TweetFollow Us on Twitter

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;

Example:

Input:  baseNumber = 107
 n = 3

Output:

 function result = 131

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

Congratulations

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;
            return;
        }
    }
    
    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)];
        else
        return;
    }
}
 
/*******************************************************/
 
/* 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 );
}

 

Community Search:
MacTech Search:

Software Updates via MacUpdate

Latest Forum Discussions

See All

Six fantastic ways to spend National Vid...
As if anyone needed an excuse to play games today, I am about to give you one: it is National Video Games Day. A day for us to play games, like we no doubt do every day. Let’s not look a gift horse in the mouth. Instead, feast your eyes on this... | Read more »
Old School RuneScape players turn out in...
The sheer leap in technological advancements in our lifetime has been mind-blowing. We went from Commodore 64s to VR glasses in what feels like a heartbeat, but more importantly, the internet. It can be a dark mess, but it also brought hundreds of... | Read more »
Today's Best Mobile Game Discounts...
Every day, we pick out a curated list of the best mobile discounts on the App Store and post them here. This list won't be comprehensive, but it every game on it is recommended. Feel free to check out the coverage we did on them in the links below... | Read more »
Nintendo and The Pokémon Company's...
Unless you have been living under a rock, you know that Nintendo has been locked in an epic battle with Pocketpair, creator of the obvious Pokémon rip-off Palworld. Nintendo often resorts to legal retaliation at the drop of a hat, but it seems this... | Read more »
Apple exclusive mobile games don’t make...
If you are a gamer on phones, no doubt you have been as distressed as I am on one huge sticking point: exclusivity. For years, Xbox and PlayStation have done battle, and before this was the Sega Genesis and the Nintendo NES. On console, it makes... | Read more »
Regionally exclusive events make no sens...
Last week, over on our sister site AppSpy, I babbled excitedly about the Pokémon GO Safari Days event. You can get nine Eevees with an explorer hat per day. Or, can you? Specifically, you, reader. Do you have the time or funds to possibly fly for... | Read more »
As Jon Bellamy defends his choice to can...
Back in March, Jagex announced the appointment of a new CEO, Jon Bellamy. Mr Bellamy then decided to almost immediately paint a huge target on his back by cancelling the Runescapes Pride event. This led to widespread condemnation about his perceived... | Read more »
Marvel Contest of Champions adds two mor...
When I saw the latest two Marvel Contest of Champions characters, I scoffed. Mr Knight and Silver Samurai, thought I, they are running out of good choices. Then I realised no, I was being far too cynical. This is one of the things that games do best... | Read more »
Grass is green, and water is wet: Pokémo...
It must be a day that ends in Y, because Pokémon Trading Card Game Pocket has kicked off its Zoroark Drop Event. Here you can get a promo version of another card, and look forward to the next Wonder Pick Event and the next Mass Outbreak that will be... | Read more »
Enter the Gungeon review
It took me a minute to get around to reviewing this game for a couple of very good reasons. The first is that Enter the Gungeon's style of roguelike bullet-hell action is teetering on the edge of being straight-up malicious, which made getting... | Read more »

Price Scanner via MacPrices.net

Take $150 off every Apple 11-inch M3 iPad Air
Amazon is offering a $150 discount on 11-inch M3 WiFi iPad Airs right now. Shipping is free: – 11″ 128GB M3 WiFi iPad Air: $449, $150 off – 11″ 256GB M3 WiFi iPad Air: $549, $150 off – 11″ 512GB M3... Read more
Apple iPad minis back on sale for $100 off MS...
Amazon is offering $100 discounts (up to 20% off) on Apple’s newest 2024 WiFi iPad minis, each with free shipping. These are the lowest prices available for new minis among the Apple retailers we... Read more
Apple’s 16-inch M4 Max MacBook Pros are on sa...
Amazon has 16-inch M4 Max MacBook Pros (Silver and Black colors) on sale for up to $410 off Apple’s MSRP right now. Shipping is free. Be sure to select Amazon as the seller, rather than a third-party... Read more
Red Pocket Mobile is offering a $150 rebate o...
Red Pocket Mobile has new Apple iPhone 17’s on sale for $150 off MSRP when you switch and open up a new line of service. Red Pocket Mobile is a nationwide MVNO using all the major wireless carrier... Read more
Switch to Verizon, and get any iPhone 16 for...
With yesterday’s introduction of the new iPhone 17 models, Verizon responded by running “on us” promos across much of the iPhone 16 lineup: iPhone 16 and 16 Plus show as $0/mo for 36 months with bill... Read more
Here is a summary of the new features in Appl...
Apple’s September 2025 event introduced major updates across its most popular product lines, focusing on health, performance, and design breakthroughs. The AirPods Pro 3 now feature best-in-class... Read more
Apple’s Smartphone Lineup Could Use A Touch o...
COMMENTARY – Whatever happened to the old adage, “less is more”? Apple’s smartphone lineup. — which is due for its annual refresh either this month or next (possibly at an Apple Event on September 9... Read more
Take $50 off every 11th-generation A16 WiFi i...
Amazon has Apple’s 11th-generation A16 WiFi iPads in stock on sale for $50 off MSRP right now. Shipping is free: – 11″ 11th-generation 128GB WiFi iPads: $299 $50 off MSRP – 11″ 11th-generation 256GB... Read more
Sunday Sale: 14-inch M4 MacBook Pros for up t...
Don’t pay full price! Amazon has Apple’s 14-inch M4 MacBook Pros (Silver and Black colors) on sale for up to $220 off MSRP right now. Shipping is free. Be sure to select Amazon as the seller, rather... Read more
Mac mini with M4 Pro CPU back on sale for $12...
B&H Photo has Apple’s Mac mini with the M4 Pro CPU back on sale for $1259, $140 off MSRP. B&H offers free 1-2 day shipping to most US addresses: – Mac mini M4 Pro CPU (24GB/512GB): $1259, $... Read more

Jobs Board

All contents are Copyright 1984-2011 by Xplain Corporation. All rights reserved. Theme designed by Icreon.