TweetFollow Us on Twitter

Apr 94 Challenge
Volume Number:10
Issue Number:4
Column Tag:Programmers’ Challenge

Related Info: Memory Manager

Programmers’ Challenge

By Mike Scanlin, MacTech Magazine Regular Contributing Author

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:, 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.


This month’s challenge is to swap two adjacent blocks of memory using a finite amount of temporary swap space. This is something the Memory Manager has to do quite often as it shuffles blocks around in the heap.

The prototype of the function you write is:

/* 1 */
void SwapBlocks(p1, p2, swapPtr size1,
  size2, swapSize)
void    *p1;
void    *p2;
void    *swapPtr;
unsigned long  size1;
unsigned long  size2;
unsigned long  swapSize;

p1 and p2 point to the beginnings of the two blocks to swap. size1 and size2 are their respective sizes (in bytes). Both blocks begin on addresses divisible by 4 and have sizes that are divisible by 4. swapPtr points to the scratch area you can use (if you need to) and swapSize is the size of that area (between 256 and 4096 bytes, inclusive). swapPtr and swapSize are also each divisible by 4. If the two blocks look like this on entry:

/* 2 */
^       ^
p1      p2    size1 = 8   size2 = 12

then the same memory locations will look like this on exit:

/* 3 */

When measuring performance I will be calling your routine many times. The distribution of the sizes of the blocks is as follows:

4 to 16 bytes 20% of the time

20 to 32 bytes 20% of the time

36 to 64 bytes 20% of the time

68 to 256 bytes 20% of the time

260 to 4096 bytes 10% of the time

4100 or more bytes 10% of the time

You would normally write this kind of routine in assembly, but let’s see how well you can do in pure C (remember, everyone has the same handicap). If you want to submit a pure assembly solution along with your pure C solution then please do so (but the assembly version will NOT be counted as an entry in the challenge and it will not win anything other than a mention in this column).


Of the 11 entries I received for the We Pry Any Heap (Happy New Year) anagram challenge, only 5 worked correctly. Congrats to Larry Landry (Rochester, NY) for the dual honor of coming in 1st both in terms of speed and smallest code size.

The times for anagramming “programmer” (462 anagrams) and “mactech magazine” (3365 anagrams) with a 19,335 word English dictionary are given here (more weight was given to longer inputs (15-30 characters) when ranking contestants). 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 code time 1 time 2

Larry Landry (1) 830 20 1048

Stepan Riha (5) 2352 45 1166

Bob Boonstra (6) 1370 52 1688

Allen Stenger (3) 1044 23 1701

Mark Nagel 1134 81 51407

Most of the entrants figured out that the key to speeding up the anagram process was to pare down the size of the dictionary first. Once you have the input characters you can eliminate any word in the dictionary that: (1) contains more characters than the input, (2) contains at least one letter not in the input set or, (3) contains more of any particular character than the input. For instance, if your input is “programmer” then you can remove any word in the dictionary that (1) is more than 10 characters long, (2) is not made up entirely of the letters [p, r, o, g, a, m, e] and, (3) contains more than any one of: 1 p, 3 r’s, 1 o, 1 g, 1 a, 2 m’s, or 1 e.

Stepan Riha (Austin, TX) took this “reduce the dictionary” idea one step further and came up with a way to store words that are permutations of each other (like ‘stop’, ‘post’ and ‘pots’) as one entry in the dictionary (and when it’s time to output an anagram he outputs all permutations for each word in the output).

Several people wrote to me and asked if reordering the words in each output anagram was necessary (i.e. ‘pale rain’ and ‘rain pale’). I admit that it wasn’t clear in the puzzle specification exactly what qualified as a ‘unique’ anagram so I allowed either interpretation. The only one of the 5 correct entries that did count word reorderings as unique anagrams is Mark Nagel (Irvine, CA) and his times above reflect that fact.

Here’s Larry’s winning solution:

Anagram Programmer's Challenge

by Larry Landry

This implementation uses a large amount of memory to optimize the CPU utilization. To guarantee that we have enough memory for all matching words, we actually allocate an array of pointers for 30,000 words. Since the rules stated that there would be about 20,000 words in the dictionary, even if every word matched, we would still have enough storage. In reality this number could probably be less than 1-200 for all but the most rare of scenarios.

The basic algorithm is: 1) Convert the input string into a table of counts for each character from a-z. So "sammy" would have a count of 2 for "m" and 1 for each of "s", "a", and "y". This makes testing for the presence of a character as simple as checking and indexed value in an array. 2) Parse through the dictionary and find the words that can be composed of some portion of characters from the input characters. Build a list of pointers to each word. The number of words in this list will be in the tens instead of thousands. 3) Recursively process the words in this list and find strings of words that use up all of the characters. For each matching sequence, output the words to the file. The processing required by this algorithm is then

D * C1 + M * log2(M) * C2


D = size of input dictionary

M = number of matching words

C1 & C2 are constants

This algorithm works very well for cases where there are few words that match the input letters. The worst case scenario where all words can be made from the input letters will still take a very long time. I expect that matching words will typically be less than 100.

/* 4 */
#include   <stdio.h>

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

#define MAX_WORDS   30000L
#define OUTPUT_BUFFER_SIZE  10000L
#define RETURN  '\n'

typedef struct {
   short   fWordLength;
} WordLoc;

/* Usage counts for each character (only indexes 'a' to 'z' are actually 
used) */
typedef uchar   CharData[256];

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

ulong findInputWords(register char *wordBuffer,
   WordLoc *validWords);
ulong findAnagrams(short numValidChars, ulong wordCount,
   WordLoc *validWords, short prevWordCount);

/* I use some global variables here to avoid passing them down into the 
recursive routine findAnagrams().  These values are constant once findAnagrams() 
is called. */

char     gOutputBuffer[OUTPUT_BUFFER_SIZE];
char    *gOutputBufferEnd = gOutputBuffer + 
char    *gOutputPtr;
CharData  gValidChars;
WordLoc  *gWordsInUse[255];
FILE    *gOutputFile;

unsigned long Anagram(Str255 inputText, FILE *wordList,
   FILE *outputFile)
   fpos_t  wordBufferLength;
   char*  wordBuffer;
   short   index;
   short   numValidChars;
   WordLoc validWords[MAX_WORDS];
   char   ch;
   ulong   wordCount;

   gOutputFile = outputFile;
   gOutputPtr = gOutputBuffer;

/* To save on file I/O time, read the whole file all at once.  First, 
find the length of the file by seeking the end and finding the file pos. 
 Then allocate a buffer of that size, plus 2 bytes  (for a return and 
NULL char) and read the data into it.  Finally put the return and NULL 
char at the end. */

   fseek(wordList, 0L, SEEK_END);
   fgetpos(wordList, &wordBufferLength);
   wordBuffer = (char*) NewPtr((Size) wordBufferLength + 2);
   if (wordBuffer == NULL)
   return 0L;  /* real error handling here */
   fread(wordBuffer, (size_t) 1,
   (size_t) wordBufferLength, wordList);
   if (wordBuffer[wordBufferLength-1] != RETURN)
   wordBuffer[wordBufferLength++] = RETURN;
   wordBuffer[wordBufferLength] = '\0';

/* To save time ruling out words, we build a list of the valid characters 
in the words.  We start with no valid characters. */
   for (index='a'; index<'z'; index++)
   gValidChars[index] = 0;

/* Now build the list of valid characters.  Each array entry will be 
a count of how many times that character is present. */
   numValidChars = *inputText++;
   for (index=numValidChars; index>0; index--)
   if ((ch = *inputText++) != ' ')
/* Find the list of words that can be made up from the letters in the 
input word */
   wordCount = findInputWords(wordBuffer,
/* Now find the list of full anagrams that can be created from these 
words */
   wordCount = findAnagrams(numValidChars, wordCount,
   &validWords[MAX_WORDS-wordCount], 0);
/* Write the results to the output */
   *gOutputPtr = 0;/* Terminate the string */
   fprintf(outputFile, gOutputBuffer);
   return wordCount;
} /* Anagram */

ulong findInputWords(register char *wordBuffer,
   WordLoc *validWords)
   char*saveStart = wordBuffer;
   ulong   numberWords = 0;
   char ch;

   while (*wordBuffer)
   ch = *wordBuffer++;
   if (ch == RETURN)
/* Record this entry as a valid word */
   validWords->fWordStart = saveStart;
   validWords->fWordLength = (short)(wordBuffer -
   saveStart - 1);

   while (saveStart < wordBuffer)

/* Save the new start of word pointer */
   saveStart = ++wordBuffer;
   } else if (gValidChars[ch])
/* This word didn't match so reset and go to the next word */
   while (saveStart < wordBuffer)
   while (*wordBuffer++ != RETURN)
/* Save the new start of word pointer */
   saveStart = wordBuffer;
   } /* else */
   } /* while */
   return numberWords;
} /* findInputWords */

ulong findAnagrams(short numValidChars, ulong wordCount,
   WordLoc *validWords, short prevWordCount)
   ulong   wordIndex;
   ulong   usedIndex;
   short   chIndex;
   ulong   matchCount = 0;
   Boolean wordFits;
   char ch;
   WordLoc *theWord;

/* Try each word we have against the list of characters. */
   for (wordIndex=0; wordIndex<wordCount; wordIndex++)
/* If there aren't enough characters left,  it can't be a match */
   if (validWords->fWordLength <= numValidChars)
/* Go through the chars in this word testing to make sure that there 
is at least one of each char  available */
   wordFits = TRUE;
   for (chIndex=0; chIndex<validWords->fWordLength; chIndex++)
   ch = validWords->fWordStart[chIndex];
   if (gValidChars[ch])
/* Found an unavailable character, so this can't be part of the anagram. 
 Reset the character usage array and go to the next word. */
   wordFits = FALSE;
   while (--chIndex >= 0)
   break;  /* get out of the for loop */
   } /* else */
   } /* for */

   if (wordFits)
/* This word fit, so see if it uses all the characters.   If so, then 
we have found an anagram.  Output the  anagram and increment the anagram 
count. */
   if (validWords->fWordLength == numValidChars)
/* Copy the previous words for this anagram separated by spaces. */
   for (usedIndex=0; usedIndex<prevWordCount; usedIndex++)
   theWord = gWordsInUse[usedIndex];
   memcpy(gOutputPtr, theWord->fWordStart,
   (size_t) theWord->fWordLength);
   gOutputPtr += theWord->fWordLength;
   *gOutputPtr++ = ' ';
   } /* for */
/* Now copy this new word and a return character */
   memcpy(gOutputPtr, validWords->fWordStart,
   (size_t) validWords->fWordLength);
   gOutputPtr += validWords->fWordLength;
   *gOutputPtr++ = RETURN;

/* To ensure that we don't overrun the output buffer check against the 
end of the buffer.  If the end pointer has been passed, write the data 
to the file  and reset the output pointer to the beginning of the buffer. 
   if (gOutputPtr > gOutputBufferEnd)
 *gOutputPtr = 0;/* Terminate the string */
   fprintf(gOutputFile, gOutputBuffer);
   gOutputPtr = gOutputBuffer;
   } /* if */
   }  /* if */
/* This word did fit, but didn't use all of the characters so add it 
to the list of previous words  in the anagram and then call this procedure 
recursively to find if there are more words that can be added to make 
an anagram with this base. */
   gWordsInUse[prevWordCount] = validWords;
   matchCount += findAnagrams(
   numValidChars - validWords->fWordLength,
   wordCount - wordIndex, validWords,
   prevWordCount + 1);
   } /* else */

/* Now undo the characters we took out of the validChar array */
   for (chIndex=0;chIndex<validWords->fWordLength;chIndex++)
   } /* if */
   } /* if */

 } /* for */

   return matchCount;
} /* findAnagrams */


Community Search:
MacTech Search:

Software Updates via MacUpdate

Planet Diver guide - How to survive long...
Planet Diver is an endless arcade game about diving through planets while dodging lava, killing bats, and collecting Starstuff. Here are some tips to help you go the distance. [Read more] | Read more »
KORG iDS-10 (Music)
KORG iDS-10 1.0.0 Device: iOS iPhone Category: Music Price: $9.99, Version: 1.0.0 (iTunes) Description: ** Debut Discount: 50% OFF! Sale Price US$9.99 (Regular price US$19.99). Other all Korg apps are also 50% OFF until Dec 28! **... | Read more »
World of Tanks Generals guide - Tips and...
World of Tanks Generals is a brand new card game by the developer behind the World of Tanks shooter franchise. It plays like a cross between chess and your typical card game. You have to keep in consideration where you place your tanks on the board... | Read more »
TruckSimulation 16 guide: How to succeed...
Remember those strangely enjoyable truck missions in Grand Theft Auto V whereit was a disturbing amount of fun to deliver cargo? TruckSimulation 16 is reminiscent of that, and has you play the role of a truck driver who has to deliver various... | Read more »
The best GIF making apps
Animated GIFs have exploded in popularity recently which is likely thanks to a combination of Tumblr, our shorter attention spans, and the simple fact they’re a lot of fun. [Read more] | Read more »
The best remote desktop apps for iOS
We've been sifting through the App Store to find the best ways to do computer tasks on a tablet. That gave us a thought - what if we could just do computer tasks from our tablets? Here's a list of the best remote desktop apps to help you use your... | Read more »
Warhammer 40,000: Freeblade guide - How...
Warhammer 40,000: Freebladejust launched in the App Store and it lets you live your childhood dream of blowing up and slashing a bunch of enemies as a massive, hulking Space Marine. It's not easy being a Space Marine though - and particularly if... | Read more »
Gopogo guide - How to bounce like the be...
Nitrome just launched a new game and, as to be expected, it's a lot of addictive fun. It's called Gopogo, and it challenges you to hoparound a bunch of platforms, avoiding enemies and picking up shiny stuff. It's not easy though - just like the... | Read more »
Sago Mini Superhero (Education)
Sago Mini Superhero 1.0 Device: iOS Universal Category: Education Price: $2.99, Version: 1.0 (iTunes) Description: KAPOW! Jack the rabbit bursts into the sky as the Sago Mini Superhero! Fly with Jack as he lifts impossible weights,... | Read more »
Star Wars: Galaxy of Heroes guide - How...
Star Wars: Galaxy of Heroes is all about collecting heroes, powering them up, and using them together to defeat your foes. It's pretty straightforward stuff for the most part, but increasing your characters' stats can be a bit confusing because it... | Read more »

Price Scanner via

World’s First USB-C Adapter For MacBook Suppo...
Innergie, a brand of Delta Electronics, has announced its official release of the world’s first USB-C adapter supporting four DC output voltages, the PowerGear USB-C 45. This true Type C adapter... Read more
13-inch and 11-inch MacBook Airs on sale for...
B&H Photo has 13″ and 11″ MacBook Airs on sale for up to $120 off MSRP as part of their Holiday sale including free shipping plus NY sales tax only: - 11″ 1.6GHz/128GB MacBook Air: $819 $90 off... Read more
13-inch MacBook Pros on sale for up to $150 o...
Take up to $150 off MSRP on the price of a new 13″ MacBook Pro at B&H Photo today as part of their Holiday sale. Shipping is free, and B&H charges NY tax only. These prices are currently the... Read more
13-inch 128GB MacBook Air now on sale for $79...
Best Buy has just lowered their price on the 2015 13″ 1.6GHz/128GB MacBook Air to $799.99 on their online store for Cyber Monday. Choose free shipping or free local store pickup (if available). Sale... Read more
Best Buy lowers 13-inch MacBook Pro prices, n...
Best Buy has lowered prices on select 13″ MacBook Pros this afternoon. Now save up to $200 off MSRP for Cyber Monday on the following models. Choose free shipping or free local store pickup (if... Read more
Cyber Monday: Apple MacBooks on sale for up t...
Apple resellers have MacBook Pros, MacBook Airs, and MacBooks on sale for up to $250 off MSRP for Cyber Monday 2015. The following is a roundup of the lowest prices available for new models from any... Read more
Cyber Monday: Apple Watch on sale for up to $...
B&H Photo has the Apple Watch on sale for Cyber Monday for $50-$100 off MSRP. Shipping is free, and B&H charges NY sales tax only: - Apple Watch Sport: $50 off - Apple Watch: $50-$100 off B... Read more
Cyber Monday: 15% off Apple products, and sto...
Use code CYBER15 on Cyber Monday only to take 15% on Apple products at Target, and store-wide. Choose free shipping or free local store pickup (if available). Sale prices for online orders only, in-... Read more
iPad Air 2 And iPad mini Among Top Five Black...
Adobe has released its 2015 online shopping data for Black Friday and Thanksgiving Day. The five best selling electronic products on Black Friday were Samsung 4K TVs, Apple iPad Air 2, Microsoft Xbox... Read more
All-in-one PC Shipments Projected To Drop Ove...
Digitimes’ Aaron Lee and Joseph Tsai report that all-in-one (AIO) PC shipments may drop a double-digit percentage on-year in 2015 due to weaker-than-expected demand, although second-largest AIO make... Read more

Jobs Board

*Apple* New Products Tester Needed - Apple (...
…we therefore look forward to put out products to quality test for durability. Apple leads the digital music revolution with its iPods and iTunes online store, continues Read more
Software Engineer, *Apple* Watch - Apple (U...
# Software Engineer, Apple Watch Job Number: 33362459 Santa Clara Valley, Califo ia, United States Posted: Jul. 28, 2015 Weekly Hours: 40.00 **Job Summary** Join the Read more
SW Engineer - *Apple* Music - Apple (United...
# SW Engineer - Apple Music Job Number: 40899104 San Francisco, Califo ia, United States Posted: Aug. 18, 2015 Weekly Hours: 40.00 **Job Summary** Join the Android Read more
Sr Software Engineer *Apple* Pay - Apple (U...
# Sr Software Engineer Apple Pay Job Number: 44003019 Santa Clara Valley, Califo ia, United States Posted: Nov. 13, 2015 Weekly Hours: 40.00 **Job Summary** Apple Read more
*Apple* Site Security Manager - Apple (Unite...
# Apple Site Security Manager Job Number: 42975010 Culver City, Califo ia, United States Posted: Oct. 2, 2015 Weekly Hours: 40.00 **Job Summary** The Apple Site Read more
All contents are Copyright 1984-2011 by Xplain Corporation. All rights reserved. Theme designed by Icreon.