TweetFollow Us on Twitter

Dec 95 Challenge
Volume Number:11
Issue Number:12
Column Tag:Programmer’s Challenge

Programmer’s Challenge

By Bob Boonstra, Westford, Massachusetts

Note: Source code files accompanying article are located on MacTech CD-ROM or source code disks.

Find Again And Again

This month the Challenge is to write a text search engine that is optimized to operate repeatedly on the same text. You will be given a block of text, some storage for data structures, and an opportunity to analyze the text before being asked to perform any searches against that text. Then you will repeatedly be asked to find a specific occurrence of a given word in that block of text. The prototypes for the code you should write are:

void InitFind(
 char *textToSearch, /* find words in this block of text  */
 long textLength,/* number of chars in textToSearch   */
 void *privateStorage,    /* storage for your use              */
 long storageSize/* number of bytes in privateStorage */
);
long FindWordOccurrence( 
    /* return offset of wordToFind in textToSearch   */
 char *wordToFind, /* find this word in textToSearch    */
 long wordLength,/* number of chars in wordToFind     */
 long occurrenceToFind, /* find this instance of wordToFind  */
 char *textToSearch, /* same parameter passed to InitFind */
 long textLength,/* same parameter passed to InitFind */
 void *privateStorage,  /* same parameter passed to InitFind */
 long storageSize/* same parameter passed to InitFind */
);

The InitFind routine will be called once for a given block of textLength characters at textToSearch to allow you to analyze the text, create data structures, and store them in privateStorage. When InitFind is called, storageSize bytes of memory at privateStorage will have been preallocated and initialized to zero.

FindWordOccurrence is to search for words, where a word is defined as a continuous sequence of alphanumeric characters delimited by a non-alphanumeric character (e.g., space, tab, punctuation, hyphen, CR, NL, or other special character). Your code should look for complete words - it would be incorrect, for example, to return a value pointing to the word “these” if the wordToFind was “the”. The wordToFind will be a legal word (i.e., no embedded delimiters). FindWordOccurrence should return the offset in textToSearch of the occurrenceToFind-th instance of wordToFind. It should return -1 if wordToFind does not occur in textToSearch, or if there are fewer than occurrenceToFind instances of wordToFind.

Both the InitFind and the FindWordOccurrence routines will be timed in determining the winner. In designing your code, you should assume that FindWordOccurrence will be called approximately 1000 times for each call to InitFind (with the same textToSearch, but possibly differing values of wordToFind and occurrenceToFind).

There is no predefined limit on textLength - you should handle text of arbitrary length. The amount of privateStorage available could be very large, but is guaranteed to be at least 64K bytes. While the test cases will include at least one large textToSearch with a small storageSize, most test cases will provide at least 32 bytes for each occurrence of a word in textToSearch, so you might want to optimize for that condition.

Other fine print: you may not change the input pointed to by textToSearch or wordToFind, and you should not use any static storage other than that provided in privateStorage.

This will be a native PowerPC Challenge, scored using the latest CodeWarrior compiler. Good luck, and happy searching.

Programmer’s Challenge Mailing List

We are pleased to announce the creation of the Programmer’s Challenge Mailing List. The list will be used to distribute the latest Challenge, provide answers to questions about the current Challenge, and discuss suggestions for future Challenges. The Challenge problem will be posted to the list each month, sometime between the 20th and the 25th of the month. This should alleviate problems caused by variations in the publication and mailing date of the magazine, and provide a predictable amount of time to work on each Challenge.

To subscribe to the list, send a message to autoshare@mactech.com with the SUBJECT line “sub challenge YourName”, substituting your real name for YourName. To unsubscribe from the list, send a message to autoshare@mactech.com with the SUBJECT line “unsub challenge”.

Note: the list server, autoshare, is set to accept commands in the SUBJECT line, not the body of the message. If you have any problems, please contact online@mactech.com.

Two Month’s Ago Winner

The Master Mindreader Challenge inspired ten readers to enter, and all ten solutions gave correct results. Congratulations to Xan Gregg (Durham, N.C.) for producing the fastest entry and winning the Challenge.

The problem required you to write code that would correctly guess a sequence of colors using a callback routine provided in the problem statement that returned two values for each guess: the number of elements of the guess where the correct color is located in the correct place in the sequence, and the number of elements where the correct color is in an incorrect place in the sequence. The number of guesses was not an explicit factor in determining the winner, but the time used by the callback routine was included in determining the winner. Participants correctly noted that this made the relative execution time of the guessing routine and the callback routine a factor in designing a fast solution. A couple of entries went so far as to offer their own, more efficient, callbacks. Nice try, but I didn’t use them - the callback in the problem was designed to provide a known time penalty for making a guess, and that was the callback I used in evaluating solutions.

The callback I supplied had one unanticipated side effect - it permitted callers to supply an out-of-range value for positions in the sequence that they didn’t care about for that guess, and six of the entries took advantage of this loophole. This wasn’t what I had intended, and I gave some thought to giving priority to solutions that did not use the loophole. In the end, however, I decided not to treat these entries any differently, because the solution statement permitted and provided a defined callback behavior for out-of-range guesses. As it turned out, the winning entry and three of the fastest four entries did not use out-of-range guesses.

Xan’s winning code first makes a sequence of guesses to determine how many positions are set to each of the possible colors. He then starts with an initial guess corresponding to these colors and begins swapping positions to determine how the number of correctly placed colors is affected. Separate logic handles the cases where the number of correctly placed colors increased or decreased by 0, 1, or 2, all the while keeping track of which color possibilities have been eliminated for each position. These and other details of Xan’s algorithm are documented in the comments to his code.

The table of results below indicates, in addition to execution time, the cumulative number of guesses used by each entry for all test cases. In general, it shows the expected rough correlation between execution time and the number of guesses, with a significant exception for the second-place entry from Ernst Munter, which took significantly fewer guesses. Ernst precalculated tables to define the guessing strategy for problems of length 5 or less and devised a technique for partitioning larger problems to use these tables. Normally I try to discourage the use of extensive precalculated data, but I decided to allow this entry because the amount of data was not unreasonable, because the tables guided the algorithm but did not precalculate a solution, and because I thought the approach was innovative and interesting. Although including the second-place entry in the article is not possible because of length restrictions, I have included the preamble from Ernst’s solution describing his approach.

Here are the times and code sizes for each of the entries. Numbers in parentheses after a person’s name indicate that person’s cumulative point total for all previous Challenges, not including this one.

Name time guesses code data out-of-range

values used?

Xan Gregg (61) 102 4123 1360 16 no

Ernst Munter (90) 109 2880 6264 5480 limited

Gustav Larsson (60) 116 3700 712 40 no

Greg Linden 127 5002 576 16 no

M. Panchenko (4) 146 5391 344 16 yes

Eric Lengyel (20) 176 6456 312 16 yes

Peter Hance 206 6557 336 16 yes

J. Vineyard (42) 228 9933 328 16 no

Ken Slezak (10) 251 6544 808 16 yes

Stefan Sinclair 259 11058 200 16 yes

Top 20 Contestants of All Time

Here are the Top 20 Contestants for the Programmer’s Challenges to date. The numbers below include points awarded for this month’s entrants. (Note: ties are listed alphabetically by last name - there are more than 20 people listed this month because of ties.)

Rank Name Points

1. [Name deleted] 176

2. Munter, Ernst 100

3. Gregg, Xan 81

4. Karsh, Bill 78

5. Larsson, Gustav 67

6. Stenger, Allen 65

7. Riha, Stepan 51

8. Goebel, James 49

9. Nepsund, Ronald 47

10. Cutts, Kevin 46

11. Mallett, Jeff 44

12. Kasparian, Raffi 42

13. Vineyard, Jeremy 42

14. Darrah, Dave 31

15. Landry, Larry 29

16. Elwertowski, Tom 24

17. Lee, Johnny 22

18. Noll, Robert 22

19. Anderson, Troy 20

20. Beith, Gary 20

21. Burgoyne, Nick 20

22. Galway, Will 20

23. Israelson, Steve 20

24. Landweber, Greg 20

25. Lengyel, Eric 20

26. Pinkerton, Tom 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 Xan’s winning solution:

MindReader

By Xan Gregg,Durham, N.C.

/*  
  I try to minimize the number of guesses without adding too much complexity to the
  code.  First I figure out how many of each color are present in the answer by
  essentially repeatedly guessing all of each color.
  
  Then I figure out the correct positions one at a time starting at slot 0.  I exchange it
  with each other slot (one at a time) until the correct color is found.  When there is a
  change in the numCorrect response from checkGuess I can tell which of the two
  slots caused the change by looking at my remembered information or, if necessary,
  by performing a second guess with one of the colors in both slots.
  
  The “remembered information” includes keeping track of colors that were
  determined (via the numCorrectchange) to be wrong before and/or a swap is made. 
  This doesn’t help out too often, but it doesn’t take much time to record compared to
  calling checkGuess.
  
  While the outer loop determines the color of each slot “left-to-right” (0 to n-1), I
  found that indexing the inner loop right-to-left instead of left-to-right increased the
  speed by 30% - 40%.  I wish I understood why!
  
  Oddly, the checkGuess function spends most of its time figuring out the numWrong
  value, which we generally ignore.
*/

typedef void (*CheckGuessProcPtr)(
        unsigned char  *theGuess,
        unsigned short *numInCorrectPos,
        unsigned short *numInWrongPos);

#define kMaxLength 16

#define Bit(color) (1L << (long) (color))


MindReader

void MindReader(unsigned char guess[],
    CheckGuessProcPtr checkGuess,
        unsigned short answerLength,
        unsigned short numColors)
{
  long    prevColorsFound;
  long    colorsFound;
  long    curColor;
  long    i, j;
  long    curCorrect;
  long    numOfColor[kMaxLength + 1];  /* 1-based */
  Boolean isCorrect[kMaxLength];
  long    possibilities[kMaxLength];   /* bit fields */
  long    colorBit1;
  long    colorBit2;
  char    color1;
  char    color2;
  long    delta;
  unsigned short  newCorrect;
  unsigned short  newWrong;
  
  /* first find the correct set of colors */
  colorsFound = 0;
  curColor = 1;
  while (colorsFound < answerLength)
   {
    for (i = colorsFound; i < answerLength; i++)
      guess[i] = curColor;
    (*checkGuess)(guess, &newCorrect, &newWrong);
    prevColorsFound = colorsFound;
    colorsFound = newCorrect + newWrong;
    numOfColor[curColor] = colorsFound - prevColorsFound;
    curColor++;
   }
  
  /* now work on the order */
  for (i = 0; i < answerLength; i++)
   {
    isCorrect[i] = false;
    possibilities[i] = -1;  /* all colors */
   }
  curCorrect = newCorrect;
  /* step through every slot, starting at 0 */
  for (i = 0; curCorrect < answerLength; i++)
   {
    if (isCorrect[i])
      continue;
    color1 = guess[i];
    colorBit1 = Bit(color1);
    /* try swapping slot i with every other open */
    /* slot, starting with the last one */
    j = answerLength;
    nextSubSlot:
    j--;
    if (guess[i] == guess[j])
      goto nextSubSlot;
    if (isCorrect[j])
      goto nextSubSlot;
    color2 = guess[j];
    colorBit2 = Bit(color2);
    if ((possibilities[i] & colorBit2) == 0)
      goto nextSubSlot;  /* no hope here */
    /* swap slots i & j and check result */
    guess[i] = color2;
    guess[j] = color1;
    (*checkGuess)(guess, &newCorrect, &newWrong);
    delta = newCorrect - curCorrect;
    if (delta >= 0)
      if (delta == 0)
       {  /* either both are incorrect OR */
                           /* one is correct and answer[i]==answer[j] */
        guess[i] = color1;
        guess[j] = color2;
        if (numOfColor[color1] == 1)
         {  /* color1 can’t be in both places */
          possibilities[i] &= ~colorBit1;
          possibilities[j] &= ~colorBit1;
         }
        if (numOfColor[color2] == 1)
         {  /* color2 can’t be in both places */
          possibilities[i] &= ~colorBit2;
          possibilities[j] &= ~colorBit2;
         }
       }
      else if (delta == 1)
       {  /* both were wrong, now one is correct */
                        /* find out which is correct */
        curCorrect = newCorrect;
        if ((possibilities[j] & colorBit1) == 0)
         {  /* i must be color2 */
          possibilities[j] &= ~colorBit2;
          numOfColor[color2] -= 1;
          goto nextSlot;
         }
        else if ((possibilities[i] & colorBit2) == 0)
         {  /* j must be color1 */
          isCorrect[j] = true;
          possibilities[i] &= ~colorBit1;
          numOfColor[color1] -= 1;
          color1 = color2;
          colorBit1 = colorBit2;
         }
        else
         {  /* we’ll have to make another guess to */
                        /* see which is correct */
          guess[i] = color1;
          (*checkGuess)(guess, &newCorrect, &newWrong);
          if (newCorrect == curCorrect)
           {  /* j must be color1 */
            possibilities[i] &=
                  (~(colorBit1 | colorBit2));
            isCorrect[j] = true;
            guess[i] = color2;
            numOfColor[color1] -= 1;
            color1 = color2;
            colorBit1 = colorBit2;
           }
          else
           {  /* i must be color2 */
            possibilities[j] &=
                  (~(colorBit1 | colorBit2));
            guess[i] = color2;
            numOfColor[color2] -= 1;
            goto nextSlot;
           }
         }
       }
      else  /* delta == 2 */
       {  /* both were wrong, now both correct */
        isCorrect[j] = true;
        numOfColor[color1] -= 1;
        numOfColor[color2] -= 1;
        curCorrect = newCorrect;
        goto nextSlot;
       }
    else  /* delta < 0 */
      if (delta == -1)
       {  /* one was correct before swap, now neither is */
        guess[i] = color1;
        guess[j] = color2;
        if ((possibilities[i] & colorBit1) == 0)
         {  /* color2 in slot j was correct */
          isCorrect[j] = true;
          numOfColor[color2] -= 1;
          possibilities[i] &= ~colorBit2;
         }
        else if ((possibilities[j] & colorBit2) == 0)
         {  /* color1 in slot i was correct */
          possibilities[j] &= ~colorBit1;
          numOfColor[color1] -= 1;
          goto nextSlot;
         }
        else
         {  /* we’ll have to make another guess to */
                        /* see which was correct */
          guess[j] = color1;
          (*checkGuess)(guess, &newCorrect, &newWrong);
          if (newCorrect == curCorrect)
           {  /* color1 in slot i was correct */
            possibilities[j] &=
                  (~(colorBit1 | colorBit2));
            guess[j] = color2;
            numOfColor[color1] -= 1;
            goto nextSlot;
           }
          else
           {  /* color2 in slot j was correct */
            possibilities[i] &=
                  (~(colorBit1 | colorBit2));
            guess[j] = color2;
            isCorrect[j] = true;
            numOfColor[color2] -= 1;
           }
         }
       }
      else  /* delta == -2 */
       {  /* both were already correct */
        guess[i] = color1;
        guess[j] = color2;
        isCorrect[j] = true;
        numOfColor[color1] -= 1;
        numOfColor[color2] -= 1;
        goto nextSlot;
       }
    goto nextSubSlot;
    nextSlot: ;
   }
  done: ;
}

Alternative Approach (Description Only)

Copyright 1995, Ernst Munter, Kanata, ON, Canada.

/*
  Problem:
    Find the value of a multidigit code, by a question and answer method.  Each
    question is a guess of the code, the answer is the number of digits that are correct,
    reported as either in correct or wrong positions.

    The challenge is to minimize total time, that is in the first order, keep the number of
    guesses small, since the time to check the guess is included in total time.  But
    spending too much time minimizing the number of guesses is counterproductive.

  Assumptions:
    1. It is OK to guess a color that is not within the range 1 to numColors.  It will not
        be “correct” or “wrong”, but it will also not corrupt the CheckGuess function.

    2. The “opponent” will call with randomly generated correctAnswer codes, and not
        try to defeat the MindReader by learning the solution strategy.

    3. The objective is not to be a true Mindreader, as this could be done by reading into
         the (*checkGuess) code, the address of which is handy.  One would then
         disassemble PowerPC instructions to discover the hidden address of
         correctAnswer.

  Solution:
    It is relatively simple to manually construct solution trees for small N
    (N=answerLength), and make them into a lookup table.

    I have made a table for N=4, and hardcoded the trees for N=2 and N=3.

    The table for N=5 was too large to be done easily by hand, and I wrote a Tree
    Builder program to construct its 246 nodes.  I then hand tuned the 2 smallest parts
    of it.

    I felt, a 246 node tree is about at the limit of what might be tolerable in a static
    array.  The tree for N=6 would have 1400 or so nodes.  There are diminishing
    returns.  Adding the N=5 tree improved the higher splits (2 or 3 splits instead of 3 or
    4), but gained only a few percentage points on the callBack frequency overall;

    To keep the trees manageable, the permutation patterns and the color schemes are
    normalized.

    Now the details:

    Even if numColors > N, there can be at most N distinct colors in the answer, for
    example 5, if answerLength=5.

    And we can arrange a color mapping so that all colors are refered to by index 1, 2,
    3, etc, with the most frequently occurring color labeled #1.

    For N=5, this reduces the possible answers to 7 color schemes, 11111, 11112,
    11122, 11123, 11223, 11234, 12345.

    To solve for N<=5, the function “ProcessSlice()” only needs the color mapping, and
    a list of the colors, suitably sorted.

    For example, the real answer “73646” can be solved by walking the solution tree in
    4 steps, given the color list 6,3,4,7 and the pattern to be found is 42131.  The
    pattern at the root of the tree T11234, is 11234.

    To obtain the pattern information, I “scan” the answer with successive guesses
    (somewhat optimized for answer lengths of 2 to 4, to eliminate some obviously
    unneeded calls to checkGuess).  The basic idea is:

    correctAnswer
    7 3 6 4 6

    Six or seven calls to checkGuess, to build the color and color-frequency lists:

    guess       correct wrong   yetToFind colorList
    1 1 1 1 1   0       0       5         -
    2 2 2 2 2   0       0       5         -
    3 3 3 3 3   1       0       4         3
    4 4 4 4 4   1       0       3         3,4
    5 5 5 5 5   0       0       3         3,4
    6 6 6 6 6   2       0       1         6,3,4
    7 7 7 7 7   1       0       0         6,3,4,7

    The last call back is avoided if the color==numColors occurs in the code.

    Then, using the tree, the correct answer is found with four more calls to checkGuess:
                                 goal   42131
    6 6 3 4 7   1       x   tree code   11234
    6 3 6 7 4   2       x               12143
    4 3 6 6 7   2       x               32114
    4 6 6 7 3   1       x               31142

    7 3 6 4 6                           42131 (no other choice)

    This results in a total of 10 or 11 calls or less to the checkGuess function.

    On average, 10 calls are needed to solve 5-wide answers, when numColors is
    randomly set to a value from 1 to 16.

    For N>5, the size of tree grows very rapidly.  So I decided to split the answer into
    multiple slices, and treat each as separate problems of width 3, 4, or 5:

    6 = 3 + 3
    7 = 4 + 3
    8 = 4 + 4
    9 = 5 + 4
    10 = 5 + 5
    11 = 5 + 6 = 5 + (3 + 3)
    12 = 5 + 7 = 5 + (4 + 3)
    13 = 5 + 8 = 5 + (4 + 4)
    14 = 5 + 9 = 5 + (5 + 4)
    15 = 5 + 10 = 5 + (5 + 5)
    16 = 8 + 8 = (4 + 4) + (4 + 4)

    To create a split, we call checkGuess with guesses of a solid color for the left side,
    and 0s for the right. (e.g. first guess 1 1 1 1 0 0 0 0, to split 8).  As a result,
    correctPos gives the number of 1s in the left slice, and wrongPos, the number of 1s
    in the right slice.  If we get correctPos+wrongPos=4 as an answer, we must call
    again because there might be more than four 1s in the answer;  the guess 0 0 0 0 1 1
    1 1 will do it.

    Performance:

    Overall, I find an almost linear relationship between the total number of call backs
    (CB) and the value of answerLength (AL), approximately CB = AL * 1.26 + 2.84
    when numColors varies randomly from 1 to 16.
*/

 
AAPL
$111.78
Apple Inc.
-0.87
MSFT
$47.66
Microsoft Corpora
+0.14
GOOG
$516.35
Google Inc.
+5.25

MacTech Search:
Community Search:

Software Updates via MacUpdate

calibre 2.13 - Complete e-library manage...
Calibre is a complete e-book library manager. Organize your collection, convert your books to multiple formats, and sync with all of your devices. Let Calibre be your multi-tasking digital librarian... Read more
Mellel 3.3.7 - Powerful word processor w...
Mellel is the leading word processor for OS X and has been widely considered the industry standard since its inception. Mellel focuses on writers and scholars for technical writing and multilingual... Read more
ScreenFlow 5.0.1 - Create screen recordi...
Save 10% with the exclusive MacUpdate coupon code: AFMacUpdate10 Buy now! ScreenFlow is powerful, easy-to-use screencasting software for the Mac. With ScreenFlow you can record the contents of your... Read more
Simon 4.0 - Monitor changes and crashes...
Simon monitors websites and alerts you of crashes and changes. Select pages to monitor, choose your alert options, and customize your settings. Simon does the rest. Keep a watchful eye on your... Read more
BBEdit 11.0.2 - Powerful text and HTML e...
BBEdit is the leading professional HTML and text editor for the Mac. Specifically crafted in response to the needs of Web authors and software developers, this award-winning product provides a... Read more
ExpanDrive 4.2.1 - Access cloud storage...
ExpanDrive builds cloud storage in every application, acts just like a USB drive plugged into your Mac. With ExpanDrive, you can securely access any remote file server directly from the Finder or... Read more
Adobe After Effects CC 2014 13.2 - Creat...
After Effects CC 2014 is available as part of Adobe Creative Cloud for as little as $19.99/month (or $9.99/month if you're a previous After Effects customer). After Effects CS6 is still available... Read more
Evernote 6.0.5 - Create searchable notes...
Evernote allows you to easily capture information in any environment using whatever device or platform you find most convenient, and makes this information accessible and searchable at anytime, from... Read more
Command-C 1.1.7 - Clipboard sharing tool...
Command-C is a revolutionary app which makes easy to share your clipboard between iOS and OS X using your local WiFi network, even if the app is not currently opened. Copy anything (text, pictures,... Read more
Tidy Up 4.0.2 - Find duplicate files and...
Tidy Up is a complete duplicate finder and disk-tidiness utility. With Tidy Up you can search for duplicate files and packages by the owner application, content, type, creator, extension, time... Read more

Latest Forum Discussions

See All

Make your own Tribez Figures (and More)...
Make your own Tribez Figures (and More) with Toyze Posted by Jessica Fisher on December 19th, 2014 [ permalink ] Universal App - Designed for iPhone and iPad | Read more »
So Many Holiday iOS Sales Oh My Goodness...
The holiday season is in full-swing, which means a whole lot of iOS apps and games are going on sale. A bunch already have, in fact. Naturally this means we’re putting together a hand-picked list of the best discounts and sales we can find in order... | Read more »
It’s Bird vs. Bird in the New PvP Mode f...
It’s Bird vs. Bird in the New PvP Mode for Angry Birds Epic Posted by Jessica Fisher on December 19th, 2014 [ permalink ] Universal App - Designed for iPhone and iPad | Read more »
Telltale Games and Mojang Announce Minec...
Telltale Games and Mojang Announce Minecraft: Story Mode – A Telltale Games Series Posted by Jessica Fisher on December 19th, 2014 [ permalink ] | Read more »
WarChest and Splash Damage Annouce Their...
WarChest and Splash Damage Annouce Their New Game: Tempo Posted by Jessica Fisher on December 19th, 2014 [ permalink ] WarChest Ltd and Splash Damage Ltd are teaming up again to work | Read more »
BulkyPix Celebrates its 6th Anniversary...
BulkyPix Celebrates its 6th Anniversary with a Bunch of Free Games Posted by Jessica Fisher on December 19th, 2014 [ permalink ] BulkyPix has | Read more »
Indulge in Japanese cuisine in Cooking F...
Indulge in Japanese cuisine in Cooking Fever’s new sushi-themed update Posted by Simon Reed on December 19th, 2014 [ permalink ] Lithuanian developer Nordcurrent has yet again updated its restaurant simulat | Read more »
Badland Daydream Level Pack Arrives to C...
Badland Daydream Level Pack Arrives to Celebrate 20 Million Downloads Posted by Ellis Spice on December 19th, 2014 [ permalink ] | Read more »
Far Cry 4, Assassin’s Creed Unity, Desti...
Far Cry 4, Assassin’s Creed Unity, Destiny, and Beyond – AppSpy Takes a Look at AAA Companion Apps Posted by Rob Rich on December 19th, 2014 [ permalink ] These day | Read more »
A Bunch of Halfbrick Games Are Going Fre...
A Bunch of Halfbrick Games Are Going Free for the Holidays Posted by Ellis Spice on December 19th, 2014 [ permalink ] Universal App - Designed for iPhone and iPad | Read more »

Price Scanner via MacPrices.net

The Apple Store offering free next-day shippi...
The Apple Store is now offering free next-day shipping on all in stock items if ordered before 12/23/14 at 10:00am PT. Local store pickup is also available within an hour of ordering for any in stock... Read more
It’s 1992 Again At Sony Pictures, Except For...
Techcrunch’s John Biggs interviewed a Sony Pictures Entertainment (SPE) employee, who quite understandably wished to remain anonymous, regarding post-hack conditions in SPE’s L.A office, explaining “... Read more
Holiday sales this weekend: MacBook Pros for...
 B&H Photo has new MacBook Pros on sale for up to $300 off MSRP as part of their Holiday pricing. Shipping is free, and B&H charges NY sales tax only: - 15″ 2.2GHz Retina MacBook Pro: $1699... Read more
Holiday sales this weekend: MacBook Airs for...
B&H Photo has 2014 MacBook Airs on sale for up to $120 off MSRP, for a limited time, for the Thanksgiving/Christmas Holiday shopping season. Shipping is free, and B&H charges NY sales tax... Read more
Holiday sales this weekend: iMacs for up to $...
B&H Photo has 21″ and 27″ iMacs on sale for up to $200 off MSRP including free shipping plus NY sales tax only. B&H will also include a free copy of Parallels Desktop software: - 21″ 1.4GHz... Read more
Holiday sales this weekend: Mac minis availab...
B&H Photo has new 2014 Mac minis on sale for up to $80 off MSRP. Shipping is free, and B&H charges NY sales tax only: - 1.4GHz Mac mini: $459 $40 off MSRP - 2.6GHz Mac mini: $629 $70 off MSRP... Read more
Holiday sales this weekend: Mac Pros for up t...
B&H Photo has Mac Pros on sale for up to $500 off MSRP. Shipping is free, and B&H charges sales tax in NY only: - 3.7GHz 4-core Mac Pro: $2599, $400 off MSRP - 3.5GHz 6-core Mac Pro: $3499, $... Read more
Save up to $400 on MacBooks with Apple Certif...
The Apple Store has Apple Certified Refurbished 2014 MacBook Pros and MacBook Airs available for up to $400 off the cost of new models. An Apple one-year warranty is included with each model, and... Read more
Save up to $300 on Macs, $30 on iPads with Ap...
Purchase a new Mac or iPad at The Apple Store for Education and take up to $300 off MSRP. All teachers, students, and staff of any educational institution qualify for the discount. Shipping is free,... Read more
iOS and Android OS Targeted by Man-in-the-Mid...
Cloud services security provider Akamai Technologies, Inc. has released, through the company’s Prolexic Security Engineering & Research Team (PLXsert), a new cybersecurity threat advisory. The... Read more

Jobs Board

*Apple* Store Leader Program (US) - Apple, I...
…Summary Learn and grow as you explore the art of leadership at the Apple Store. You'll master our retail business inside and out through training, hands-on experience, Read more
Project Manager, *Apple* Financial Services...
**Job Summary** Apple Financial Services (AFS) offers consumers, businesses and educational institutions ways to finance Apple purchases. We work with national and Read more
*Apple* Retail - Multiple Positions (US) - A...
Sales Specialist - Retail Customer Service and Sales Transform Apple Store visitors into loyal Apple customers. When customers enter the store, you're also the Read more
*Apple* Retail - Multiple Positions (US) - A...
Sales Specialist - Retail Customer Service and Sales Transform Apple Store visitors into loyal Apple customers. When customers enter the store, you're also the Read more
*Apple* Retail - Multiple Positions (US) - A...
Job Description: Sales Specialist - Retail Customer Service and Sales Transform Apple Store visitors into loyal Apple customers. When customers enter the store, Read more
All contents are Copyright 1984-2011 by Xplain Corporation. All rights reserved. Theme designed by Icreon.