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Oct 92 Challenge
Volume Number:8
Issue Number:6
Column Tag: Programmers' Challenge

Programmers' Challenge

By Mike Scanlin, MacTutor Regular Contributing Author

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

Programming Challenge of the Month - NAME NO ONE MAN

This month’s challenge involves palindromes -- things that read the same backward and forward (like the letters in “name no one man” or “a toyota”). The goal is to write a routine that finds the nth palindrome greater than a given baseNumber (when it’s displayed as a base 10 integer without leading zeros). Our numeric palindromes will only consist of digits from 0 to 9 and will not be larger than 9 digits long (return -1 if the palindrome requested is larger than 999,999,999). The prototype is:

long FindNthPalindrome(baseNumber, n)
 long baseNumber;
 short  n;


Input:  baseNumber = 107
 n = 3


 function result = 131

Remember, speed is more important than size. This is a fairly simple programming challenge -- but how fast can you make it?


To Aaron Zick (San Francisco, CA) for winning the very first MacTutor Programming Challenge (rubber banded pegs). Among the submitted solutions yielding correct results, his was the fastest and the second smallest. He will be receiving a cool t-shirt as soon as they are available.

The key to writing a fast routine was knowing that you don’t have to use trig functions to calculate the area of a convex polygon. As William Karsh (Manteno, IL) explained in his well commented solution, the area of a “simply connected, piecewise differentiable” region can be computed as follows: For each segment going around the perimeter, bounded by points p1 to p2, calculate p1.h * (p2.v - p1.v) - p1.v * (p2.h - p1.h). The area is the sum of all of these pieces (you may need to multiply by -1 for orientation). Sorry, William, you had the right idea but your code was twice as large and 5% slower than the winning solution.

Jim Walker (Columbia, SC) deserves mention for the smallest code (half the size of the winning solution) and for reminding us that you can calculate the area of a triangle by using the following macro (which might come in handy in one of your own applications, so keep it in mind): AREA(x, y, z) = ((z.h-y.h) * (y.v-x.v) - (z.v-y.v) * (y.h-x.h)) (the sign will be negative if going from x to y to z involves a left turn). Unfortunately Jim’s easy-to-read and elegant routine was 5% to 25% slower than Aaron’s.

Here’s Aaron’s winning solution to the August Challenge (some comments have been removed for space reasons. Aaron’s complete source is on the source code disk):

/* Max holes per side of the peg board. */
#define HOLES 13
void GetPerimeter( Point thePegs[], short 
 numPegs, Point outerPegs[], short 
 sideLast[] );
void GetEdgePegs( Point outerPegs[], short 
 test, short last, Point edgePegs[], 
 short *numEdgePegs );
void CheckEdgePegs( Point edgePegs[], short 
 *numEdgePegs, Point newPeg, short first);
void IntegrateArea( Point edgePegs[], short 
 numEdgePegs, Fixed *area ); 
/* BandedPegs takes an array of points 
 * representing pegs on a pegboard and 
 * returns an array of points representing 
 * the pegs that would be touched by a 
 * rubber band surrounding as many pegs as  
 * possible. It also returns the area thus               surrounded. 
void BandedPegs( short numPegs, Point thePegs[],
 short *numEdgePegs, Point edgePegs[], Fixed *area )
    Point   outerPegs[4*(HOLES-1)+1];
    short   sideLast[4], first, last, i;
    if( numPegs > 3 ) {
        GetPerimeter( thePegs, numPegs, outerPegs, sideLast );
 /* Initialize some variables and march around
  * the sides of the board. */
        *numEdgePegs = first = i = 0;
        do {
 /* If there's at least one new peg along the
  * column tops (bottoms), see which ones contact
  * the rubber band. */
            last = sideLast[i++];
            if( first < last ) {
                GetEdgePegs( outerPegs, first, last, edgePegs,
                 numEdgePegs );
                first = last;
 /* Count all pegs from the last (first) column
  * as edge pegs. */
            last = sideLast[i++];
            while( first < last )
                edgePegs[(*numEdgePegs)++] = outerPegs[first++];
        } while( i < 4 ); /* Repeat for four sides. */
    else { 
      /* With 3 or fewer pegs, all will touch the rubber band. */
        *numEdgePegs = numPegs;
        for( i = 0; i < numPegs; i++ ) edgePegs[i] = thePegs[i];
        if( numPegs < 3 ) {
        /* With less than 3 pegs, area must be 0. */
            *area = 0;
    IntegrateArea( edgePegs, *numEdgePegs, area );
 /* If there are more than 3 pegs, and they are all
  * in a straight line (indicated by a zero area),
  * the above algorithm will have counted the interior
  * points twice.  The following will remove the
  * redundant set of interior points.  Note that
  * it's also okay for 3 pegs, but no fewer. */
    if( *area == 0 )
      *numEdgePegs = (*numEdgePegs + 3)/2;
/* This function finds the pegs which roughly
 * define the four sides of the rubber band. */
void GetPerimeter( Point thePegs[], short numPegs,
 Point outerPegs[], short sideLast[] )
    short   colmin[HOLES], colmax[HOLES],
            rowmin[HOLES], rowmax[HOLES],
            col, row, col1, col2, n;
    for( n = 0; n < HOLES; n++  ) {
        colmin[n] = rowmin[n] = HOLES;
        colmax[n] = rowmax[n] = -1;
 /* Check each peg to see if it sets a new extreme
 * in any row or column. */
    for( n = 0; n < numPegs; n++ ) {
        row = thePegs[n].v;
        col = thePegs[n].h;
        if( col < colmin[row] ) colmin[row] = col;
        if( col > colmax[row] ) colmax[row] = col;
        if( row < rowmin[col] ) rowmin[col] = row;
        if( row > rowmax[col] ) rowmax[col] = row;
 /* Collect the pegs at the tops of each column. */
    n = -1;
    for( col = 0; col < HOLES; col++ ) {
        if( (row = rowmin[col]) < HOLES ) {
            outerPegs[++n].v = row;
            outerPegs[n].h = col;
    sideLast[0] = n;
    col1 = outerPegs[0].h;
    col2 = outerPegs[n].h;
 /* Collect all but the top peg of the last column,
  * from top to bottom. */
    for( row = rowmin[col2] + 1; row <= rowmax[col2]; row++ ) {
        if( colmax[row] == col2 ) {
            outerPegs[++n].v = row;
            outerPegs[n].h = col2;
    sideLast[1] = n;
 /* From last to first, collect the pegs at the
  * bottoms of all but the last column. */
    for( col = col2 - 1; col >= col1; col-- ) {
        if( (row = rowmax[col]) >= 0 ) {
            outerPegs[++n].v = row;
            outerPegs[n].h = col;
    sideLast[2] = n;
 /* Collect all but the bottom peg of the first column,
  * from bottom to top. */
    for( row = rowmax[col1] - 1; row >= rowmin[col1]; row-- ) {
        if( colmin[row] == col1 ) {
            outerPegs[++n].v = row;
            outerPegs[n].h = col1;
    sideLast[3] = n;
/* This function finds the pegs which would push
 * a rubber band to the left of a line between a
 * given starting point and a given ending point.
 * It counts the starting point (but not the
 * ending point) as such a peg. */
void GetEdgePegs( Point outerPegs[], short test, short last,
                  Point edgePegs[], short *numEdgePegs )
    Point   testPeg, backPeg, nextPeg;
    short   convex, first;
    first = *numEdgePegs;

    backPeg = edgePegs[(*numEdgePegs)++] = outerPegs[test];
    nextPeg = outerPegs[last];
 /* Loop through the array of outerPegs from the
  * one after the starting point to the one just
  * before the ending point. */
    while( ++test < last ) {
        testPeg = outerPegs[test];
 /* See if the path connecting backPeg, testPeg,
  * and nextPeg is convex, straight, or concave. */
        if( (convex = (nextPeg.v-backPeg.v)*(testPeg.h-backPeg.h)
 -(testPeg.v-backPeg.v)*(nextPeg.h-backPeg.h)) >= 0 ) {
 /* If convex or straight, count the test
  * peg as an edge peg. */
            edgePegs[(*numEdgePegs)++] = backPeg = testPeg;
 /* If convex, the rubber band's path will change,
  * so we need to check previous edge pegs to see
  * if they are still edge pegs. */
            if( convex > 0 )
              CheckEdgePegs( edgePegs, numEdgePegs, testPeg, first );

/* If a peg just added to the list of edge pegs
 * has extended the rubber band, this routine will
 * search backward through the list, throwing out pegs
 * that are no longer contacted, until it finds one
 * that still is. */
void CheckEdgePegs( Point edgePegs[], short *numEdgePegs,
                    Point newPeg, short first )
    Point   testPeg, backPeg;
    short   test;
    test = *numEdgePegs - 1;
 /* Loop backward through the list of edge pegs,
  * starting with the one before that just added,
  * stopping before the first that can't be removed. */
    while( --test > first ) {
        testPeg = edgePegs[test];
        backPeg = edgePegs[test-1];
 /* If the path between newPeg, testPeg,
  * and backPeg is concave, remove the peg. */
        if( (newPeg.v-backPeg.v)*(testPeg.h-backPeg.h)
 -(testPeg.v-backPeg.v)*(newPeg.h-backPeg.h) < 0 )
            edgePegs[test] = edgePegs[--(*numEdgePegs)];
/* This function integrates the area enclosed
 * by a rubber band. */
void IntegrateArea( Point edgePegs[],
 short numEdgePegs, Fixed *area ) 
    Point   thePeg, lastPeg;
    long    integral = 0;
    short   i;
 /* Starting and ending with the last peg,
  * integrate double the area under the closed path. */
    lastPeg = edgePegs[numEdgePegs-1];
    for( i = 0; i < numEdgePegs; i++ ) {
        thePeg = edgePegs[i];
        integral += (thePeg.h + lastPeg.h)*(thePeg.v - lastPeg.v);
        lastPeg = thePeg;
 /* Correct a negative integral if the path was
  * counterclockwise. */
    if( integral < 0 ) integral = -integral;
 /* By shifting, simultaneously halve the integral
  * and convert it to a fixed. */
    *area = (Fixed)( integral << 15 );


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