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Programmers’ Challenge

By Mike Scanlin, MacTech Magazine Regular Contributing Author

Rotated Bits

This month’s challenge is to write a routine that quickly rotates a 1-bit deep bitMap 90 degrees clockwise. You are given pointers to a source bitMap and destination bitMap. Space for the destination pixels has already been allocated (the baseAddr field is valid) and the rowBytes and bounds fields have been set up for you. All you have to do is rotate the bits.

The function prototype you write is:

void RotateBitMapClockwise(srcBitMapPtr, dstBitMapPtr)
BitMap  *srcBitMapPtr;
BitMap  *dstBitMapPtr;

To check for correctness, you might want to write a test program that rotates the same bitMap four times and then diffs the result with the original (they should be identical). Have fun.

February Winner

The winner of the “Insane Anglo Warlord” challenge is Jeremy Vineyard (Lawrence, KS) whose solution was the only one of the two I received which gave valid results (the other one gave intermediate strings that had a character not in the original strings). Remember kids: you have to play the game to have a chance at winning.

Here’s the output of Jeremy’s Descramble routine when the input strings are “INSANE ANGLO WARLORD” and “RONALD WILSON REAGAN” when the number of intermediate steps is set to 10:

INSANE ANGLO WARLORD
OIDSLNE N NGWRALORAA
 ORDLINE AWNSNLORAAG
R AODSILEWANN ORALGN
R WADOSGLNIAA RLOENN
R IWODNSALRAAL OEGNN
ROD INLASL RAWOEAGNN
RONDIL A SLORWAEAGNN
RONLDLI S AORWAEAGNN
RONDL LWASIOR AEAGNN
RONWLD AILSOR AEAGNN
RONALD WILSON REAGAN

And here is Jeremy’s code to generate those strings:

/***********************************
 * Descramble.c
 * A procedure that returns a smooth 
 * transition from one string to 
 * another with the same characters, 
 * but in a different order.
 ***********************************/

#include <stdlib.h>
#include <string.h>

void Descramble(
 Str255 startString,
 Str255 endString,
 unsigned short  numSteps,
 Str255 *stepStringPtrs[20] )
{
 short  i, j, k;
 short  strLength = startString[0];
 div_t  divResult;
 short  destination, memSwitch;
 BooleanswitchPossible;

 short  distances[256];
 short  destMap[256];
 BooleandestSwitchMap[256];
 BooleancheckList[256];
 
 /* Clear the destination switching 
  * map for use. Only clear it up to 
  * strLength to save time. */
 for (i = 1; i <= strLength; i++)
 destSwitchMap[i] = false;
 
 /* This loop searches for a duplicate 
  * char 'j' for the char 'i'. Once it  * finds a duplicate, it checks 
to 
  * see whether it is already being 
  * used as another char's 
  * destination. If not, it shows that 
  * char 'j' is char 'i's destination, 
  * sets the destMap to show that char 
  * 'j' is being used, and
  * calculates how far the char 'i' needs to travel on 
  * each step to get to its destination. */
 for (i = 1; i <= strLength; i++)
 for (j = 1; j <= strLength; j++)
 if ( startString[i] == endString[j] &&
  ! destSwitchMap[j] )
 {
 /* If destSwitchMap[j] is set, then it already has a 
  * char which is using it for a destination. */
 destSwitchMap[j] = true;
 
 /* Remember this char's dest. */
 destMap[i] = j;
 
 /* Find out how far this char should move 
  * on each step. */
 if (i == j)
 distances[i] = 0;
 else
 {
    divResult = div(j - i, numSteps);
    distances[i] = divResult.quot;
 
 /* Use the remainder to make
  * sure this char moves each step.*/ 
    if (divResult.rem != 0)
    if (divResult.rem < 0)
       distances[i] -= 1;
    else
   distances[i] += 1;
 
    /* Increment to exit loop. */
    j = 512;
 }
 }
 
 /* In this loop, each character tries to move towards its 
  * destination. Its distance is then recalculated to 
  * compensate for it being switched. This creates a 
  * 'morphing' effect, where the letters in the start string 
  * gradually change to their positions in the end string. */
 for (i = 0; i < numSteps; i++)
 {
 /* Copy the appropriate string for switching. */
 if (i > 0)
 memcpy(*stepStringPtrs[i],
  *stepStringPtrs[i - 1],
  strLength + 1);
 else
 memcpy(*stepStringPtrs[0], startString,
   strLength + 1);
 
 /* Clear the check list for use. */
 for (k = 1; k <= strLength; k++)
 checkList[k] = false;
 
 /* This loop switches characters until
  * switchPossible = false. */
 do
 {
  switchPossible = false;

  for (j = 1; j <= strLength; j++)
  {
   if (distances[j] != 0 &&
  ! checkList[j])
   {
      /* Calculate this char's intended
 * destination for this step. */
      destination = j + distances[j];
        
      if (checkList[destination])
      {
 /* If the destination has already been used, find 
  * the nearest one to it to switch to. */
          destination = -1;
          
         if (distances[j] > 0)
          for (k = destination - 1; k > j;
  k -= 1)
          if (! checkList[k] &&
  distances[k] != 0)
          {
          destination = k;
          k = 512;
          }
          else;
         else
          for (k = destination + 1; k < j;
  k++)
          if (! checkList[k] &&
  distances[k] != 0)
          {
          destination = k;
          k = 512;
          }
      }
      
      if (destination > 0)
      {
       /* If destination is a valid number, do the 
   * neccessary switching. Switch the characters
   * in the string */
        memSwitch =
   stepStringPtrs[i][0][destination];
        stepStringPtrs[i][0][destination]
   = stepStringPtrs[i][0][j];
        stepStringPtrs[i][0][j] =
   memSwitch;
        
        /* Switch the character's mapped destinations. */
        memSwitch = destMap[destination];
        destMap[destination] = destMap[j];
        destMap[j] = memSwitch;
        
        /* Show that the destination has been switched. */
        checkList[destination] = true;
        
        /* Recalculate the switched char's distance. */
        if (destMap[j] == j)
        distances[j] = 0;
        else if (i + 1 < numSteps)
        {
     divResult = div(destMap[j] - j,
  numSteps - i - 1);
     distances[j] = divResult.quot;
     if (divResult.rem != 0)
     if (divResult.rem < 0)
      distances[j] -= 1;
     else
     distances[j] += 1;
  }
  
    switchPossible = true;
      }
   }
   }
 } while (switchPossible);
 
 if (i + 1 == numSteps)
 return;
 
 /* Recalculate all distances to compensate
  * for switched char's. */
 for (k = 1; k <= strLength; k++)
 if (distances[k] != 0)
 {
 if (destMap[k] == k)
 distances[k] = 0;
 else if (i + 1 < numSteps)
 {
    /* Recalculate how far this char needs to move 
 * with the number of steps that are left. */
    divResult = div(destMap[k] - k,
  numSteps - i - 1);
    distances[k] = divResult.quot;
    if (divResult.rem != 0)
    if (divResult.rem < 0)
    distances[k] -= 1;
    else
    distances[k] += 1;
 }
 }
 }
}

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).