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Jan 01 Challenge Volume Number: 17 (2001)
Issue Number: 1
Column Tag: Programmer's Challenge

By Bob Boonstra, Westford, MA

Tetris

When George Warner first suggested that I base a Challenge on Tetris, I was skeptical. I hadn't played Tetris in a long time, and my recollection was that Tetris is a game not only of strategy, but of manual dexterity as well. After some email conversation, however, I think we've found a way to formulate Tetris info a meaningful Challenge.

You remember how Tetris is played, right? The game is played on a board sized perhaps 10 cells wide and 20 cells high into which pieces of varying sizes are dropped. The player is able to move the pieces left or right as they drop, or to rotate them clockwise or counter-clockwise, or to drop them to the bottom of the board. The player accumulates points as each piece is positioned into its final location. As one or more rows become completely filled, those rows are removed, the other rows are shifted down, and more points are accumulated. As rows are deleted, the game progresses to higher and higher levels where the pieces drop faster and faster. The game continues until there is no room for the next piece to drop into the board.

In this month's Challenge version of Tetris, the board size is generalized to something potentially larger than 10x20, the game speed remains constant, and the game continues until a specified time limit expires. You can make one move of the current game piece, a translation or a rotation, for each tick of the game clock. You can take as long as you like to figure out the best move, but every microsecond you use to calculate your move subtracts from the time limit and reduces your opportunity to score additional points. So you need to plan your moves both carefully and quickly.

The prototype for the code you should write is:

typedef char Piece[7][7];
   /* aPiece[row][col] is 1 if aPiece occupies cell (row,col), 0 otherwise */

typedef char Board[256][256];
   /* aBoard[row][col] is -1 if cell (row,col) is empty, otherwise it is the Piece index */
   /* row 0 is the top row, col 0 is the leftmost column */

typedef enum {
   kNoMove=0,                         /* allow piece to fall normally */
   kMoveLeft, kMoveRight,         /* move piece left/right one column */
   kDrop,                              /* drop piece to bottom of board */
   kRotateClockwise,             /* rotate piece clockwise 90 degrees */
   kRotateCounterClockwise      /* rotate piece counterclockwise 90 degrees */
} MoveType;

void InitTetris(
   short boardWidth,               /* width of board in cells */
   short boardHeight,            /* height of board in cells */
   short numPieceTypes,         /* number of types of pieces */
   const Piece gamePieces[],   /* pieces to play */
   long timeToPlay                  /* game time, in milliseconds */
);

MoveType /* move active piece */ Tetris(
   const Board gameBoard,      
            /* current state of the game board, 
            bottom row is [boardHeight-1], left column is [0] */
   short activePieceTypeIndex,   /* index into gamePieces of active piece */
   short nextPieceTypeIndex,      /* index into gamePieces of next piece */
   long pointsEarned,               /* number of points earned thus far */
   long timeToGo                        /* time remaining, in milliseconds */
);

void TermTetris(void);

The Tetris Challenge will work like this. Your InitTetris routine will be called first, to allow you to set up the problem based on the board configuration and the Piece shapes to be used in the problem. Next, your Tetris routine will be called multiple times, allowing you to manipulate the falling Tetris pieces, with the objective of accumulating as many points as possible. Your Tetris routine will continue to be called until the specified timeToPlay has expired, or until no more Pieces can be placed on the board. Then your TermTetris routine will be called, allowing you to deallocate any dynamically allocated storage.

InitTetris will be provided with the width (boardWidth) and height (boardHeight) of the game Board. It will also be given the numPieceTypes gamePieces to be used in the game, and the length of the game (timeToPlay) in milliseconds.

Each call to Tetris gives you the current state of the gameBoard; you should determine what you want to do with the active game Piece and return the corresponding MoveType. The test code will attempt to translate or rotate the active game Piece according to the MoveType you specify prior to dropping the Piece down one row. If the Piece cannot be moved or rotated as you request, the Piece will simply drop down one row. When the active Piece drops as far as it can, it will remain the active piece for one more game cycle, so that you can move it left or right by one position should you so choose. The Tetris parameters also provide you with the type of the next piece to be played (nextPieceTypeIndex), the pointsEarned so far, and the timeToGo still remaining in the game.

The gameBoard will contain the value -1 in each empty cell, and the index of the Piece occupying that cell for nonempty cells. Game Piece shapes are specified in a 7x7 array, with a 1 indicating that the corresponding cell is occupied. Pieces rotations occur about the central cell in that array. When Tetris is called with a new active game Piece, the Piece will be positioned just above the gameBoard, with the bottom row of the Piece due to appear on the gameBoard the next time Tetris is called.

The winner will be the solution that scores the most Tetris points within the specified timeToPlay. You score 1 point for each row that a Piece falls when it is dropped. When you eliminate a row, you earn 100 points. If multiple rows are completed at the same time, you win additional points: the second row completed by dropping a block is worth 200 points, the third 300 points, and the fourth 400 points. If you should eliminate all blocks on the board, you earn an additional 1000 points.

The Challenge prize will be divided between the overall winner and the best scoring entry from a contestant that has not won the Challenge recently. If you have wanted to compete in the Challenge, but have been discouraged from doing so, perhaps this is your chance at some recognition and a share of the Challenge prize.

This will be a native PowerPC Challenge, using the CodeWarrior Pro 6 environment. Solutions may be coded in C, C++, or Pascal. You can also provide a solution in Java, provided you also provide a test driver equivalent to the C code provided on the web for this problem.

Three Months Ago Winner

Congratulations to Ernst Munter (Kanata, Ontario) for another Programmer's Challenge victory. Ernst submitted the winning entry to the October "Which Bills Did They Pay" Challenge. This Challenge required contestants to sort out a set of invoices and a set of payments, matching payments to invoices. The solutions had to deal with payments that settled multiple invoices and partial payments of an invoice. Scoring was based on minimizing a "late-dollar-days" value that was the product of the amount of the invoice (or part of an invoice) times the number of days the amount went unpaid, thus encouraging the solution to apply payments to the oldest applicable invoice.

Ernst beat out the second-place entry by new contestant Sue Flowers. Sue's solution generated the same late-dollar-days value that Ernst's entry generated, but required significantly more execution time to do so. In several of the larger test cases, Sue's entry generated the same bill reconciliation log as Ernst's entry did, although this was not true in all cases. Besides Ernst's and Sue's entries, two additional entries for this Challenge were submitted, but neither solved the test cases correctly.

As always, Ernst's code is well commented. His entry makes two passes through the payment list, the first time looking for either a perfect match with a single invoice or a perfect match with multiple invoices. Ernst's CombineInvoices routine uses a stack-based technique to find the combination of invoices that exactly matches the payment amount. Although I didn't specifically analyze the performance of the individual routines in Ernst's code, I believe that the CombineInvoices routine is key to the performance Ernst achieved. In the second pass, his code looks for the "best" possible partial payment match, where "best" is defined as the invoice with the invoiced amount closest to the payment amount. Finally, the code makes a third pass through any remaining unpaid invoices to calculate the late-dollar-days statistic.

The table below lists, for each of the solutions submitted, the late-dollar-days value produced by the combined test cases, the cumulative execution time, and the number of reconciliation records generated. It also provides the code size, data size, and programming language used for each entry. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges prior to this one.

Name$DaysTime (msecs)RecordsErrors?Code SizeData SizeLang
Ernst Munter (661)18297920.43679no2756180C++
Sue Flowers182979246.54712no4156233C
A.D.446926800.7811yes127224C++
R. S.crash138361775C++

Top Contestants...

Listed here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated 10 or more points during the past two years. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants.

Rank Name Points
1. Munter, Ernst 251
2. Saxton, Tom 96
3. Maurer, Sebastian 68
4. Rieken, Willeke 65
5. Boring, Randy 52
6. Shearer, Rob 48
7. Taylor, Jonathan 36
8. Wihlborg, Charles 29

... and the Top Contestants Looking For a Recent Win

Starting this month, in order to give some recognition to other participants in the Challenge, we are also going to list the high scores for contestants who have accumulated points without taking first place in a Challenge. Listed here are all of those contestants who have accumulated 6 or more points during the past two years.

9. Downs, Andrew 12
10. Jones, Dennis 12
11. Day, Mark 10
12. Duga, Brady 10
13. Fazekas, Miklos 10
14. Flowers, Sue 10
15. Selengut, Jared 10
16. Strout, Joe 10
17. Hala, Ladislav 7
18. Miller, Mike 7
19. Nicolle, Ludovic 7
20. Schotsman, Jan 7
21. Widyyatama, Yudhi 7
22. Heithcock, JG 6

There are three ways to earn points: (1) scoring in the top 5 of any Challenge, (2) being the first person to find a bug in a published winning solution or, (3) being the first person to suggest a Challenge that I use. The points you can win are:

1st place 20 points
2nd place 10 points
3rd place 7 points
4th place 4 points
5th place 2 points
finding bug 2 points
suggesting Challenge 2 points

Here is Ernst's winning "What Bills Did They Pay" solution:

WhichBills.cp
Copyright © 2000
Ernst Munter

/*
October 4, 2000.
Submission to MacTech Programmer's Challenge for October 2000.
Copyright © 2000, Ernst Munter, Kanata, ON, Canada.
  
      "What Bills Did They Pay"

The Problem
------
Reconcile a set of invoices and payments where invoices may be paid with more than one
instalment,  and where payments may cover several invoices. 

The objective is to minimize the overall "late dollar days" subject to a set of rules. 

The Solution
------
I tackle the problem in three passes over the lists of invoices and payments.  During the
first  and second phases, reconciled invoice and payment records are removed from the lists.

In the first pass, each payment may  
   - be perfectly matched to a single invoice,
   - exactly match the sum of a number of invoices,
   - not match.
   
In the second pass, any remaining payments are applied to the smallest available invoice.
This may still leave unmatched invoices or payments unreconciled.

The amount of "late dollar days" is the sum of each reconciled amount multiplied by the delay
between the invoice and the payment.

The "late dollar days" figure includes each of the remaining unmatched invoices, multiplied
by  the delay from the invoice date to the last payment date.  

Any left-over payment that cannot be reconciled is disregarded.

Minimization of "late dollar days"
-----------------
If all payments have been accounted for, the resulting "late dollar days" number is
independent of the how the payments are applied to the invoices.  This is true because all
unpaid invoices  are effectively "paid" off on the last payment day.

The objective (to minimize "late dollar days") is thus achieved when all payments are matched
against invoices.  If this is not possible, it is best to 
   - minimize the number or amount of unmatched payments,
   - leave the invoices that cannot be matched to as late a date as possible.
   
No exhaustive attempt is made of legal permutations of multiple and partial matches, to
eliminate ALL unmatched payments.  It is hoped that the heuristic approach comes fairly
close.
*/

#include "ReconcilePayments.h"

struct Calendar
// Linear calendar for calculating differences in days 
const char months[12]={0,31,28,31,30,31,30,31,31,30,31,30};
static struct Calendar {
   short startMonth[4][16];
   Calendar()
   {
      for (long y=0;y<4;y++)
      {
         startMonth[y][1]=0;
         startMonth[y][2]=months[1];
         startMonth[y][3]=startMonth[y][2]+months[2];
         if (y==0) startMonth[y][3]++;
         for (long m=4;m <= 12;m++)
            startMonth[y][m]=startMonth[y][m-1]+months[m-1];
      }
   }
} gCalendar;

// Converts a Macintosh 14-byte DateTimeRec into a long (resolution to days only).
inline long DayNumber(const DateTimeRec & x)
{
   return
      (1461*x.year-1)/4+
      gCalendar.startMonth[3 & x.year][x.month] + 
      x.day;
}

struct MyInvoice
// My version of the invoice, using linear dates
struct MyInvoice {
   long   amount;
   long   day;
   MyInvoice*   next;   // invoices are stored in a doubly linked list
   MyInvoice*   prev;
   long   number;
   MyInvoice(){}
   MyInvoice(MyInvoice* nxt,MyInvoice* prv) :
      amount(0),   day(0),
      next(nxt),   prev(prv),   
      number(0){}
   MyInvoice(const Invoice & inv,MyInvoice* nxt,MyInvoice* prv) :
      amount(inv.invoiceAmount),
      day(DayNumber(inv.invoiceDate)),
      next(nxt),
      prev(prv),
      number(inv.invoiceNumber){}
   void Remove()
   // bridges link pointers to remove invoice from the list it is on
   {
      if (next) next->prev=prev;
      prev->next=next;      
         // prev is never 0 except in the list root which is never removed
   }
};
typedef MyInvoice* MyInvoicePtr;

struct MyPayment
// My version of the payment, using linear dates
struct MyPayment {
   long   amount;
   long   day;
   MyPayment*   next;         // payments are stored in a doubly linked list
   MyPayment*   prev;
   long   number;
   MyInvoice*   mostRecentInvoice;   // .. relative to this payment
   MyPayment(){}
   MyPayment(MyPayment* nxt,MyPayment* prv) :
      amount(0),   day(0),
      next(nxt),   prev(prv),   
      number(0),   mostRecentInvoice() {}
   MyPayment(const Payment & pay,MyPayment* nxt,MyPayment* prv) :
      amount(pay.paymentAmount),
      day(DayNumber(pay.paymentDate)),
      next(nxt),   prev(prv),
      number(pay.paymentNumber),   
      mostRecentInvoice() {}
   void Remove()
   {
      if (next) next->prev=prev;
      prev->next=next;
   }
};
typedef MyPayment* MyPaymentPtr;

Count
inline 
long Count(const Invoice theInvoices[])
// Counts the number of invoices, up to the sentinel with invoiceNumber 0
{
   const Invoice* I=theInvoices-1;
   long n=0;
   while ((++I)->invoiceNumber) {n++;}
   return n;
}
 
inline
long Count(const Payment thePayments[])
// Counts the number of payments, up to the sentinel with paymentNumber 0
{
   const Payment* P=thePayments-1;
   long n=0;
   while ((++P)->paymentNumber) {n++;}
   return n;
}

Reconcile
inline
long Reconcile(long amount,Reconciliation & R,
                        MyInvoice & I,MyPayment & P)
// Copies info into a reconciliation record and returns late dollar days 
{
   long iAmount=I.amount;
//   long pAmount=P.amount;      // there is no need to track partially used payments
   R.paymentNumber=P.number;
   R.appliedAmount=amount;
   R.invoiceNumber=I.number;
   I.amount = iAmount - amount;         
//   P.amount = pAmount - amount;// there is no need to track partially used payments
   return amount*(P.day-I.day);
}

CombineInvoices
inline
MyInvoicePtr* 
CombineInvoices(long balance,MyInvoice* I,
                  MyInvoice* mostRecentInvoice,MyInvoicePtr* stack)
// Searches invoices, from the range of invoices up to "mostRecentInvoice" to add up 
// to "balance".  
// The result is a sublist on the stack.
// Returns the top of the stack, or 0 if no match was found.
{
   *stack=0; 
   MyInvoicePtr* SP=stack+1;
   while (I && (I<=mostRecentInvoice))
   {
      if (balance < I->amount)      // invoice too large: try next in list
      {
         I=I->next;               
         if ((I>mostRecentInvoice) || (I==0))   // reached end of range
         {
            I=*-SP;                         // unwind stack
            if (I==0) return 0;      // reached bottom: no multiple found
            balance += I->amount;   // restore previous balance
            I=I->next;                      // and try next in list instead
         }
         continue;
      }
      
      if (balance > I->amount)      // include this one ... perhaps
      {
         if (I<mostRecentInvoice)   
                  // it's in range, and there is at least one other
         {
            *SP++=I;            // push this invoice on stack
            balance -= I->amount;
            I=I->next;
         } else                  // go back to previously found and skip it
         {
            I=*-SP;                         // unwind stack
            if (I==0) return 0;    // reached bottom: no multiple found
            balance += I->amount;   // restore previous balance
            I=I->next;                      // and try next in list instead
         }
         continue;
      }
      
      if (balance == I->amount)      // success
      {
         *SP++=I;                  // push this invoice on stack
         return SP;               // and return stack
         
      }
   }
   return 0;
}

PartialPayment
inline
MyInvoice* PartialPayment(long balance,MyInvoice* I,
                              MyInvoice* mostRecentInvoice)
// Returns the first invoice that exactly matches the balance,
//    failing an exact match, returns the invoice with the closest higher amount.  
// Returns 0 if no exact or partial match is found.
{
   MyInvoice* bestI=0;
   long bestAmount=0x7FFFFFFF;
   while (I && (I<=mostRecentInvoice)) {
      if (I->amount == balance)
      {
         return I;
      }
      
      if (I->amount > balance)
      {
         if (I->amount < bestAmount)
         {
            bestI=I;
            bestAmount=I->amount;
         }
      }
      I=I->next;
   }
   return bestI;
}

ReconcilePayments
long ReconcilePayments (
   const Invoice theInvoices[],
   const Payment thePayments[],
   Reconciliation theReconciliation[],
   long numberOfReconciliationRecords,
   long *lateDollarDays )
// Attempts to reconcile all invoices with the payments and computes the 
//   lateDollarDays.
// Returns the number of ReconciliationRecords actually filled out.
{
                              
   // Prepare private copies of invoices                              
   long numInvoices=Count(theInvoices);
   MyInvoice* INV = new MyInvoice[numInvoices];
   
   MyInvoice* I = INV;
   const Invoice*   theI = theInvoices;
   MyInvoice iList=MyInvoice(INV,0);
   MyInvoice* mostRecentInvoice = &iList;
   
   // Link all invoices into a list
   for (int i=0;i<numInvoices;i++,theI++)
   {
      MyInvoice* nextInvoice=I+1;
      *I=MyInvoice(*theI,nextInvoice,mostRecentInvoice);
      mostRecentInvoice=I;
      I=nextInvoice;
   }
   mostRecentInvoice->next=0;
         
   // Prepare private copies of payments   
   long numPayments=Count(thePayments);
   MyPayment* PAY = new MyPayment[numPayments];
   MyPayment* P = PAY;
   const Payment*   theP = thePayments;
   MyPayment pList=MyPayment(PAY,0);
   MyPayment* lastPayment = &pList;
   
   // Link all payments into a list
   for (int p=0;p<numPayments;p++,theP++)
   {
      MyPayment* nextPayment=P+1;
      *P = MyPayment(*theP,nextPayment,lastPayment);
      lastPayment=P;
      P=nextPayment;
   }
   lastPayment->next=0;
   long lastPayday=(-P)->day;
   
   Reconciliation* R = theReconciliation;
                  // shorthand for a long variable name
   long maxR = numberOfReconciliationRecords;
                  // shorthand for a long variable name

   // Get memory for a stack for the non-recursive search
   long stackSize=1 + numInvoices;
   MyInvoicePtr*    iStack=new MyInvoicePtr[stackSize];
   
   long numR=0;
   long lateDD=0;
   
/* PHASE 1 
   Scan all payments, trying to match single or multiple invoices against them
*/
   for (P=pList.next; P; P=P->next)
   {
      I=iList.next;
      if (I==0) break;         // invoice list is empty: we're done 
      long amount=P->amount;
      long theDay=P->day;
      
      // scan forward through invoices to find a perfect match for a single invoice
      MyInvoice* perfectMatch=0;
      mostRecentInvoice=0;
      while (I && (I->day <= theDay))   
      {
         if (I->amount == amount)
         {
            perfectMatch=I;      // break on earliest match found
            break;
         }
         mostRecentInvoice=I;
         I=I->next;
      }
      P->mostRecentInvoice=mostRecentInvoice;
                  // side effect: gets invoice range for each payment
      
      // use earliest perfect match if there is one
      if (perfectMatch)   
      {
         lateDD += Reconcile(amount,R[numR++],*perfectMatch,*P);
         P->Remove();         // remove payment from list   
         perfectMatch->Remove();   // this invoice can be removed
         
         if (numR >= maxR)      // all available recRecords used: must quit
            goto phase3;
         continue;   
      }        
      
      // try combining multiple invoices 
      MyInvoicePtr* SP=
         CombineInvoices(amount,iList.next,mostRecentInvoice,iStack);
      if (SP)
      {   
         P->Remove();         // remove payment from the list   
         I=*-SP;                  // pop one invoice from the stack
         do {
            lateDD += Reconcile(I->amount,R[numR++],*I,*P);
            I->Remove();      // this invoice can be removed
            I=*-SP;               // pop next invoice from the stack
            if (numR >= maxR)   // all available recRecords used: must quit
               goto phase3;
         } while (I);
      } 
   }

/* PHASE 2 
   Scan remaining payments, trying to match them as partial payments against the remaining invoices
*/   
   for (P=pList.next; P; P=P->next)
   {
      I=iList.next;
      if (I==0) break;         // no invoices left to balance: next phase
      long amount=P->amount;
      long theDay=P->day;
      
      // find "best" partial payment
      I=PartialPayment(amount,iList.next,P->mostRecentInvoice);   
      if (I)
      {
         P->Remove();
         lateDD += Reconcile(amount,R[numR++],*I,*P);
         if (I->amount==0)
            I->Remove();
         if (numR >= maxR)      // all available recRecords used: must quit
            break;
      }
   }
      
/* PHASE 3 
   Scan remaining invoices, adding their unpaid balances to the late dollar days
*/
phase3:
   for (I = iList.next; I; I=I->next)         
      lateDD+=(lastPayday-I->day)*I->amount;

   // free dynamic memory
   delete [] iStack;
   delete [] PAY;
   delete [] INV;
   
   *lateDollarDays = lateDD;
   return numR;
}

 

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Despicable Me: Minion Rush, by Gameloft, is introducing a new runner to the mix in their latest update. Now you can play as Carl, the prankster minion. Carl has a few new abilities to play with, including running at a higher speed from the start.... | Read more »

Price Scanner via MacPrices.net

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
Save up to $80 on iPad Air 2s, NY tax only, f...
 B&H Photo has iPad Air 2s on sale for $80 off MSRP including free shipping plus NY sales tax only: - 16GB iPad Air 2 WiFi: $469.99 $30 off - 64GB iPad Air 2 WiFi: $549.99 $50 off - 128GB iPad... Read more
iMacs on sale for up to $205 off MSRP
B&H Photo has 21″ and 27″ iMacs on sale for up to $205 off MSRP including free shipping plus NY sales tax only: - 21″ 1.4GHz iMac: $1019 $80 off - 21″ 2.7GHz iMac: $1189 $110 off - 21″ 2.9GHz... Read more
Färbe Technik Offers iPhone Battery Charge LI...
Färbe Technik, which manufactures and markets of mobile accessories for Apple, Blackberry and Samsung mobile devices, is offering tips on how to keep your iPhone charged while in the field: •... Read more
Electronic Recyclers International CEO Urges...
Citing a recent story on CNBC about concerns some security professionals have about the forthcoming Apple Watch, John Shegerian, Chairman and CEO of Electronic Recyclers International (ERI), the... Read more
Save up to $380 with Apple refurbished iMacs
The Apple Store has Apple Certified Refurbished iMacs available for up to $380 off the cost of new models. Apple’s one-year warranty is standard, and shipping is free: - 27″ 3.5GHz 5K iMac – $2119 $... Read more
Mac minis on sale for up to $75 off, starting...
MacMall has Mac minis on sale for up to $75 off MSRP including free shipping. Their prices are the lowest available for these models from any reseller: - 1.4GHz Mac mini: $459.99 $40 off - 2.6GHz Mac... Read more
College Student Deals: Additional $50 off Mac...
Take an additional $50 off all MacBooks and iMacs at Best Buy Online with their College Students Deals Savings, valid through April 11, 2015. Anyone with a valid .EDU email address can take advantage... Read more
Mac Pros on sale for up to $260 off MSRP
B&H Photo has Mac Pros on sale for up to $260 off MSRP. Shipping is free, and B&H charges sales tax in NY only: - 3.7GHz 4-core Mac Pro: $2799, $200 off MSRP - 3.5GHz 6-core Mac Pro: $3719.99... Read more
13-inch 2.5GHz MacBook Pro on sale for $100 o...
B&H Photo has the 13″ 2.5GHz MacBook Pro on sale for $999 including free shipping plus NY sales tax only. Their price is $100 off MSRP. Read more

Jobs Board

DevOps Software Engineer - *Apple* Pay, iOS...
**Job Summary** Imagine what you could do here. At Apple , great ideas have a way of becoming great products, services, and customer experiences very quickly. Bring 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
Sr. Technical Services Consultant, *Apple*...
**Job Summary** Apple Professional Services (APS) has an opening for a senior technical position that contributes to Apple 's efforts for strategic and transactional Read more
Lead *Apple* Solutions Consultant - Retail...
**Job Summary** Job Summary The Lead ASC is an Apple employee who serves as the Apple business manager and influencer in a hyper-business critical Reseller's store Read more
*Apple* Pay - Site Reliability Engineer - Ap...
**Job Summary** Imagine what you could do here. At Apple , great ideas have a way of becoming great products, services, and customer experiences very quickly. Bring Read more
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