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May 00 Challenge

Volume Number: 16 (2000)
Issue Number: 5
Column Tag: Programmer's Challenge

Programmer's Challenge

by Bob Boonstra, Westford, MA

BigNum Math

Back in September, 1995, we conducted an RSA Challenge that involved raising large integers to integral powers, modulo a third integer. The representation we used for those large integers was a BigNum type, where each digit of the large integer was stored in a byte. That representation and the operations on it were not particularly efficient, and this month we will belatedly recitfy that situation. Your Challenge is to implement a new BigNum type, of your own design, along with a number of arithmetic operations on these BigNums..

The prototype for the code you should write is:

typedef struct BigNum {
	long lengthInDigits;	/* length of the BigNum in digits */
	void *bigNumData;			/* pointer to BigNum data */
} BigNum;

BigNum NewBigNum (			/* create a BigNum */
	char sign,						/* +1 or -1 */
	char digits[],				/* digits to be made into a BigNum */
	long numDigits				/* number of digits */
);

void DisposeBigNum (		/* dispose of a BigNum */
	BigNum theBigNum			/* the BigNum to be disposed of */
);

BigNum AddBigNums (			/* sum two BigNums, returning a new one */
	BigNum bigNumA,				/* return the sum A+B */
	BigNum bigNumB
);

BigNum SubtractBigNums (	/* subtract two BigNums, returning a new one */
	BigNum bigNumA,				/* return the difference A-B */
	BigNum bigNumB
);

BigNum MultiplyBigNums (	/* multiply two BigNums, returning a new one */
	BigNum bigNumA,				/* return the product A*B */
	BigNum bigNumB
);

BigNum DivideBigNums (		/* divide two BigNums, returning a new one */
	BigNum bigNumA,				/* return the quotient A/B, discarding the remainder */
	BigNum bigNumB
);

BigNum ModBigNums (			/* divide two BigNums, returning a new one */
	BigNum bigNumA,				/* return the remainder A%B, discarding the quotient */
	BigNum bigNumB
);

BigNum PowerBigNums (		/* calculate one Bignum to the power of another, returning a new one */
	BigNum bigNumA,				/* return A raised to the power B, discarding the quotient */
	BigNum bigNumB
);

BigNum SqrtBigNum (			/* find the sqrt of a BigNum, returning a new one */
	BigNum bigNumA				/* return the square root of A */
);

long /* numDigits */ BigNumToDigits( /* convert a bigNum to decimal digits */
	BigNum bigNumA,				/* bigNum to be converted to decimal digits 0-9 */
	char *sign,						/* return +1 or -1 */
	char digits[]					/* decimal digits of bigNumA, preceeded by '-' if negative */
									/* storage for digits preallocated based on bigNumA.lengthInDigits */
);

The first thing you need to do is decide on an internal representation for BigNums. Then you need to write a NewBigNum routine that will create a BigNum from a sequence of numDigits digits and a sign value. Your NewBigNum code is responsible for allocating memory for the BigNumData. The DisposeBigNum routine is responsible for deallocating that memory. The caller of your code is responsible for pairing every NewBigNum call with a DisposeBigNum call, and the two routines should be implemented so as not to create any memory leaks. In addition to these allocation and deallocation routines, you need to write code to perform addition (AddBigNums), subtraction (SubtractBigNums), multiplication (MultiplyBigNums), division (DivideBigNums), remainders (ModBigNums), and exponentiation (PowerBigNums). Each of these routines takes two arguments, calculates the result, and returns the result in a new BigNum allocated by your code. Each of these returned BigNums will also be disposed of by a call to DisposeBigNum before the test is over, although they might be used for calculations in the interim.

Just to spice things up, you also need to provide a SqrtBigNum routine that calculates and returns the integer square root of a BigNum, the largest BigNum whose square is no larger than the original number.

And finally, to help me decipher your BigNums, you need to provide a BigNumToDigits conversion routine that converts your private BigNum data structure into a sequence of digits, along with a sign, and returns the number of digits in the decimal representation of the BigNum.

I'm not providing information on the distribution of calls to the various routines, except to say that the arithmetic routines will significantly outnumber the allocation and deallocation routines. The winner will be the solution that correctly completes a sequence of arithmetic operations on BigNums in the least amount of time. You are strongly encouraged to adequately comment the code in your submissions. Not only does that make your code more understandable if it is published as the winning solution, but it also helps me track down any minor problems that might occur.

I'll close with a plug for the Challenge mailing list, where you can receive notice of the problems before the hard-copy magazine reaches your mailbox, and where any post-publication clarifications are distributed. Subscription instructions can be found at www.mactech.com/progchallenge/. This will be a native PowerPC Challenge, using the CodeWarrior Pro 5 environment. Solutions may be coded in C, C++, or Pascal.

Three Months Ago Winner

The February Challenge required readers to calculate a minimal Latin square of a given order. Latin Squares are nxn arrays of integers, where each row and each column contains each integer from 1 to n exactly once. Congratulations to Willeke Rieken (The Netherlands) for coming up with the winning solution to the Latin Squares Challenge.

Eleven readers submitted entries to this Challenge, and their performance varied widely in efficiency. My test scenario was based on 28 test cases, consisting of the Latin Squares of orders 4, 5, 8, 9, 12, 13, 16, 17, 20, 21, 24, 25, 28, 29, 32, 33, 36, 37, 40, 41, 44, and 45. I selected those numbers because they formed a regular pattern that could be continued as far as the solutions would allow, and because they contained a mix of odd numbers, even numbers, perfect squares, prime numbers, and powers of two. My original intent was to test even larger numbers, but even the best solutions took too long to calculate some of the larger numbers.

Even limiting the tests to these cases, some of the solutions took a long time to execute, so I divided the tests into three sets. The first set consisted of the first ten test cases, and I ran all of the entries against that set. Three of the entries either did not complete all of the cases, or calculated a Latin Square that was larger than the squares calculated by other solutions. Three of the entries had fast execution times for the ten cases, and one more had an execution time within roughly two orders of magnitude of the best ones. So I ran the top four solutions against the next six test cases. Two of the entries completed those cases correctly, so I ran those cases against the final six test cases. The second place solution by Ernst Munter was by far the faster of the two, but unfortunately, it did not compute the minimal solution for the square of order 37. Where Ernst calculated a solution that included the following as the 28th row:

 28 27 26 25 32 31 30 35 36 37 33 34 19 20 17 21 7 8 9 5 6 4 ...

... Willeke's entry produced the following smaller value:

 28 27 26 25 32 31 30 35 36 37 33 34 19 20 17 21 7 8 9 5 6 3 ...

I decided not to disqualify solutions that produced suboptimal Latin Squares, or that failed to produce a result in a reasonable time. Instead, I ranked solutions by how many test cases they were able to complete, then how many they completed correctly, and then in order of increasing execution time. The problem statement called for the use of execution time only for correct solutions, but I felt that it was fairest to allow solutions that produced a suboptimal result to compete based on how well they did.

Willeke's algorithm takes advantage of the fact that squares whose size is a power of two can be generated with a systematic pattern of switching pairs of numbers in row n to create rows a power of 2 away from row n. He accomplishes this in his FillSquare2 routine. Squares of other sizes are filled by first filing the largest subsquare of size k (k a power of 2), filling the top right n-k square optimally, filling the diagonal, and then completing the square by trial and error. Ernst's entry makes more efficient use of information about which digits are forced into use before a particular column in a given row because the digit has already been used in subsequent columns. Ernst observes in his entry that execution time does not grow with problem size, and that problems of certain sizes (e.g., 41) take much longer to execute than one might expect based on the time required for squares of dimensions close in value.

The first table below lists, for each of the entries submitted, the final ranking based on all test cases completed, total execution time for the first ten cases, the number of test cases completed, the number completed incorrectly, and the code size, data size, and language parameters. 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. The second and third tables provide the results for the remaining twelve test cases.

Note that while the top four positions in this Challenge were won by four of our top contestants in the points standing (the fifth did not compete), there are a number of new names in the list of contestants. Keep trying, folks, I know from personal experience that it takes a while to become good at this, but it is possible to knock the leaders from their perches.

Cases 1-10

Name Rank Time (msec) Completed Cases Incorrect Cases Code Size Data Size Lang
Willeke Rieken 68) 1 4.1 10 0 3976 8 C++
Ernst Munter (557) 2 2.4 10 0 3224 96 C++
Randy Boring (116) 3 3.7 10 0 3828 42 C++
Sebastian Maurer (97) 4 524.5 10 0 1336 52 C++
Claes Wihlborg 5 5271.1 10 0 2596 73 C
Bjorn Davidsson (6) 6 141740.7 10 0 2232 120 C++
Michael Lewis 7 155346.4 10 0 5112 207 C++
Paul Russell 8 1436033.6 10 0 1660 8 C
Jonny Taylor (24) 9 4.3 9 0 5788 156 C
Derek Ledbetter (4) 10 1917.3 10 2 13088 312 C++
S. S. (withdrawn) 11 2.4 7 0 592 8 C++

Cases 11-16

Name Time (msec) Completed Cases Incorrect Cases
Ernst Munter 6.1 6 0
Willeke Rieken 1968.2 6 0
Randy Boring 40604.8 3 0
Sebastian Maurer N/A 0 0

Cases 17-22

Name Time (msec) Completed Cases Incorrect Cases
Ernst Munter 3200253.1 6 1
Willeke Rieken 13013297.2 6 0

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 215
2. Saxton, Tom 139
3. Maurer, Sebastian 91
4. Rieken, Willeke 61
5. Boring, Randy 50
6. Heathcock, JG 43
7. Shearer, Rob 43
8. Taylor, Jonathan 24
9. Brown, Pat 20
9. Hostetter, Mat 20
10. Downs, Andrew 12
11. Jones, Dennis 12
12. Hart, Alan 11
13. Duga, Brady 10
14. Hewett, Kevin 10
15. Murphy, ACC 10
16. Selengut, Jared 10
17. Strout, Joe 10

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 Willeke's winning Latin Squares solution:

LatinSquares.cp
Copyright © 2000
Welleke Rieken

/*
	After generating several squares a pattern emerged.
	If n is even, every second row can be generated by
	switching pairs of numbers of the row above.
	If n can be divided by 4, every third and fourth
	row can be generated by switching squares of 2 by 2
	numbers of the 2 rows above.
	Example: n = 12 is generated by generating
	n = 3 and replacing every number by a square with
	n = 4.
	Other n's are generated by generating the biggest
	power of n that fits in the square and generating
	a square of n - 2^x at the top right. This square can
	be repeated to the bottom left till the first square ends.
	the numbers in the first column are in ascending order.
	the diagonal from top rigth to bottom left is filled with n.
	Example: n = 7
	1234567
	2143675
	3412756
	4567xxx
	5x7xxxx
	67xxxxx
	7xxxxxx
	The remaining numbers are generated by trial and error.
*/

#include "LatinSquares.h"

FillSquare2
static void FillSquare2(long n, short *latinSquare,
												long theDim,
												long theStartRow, long theStartCol,
												long theStartVal, long theNrOfRows)
// n is a power of 2. fill the first row with ascending numbers
// and switch them around to generate the other rows.
{
	short *aFrom1, *aTo1, *aFrom2, *aTo2;
	long 	aValue = theStartVal + 1, aRowsDone, aMultiple;
	short	*aStartSquare = latinSquare + (theStartRow * n) + 
			theStartCol;
	
	// fill first row
	aFrom1 = aStartSquare;
	for (long aCol = 0; aCol < theDim; aCol++)
	{
		*aFrom1  = aValue;
		aValue++;
		aFrom1++;
	}
	aRowsDone = 1;
	aMultiple = 1;
	while (aRowsDone < theNrOfRows)
	{
		for (long aRow = 0; aRow < aMultiple; aRow++)
		{
			if (aRow >= theNrOfRows)
				break;
			for (long anOffset = 0; anOffset < theDim; anOffset += 
							(aMultiple * 2))
			{
				aFrom2 = aStartSquare + (aRow * n) + anOffset;
				aFrom1 = aFrom2 + aMultiple;
		aTo1 = aStartSquare + ((aMultiple + aRow) * n) + anOffset;
				aTo2 = aTo1 + aMultiple;
				for (long aCol = 0; aCol < aMultiple; aCol++)
				{
					*aTo1 = *aFrom1;
					aFrom1++;
					aTo1++;
					*aTo2 = *aFrom2;
					aFrom2++;
					aTo2++;
				}
			}
		}
		aRowsDone += aMultiple;
		aMultiple <<= 1;
	}
}

CopySquare
static inline void CopySquare(long n, short *theFrom, short *theTo,
															long theDim)
{
// copy a square of size theDim from theFrom to theTo
	short *aFrom, *aTo;
	
	for (long aRow = 0; aRow < theDim; aRow++)
	{
		aFrom = theFrom + (aRow * n);
		aTo = theTo + (aRow * n);
		for (long aCol = 0; aCol < theDim; aCol++)
		{
			*aTo = *aFrom;
			aFrom++;
			aTo++;
		}
	}
}

CantFillRow
static short CantFillRow(long theDim, short *theValInCol,
													short *theValInRow, long theCol,
													long *theValue)
// check if there are numbers that can't be placed and if there
// are enough columns for the bigger numbers
{
	long	aGreaterPlacesNeeded = 0;
	short	aValOK = 0;
	for (long i = *theValue + 1; i < theDim; i++)
		if (!theValInRow[i])
		{
			aGreaterPlacesNeeded++;
			aValOK = 0;
			for (long j = theCol + 1; j < theDim; j++)
				if (!theValInCol[j * theDim + i])
				{
					aValOK = 1;
					break;
				}
			if (!aValOK)
			{
				*theValue = i - 1;
				return 1;
			}
		}
	for (long j = theCol + 1; j < theDim; j++)
	{
		aValOK = 0;
		for (long i = *theValue + 1; i < theDim; i++)
			if (!(theValInRow[i] || theValInCol[j * theDim + i]))
			{
				aValOK = 1;
				break;
			}
		if (aValOK)
			aGreaterPlacesNeeded-;
	}
	if (aGreaterPlacesNeeded > 0)
		return 1;
	return 0;	
}

CompleteSquare
static void CompleteSquare(long n, short *latinSquare,
						long theDim, long theSubDim,
						long theStartRow, long theStartCol,
						long theStartVal)
// fill remaining numbers by trial and error
{
	short	*aStartSquare = latinSquare +
												((theStartRow * n) << theSubDim) +
												(theStartCol << theSubDim);
	short	*aValInRow = new short[theDim];
	short	*aValInCol = new short[theDim * theDim];
	short	*aToBeFilled = new short[theDim * theDim];
	long	aRow, aCol, aValue, aSubDimvalue;
	short	*p, *q;

	aSubDimvalue = 1 << theSubDim;
	// fill left row and diagonal
	p = aStartSquare + ((n + theDim - 2) << theSubDim);
	q = aStartSquare + (n << theSubDim);
	for (aCol = 1; aCol < theDim; aCol++)
	{
		CopySquare(n, aStartSquare + ((theDim - 1) << theSubDim),
								p, aSubDimvalue);
		p += ((n - 1) << theSubDim);
		CopySquare(n, aStartSquare + (aCol << theSubDim),
								q, aSubDimvalue);
		q += (n << theSubDim);
	}
	// which numbers are used and which numbers have to be filled in
	for (aCol = 0; aCol < theDim * theDim; aCol++)
	{
		aValInCol[aCol] = 0;
		aToBeFilled[aCol] = 1;
	}
	for (aRow = 0; aRow < theDim; aRow++)
	{
		p = aStartSquare + ((aRow * n) << theSubDim);
		for (aCol = 0; aCol < theDim; aCol++)
		{
			if (*p)
			{
				aValue = aCol * theDim +
									(((*p - 1) >> theSubDim) - theStartVal);
				aValInCol[aValue] = 1;
				aToBeFilled[aValue] = 0;
			}
			p += (aSubDimvalue);
		}
	}

	// which numbers are in this row
	for (aValue = 0; aValue < theDim; aValue++)
		aValInRow[aValue] = 0;
	aValue = 0;
	aRow = 1;
	aCol = 0;
	p = aStartSquare + (n << theSubDim);
	while (1)
	{
		// find next place to ve filled
		while (*p)
		{
			aValInRow[((*p - 1) >> theSubDim) - theStartVal] = 1;
			aCol++;
			p += (aSubDimvalue);
			if (aCol >= theDim)
			{
				aCol = 0;
				aRow++;
				p = aStartSquare + ((aRow * n) << theSubDim);
				for (aValue = 0; aValue < theDim; aValue++)
					aValInRow[aValue] = 0;
				aValue = 0;
			}
		}
		// find next posible value
		while ((aValue < theDim) &&
						(aValInCol[aCol * theDim + aValue] ||
							aValInRow[aValue] ||
							CantFillRow(theDim, aValInCol, aValInRow,
													aCol, &aValue)))
			aValue++;
		if (aValue < theDim)
		{
			// place value
			aValInCol[aCol * theDim + aValue] = 1;
			aValInRow[aValue] = 1;
			CopySquare(n, aStartSquare + (aValue << theSubDim),
									p, aSubDimvalue);
			
			// next column
			aCol++;
			p += (aSubDimvalue);
			if (aCol >= theDim)
			{
				// next row
				aRow++;
				if (aRow < theDim)
				{
					p = aStartSquare + ((aRow * n) << theSubDim);
					for (aValue = 0; aValue < theDim; aValue++)
						aValInRow[aValue] = 0;
					for (aCol = 0; aCol < theDim; aCol++)
					{
						aValInRow[aValue] = 0;
						if (*p)
					aValInRow[((*p - 1) >> theSubDim) - theStartVal] = 1;
						p += (aSubDimvalue);
					}
					aCol = 0;
					p = aStartSquare + ((aRow * n) << theSubDim);
				}
				else
				{
					return;
				}
			}
			aValue = 0;
		}
		else
		{
			// undo
			aCol-;
			p -= (aSubDimvalue);
			aValue = ((*p - 1) >> theSubDim) - theStartVal;
			while (aCol >= 0 && !aToBeFilled[aCol * theDim + aValue])
			{
				aCol-;
				if (aCol >= 0)
				{
					p -= (aSubDimvalue);
					aValue = ((*p - 1) >> theSubDim) - theStartVal;
				}
			}
			if (aCol < 0)
			{
				aRow-;
				p = aStartSquare +
						(((aRow * n) + theDim - 1) << theSubDim);
				aCol = theDim - 1;
				for (aValue = 0; aValue < theDim; aValue++)
					aValInRow[aValue] = 1;
			}
			aValue = ((*p - 1) >> theSubDim) - theStartVal;
			*p = 0;
			aValInCol[aCol * theDim + aValue] = 0;
			aValInRow[aValue] = 0;
			aValue++;
		}
	}
	delete[] aValInCol;
	delete[] aValInRow;
	delete[] aToBeFilled;
}

FillSquare
static void FillSquare(long n, short *latinSquare,
						long theDim, long theSubDim,
						long theStartRow, long theStartCol,
						long theStartVal, long theNrOfRows)
// fill latin square
// if n can be divided by a power of 2,
// theSubDim is 2^x, theDim is n/(2^x)
{
	if (theDim == 1)	// n is a power of 2
		FillSquare2(n, latinSquare, 1 << theSubDim,
				theStartRow << theSubDim, theStartCol << theSubDim,
				theStartVal << theSubDim, theNrOfRows << theSubDim);
	else
	{
		long	aMaxPower2, aNrOfRows, aStartCol, aStartRow;
		short	*aStartSquare = latinSquare +
												((theStartRow * n) << theSubDim) +
															(theStartCol << theSubDim);
		aMaxPower2 = 1;
		while (aMaxPower2 <= theDim) aMaxPower2 <<= 1;
		aMaxPower2 >>= 1;
		// fill top left of the square with a square with n = 2^aMaxPower2
		FillSquare2(n, latinSquare, aMaxPower2 << theSubDim,
				theStartRow << theSubDim, theStartCol << theSubDim,
				theStartVal << theSubDim, aMaxPower2 << theSubDim);
		aNrOfRows = theDim - aMaxPower2;
		if (aNrOfRows > theNrOfRows) aNrOfRows = theNrOfRows;
		// fill top right of the square with a square with n = theDim - 2^aMaxPower2
		FillSquare(n, latinSquare, theDim - aMaxPower2, theSubDim,
								theStartRow, theStartCol + aMaxPower2,
								theStartVal + aMaxPower2, aNrOfRows);
		// copy the square from the top right along the diagonal to the bottom left
		aStartCol = aMaxPower2 - aNrOfRows;
		aStartRow = aNrOfRows;
		while (aStartCol >= 0 && aStartRow < theNrOfRows)
		{
			if (aStartRow + aNrOfRows > theNrOfRows)
				aNrOfRows = theNrOfRows - aStartRow;
			if (aNrOfRows > aStartCol && aStartCol > 0)
				aNrOfRows = aStartCol;
	for (long aRow = 0; aRow < (aNrOfRows << theSubDim); aRow++)
			{
				short	*aFrom = aStartSquare + (aRow * n) +
												(aMaxPower2 << theSubDim);
				short	*aTo = aStartSquare +
									(((aStartRow << theSubDim) + aRow) * n) +
											(aStartCol << theSubDim);
				for (long aCol = 0; aCol < ((theDim - aMaxPower2) << 
							theSubDim); aCol++)
				{
					*aTo = *aFrom;
					aFrom++;
					aTo++;
				}
			for (long aCol = ((aStartCol + (theDim - aMaxPower2)) <<
				theSubDim); aCol < (aMaxPower2 << theSubDim); aCol++)
				{
					*aTo = 0;
					aTo++;
				}
			}
			aStartCol -= (theDim - aMaxPower2);
			aStartRow += (theDim - aMaxPower2);
		}
		// generate the remaning numbers
		CompleteSquare(n, latinSquare, theDim, theSubDim,
										theStartRow, theStartCol, theStartVal);
	}
}

LatinSquares
void LatinSquares(
  short n, /* dimension of the latin square to be generated */
  short *latinSquare /* set latinSquare[c + r*n] to square value row r, col c */
) {
	short	*p = latinSquare;
	long	aSubDim = 0;
	// init
	for (long i = 0; i < n * n; i++, p++)
		*p = 0;
	// can n be divided by a power of 2
	while (!(n & (1 << aSubDim))) aSubDim++;
	FillSquare(n, latinSquare, n >> aSubDim, aSubDim,
							0, 0, 0, n >> aSubDim);
}
 

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OmniFocus helps you manage your tasks the way that you want, freeing you to focus your attention on the things that matter to you most. Capturing tasks and ideas is always a keyboard shortcut away in... Read more
TinkerTool 5.4 - Expanded preference set...
TinkerTool is an application that gives you access to additional preference settings Apple has built into Mac OS X. This allows to activate hidden features in the operating system and in some of the... Read more
Tinderbox 6.3.1 - Store and organize you...
Tinderbox is a personal content management assistant. It stores your notes, ideas, and plans. It can help you organize and understand them. And Tinderbox helps you share ideas through Web journals... Read more

Gallery Doctor (Photography)
Gallery Doctor 1.0 Device: iOS iPhone Category: Photography Price: $2.99, Version: 1.0 (iTunes) Description: Free up valuable iCloud and iPhone storage with Gallery Doctor, the only iPhone cleaner that automatically identifies the... | Read more »
You Against Me (Games)
You Against Me 1.0 Device: iOS Universal Category: Games Price: $.99, Version: 1.0 (iTunes) Description: A simple game… You. Me. Claim, steal, lock, score, win! | Read more »
Yep, it's True - Angry Birds 2 is O...
The not exactly rumors were true and the birds are back. Angry Birds 2 has come to the App Store and the world will... well I suppose it'll still be the same, but now we have more bird-flinging options! [Read more] | Read more »
You Could Design Your Own Card for Chain...
If you've ever wanted to create your own item, weapon, trap, or even monster for Chainsaw Warrior: Lords of the Night, this is your chance. Auroch Digital is currently holding a contest so that fans can fight to the death (not really) to see which... | Read more »
Bitcoin Billionaire is Going Back in Tim...
If you thought you managed to buy everything there is to buy in Bitcoin Billionaire and make all the money, well you though wrong. Those of you who made it far enough might remember investing in time travel - and it looks like that investment is... | Read more »
Domino Drop (Games)
Domino Drop 1.0 Device: iOS Universal Category: Games Price: $1.99, Version: 1.0 (iTunes) Description: Domino Drop is a delightful new puzzle game with dominos and gravity!Learn how to play it in a minute, master it day by day.Your... | Read more »
OPERATION DRACULA (Games)
OPERATION DRACULA 1.0.1 Device: iOS Universal Category: Games Price: $5.99, Version: 1.0.1 (iTunes) Description: 25% off launch sale!!! 'Could prove to be one of the most accurate representations of the Japanese bullet hell shmup... | Read more »
Race The Sun (Games)
Race The Sun 1.01 Device: iOS iPhone Category: Games Price: $4.99, Version: 1.01 (iTunes) Description: You are a solar craft. The sun is your death timer. Hurtle towards the sunset at breakneck speed in a futile race against time.... | Read more »
Tap Delay (Music)
Tap Delay 1.0.0 Device: iOS Universal Category: Music Price: $4.99, Version: 1.0.0 (iTunes) Description: Back in the “old days”, producers and engineers created delay and echo effects using tape machines. Tap Delay combines the warm... | Read more »
This Week at 148Apps: July 20-24, 2015
July is Heating Up With 148Apps How do you know what apps are worth your time and money? Just look to the review team at 148Apps. We sort through the chaos and find the apps you're looking for. The ones we love become Editor’s Choice, standing out... | Read more »

Price Scanner via MacPrices.net

Apple’s Education discount saves up to $300 o...
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
12-inch MacBooks in stock for $20 off, save o...
Adorama has 12″ Retina MacBooks in stock for $20 off MSRP including free shipping plus NY & NJ sales tax only. For a limited time, Adorama will include a free Apple USB-C to USB Adapter, free 4-... Read more
College Student Deals: Additional $100 off Ma...
Take an additional $100 off all MacBooks and iMacs at Best Buy Online with their College Students Deals Savings, valid through August 8, 2015. Anyone with a valid .EDU email address can take... Read more
2015 13-inch 2.7GHz Retina MacBook Pro on sal...
B&H Photo has the new 2015 13″ 2.7GHz/128GB Retina MacBook Pro on sale today for $1199 including free shipping plus NY sales tax only. Their price is $100 off MSRP. Read more
2.8GHz Mac mini available for $988, includes...
Adorama has the 2.8GHz Mac mini available for $988, $11 off MSRP, including a free copy of Apple’s 3-Year AppleCare Protection Plan. Shipping is free, and Adorama charges sales tax in NY & NJ... Read more
Updated Mac Price Trackers
We’ve updated our Mac Price Trackers with the latest information on prices, bundles, and availability on systems from Apple’s authorized internet/catalog resellers: - 15″ MacBook Pros - 13″ MacBook... Read more
High-Precision Battery Fuel Gauge IC Extends...
Renesas Electronics Corporation has announced its new lithium-ion (Li-ion) battery fuel gauge IC, the RAJ240500, designed to extend battery life for connected mobile devices such as tablets, notebook... Read more
27-inch 3.3GHz 5K iMac on sale for $1799, $20...
B&H Photo has the 27″ 3.3GHz 5K iMac on sale for $1799 including free shipping plus NY tax only. Their price is $200 off MSRP, and it’s the lowest price available for this model from any Apple... Read more
Twelve South Free Dual Screen Backgrounds Co...
Twelve South has posted a second collection of travel Desktop photos, noting: For the Twelve South team, a vacation is never just a vacation. It’s a time to try out new prototypes on the road, visit... Read more
Apple Refurbished iMacs available for up to $...
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 – $1949 $... Read more

Jobs Board

Engineering Manager, Search Relevance, *Appl...
**Job Summary** Apple 's new Spotlight Suggestions service provides fast, relevant search results from the Inte et in Spotlight and Safari on iOS and OS X. We are looking Read more
Lead Infrastructure Engineer - *Apple* /Mac P...
…of a team * Requires proven problem solving skills Preferred Additional: * Apple Certified System Administrator (ACSA) * Apple Certified Technical Coordinator (ACTC) 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
*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
*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
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