genre. Between the Blackjack-style combat and strange gear system alone, you’re left to your own devices to figure out how best to use everything to your... | Read more »
Webzen’s latest RPG First Hero is offici...
You might be busy sending your hulking Dark Knight into the midst of battle in Webzen’s other recent release: the long-anticipated MU Origin 2. But for something a little different, the South Korean publisher has launched First Hero. Released today... | Read more »

Price Scanner via MacPrices.net

Amazon drops prices, now offers clearance 13″...
Amazon has new dropped prices on clearance 13″ 2.3GHz Dual-Core non-Touch Bar MacBook Pros by $200 off Apple’s original MSRP, with prices now available starting at $1099. Shipping is free. Be sure to... Read more
2018 15″ MacBook Pros now on sale for $500 of...
Amazon has dropped prices on select clearance 2018 15″ 6-Core MacBook Pros to $500 off Apple’s original MSRP. Prices now start at $1899 shipped: – 2018 15″ 2.2GHz Touch Bar MacBook Pro Silver: $1899.... Read more
Price drop! Clearance 12″ 1.2GHz Silver MacBo...
Amazon has dropped their price on the recently-discontinued 12″ 1.2GHz Silver MacBook to $849.99 shipped. That’s $450 off Apple’s original MSRP for this model, and it’s the cheapest price available... Read more
Apple’s 21″ 3.0GHz 4K iMac drops to only $936...
Abt Electronics has dropped their price on clearance, previous-generation 21″ 3.0GHz 4K iMacs to only $936 shipped. That’s $363 off Apple’s original MSRP, and it’s the cheapest price we’ve seen so... Read more
Amazon’s Prime Day savings on Apple 11″ iPad...
Amazon has new 2018 Apple 11″ iPad Pros in stock today and on sale for up to $250 off Apple’s MSRP as part of their Prime Day sale (but Prime membership is NOT required for these savings). These are... Read more
Prime Day Apple iPhone deal: $100 off all iPh...
Boost Mobile is offering Apple’s new 2018 iPhone Xr, iPhone Xs, and Xs Max for $100 off MSRP. Their discount reduces the cost of an Xs to $899 for the 64GB models and $999 for the 64GB Xs Max. Price... Read more
Clearance 13″ 2.3GHz Dual-Core MacBook Pros a...
Focus Camera has clearance 2017 13″ 2.3GHz/128GB non-Touch Bar Dual-Core MacBook Pros on sale for $169 off Apple’s original MSRP. Shipping is free. Focus charges sales tax for NY & NJ residents... Read more
Amazon Prime Day deal: 9.7″ Apple iPads for $...
Amazon is offering new 9.7″ WiFi iPads with Apple Pencil support for $80-$100 off MSRP as part of their Prime Day sale, starting at only $249. These are the same iPads found in Apple’s retail and... Read more
Amazon Prime Day deal: 10% (up to $20) off Ap...
Amazon is offering discounts on new 2019 Apple AirPods ranging up to $20 (10%) off MSRP as part of their Prime Day sales. Shipping is free: – AirPods with Charging Case: $144.99 $15 off MSRP –... Read more
Amazon Prime Day deal: $50-$80 off Apple Watc...
Amazon has Apple Watch Series 4 and Series 3 models on sale for $50-$80 off Apple’s MSRP as part of their Prime Day deals with prices starting at only $199. Choose Amazon as the seller rather than a... Read more

Jobs Board

*Apple* Systems Architect/Engineer, Vice Pre...
…its vision to be the world's most trusted financial group. **Summary:** Apple Systems Architect/Engineer with strong knowledge of products and services related to Read more
*Apple* Graders/Inspectors (Seasonal/Hourly/...
…requirements. #COVAentryleveljobs ## Minimum Qualifications Some knowledge of agricultural and/or the apple industry is helpful as well as the ability to comprehend, Read more
Best Buy *Apple* Computing Master - Best Bu...
**710003BR** **Job Title:** Best Buy Apple Computing Master **Job Category:** Store Associates **Location Number:** 000171-Winchester Road-Store **Job Description:** Read more
Best Buy *Apple* Computing Master - Best Bu...
**709786BR** **Job Title:** Best Buy Apple Computing Master **Job Category:** Sales **Location Number:** 000430-Orange Park-Store **Job Description:** **What does a Read more
Geek Squad *Apple* Master Consultation Agen...
**709918BR** **Job Title:** Geek Squad Apple Master Consultation Agent **Job Category:** Services/Installation/Repair **Location Number:** 000106-Palmdale-Store Read more
All contents are Copyright 1984-2011 by Xplain Corporation. All rights reserved. Theme designed by Icreon.