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Nov 95 Challenge
Volume Number:11
Issue Number:11
Column Tag:Programmer’s Challenge

Programmer’s Challenge

By Bob Boonstra, Westford, Massachusetts

Note: Source code files accompanying article are located on MacTech CD-ROM or source code disks.

Enclosing Bounds

The Challenge this month is based on a suggestion by Mike Scanlin, who remains a fan of the column. (We’re still waiting for Mike’s first Challenge entry, however.) The problem is to write a routine that will return a rectangle enclosing all non-white pixels in a selected area of an image. This code might be useful in a drawing or painting program, where the user would be allowed to select a subset of the image by clicking and dragging, and the software would select all of the elements of the image contained within that selection. The prototype of the code you will write is:

void EnclosingBounds(
 PixMapHandle pm,  /* handle to PixMap containing image */
 Rect selection, /* subset of image to enclose */
 Rect *enclosingRect /* enclosing rect return value */

Your code should examine all of the pixels within the selection rectangle of the PixMap and return the smallest rectangle containing all of the non-white pixels. Pixels outside the selection rectangle should be ignored. The bounds rectangle of the PixMap will be no larger than 2048 pixels in each dimension, the baseAddr pointer will be longword aligned, and rowBytes will be a multiple of 4. You should deal with pixelSize values of 1, 8, or 32, with values of 8 and 32 being weighted most heavily in measuring performance. For PixMaps with indexed pixels (cmpCount==1), the color table will contain white as the first table entry (as all good color tables are supposed to). For PixMaps with direct pixels, the unused (alpha) bits of each pixel will be zero.

You may use either the Metrowerks or the Symantec compilers for this native PowerPC Challenge. If you have any questions, or would like some test data for your code, please send me e-mail at one of the Programmer’s Challenge addresses, or directly to

Two Months Ago Winner

Congratulations to Eric Lengyel (Blacksburg, VA) for submitting the fastest and smallest entry to the Reversible Scrambling Algorithm Challenge. Despite an unfortunate delay in publication of the magazine that left participants with less time than usual to complete the Challenge, three of the four entries I received by the extended deadline worked correctly, at least in part.

You might recall that the Challenge was to write code that would raise a large integer message to a power and compute the remainder modulo another large integer. The name of the Challenge comes from the fact that this technique is reversible, given properly chosen integers. Eric is a graduate student in Mathematics at Virginia Tech, and he took advantage of a highly optimized multiple precision integer arithmetic library that he had written as part of a number theory project involving the factorization of very large numbers.

Each of the working entries converted the BigNum representation provided in the problem into one that right-justified numbers into a fixed-length data structure. While this imposes a restriction on the maximum size integer that the code can handle, this assumption was permitted by the problem statement. In Eric’s code, the restriction is controlled by a single #define statement.

Eric uses a binary exponentiation algorithm to raise the message to the specified power, and takes advantage of facts from number theory that allow the remainder to be computed at each step of the exponentiation. The time to perform the exponentiation is therefore proportional to the logarithm of the exponent. Eric’s multiplication and division routines use the 68020’s capability to compute the 64-bit product of two longwords and to divide a 64-bit dividend by a longword. The multiplication, division, exponentiation, and compare routines in Eric’s code are general purpose and could be used in any 68K application that needs large integers.

Honorable mention goes to Ernst Munter, who submitted an entry in pure C that was actually the fastest code for the short modulus test cases. Unfortunately, his entry did not produce correct results for the longer moduli.

Here are the times and code sizes for the entries that worked correctly (or partially correctly). Execution time is presented for two specific test cases, with modulus lengths of 22 and 88 bytes, respectively, as well as the total time for all of the test cases I ran. Cases that produced incorrect results are indicated with an asterisk. Numbers in parens after a person’s name indicate that person’s cumulative point total for all previous Challenges, not including this one.

Name time1 time2 Total time code data

(22) (88)
bytes bytes

Eric Lengyel 47 463 2083 1190 0

Xan Gregg (51) 35 967 3175 1558 0

Ernst Munter (C entry) (90) 17 * * 4266 11788

Top 20 Contestants of All Time

Here are the Top 20 Contestants for the Programmer’s Challenges to date. The numbers below include points awarded for this month’s entrants. (Note: ties are listed alphabetically by last name - there are more than 20 people listed this month because of ties.)

Rank Name Points

1. [Name deleted] 176

2. Munter, Ernst 90

3. Karsh, Bill 78

4. Stenger, Allen 65

5. Gregg, Xan 61

6. Larsson, Gustav 60

7. Riha, Stepan 51

8. Goebel, James 49

9. Nepsund, Ronald 47

10. Cutts, Kevin 46

11. Mallett, Jeff 44

12. Kasparian, Raffi 42

13. Vineyard, Jeremy 42

14. Darrah, Dave 31

15. Landry, Larry 29

16. Elwertowski, Tom 24

17. Lee, Johnny 22

18. Noll, Robert 22

19. Anderson, Troy 20

20. Beith, Gary 20

21. Burgoyne, Nick 20

22. Galway, Will 20

23. Israelson, Steve 20

24. Landweber, Greg 20

25. Lengyel, Eric 20

26. Pinkerton, Tom 20

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 Eric’s winning solution:


Copyright © 1995 Eric Lengyel
I call my fixed length numbers “BigFixed” and translate from BigNum’s to BigFixed’s in the PowerAndRemainder 
routine.  These are the assembly language routines which are the guts of my program:

(1) PowerMod - raises a number to a power and reduces it by a modulus.  It uses a fast binary exponentiation 
algorithm, reducing by the modulus at each step.
(2) Multiply - multiplies 2 BigNum’s together.
(3) MultQ - mutliplies a BigNum by a long int.
(4) Divide - divides one BigNum by another and supplies the quotient and remainder.
(5) Compare - determines the ordering of 2 BigNum’s.

Some of the loops have been expanded to make more efficient use of the instruction cache.

#define NumSize 72

typedef struct BigNum
   short           numDig;
   unsigned char   *dig;
} BigNum;

typedef struct BigFixed
   unsigned char   dig[NumSize*4];
} BigFixed;

/* We need 72 longs because the division routine needs the most significant longword to be zero and the 
speed optimization requires that NumSize be a multiple of four. */

void PowerAndRemainder(BigNum *msg, BigNum *exp, BigNum *n,
   BigNum *res);
void PowerMod(BigFixed *msg, BigFixed *exp, BigFixed *n,
   BigFixed *res);
void Multiply(BigFixed *src1, BigFixed *src2, BigFixed *dst);
void MultQ(BigFixed *src1, long src2, BigFixed *dst);
void Divide(BigFixed *end, BigFixed *sor, BigFixed *dst);
short Compare(BigFixed *src1, BigFixed *src2);


void PowerAndRemainder(BigNum *msg, BigNum *exp, BigNum *n,
   BigNum *res)
   short      a, b, numDigits;
   BigFixed   msg0, exp0, n0, res0;
   for (a = 0; a < NumSize*4; a++)
      b = NumSize*4 - msg->numDig;
      if (a < b) msg0.dig[a] = 0;
      else msg0.dig[a] = msg->dig[a - b];
      b = NumSize*4 - exp->numDig;
      if (a < b) exp0.dig[a] = 0;
      else exp0.dig[a] = exp->dig[a - b];
      b = NumSize*4 - n->numDig;
      if (a < b) n0.dig[a] = 0;
      else n0.dig[a] = n->dig[a - b];
   PowerMod(&msg0, &exp0, &n0, &res0);
   a = 0;
   while (res0.dig[a] == 0) a++;
   numDigits = res->numDig = NumSize*4 - a;
   for (b = 0; b < numDigits; b++)
      res->dig[b] = res0.dig[a++];


void PowerMod(BigFixed *msg, BigFixed *exp, BigFixed *n,
   BigFixed *res)
   BigFixed   acc, scrap;

   LEA      acc, A0            ;   Start with one in
   MOVEQ    #NumSize/4-2, D0   ;   accumulator
1} ; Test a bit in current
                               ;   longword of exponent
   BEQ      @1                 ;   If zero, skip multiply
   PEA      acc                ;   Multiply accumulator
   PEA      acc                ;   by base
   MOVE.L   msg, -(A7)
   JSR      Multiply
   ADDA.W   #12, A7
   MOVE.L   n, -(A7)           ;   Compare accumulator
   PEA      acc                ;   to modulus
   JSR      Compare
   ADDQ.W   #8, A7
   TST.B    D0
   BMI      @1                 ;   If it’s less, skip
                               ;   reduction
   PEA      scrap              ;   Reduce modulo “n”
   MOVE.L   n, -(A7)
   PEA      acc
   JSR      Divide
   ADDA.W   #12, A7

/* Multiply src1 by src2 and put product in dst */

void Multiply(BigFixed *src1, BigFixed *src2, BigFixed *dst)
   short      topStop, botStop;
   BigFixed   acc, line;
   MOVEM.L  D0-D7/A0-A4, -(A7)
   LEA      acc, A0            ;   Clear accumulator
   MOVEQ    #NumSize/4-1, D0
D5       ;   Do 64-bit multiply
   ADD.L    D2, D5             ;   Add carry to low
                               ;   longword of product
   CLR.L    D2                 ;   Use D2 as dummy to
                               ;   extend carry
   ADDX.L   D2, D6             ;   Add zero to high
                               ;   longword with carry
   MOVE.L   D6, D2             ;   Anything in high
                               ;   longword gets carried
   MOVE.L   D5, 00(A2, D3.W*4) ;   Store low longword in
                               ;   partial product
   SUBQ.W   #1, D3             ;   Loop through all
   CMP.W    topStop, D3        ;   longwords in top number
   BGE      @1
   MOVEA.L  A2, A0             ;   Now add partial product
                               ;   to accumulator
   MOVE.L   D4, D0             ;   Calculate correct
                               ;   position in product
   LEA      acc, A1            ;   Get accumulator’s addr
   ADDQ.W   #1, D0
   ADDA.W   #NumSize * 4, A0
   LSL.W    #2, D0
   ADDA.W   D0, A1
   MOVE.W   D4, D1
   MOVE.L   -(A1), D0          ;   Get longword of product
   SUBQ     #1, D1
   ADD.L    -(A0), D0          ;   Add longword of
   MOVE.L   D0, (A1)           ;   partial product
   TST.W    D1                 ;   If no more longwords,
   BMI      @2                 ;   then branch
/* Multiply src1 by src2 and put product in dst */

void MultQ(BigFixed *src1, long src2, BigFixed *dst)
   BigFixed   pro;
   MOVEM.L  D0-D7/A0/A1, -(A7)
   LEA      pro, A0            ;   Clear product
   MOVEQ    #NumSize/4-1, D0
D4          ;   Do 64-bit multiply
                               ;   by bottom number
   ADD.L    D2, D4             ;   Add carry
   CLR.L    D2                 ;   Use D2 as dummy to
                               ;   extend carry
   ADDX.L   D2, D5             ;   Add zero with carry
   MOVE.L   D5, D2             ;   High longword
                               ;   becomes carry
   MOVE.L   D4, 00(A1, D0.W*4) ;   Put partial product
                               ;   into result
   SUBQ.W   #1, D0             ;   Loop through all
   CMP.W    D1, D0             ;   longwords in top #
   BGE      @1

/* Divide end (dividend) by sor (divisor) and put quotient in dst.  Remainder will end
    up in end */

void Divide(BigFixed *end, BigFixed *sor, BigFixed *dst)
   long      pq;
   BigFixed  quo, line;
   MOVEM.L  D0-D7/A0-A4, -(A7)
   LEA      quo, A0            ;   Clear quotient
   MOVEQ    #NumSize/4-1, D0
D4          ;   Do 64-bit division

/* Compare src1 and src2.  Returns 1 if src1 > src2, 0 if they’re equal, and -1 if src1 < 
    rc2. */

short Compare(BigFixed *src1, BigFixed *src2)
   MOVEM.L  D1/D2/A0/A1, -(A7)
   MOVEA.L  src1, A0           ;   Get src1’s address
   MOVEA.L  src2, A1           ;   Get src2’s address
   MOVEQ    #1, D0             ;   Start with +1
   MOVE.L   (A0)+, D2
   CMP.L    (A1)+, D2          ;   Compare 1st longwords
   BLT      @1                 ;   If src1 less, branch
   BNE      @2                 ;   If !=, src1 must
   MOVE.L   (A0)+, D2          ;   be greater
   CMP.L    (A1)+, D2          ;   Cmp 3 more longwords
   BCS      @1                 ;   (Unsigned)
   BNE      @2
   MOVE.L   (A0)+, D2
   CMP.L    (A1)+, D2
   BCS      @1
   BNE      @2
   MOVE.L   (A0)+, D2
   CMP.L    (A1)+, D2
   BCS      @1
   BNE      @2
   MOVEQ    #NumSize/4-2, D1   ;   Number of longwords
                               ;   remaining / 4


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