Jun 95 Challenge
 Volume Number: 11 Issue Number: 6 Column Tag: Programmer’s Challenge

# Programmer’s Challenge

By Bob Boonstra and Mike Scanlin

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

## Goodbye From Mike

This month is a special month. I have decided to turn the Programmer Challenge over to Bob Boonstra. Readers of this column will recognize Bob’s name from his many excellent winning solutions to the Challenges I’ve posed. In fact, Bob is the proud owner of six 1st-place showings. I can think of no-one I’d rather turn this column over to than Bob. I will judge the last two puzzles I have posed and Bob will judge the puzzles he poses (the first of which, Check Checkmate, is in this issue).

It’s been almost three years since I first started this column. While it has been incredibly rewarding to see so much optimal code and meet so many like-minded efficiency nuts, my life has changed during those three years and I no longer have the time that this column deserves. Even though I won't be writing it any more I'm sure this column will remain my favorite part of MacTech (I may even submit an entry once in a while...).

Thanks to everyone who took the time to write to me over the years with suggestions for Challenges. I have given Bob the compiled list of ideas. I’m sure he’d love to hear from you if you have any other ideas or comments on what you want to see in this column. I leave you in Bob’s capable hands...

Optimally yours,

Mike Scanlin
scanlin@genmagic.com

## Check Checkmate

This month’s Challenge deals with the game of chess. You will be given a chess position and asked to produce the list of moves and captures that it is legal for one of the sides to make in accordance with the rules of chess. The objective is to produce the legal move list in minimum time.

Here are the prototypes for the routines you should write:

```typedef enum {rowA=0,rowB,rowC,rowD,rowE,rowF,rowG,rowH} Row;
typedef enum {col1=0,col2,col3,col4,col5,col6,col7,col8} Col;
typedef enum {whiteSide=0,blackSide} Side;
typedef enum {king=0,queen,rook,bishop,knight,pawn} ChessPiece ;

typedef struct Square {
Row row;
Col col;
} Square;

typedef struct PiecePosition {
Square sq;
Side side;
ChessPiece piece;
} PiecePosition;

typedef struct ChessMove {
Square fromSq;
Square toSq;
Boolean moveIsCapture;
Boolean opponentPlacedInCheck;
} ChessMove;

short /*numberOfMoves*/ LegalChessMoves(
PiecePosition piecePositionArray[],
short numberOfPieces,
Side sideToMove,
ChessMove legalMoveArray[],
void *privateDataPtr
);

void InitChess(void *privateDataPtr);

```

This Challenge will consist of one call to InitChess, followed by multiple calls to LegalChessMoves. The InitChess routine will not be timed, and may initialize up to 32K bytes of storage preallocated by my testbench and pointed to by privateDataPtr. Your program should not use any static storage besides that pointed to by privateDataPtr.

Each call to LegalChessMoves will provide a piecePositionArray containing numberOfPieces chess pieces. Each piece is described in a PiecePosition struct containing the ChessPiece, Side, and position. The position is provided as a Square struct containing a Row and a Col, where rowA represents the starting row for the White major pieces, and rowH the starting row for Black. The columns are numbered from left to right as viewed from the White side, so that (rowH, col4) is the starting square of the Black queen in a new game. The piecePositionArray will describe a legal set of chess piece positions, anything from the initial positions in a new game to an end-game position. No position will be provided that could not be reached during a chess game. Remember that due to past pawn promotions, a side may have (for example) more than one queen. You should generate the moves for the side provided in sideToMove. The privateDataPtr parameter will be the same pointer provided to InitChess.

LegalChessMoves should store in legalMoveArray the complete list of legal ChessMoves for the sideToMove. The legalMoveArray will be allocated by my testbench and will be large enough to hold all of the legal moves for the given chess position. You should describe each legal move/capture by placing the row/col location of the moving piece in fromSq and the rol/col of the destination in toSq. In addition, for each move, the Boolean moveIsCapture should be set to true if the move captures an opponent's piece (and set to false otherwise). Similarly, the Boolean opponentPlacedInCheck should be set to true if the move places the opponent in check (and set to false otherwise). The return value of the function should be the number of moves stored in legalMoveArray. In generating the list of legal moves, you should assume that castling moves or en passant pawn captures cannot occur. Remember that it is not legal to make a chess move which places or leaves the moving side's king in check. If the sideToMove has been checkmated and cannot move, LegalChessMoves should return zero.

This month brings one change to the rules of the contest. From now on, with apologies (and sympathies) to any MacPlus users still out there, the 68020 code generation option will be turned ON. This Challenge will be scored using the THINK C compiler and 68K instruction set, but future Challenges will include occasional use of the CodeWarrior compiler and/or the PowerPC instruction set. This month, you should also include the following pragmas in your code:

```#pragma options (mc68020, !mc68881, require_protos)
#pragma options (pack_enums, align_arrays)

```

Have fun, and e-mail me if there are any questions.

## Two Months Ago Winner

I guess most people spent the beginning of April working on their taxes instead of the Stock Market Database Challenge because I only received 3 entries. Of those, only two worked correctly. Despite the lack of competition, I’m happy to say that the winning solution is quite good. Congratulations to Xan Gregg (Durham, NC) for his efficient solution. And thanks to Ernst Munter for taking the time to enter. Ernst was at a bit of a disadvantage, considering he had to develop his code on a DOS-based machine that didn’t have the RAM or disk space specified in the problem.

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

Name time code

Xan Gregg (4) 3523 1766

Ernst Munter (53) 61714 3470

The key to implementing a fast database where not everything fits into RAM is to have fast RAM-based indexes to your disk-based data and to minimize disk accesses. Also, for those times when you do read/write to the disk it’s really important to do it in even sector amounts. Xan uses 8K reads/writes. Sectors on most devices are 512 bytes each so 8K is a good choice. If your application ever needs to swap data to disk then you should study Xan’s code. It could save your users a lot of time.

## Top 20 Contestants Of All Time

Here are the Top 20 Contestants for the 34 Programmer’s Challenges to date. The numbers below include points awarded for this months’ entrants. (Note: ties are listed alphabetically by last name - there are 24 people listed this month because 7 people have 20 points each.)

1. Boonstra, Bob 176

2. Karsh, Bill 71

3. Stenger, Allen 65

4. Munter, Ernst 63

6. Riha, Stepan 51

7. Goebel, James 49

8. Cutts, Kevin 46

9. Nepsund, Ronald 40

10. Vineyard, Jeremy 40

11. Darrah, Dave 34

12. Mallett, Jeff 34

13. Landry, Larry 29

14. Elwertowski, Tom 24

15. Gregg, Xan 24

16. Kasparian, Raffi 24

17. Lee, Johnny 22

18. Anderson, Troy 20

19. Burgoyne, Nick 20

20. Galway, Will 20

21. Israelson, Steve 20

22. Landweber, Greg 20

23. Noll, Bob 20

24. Pinkerton, Tom 20

There are three ways to earn points: (1) by scoring in the top 5 of any challenge, (2) by 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 5 points

suggesting challenge 2 points

Here is Xan’s winning solution:

```#include <OSUtils.h>
#include <Memory.h>
#include <Files.h>

/* OVERVIEW  market.c - by Xan Gregg
I cache all symbol names in memory and as many blocks
of trade data from disk as possible are cached in the rest
of the allowed memory.  For each symbol I keep a “pointer”
(block# and entry#) to the most recent trade involving that
symbol.

The header includes an array of 26K entries in which each
entry points to a list of symbols.  All symbols with the
same first three letters are on the same list.  So the array
of symbol lists is indexed by the first three letters.

The trade data are stored on disk as a summary of the trade
information passed to NewTrade().  The time is stored in
a long, and the volume is stored as a daily cummulative
value.  The data is accumulated into a 8K block which
is written to disk when it gets full.

I know that there will be under 500,000 trades (since
the trade is at most 10MB and each Trade struct is 22
bytes).  And I have been told there will be about 10,000
symbols and at most 16K.  So that puts each symbol at
an average of 50 trades each.  That’s good for my approach
of indexing by the symbol name.  However, since I iterate
through the list of trades in reverse order, I am at a
disadvantage if a lot of older data is requested.

*/

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

typedef struct TimeStamp {
uchar  yearsFrom1900;
uchar  month;
uchar  day;
uchar  hour;
uchar  minute;
uchar  second;
} TimeStamp;

typedef char Str7[8];

Str7   symbol;
TimeStamptime;
Fixed  price;
ulong  numShares;

Fixed PriceIs(void *privateDataPtr, Str7 symbol, TimeStamp time);
ulong VolumeIs(void *privateDataPtr, Str7 symbol, TimeStamp time);

#define kBlockSize  8192L
#define kBlockEntries   (kBlockSize/16)
#define kFileSize   (8*1024L*1024L)
/* enough to store 500,000 trades */
#define kNumBlocks  1024L

#define kNumSymbolIndices (26*32*32L)
#define kSecondsPerDay  (60L*60L*24)

typedef struct SymbolEntry SymbolEntry;

struct SymbolEntry
{
long        restOfSymbol;   /* last 4 chars */
SymbolEntry *next;          /* with same 1st 3 chars */
short       latestBlockNum; /* location on disk */
short       latestEntryNum; /* location with block */
};

/* 16-byte struct stored on disk for each trade */
typedef struct
{
ulong   seconds;      /* seconds since 1904 */
Fixed   price;        /* same as from Trade record */
ulong   totalShares;  /* current and previous */
short   prevBlockNum;
short   prevEntryNum;

typedef struct
{
short      blockNum;   /* where it is on disk */
short   filler;
} CachedBlock;

typedef struct
{
CachedBlock  *blockLocation[kNumBlocks]; /* 4KB */
Ptr          symbolDataEnd;
CachedBlock  *lowestBlockP;
CachedBlock  *highestBlockP;
CachedBlock  *oldestBlockP;
CachedBlock  *currentOutBlockP;
short        currentEntryNum;
short        refNum;
ParamBlockRec pb;
ushort       symbolIndex[kNumSymbolIndices]; /* 26KB */
SymbolEntry  symbols[1]; /* variable length */

/* the above struct takes less than 32K, and the
symbol data takes less than 192K (assuming a max of
16K symbols), so the cached blocks can use everything
after the first 224K */

#define gBlockLocation      dataP->blockLocation
#define gSymbolDataEnd      dataP->symbolDataEnd
#define gLowestBlockP       dataP->lowestBlockP
#define gHighestBlockP      dataP->highestBlockP
#define gOldestBlockP       dataP->oldestBlockP
#define gCurrentOutBlockP   dataP->currentOutBlockP
#define gCurrentEntryNum    dataP->currentEntryNum
#define gRefNum             dataP->refNum
#define gPB                 dataP->pb
#define gSymbolIndex        dataP->symbolIndex
#define gSymbols            dataP->symbols

{
CachedBlock *blockP;

maxRAM = maxRAM & -4L;  /* make sure end is aligned */
if (dataP == 0)
DebugStr(“\p NO MEM”);

/* skip the first symbol as 0 is not a valid index */
gSymbolDataEnd = (Ptr) (gSymbols + 1);

/* Initialize memory blocks at end of data area. */
blockP = (CachedBlock *) (((Ptr) dataP) + maxRAM);
gCurrentOutBlockP = blockP - 1;
/* leaving 224K for headers and symbols */
while ((Ptr) (blockP - 1) > ((Ptr) dataP) + 224*1024L)
{
blockP-;
blockP->blockNum = -1;
}
gLowestBlockP = blockP;
gHighestBlockP = gCurrentOutBlockP;
gOldestBlockP = gHighestBlockP - 1;
gCurrentOutBlockP->blockNum = 0;
gBlockLocation[0] = gCurrentOutBlockP;

{ /* create swap file */
OSErr   err;
long    size;
err = Create(“\pXansSwapFile”, 0, ‘xxxx’, ‘DATA’);
if (err)
{
FSDelete(“\pXansSwapFile”, 0);
err = Create(“\pXansSwapFile”, 0, ‘xxxx’, ‘DATA’);
if (err)
DebugStr(“\p NO FILE”);
}
FSOpen(“\pXansSwapFile”, 0, &gRefNum);
if (gRefNum == 0)
DebugStr(“\p NO FILE”);
size = kFileSize;
err = AllocContig(gRefNum, &size);
size = kFileSize;
if (err)
err = Allocate(gRefNum, &size);
if (err)
DebugStr(“\p NO FILE”);
SetEOF(gRefNum, kFileSize);
}
return (void *) dataP;
}

/* Take a chance and use OS trap here.  If assembly
were allowed could at least cache the address of the
code and call it with a function pointer instead of
through a trap, or we could just write this whole
routine is assembly.
*/
ulong timeToSeconds(TimeStamp *timeP)
{
DateTimeRec dateTime;
ulong       seconds;

dateTime.year = 1900 + timeP->yearsFrom1900;
dateTime.month = timeP->month;
dateTime.day = timeP->day;
dateTime.hour = timeP->hour;
dateTime.minute = timeP->minute;
dateTime.second = timeP->second;
Date2Secs(&dateTime, &seconds);
return seconds;
}

{
register long           rest;
register SymbolEntry    *symbolP;
long        memRest;
short       keyIndex;
char        secondChar;
char        thirdChar;
char        *p;
short       len = symbol[0];

/* figure the keyIndex (first three letters) */
if (len < 2)
{
secondChar = 0;
thirdChar = 0;
}
else
{
secondChar = symbol[2] - ‘@’;
if (len != 2)
thirdChar = symbol[3] - ‘@’;
else
thirdChar = 0;
}
keyIndex = ((symbol[1] - ‘A’) << 10)
| (secondChar << 5) | thirdChar;

/* and the ‘rest’ (last four letters) */
memRest = 0;
p = (char *) &memRest;
while (len > 3)
*p++ = symbol[len-];

rest = memRest;
if (gSymbolIndex[keyIndex] == 0)
symbolP = 0;
else
{
symbolP = &gSymbols[gSymbolIndex[keyIndex]];
while ((symbolP != 0)
&& (symbolP->restOfSymbol != rest))
symbolP = symbolP->next;
}
return symbolP;
}

/* like findSymbol, but inserts it if necessary */
Str7 symbol)
{
register long           rest;
register SymbolEntry    *symbolP;
long        memRest;
short       keyIndex;
char        secondChar;
char        thirdChar;
char        *p;
short       len = symbol[0];

/* figure the keyIndex (first three letters) */
if (len < 2)
{
secondChar = 0;
thirdChar = 0;
}
else
{
secondChar = symbol[2] - ‘@’;
if (len != 2)
thirdChar = symbol[3] - ‘@’;
else
thirdChar = 0;
}
keyIndex = ((symbol[1] - ‘A’) << 10)
| (secondChar << 5) | thirdChar;

/* and the ‘rest’ (last four letters) */
memRest = 0;
p = (char *) &memRest;
while (len > 3)
*p++ = symbol[len-];

rest = memRest;
if (gSymbolIndex[keyIndex] == 0)
{  /* this is the first symbol with this key */
symbolP = (SymbolEntry *) gSymbolDataEnd;
gSymbolIndex[keyIndex] =
(ushort) (symbolP - gSymbols);
gSymbolDataEnd += sizeof(SymbolEntry);
symbolP->restOfSymbol = rest;
symbolP->next = 0;
symbolP->latestBlockNum = -1;
}
else
{  /* search the list for our last four letters */
symbolP = &gSymbols[gSymbolIndex[keyIndex]];
while ((symbolP->next != 0) &&
(symbolP->restOfSymbol != rest))
symbolP = symbolP->next;
if (symbolP->restOfSymbol != rest)
symbolP->next = (SymbolEntry *) gSymbolDataEnd;
symbolP = symbolP->next;
gSymbolDataEnd += sizeof(SymbolEntry);
symbolP->restOfSymbol = rest;
symbolP->next = 0;
symbolP->latestBlockNum = -1;
}
}
return symbolP;
}

/* returns a pointer to a block in memory.  If the
block is not already in memory, it is loaded from disk. */
{
CachedBlock     *blockP;

blockP = gBlockLocation[blockNum];
if (blockP == 0)
{  /* need to load it from disk */
if (gOldestBlockP->blockNum >= 0)
gBlockLocation[gOldestBlockP->blockNum] = 0;
gOldestBlockP->blockNum = blockNum;
| (1 << 5); /* don’t cache */
* kBlockSize;
gBlockLocation[blockNum] = gOldestBlockP;

/* find the next oldest block */
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
if (gOldestBlockP == gCurrentOutBlockP)
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
if ((gPB.ioParam.ioResult > 0)
&& (((Ptr) gOldestBlockP->data)
== gPB.ioParam.ioBuffer))
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
if (gOldestBlockP == gCurrentOutBlockP)
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
/* wait for read to complete */
;
}
return blockP;
}

{
CachedBlock *p;

/* make sure any previous write is done */
while (gPB.ioParam.ioResult > 0)
;
gPB.ioParam.ioRefNum = gRefNum;
gPB.ioParam.ioBuffer = (Ptr) gCurrentOutBlockP->data;
gPB.ioParam.ioReqCount = kBlockSize;
gPB.ioParam.ioPosMode = fsFromStart;
gPB.ioParam.ioPosOffset =
((long) gCurrentOutBlockP->blockNum) * kBlockSize;
PBWrite(&gPB, true);

/* find the next oldest block */
p = gOldestBlockP;
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
if (gOldestBlockP == gCurrentOutBlockP)
if (gOldestBlockP == gLowestBlockP)
gOldestBlockP = gHighestBlockP;
else
gOldestBlockP-;
gCurrentOutBlockP = p;
}

{
ulong           seconds;
SymbolEntry     *symbolP;

if (gCurrentEntryNum == kBlockEntries)
{  /* current block is full, so write it out */
short   curBlockNum = gCurrentOutBlockP->blockNum;

writeCurrentBlock(dataP);
gCurrentEntryNum = 0;
if (gCurrentOutBlockP->blockNum >= 0)
gBlockLocation[gCurrentOutBlockP->blockNum] = 0;
gCurrentOutBlockP->blockNum = curBlockNum + 1;
gBlockLocation[gCurrentOutBlockP->blockNum] =
gCurrentOutBlockP;
}
if (symbolP->latestBlockNum >= 0)
{
CachedBlock *blockP;
ulong       todaySeconds;

todaySeconds = seconds - (seconds % kSecondsPerDay);
blockP = getBlock(dataP, symbolP->latestBlockNum);
}
symbolP->latestBlockNum = gCurrentOutBlockP->blockNum;
symbolP->latestEntryNum = gCurrentEntryNum;
gCurrentEntryNum++;
}

Fixed PriceIs(void *privateP, Str7 symbol, TimeStamp time)
{
ulong       seconds;
SymbolEntry *symbolP;

seconds = timeToSeconds(&time);
symbolP = findSymbolEntry(dataP, symbol);
if (symbolP)
{
CachedBlock     *blockP;

/* look for the most recent trade that is before
or on the requested time */
blockP = getBlock(dataP, symbolP->latestBlockNum);
{
return 0;
}
}
else
return 0;
}

ulong VolumeIs(void *privateP, Str7 symbol, TimeStamp time)
{
ulong       seconds;
SymbolEntry *symbolP;
ulong       todaySeconds;

seconds = timeToSeconds(&time);
todaySeconds = seconds - (seconds % kSecondsPerDay);
symbolP = findSymbolEntry(dataP, symbol);
if (symbolP)
{
CachedBlock     *blockP;

/* look for the most recent trade that is before
the requested time and on the same day */
blockP = getBlock(dataP, symbolP->latestBlockNum);
{
return 0;
}
return 0;
else
}
else
return 0;

}
```

Community Search:
MacTech Search:

## Latest Forum Discussions

Fresh From the Land Down Under – The Tou...
After a two week hiatus, we are back with another episode of The TouchArcade Show. Eli is fresh off his trip to Australia, which according to him is very similar to America but more upside down. Also kangaroos all over. Other topics this week... | Read more »
TouchArcade Game of the Week: ‘Dungeon T...
I’m a little conflicted on this week’s pick. Pretty much everyone knows the legend of Dungeon Raid, the match-3 RPG hybrid that took the world by storm way back in 2011. Everyone at the time was obsessed with it, but for whatever reason the... | Read more »
Hello gentle readers, and welcome to the SwitchArcade Round-Up for July 19th, 2024. In today’s article, we finish up the week with the unusual appearance of a review. I’ve spent my time with Hot Lap Racing, and I’m ready to give my verdict. After... | Read more »
Draknek Interview: Alan Hazelden on Thin...
Ever since I played my first release from Draknek & Friends years ago, I knew I wanted to sit down with Alan Hazelden and chat about the team, puzzle games, and much more. | Read more »
The Latest ‘Marvel Snap’ OTA Update Buff...
I don’t know about all of you, my fellow Marvel Snap (Free) players, but these days when I see a balance update I find myself clenching my… teeth and bracing for the impact to my decks. They’ve been pretty spicy of late, after all. How will the... | Read more »
‘Honkai Star Rail’ Version 2.4 “Finest D...
HoYoverse just announced the Honkai Star Rail (Free) version 2.4 “Finest Duel Under the Pristine Blue" update alongside a surprising collaboration. Honkai Star Rail 2.4 follows the 2.3 “Farewell, Penacony" update. Read about that here. | Read more »
‘Vampire Survivors+’ on Apple Arcade Wil...
Earlier this month, Apple revealed that poncle’s excellent Vampire Survivors+ () would be heading to Apple Arcade as a new App Store Great. I reached out to poncle to check in on the DLC for Vampire Survivors+ because only the first two DLCs were... | Read more »
Homerun Clash 2: Legends Derby opens for...
Since launching in 2018, Homerun Clash has performed admirably for HAEGIN, racking up 12 million players all eager to prove they could be the next baseball champions. Well, the title will soon be up for grabs again, as Homerun Clash 2: Legends... | Read more »
‘Neverness to Everness’ Is a Free To Pla...
Perfect World Games and Hotta Studio (Tower of Fantasy) announced a new free to play open world RPG in the form of Neverness to Everness a few days ago (via Gematsu). Neverness to Everness has an urban setting, and the two reveal trailers for it... | Read more »
Meditative Puzzler ‘Ouros’ Coming to iOS...
Ouros is a mediative puzzle game from developer Michael Kamm that launched on PC just a couple of months back, and today it has been revealed that the title is now heading to iOS and Android devices next month. Which is good news I say because this... | Read more »

## Price Scanner via MacPrices.net

Amazon is still selling 16-inch MacBook Pros...
Prime Day in July is over, but Amazon is still selling 16-inch Apple MacBook Pros for \$500-\$600 off MSRP. Shipping is free. These are the lowest prices available this weekend for new 16″ Apple... Read more
Walmart continues to sell clearance 13-inch M...
Walmart continues to offer clearance, but new, Apple 13″ M1 MacBook Airs (8GB RAM, 256GB SSD) online for \$699, \$300 off original MSRP, in Space Gray, Silver, and Gold colors. These are new MacBooks... Read more
Apple is offering steep discounts, up to \$600...
Apple has standard-configuration 16″ M3 Max MacBook Pros available, Certified Refurbished, starting at \$2969 and ranging up to \$600 off MSRP. Each model features a new outer case, shipping is free,... Read more
Save up to \$480 with these 14-inch M3 Pro/M3...
Apple has 14″ M3 Pro and M3 Max MacBook Pros in stock today and available, Certified Refurbished, starting at \$1699 and ranging up to \$480 off MSRP. Each model features a new outer case, shipping is... Read more
Amazon has clearance 9th-generation WiFi iPad...
Amazon has Apple’s 9th generation 10.2″ WiFi iPads on sale for \$80-\$100 off MSRP, starting only \$249. Their prices are the lowest available for new iPads anywhere: – 10″ 64GB WiFi iPad (Space Gray or... Read more
Apple is offering a \$50 discount on 2nd-gener...
Apple has Certified Refurbished White and Midnight HomePods available for \$249, Certified Refurbished. That’s \$50 off MSRP and the lowest price currently available for a full-size Apple HomePod today... Read more
The latest MacBook Pro sale at Amazon: 16-inc...
Amazon is offering instant discounts on 16″ M3 Pro and 16″ M3 Max MacBook Pros ranging up to \$400 off MSRP as part of their early July 4th sale. Shipping is free. These are the lowest prices... Read more
14-inch M3 Pro MacBook Pros with 36GB of RAM...
B&H Photo has 14″ M3 Pro MacBook Pros with 36GB of RAM and 512GB or 1TB SSDs in stock today and on sale for \$200 off Apple’s MSRP, each including free 1-2 day shipping: – 14″ M3 Pro MacBook Pro (... Read more
14-inch M3 MacBook Pros with 16GB of RAM on s...
B&H Photo has 14″ M3 MacBook Pros with 16GB of RAM and 512GB or 1TB SSDs in stock today and on sale for \$150-\$200 off Apple’s MSRP, each including free 1-2 day shipping: – 14″ M3 MacBook Pro (... Read more
Amazon is offering \$170-\$200 discounts on new...
Amazon is offering a \$170-\$200 discount on every configuration and color of Apple’s M3-powered 15″ MacBook Airs. Prices start at \$1129 for models with 8GB of RAM and 256GB of storage: – 15″ M3... Read more

## Jobs Board

*Apple* Systems Engineer - Chenega Corporati...
…LLC,** a **Chenega Professional Services** ' company, is looking for a ** Apple Systems Engineer** to support the Information Technology Operations and Maintenance Read more
Solutions Engineer - *Apple* - SHI (United...
**Job Summary** An Apple Solution Engineer's primary role is tosupport SHI customers in their efforts to select, deploy, and manage Apple operating systems and Read more
*Apple* / Mac Administrator - JAMF Pro - Ame...
Amentum is seeking an ** Apple / Mac Administrator - JAMF Pro** to provide support with the Apple Ecosystem to include hardware and software to join our team and Read more
Operations Associate - *Apple* Blossom Mall...
Operations Associate - Apple Blossom Mall Location:Winchester, VA, United States (https://jobs.jcp.com/jobs/location/191170/winchester-va-united-states) - Apple Read more
Cashier - *Apple* Blossom Mall - JCPenney (...
Cashier - Apple Blossom Mall Location:Winchester, VA, United States (https://jobs.jcp.com/jobs/location/191170/winchester-va-united-states) - Apple Blossom Mall Read more