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Feb 94 Challenge
Volume Number:10
Issue Number:2
Column Tag:Programmers’ Challenge

Programmers’ Challenge

By Nice Silk Man

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

The Rules

Here’s how it works: Each month there will be a different programming challenge presented here. First, you must write some code that solves the challenge. Second, you must optimize your code (a lot). Then, submit your solution to MacTech Magazine (formerly MacTutor). A winner will be chosen based on code correctness, speed, size and elegance (in that order of importance) as well as the postmark of the answer. In the event of multiple equally desirable solutions, one winner will be chosen at random (with honorable mention, but no prize, given to the runners up). The prize for the best solution each month is $50 and a limited edition “The Winner! MacTech Magazine Programming Challenge” T-shirt (not to be found in stores).

In order to make fair comparisons between solutions, all solutions must be in ANSI compatible C (i.e., don’t use Think’s Object extensions). Only pure C code can be used. Any entries with any assembly in them will be disqualified (except for those challenges specifically stated to be in assembly). However, you may call any routine in the Macintosh toolbox you want (i.e., it doesn’t matter if you use NewPtr instead of malloc). All entries will be tested with the FPU and 68020 flags turned off in THINK C. When timing routines, the latest version of THINK C will be used (with ANSI Settings plus “Honor ‘register’ first” and “Use Global Optimizer” turned on) so beware if you optimize for a different C compiler. All code should be limited to 60 characters wide. This will aid us in dealing with e-mail gateways and page layout.

The solution and winners for this month’s Programmers’ Challenge will be published in the issue two months later. All submissions must be received by the 10th day of the month printed on the front of this issue.

All solutions should be marked “Attn: Programmers’ Challenge Solution” and sent to Xplain Corporation (the publishers of MacTech Magazine) via “snail mail” or preferably, e-mail - AppleLink: MT.PROGCHAL, Internet: progchallenge@xplain.com, CompuServe: 71552,174 and America Online: MT PRGCHAL. If you send via snail mail, please include a disk with the solution and all related files (including contact information). See page 2 for information on “How to Contact Xplain Corporation.”

MacTech Magazine reserves the right to publish any solution entered in the Programming Challenge of the Month and all entries are the property of MacTech Magazine upon submission. The submission falls under all the same conventions of an article submission.

WE PRY ANY HEAP

Everyone likes anagrams. If you’ve ever had an anagram program and run your friends’ names through it then you know how excited people get when they see what their name can spell when the letters are rearranged. It’s one of those little things that computers can do that impresses non-computer people like my mom more than any amount of awesome 3-D rendering or clever computer animation. This month’s challenge is to write a fast anagram routine.

The prototype of the function you write is:

/* 1 */

unsigned long Anagram(inputText, 
 wordList, outputFile)
Str255  inputText;
FILE    *wordList;
FILE    *outputFile;

InputText is a Pascal string containing the text to anagram. It will be all lowercase letters (a..z) and may contain spaces, which you should ignore (i.e. your anagram may contain more or fewer spaces; it doesn’t matter). WordList is a standard C input stream containing the dictionary of valid words you can use to make your anagrammed output. The words in the dictionary will be all lowercase and sorted from ‘a’ to ‘z’ (and there will be about 20,000 of them). There is a 0x0D byte between each word. You should keep reading words from the stream until you reach the end of file. OutputFile is a standard C output stream that you should write your anagrams to, each one separated by a 0x0D byte. The return value of the function is the number of unique anagrams that were sent to the outputFile.

Good luck and Happy New Year!

TWO MONTHS AGO WINNER

Of the 11 entries I received for the Present Packing challenge, nine worked correctly. Congrats to James Goebel (location unknown) for having the highest average number of presents packed. James previously won the ASCII85 Encode challenge and now he is tied in a 3-way tie for the most number of 1st place Challenge showings.

This challenge was judged based on the highest average number of packages packed. The times and code+data sizes are given for interest only. Numbers in parens after a person’s name indicate how many times that person has finished in the top 5 places of all previous Programmer Challenges, not including this one.

Name packages time code+data

James Goebel (2) 95.8 4007 2506

Kevin Cutts (1) 94.3 67 10806

Robert Coie 93.8 5 900

Bob Boonstra (4) 93.7 6 5684

Dave Darrah 93.6 6 1002

Paul Pedriana 93.5 5 1442

Stefan Pantke 91.4 121 20552

Allen Stenger (2) 91.0 11439 360

Jeremy Vineyard (1) 67.3 562 664

My apologies for not considering that it would be nice if you could rotate a present 90 degrees as you packed it. Several people who entered wrote to me and asked me about that before entering. Unfortunately, Donald Knipp (location unknown) didn’t ask me and just assumed that he could rotate the presents. But since the storePresProc had no way of knowing that he had rotated them he ended up putting presents on top of each other, which invalidated his entry. I’m not happy about disqualifying Donald’s otherwise clever entry but I must in order to be fair to those who were told they couldn’t rotate. In the future, I urge everyone to e-mail me if something is ambiguous or if there are questions about what assumptions you can and cannot make in your solutions.

Here’s James’ winning solution. My apologies for removing some whitespace and comments in order to fit it in this column; James’ unedited code is on the source code disk.

/* 2 */

/* PackPresents() by Clement James Goebel III.
 This code accepts a number of presents one after another and
 attempts to store as MANY as possible
 in its storage area. 
 This routine starts with an array describing the expected 
 distribution of data and then slowly changes to use an array
 that describes the sizes of observed presents as the process
 continues. These arrays are used to decide which presents 
 are too big and should not be stored as they will cause us
 to throw away smaller 
 presents latter. The routine is slow 
 and methodical as it trys to pack presents into the smallest 
   spaces it can find. It also does an ok job of guessing which 
 packages to discard, a function that might not really be 
 needed with evenly distributed data sets. But the goal was
 to pack the most, so you can't be too careful. The matrix 
 that keeps track of stored presents contains zeros where 
 presents are stored, and all other locations contain a value
 that represents the amount of free space that is contiguous
 to that location. We will always try to fill small holes 
 first, a better way might be to look for presents that are
 half the size of the hole, but that would require many more
 special cases. And when placing presents we will always try
 to get as many surfaces to touch as possible.
*/

#define WID_DIM  100
#define LEN_DIM  100
#define MIN_GIFT_DIM 5
#define MAX_GIFT_DIM 15
#define LIKES_OTHERS 3
#define LIKES_WALLS2
#define SPACE_USED 0
#define MEASURING-1

static short sgsGiftsSeen, sgsGiftsStored, sgsLeftmostPresent,
 sgsTopmostPresent;
static long  sglExpectedCount, sglItemsOnStack,
 sglTotalAreaExpected;
static short *sgasStack, *sgasAreasSeen, *sgasAreasExpected;
static long  *sgalSpace;

typedef void (*NextPresProc)
 (unsigned short *pWidth, unsigned short *pLength );
typedef void (*StorePresProc)
 (unsigned short xPos, unsigned short yPos );

void PackPresents( unsigned short usNumGifts,
 NextPresProc pNextPresProc, StorePresProc pStorePresProc );
void MyPacker( unsigned short usNumGifts,
 long sgaalStorage[WID_DIM][LEN_DIM],
 NextPresProc pNextPresProc, StorePresProc pStorePresProc );

// PackPresents()
void PackPresents( unsigned short usNumGifts,
 NextPresProc pNextPresProc, StorePresProc pStorePresProc)
{
 long w, l, lBytes;
 sgsGiftsSeen = sgsGiftsStored = 0;
 sglItemsOnStack = 0;
 sgsLeftmostPresent = WID_DIM;
 sgsTopmostPresent = LEN_DIM;
 lBytes = (MAX_GIFT_DIM+1) 
 * (MAX_GIFT_DIM+1) * sizeof( short );
 sgasAreasSeen = (void*)NewPtrClear( lBytes );
 sgasAreasExpected = (void*)NewPtrClear( lBytes );
 sgasStack = (void*)NewPtrClear( (WID_DIM * LEN_DIM ) 
 * 2 * sizeof( short ) );
 sgalSpace = (void*)NewPtrClear( WID_DIM * LEN_DIM 
 * sizeof( long ) );
// Given the range of inputs we expect to see compute
// the number of presents of each size that we 
// expect to be offered.
 sgsGiftsSeen = sglExpectedCount = 0;
 sglTotalAreaExpected = 0;
 for ( w = MIN_GIFT_DIM; w <= MAX_GIFT_DIM; w++ ) {
 for ( l = MIN_GIFT_DIM;l<=MAX_GIFT_DIM; l++) {
 sgasAreasExpected[ w * l] ++;
 sglTotalAreaExpected += w * l;
 sglExpectedCount++;
 } }
 MyPacker( usNumGifts, (void*)sgalSpace, 
 pNextPresProc, pStorePresProc );
 DisposePtr( (Ptr)sgasAreasSeen );
 DisposePtr( (Ptr)sgasAreasExpected );
 DisposePtr( (Ptr)sgasStack );
 DisposePtr( (Ptr)sgalSpace );
}

// Utility routines called by packing routine.
Boolean BestPosition( long aalSpace[WID_DIM][LEN_DIM],
 unsigned short usSpaceRemaining, unsigned short usWidth,
 unsigned short usLength, short *pusX, short *pusY );
int LargestGiftDesired( unsigned short usSpaceLeft, 
 unsigned short usTotalGifts, unsigned short usGiftsRemaining,
 unsigned short usWidth, unsigned short usLength );
void RecomputeAreas( long aalSpace[WID_DIM][LEN_DIM],
 unsigned short *pusSpaceRemaining, int iHoleSize );

// MyPacker()
// After getting each present check to see what the 
// expected sizes of the next presents will be and
// pick a largest acceptable size.  If the present
// meets the size requirement then find the best 
// location for it (presents like to sit amoung
// friends or with thier back to the wall!), and 
// store it.
void MyPacker( unsigned short usNumGifts,
 long aalGiftStorage[WID_DIM][LEN_DIM],
 NextPresProc pNextPresProc, StorePresProc pStorePresProc )
{
 unsigned short usSpaceRemaining, usGiftsRemaining;
 unsigned short usWidth, usLength;
 int  iLargestGiftDesired, iArea, i, w, l, iHoleSize;
 short X, Y;
 
 usGiftsRemaining = usNumGifts;
 usSpaceRemaining = WID_DIM * LEN_DIM;
// Fill storage array with contiguous area values.
// In the beginning all space is empty and contiguous.
 for ( w = 0; w < WID_DIM; w++ )
 for ( l = 0; l < LEN_DIM; l++ )
 aalGiftStorage[w][l] = usSpaceRemaining;    
// Get the presents.
 while ( usGiftsRemaining ) {
 (pNextPresProc)( &usWidth, &usLength );
 usGiftsRemaining--;
 iLargestGiftDesired = LargestGiftDesired( 
   usSpaceRemaining, usNumGifts, 
   usGiftsRemaining, usWidth, usLength );
 iArea = usWidth * usLength;
 if ( iArea <= iLargestGiftDesired ) {
 if ( BestPosition( aalGiftStorage, 
   usSpaceRemaining, usWidth, usLength, &X, &Y ) ) {
 iHoleSize = aalGiftStorage[X][Y];
// Store a gift.
 for ( w = 0; w < usWidth; w++ ) {
 for ( l = 0; l < usLength; l++ ) {
 aalGiftStorage[X+w][Y+l] = SPACE_USED;
 } }
 pStorePresProc( (unsigned short)X, 
 (unsigned short)Y );
 sgsGiftsStored++;
 if ( sgsLeftmostPresent > X )
 sgsLeftmostPresent = X;
 if ( sgsTopmostPresent > Y )
 sgsTopmostPresent = Y;
 usSpaceRemaining -= iArea;
 RecomputeAreas( aalGiftStorage, 
   &usSpaceRemaining, iHoleSize );
 } }    
 if ( usSpaceRemaining == 0 ) return;
}}

// Push() & Pop() implement a stack for the 
// flood fill type algorithm FloodMark to store
// data points on instead of recursing into the heap.
Push( unsigned usX, unsigned usY )
{
 sgasStack[sglItemsOnStack] = usX;
 sgasStack[sglItemsOnStack+1] = usY;
 sglItemsOnStack += 2;
}
Boolean Pop( unsigned short *pusX, unsigned short *pusY )
{
 if ( sglItemsOnStack ) {
 sglItemsOnStack -= 2;
 *pusX = sgasStack[sglItemsOnStack];
 *pusY = sgasStack[sglItemsOnStack+1];
 return( TRUE );
 }
 return( FALSE );
}

// FloodMark()
// This is an implementation of the well documented
// Floodfill alorithm for fill irregular shapes with
// paint or other such graphically stuff.  Here we
// use it to measure the number of contigious values,
// that match the iValueToMatch, variable,
// in the array.  As it encounters each value it 
// marks it with the flag MEASURING so that we can 
// then go and place the new area value back into 
// those positions.
long FloodMark( int iValueToMatch,
 long aal[WID_DIM][LEN_DIM], unsigned short X,
 unsigned short Y, Boolean *pbCanFitMinGift )
{
 long lPixelsFilled = 0;
 int l, w, iV = iValueToMatch;
 Boolean bCanFitMinGift = FALSE;
 
 sglItemsOnStack = 0;
 if ( aal[X][Y] == iV )
 Push( X, Y );
 while ( Pop( &X, &Y ) ) {
 aal[X][Y] = MEASURING;
 lPixelsFilled++;
 if ( ! bCanFitMinGift ) {
 if ( X + MIN_GIFT_DIM - 1 > WID_DIM )
 goto FAILED_MIN_TEST;
 if ( Y + MIN_GIFT_DIM - 1 > LEN_DIM )
 goto FAILED_MIN_TEST;
 for ( w = 0; w < MIN_GIFT_DIM; w++ )
 for ( l = 0; l < MIN_GIFT_DIM; l++ ) 
 if ( aal[X+w][Y+l] == SPACE_USED )
 goto FAILED_MIN_TEST;
 bCanFitMinGift = TRUE;
 }
FAILED_MIN_TEST:;
 if ( Y > 0 && aal[X][Y-1] == iV )
 Push( X, Y-1 );
 if ( Y+1 < LEN_DIM && aal[X][Y+1] == iV )
 Push( X, Y+1 );
 if ( X > 0 && aal[X-1][Y] == iV )
 Push( X-1, Y );
 if ( X+1 < WID_DIM && aal[X+1][Y] == iV )
 Push( X+1, Y );
 }
 *pbCanFitMinGift = bCanFitMinGift;
 return( lPixelsFilled );
}

// RecomputeAreas()
// The matrix that holds the current state of stored
// presents contains zeros were presents are located,
// and every other location contains a value that
// describes the area of the contigious region of 
// which it is a part.  After placing a present in an 
// empty region of size X, call this routine with
// iHoleSize = X, so that the area map can be brought
// up to date.
void RecomputeAreas( long aalSpace[WID_DIM][LEN_DIM],
 unsigned short *pusSpaceRemaining, int iHoleSize )
{
 int  i,j,k,l, iSmallest;
 long lArea;
 Boolean bCanFitMinGift;

 for ( i = 0; i < WID_DIM; i++ ) {
   for ( j = 0; j < LEN_DIM; j++ ) {
     if ( aalSpace[i][j] == iHoleSize ) {
   lArea = FloodMark( iHoleSize, aalSpace, i, j,
   &bCanFitMinGift );
   
   if ( ! bCanFitMinGift ) {
// Remove areas smaller than smallest present.
     for( k = 0; k < WID_DIM; k++ )
       for ( l = 0; l < LEN_DIM; l++ )
     if ( aalSpace[k][l] == MEASURING )
   aalSpace[k][l] = SPACE_USED;
     (*pusSpaceRemaining) -= lArea;
   } else {
     for( k = 0; k < WID_DIM; k++ )
       for ( l = 0; l < LEN_DIM; l++ )
     if ( aalSpace[k][l] == MEASURING )
   aalSpace[k][l] = lArea;
}} } } }

// NeighborCount()
// As we all know presents like lots of friends and
// are agoraphobic, so pack them in tight leaving as
// few exposed surfaces as possible.  Being next to
// a friend is better than a cold wall, but better to
// cover your rear with a wall then leave it out in 
// the open.
int NeighborCount( long aalSpace[WID_DIM][LEN_DIM],
 unsigned short usWidth, unsigned short usLength,
 unsigned short usX, unsigned short usY )
{
 unsigned short w, l;
 int iNeighbors;

 iNeighbors = 0;
 if ( usX + usWidth - 1 < sgsLeftmostPresent ) {
 if ( usX == 0 )
 iNeighbors += (LIKES_WALLS * usLength);
 if ( usX+usWidth == WID_DIM )
 iNeighbors += (LIKES_WALLS * usLength);
 } else {
 for ( l = 0; l < usLength; l++ ) {
 if ( usX == 0 )
 iNeighbors += LIKES_WALLS;
 else if ( aalSpace[usX-1][usY+l] == SPACE_USED )
 iNeighbors += LIKES_OTHERS;
 
 if ( usX+usWidth == WID_DIM )
 iNeighbors += LIKES_WALLS;
 else if ( aalSpace[usX+usWidth][usY+l] == SPACE_USED )
 iNeighbors += LIKES_OTHERS;
 } }
 if ( usY + usLength - 1 < sgsTopmostPresent ) {
 if ( usY == 0 )
 iNeighbors += (LIKES_WALLS * usWidth);
 if ( usY+usLength == LEN_DIM )
 iNeighbors += (LIKES_WALLS * usWidth);
 } else {
 for ( w = 0; w < usWidth; w++ ) {
 if ( usY == 0 )
 iNeighbors += LIKES_WALLS;
 else if ( aalSpace[usX+w][usY-1] == SPACE_USED )
 iNeighbors += LIKES_OTHERS;
 
 if ( usY+usLength == LEN_DIM )
 iNeighbors += LIKES_WALLS;
 else if ( aalSpace[usX+w][usY+usLength] == SPACE_USED )
 iNeighbors += LIKES_OTHERS;
 } }
 return( iNeighbors );
}
 
// BestPosition()
// Find the "best" position for this size present.
// If the present does not fit then return FALSE.
// Find the smallest open area that can accomadate
// this package then position it so that it is
// adjacent to as many others, or edges, as possible.
Boolean BestPosition( long aalSpace[WID_DIM][LEN_DIM],
 unsigned short usSpaceRemaining, unsigned short usWidth,
 unsigned short usLength, short *psX, short *psY )
{
 Boolean bFits = FALSE;
 short sX, sY, w, l;
 int iNeighbors, iMostNeighbors = -1;
 long lThisHole,lSmallestHole = 0x7FFFFFFF;
 
// 1st package always to the lower right corner.
 if ( sgsGiftsStored == 0 ) {
 *psX = WID_DIM - usWidth;
 *psY = LEN_DIM - usLength;
 return( TRUE );
 }

// Check all potential positions for open space.
 for ( sX = WID_DIM - usWidth; sX >= 0 ; sX-- ) {
 for ( sY = LEN_DIM - usLength; sY >= 0 ; sY-- ) {
 lThisHole = aalSpace[sX][sY];
 if ( lThisHole != SPACE_USED
 && lSmallestHole >= lThisHole ) {

 for ( w = 0; w < usWidth; w++ ) {
 for ( l = 0; l < usLength; l++ ) {
 if ( aalSpace[sX+w][sY+l] == 
    SPACE_USED ) {
 sY -= ( usLength - l );
 sY ++;
 goto SPACE_NOT_AVAILABLE;
 } }  }
 
// Count the neighbors, since presents need friends.
 iNeighbors = NeighborCount( aalSpace,
 usWidth, usLength, sX, sY );
 if ( iNeighbors > iMostNeighbors 
 || lSmallestHole > lThisHole ) {
 bFits = TRUE;
 *psX = sX;
 *psY = sY;
 iMostNeighbors = iNeighbors;
 lSmallestHole = lThisHole;
 } }
 SPACE_NOT_AVAILABLE:;
 } }
 return( bFits );
}

// LargestGiftDesired()
// To find the largest gift that we wish to accept
// we total all of the areas from smallest to largest
// of the gifts that we expect to get.  When the 
// expected total approaches the space remaining
// we choose that size as the largest gift to accept.
int LargestGiftDesired( unsigned short usSpaceLeft,
 unsigned short usTotalGifts, unsigned short usGiftsRemain,
 unsigned short usWidth, unsigned short usLength ) 
{
 long lExpected1000, lSpcLeft1000;
 long lRandomModel, lObserved;
 int    iSize, iMaxSize;
 sgsGiftsSeen++;
 if ( usWidth > MAX_GIFT_DIM 
 || usLength > MAX_GIFT_DIM )
 return( 0 );
 sgasAreasSeen[usWidth * usLength] += 1;
 if ( usGiftsRemain <= 5 )
 return( usSpaceLeft );
 iSize = MIN_GIFT_DIM * MIN_GIFT_DIM - 1;
 iMaxSize = MAX_GIFT_DIM * MAX_GIFT_DIM;
 lExpected1000 = 0;
 lSpcLeft1000 = usSpaceLeft * 1000L;
 while ( lExpected1000 + iSize * 3 * 1000L 
   <= lSpcLeft1000 && iSize <= iMaxSize ) {
 iSize++;
 lRandomModel = 1000L * 
 sgasAreasExpected[iSize] * iSize;
 lObserved = 1000L * sgasAreasSeen[iSize] * iSize;
 
 if ( lRandomModel || lObserved ) {
 lRandomModel = usGiftsRemain * lRandomModel 
 / sglExpectedCount;
 lObserved = usGiftsRemain * lObserved 
 / sgsGiftsSeen;
// Distribute weight between expected model and 
// observed sizes by the proportion of the totals
// gifts that we have already seen.
 lRandomModel *= usGiftsRemain;
 lRandomModel /= usTotalGifts;
 lObserved *= (usTotalGifts-usGiftsRemain);
 lObserved /= usTotalGifts;
 lExpected1000 += lObserved + lRandomModel;
 } }
 return( iSize );
}







  
 

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128GB MacBook Airs on sale for $100 off MSRP,...
B&H Photo has 11″ & 13″ MacBook Airs with 128GB SSDs on sale for $100 off MSRP. Shipping is free, and B&H charges NY sales tax only: - 11″ 1.6GHz/128GB MacBook Air: $799.99, $100 off MSRP... Read more
13-inch 2.5GHz MacBook Pro (refurbished) avai...
The Apple Store has Apple Certified Refurbished 13″ 2.5GHz MacBook Pros available for $829, or $270 off the cost of new models. Apple’s one-year warranty is standard, and shipping is free: - 13″ 2.... Read more
27-inch 3.2GHz iMac on sale for $1679, save $...
B&H Photo has the 27″ 3.2GHz iMac on sale for $1679.99 including free shipping plus NY sales tax only. Their price is $120 off MSRP. Read more
Apple and Cisco Partner to Deliver Fast-Lane...
Apple and Cisco have announced a partnership to create a “fast lane” for iOS business users by optimizing Cisco networks for iOS devices and apps. The alliance integrates iPhone with Cisco enterprise... Read more
Apple offering refurbished 2015 13-inch Retin...
The Apple Store is offering Apple Certified Refurbished 2015 13″ Retina MacBook Pros for up to $270 (15%) off the cost of new models. An Apple one-year warranty is included with each model, and... Read more
Apple refurbished 2015 MacBook Airs available...
The Apple Store has Apple Certified Refurbished 2015 11″ and 13″ MacBook Airs (the latest models), available for up to $180 off the cost of new models. An Apple one-year warranty is included with... Read more
21-inch iMacs on sale for up to $120 off MSRP
B&H Photo has 21″ iMacs on sale for up to $120 off MSRP including free shipping plus NY sales tax only: - 21″ 1.4GHz iMac: $999.99 $100 off - 21″ 2.7GHz iMac: $1199.99 $100 off - 21″ 2.9GHz iMac... Read more
5K iMacs on sale for up to $150 off MSRP, fre...
B&H Photo has the 27″ 3.3GHz 5K iMac on sale for $1899.99 including free shipping plus NY tax only. Their price is $100 off MSRP. They have the 27″ 3.5GHz 5K iMac on sale for $2149.99 $2199.99, $... Read more

Jobs Board

*Apple* Evangelist - JAMF Software (United S...
The Apple Evangelist is responsible for building and cultivating strategic relationships with Apple 's small and mid-market business development field teams. This 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* Desktop Analyst - KDS Staffing (Unit...
…field and consistent professional recruiting achievement. Job Description: Title: Apple Desktop AnalystPosition Type: Full-time PermanentLocation: White Plains, NYHot Read more
*Apple* Systems Engineer (Mclean, VA and NYC...
Title: Apple Systems Engineer (Mclean, VA and NYC) Location: United States-New York-New York-200 Park Ave (22005) Other Locations: United States-Virginia-Vienna-Towers Read more
*Apple* Systems Engineer (Mclean, VA and NYC...
…Assist in providing strategic direction and technical leadership within the Apple portfolio, including desktops, laptops, and printing environment. This person will Read more
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