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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 */


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