TweetFollow Us on Twitter

Mar 93 Challenge
Volume Number:9
Issue Number:3
Column Tag:Programmers’ Challenge

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.

Count Unique Words

Most word processors these days have a Count Words command. The quality in terms of accuracy and speed of these commands varies quite a bit. I tested three leading word processors with a document containing 124,829 characters and got three different answers ranging from 18446 words to 18886 words and times ranging from 4 seconds to 11 seconds. I’m not sure what the correct answer was for that document; it depends on how you define what a word is.

For purposes of this month’s challenge, a word is defined as an unbroken set of one or more letters. The input text will only contain upper and lower case letters a to z, spaces, carriage returns, periods and commas (for a total of 56 possible byte values). No digits, hyphens, tabs, other punctuation, etc. Since counting words using this simplified definition is rather trivial, you’re going to count the number of unique words instead.

The prototype of the function you write is:

unsigned short CountUniqueWords(textPtr, byteCount)
PtrtextPtr;
unsigned short byteCount;

Your function should return the number of unique words (case insensitive) in the input text. The maximum word length for individual words in the input text is 255 characters.

This is my 7th programmer’s challenge that I’ve posed to MacTech readers. I have received approximately very little feedback as to what you think of these challenges. Are they too easy, too hard, too uninteresting, or what? Do you want hard core numerical analysis puzzles (like write a fast sqrt function) or do you want Mac-specific problems (like write a fast TileAndStackWindows function) or are things okay as they are? If you have any ideas for future challenges, please send them in (credit will be given in this column if I use one of your ideas). Thanks.

Two Months Ago Winner

The winner of the “Travelling Salesman” challenge is Ronald Nepsund (Northridge, CA) whose solution was the only one of the five I received which gave correct results. The time intensive part of solutions to this class of problems is the distance between two points calculation, which involves a square root. Ronald uses a precomputed sqrt table for values 0 to 25 to eliminate much of this time.

A couple of people chose the algorithm of “find the closest city to where we currently are and move to that city; repeat until all cities have been visited” which is not correct. An example set of input data that broke everyone but Ronald’s solution is: numCities = 8, startCityIndex = 5, *citiesPtr = {1,1}, {2,1}, {3,1}, {2,2}, {1,3}, {2,3}, {3,3}, {2,4}. If you draw it and work it out by hand (through trial and error) you can see that the minimum path distance is 8.66. There is more than one correct ordering for the optimal path but all of the optimal paths will have that same length.

Here is Ronald’s winning solution to the January Challenge:

//***********************************
// Travelling Salesman 
// by Ronald M. Nepsund

#include <math.h>

#define fracBase 0x20000000

//There are two 32 by 20 arrays of longs
//which together give the distance betwean
//any two cities.
//Instead of using Array[i,j] to access
//the array Array[(i<<5)+j] is used
//and two longs are needed to accurately
//measure the distance betwean cities
//so two arrays of longs are used.
//gDistanceFrac is used to hold the
//fractional part of distance in 1/0x20000000
//of a unit.
long  gDistanceInt[640],
 gDistanceFrac[640];
//these are used to represent a path betwean
//the cities.
Byte  gNextCity[20],gOptPath[20];
//how long is the currently selected best
//path so far.
long  gBestPathLength,gFracBestPathLength;

unsignedshort  gNumCities;
unsignedshort  gStartCityIndex;

//precalculated square root for zero to 25
long  qSquTableInt[] =
 {0,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,
  4,4,4,4,4,4,4,4,4,
  5,5,5,5,5,5,5,5,5,5,5};
//precalculated square root - fractional part
//in 1/0x20000000 of a whole unit
long  qSquTableFrac[]=
 {0,0,222379212,393016784,0,126738030,
 241317968,346685095,444758425,0,
 87122155,169986639,249162657,
 325102865,398174277,468679365,0,
 66091829,130266726,192682403,
 253476060,312767944,370664138,
 427258795,482635936,0 };

void DoPath(short cityIndex, long InttPathLength,
 long fracPathLength);

//The recursive routien that actually finds
//the shortest path.

void  DoPath(registershort cityIndex,
 long InttPathLength,
 long fracPathLength)
{
 register short  i;
 BooleanlastCity;
 long   offset;
 
 if (fracPathLength > fracBase)  {
 //the fractional value variable has
 //exceeded the value of one whole
 //unit
 InttPathLength += 1;
 fracPathLength -= fracBase;
 }
 
 //Has the path has become longer than the
 //shortest path we have already found?
 if (InttPathLength > gBestPathLength ||
 ((InttPathLength == gBestPathLength) &&
  (fracPathLength >= gFracBestPathLength)))
   return;
 
 //lastCity is used to tell if all the
 //cities have been visited
 lastCity = TRUE;
 //for each city
 for(i = 0; i<gNumCities; i++)
 //if not to same city or already
 //visited city
 if ( i != cityIndex &&
  gNextCity[i] == 0xFF) {
 //not at the end of the path
 lastCity = FALSE;
 //path from city ‘cityIndex’ to ‘i’
 gNextCity[cityIndex] = i;
 //offset into distance arrays
 offset = (cityIndex << 5) + i;
 //go to next city adding the
 //distance to that city to the
 //path length
 DoPath(i,
  InttPathLength+gDistanceInt[offset],
  fracPathLength+gDistanceFrac[offset]);
 } //end if and for
 
 // if this is the last city in the chain and
 // is a shorter path than the previous best
 if ((lastCity) &&
 ((InttPathLength < gBestPathLength) ||
  ((InttPathLength == gBestPathLength) &&
   (fracPathLength < gFracBestPathLength))
   ) ) {
 // make this the current best path
 register long *LPnt1,*LPnt2;
 //this is the current shortest path
 //length now
 gBestPathLength = InttPathLength;
 gFracBestPathLength = fracPathLength;
 //copy path to ‘optPath’
 LPnt1 = (long *)&gNextCity;
 LPnt2 = (long *)&gOptPath;
 for (i= ((3+gNumCities) >> 2); i>0; i-)
 *LPnt2++ = *LPnt1++;
 } else 
 //this city is no longer connected to
 //the next city
 gNextCity[cityIndex] = 0xFF;
}

void InitDistances(unsigned short numCities
 Point *citiesPtr);

//initialize two arrays which will give the
//distance betwean any two cities.
void InitDistances(
 unsigned short numCities,
 Point  *citiesPtr)
{
 short  i,j,offset;
 register long *LPntl1,*LPntF1,
 *LPntI2,*LPntF2;
 long   dist;
 short  deltax,deltay;
 double X;
//The distance from city i to j is the same
//as from city j to i.
//Use pointers into the arrays
//We will add a constant to the pointers to
//step through the array
//instead of doing a multiplication to find
//the wanted entries in the array

 //how far is it betwean any two cities
 for (i=0; i<numCities; i++){
 LPntl1 = gDistanceInt  + i;
 LPntF1 = gDistanceFrac + i;
 offset = i << 5;
 LPntI2 = gDistanceInt  + offset;
 LPntF2 = gDistanceFrac + offset;
 for (j=0; j<=i; j++)
 if (i==j){
 //both pointers are pointing
 //to the same locations in the array
 //distance to the same city is zero
 *LPntI2++ = 0;
 *LPntl1 = 0;  LPntl1 += 32;
 *LPntF2++;
 LPntF1 += 32;
 } else {
 //calculate horizontal and vertical
 //distance betwean city ‘i’ and ‘j’
 deltax = citiesPtr[i].h-
 citiesPtr[j].h;
 deltay = citiesPtr[i].v-
 citiesPtr[j].v;
 //The distance betwean the cities is
 // squareRoot( deltax*deltax +
 // deltay*deltay)
 //Where you can, do multiplications
 //using shorts instead of long’s -
 //They are faster.
 if (-255< deltax && deltax<256)
 if (-255< deltay && deltay<256)
 dist = ((long)(deltax*deltax) + 
 (long)(deltay*deltay));
 else
 dist = ((long)(deltax*deltax) + 
 (long)deltay*deltay);
 else
 if (-255< deltay && deltay<256)
 dist = ((long)deltax*deltax + 
 (long)(deltay*deltay));
 else
 dist = ((long)deltax*deltax + 
 (long)deltay*deltay);

 //do squareRoot
 if (dist <= 25) {
 //use sqrt lookup tables for
 //0 to 25
 *LPntI2++  = *LPntl1 =
 qSquTableInt[dist];
  LPntl1 += 32;
 *LPntF2++  = *LPntF1 =
 qSquTableFrac[dist];
  LPntF1 += 32;
 } else {
        X = sqrt(dist);
 //gDistanceInt[(i<<5) + j] = X;
 //gDistanceInt[(j<<5) + i] = X;
 //integer part of distance
 // between points
 dist = X;
 *LPntl1 = *LPntI2++  = dist;
 LPntl1 += 32;
 
 //gDistanceFrac[i<<5 + j] =
 // (X - dist) * $20000000;
 //gDistanceFrac[j<<5 + i] =
 // (X - dist) * $20000000;
 // fractional part
 dist = (X - dist) * fracBase;
 *LPntF2++  = *LPntF1 = dist;
 LPntF1 += 32;
 }
 }
 }
}

void OptimalPath(unsigned short numCities
 unsigned short startCityIndex,
 Point *citiesPtr,Point *optimalPathPtr);

void OptimalPath(numCities,startCityIndex,citiesPtr,
 optimalPathPtr)
unsigned short numCities;
unsigned short startCityIndex;
Point *citiesPtr;
Point *optimalPathPtr;
{
 register short  i,j;
 long   time,index;
 double X;
 
 //generates the tables for the distances
 //betwean any two cities.
 //This routien takes up most of the time.
 InitDistances(numCities,citiesPtr);
 
 //OxFF means that there is no path from
 //this city to another
 for (i=0; i<numCities; i++)
 //no paths betwean cities
 gNextCity[i] = 0xFF;
 
 gNumCities = numCities;
 gStartCityIndex = startCityIndex;
 //any path done by DoPath will be shorter
 //than this
 gBestPathLength = 0x7FFFFFFF;
 gFracBestPathLength = 0;

 //find the best path
 DoPath(startCityIndex,0,0);  
 
 //put the best path into the form
 //desired for ‘optimalPath’
 j=startCityIndex;
 for(i=0; i<numCities; i++) {
 optimalPathPtr[i] = citiesPtr[j];
 j = gOptPath[j];
 }
}

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. 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.

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: progchallenge@xplain.com, and CompuServe: 71552,174. 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.

 

Community Search:
MacTech Search:

Software Updates via MacUpdate

How to get started with live.ly
One could be forgiven for thinking that there are already plenty of streaming video apps out there. It's just that the App Store charts would insist that you're mistaken. [Read more] | Read more »
Rodeo Stampede: Guide to all Savannah an...
A "gotta catch 'em all" joke seems appropriate here, even though we're talking animals in Rodeo Stampede and not pocket monsters. By now you've probably had plenty of rides, tamed some animals and built yourself a pretty nice zoo | Read more »
Is there cross-platform play in slither....
So you've sunken plenty of hours into crawling around in slither.io on your iPhone or iPad. You've got your stories of tragedy and triumph, the times you coiled four snakes at one time balanced out by the others when you had a length of more than... | Read more »
Rodeo Stampede guide to running a better...
In Rodeo Stampede, honing your skills so you can jump from animal to animal and outrun the herd as long as possible is only half the fun. Once you've tamed a few animals, you can bring them home with you. [Read more] | Read more »
VoxSyn (Music)
VoxSyn 1.0 Device: iOS Universal Category: Music Price: $6.99, Version: 1.0 (iTunes) Description: VoxSyn turns your voice into the most flexible vocal sound generator ever. Instantly following even subtle modulations of pitch and... | Read more »
Catch Battleplans on Google Play from Ju...
Real-time strategy title Battleplans is due for release on Google Play on June 30th, following its release for iOS systems last month. With its simple interface and pretty graphics, the crowd-pleaser brings a formerly overlooked genre out for the... | Read more »
iDoyle: The interactive Adventures of Sh...
iDoyle: The interactive Adventures of Sherlock Holmes - A Scandal in Bohemia 1.0 Device: iOS Universal Category: Books Price: $1.99, Version: 1.0 (iTunes) Description: Special Release Price $1.99 (Normally $3.99) | Read more »
Five popular free apps to help you slim...
Thanks to retail and advertising, we're used to thinking one season ahead. Here we are just a week into the summer and we're conditioned to start thinking about the fall. [Read more] | Read more »
How to ride longer and tame more animals...
It's hard to accurately describe Rodeo Stampede to people who haven't seen it yet. It's like if someone took Crossy Roadand Disco Zoo and put them in a blender, yet with a unique game mechanic that's still simple and fun for anyone. [Read more] | Read more »
Teeny Titans - A Teen Titans Go! Figure...
Teeny Titans - A Teen Titans Go! Figure Battling Game 1.0.0 Device: iOS Universal Category: Games Price: $3.99, Version: 1.0.0 (iTunes) Description: Teeny Titans, GO! Join Robin for a figure battling RPG of epic proportions! TEENY... | Read more »

Price Scanner via MacPrices.net

13-inch 256GB MacBook Air on sale for $1149,...
Amazon has the 2016 13″ 1.6GHz/256GB MacBook Air (model MMGG2LL/A) on sale for $1149.99 including free shipping. Their price is also $50 off MSRP. Read more
Haven App Launches New Age Of Wirless 911 Eme...
Haven from RapidSOS represents a transformation in access to emergency services from a phone call solely dependent on voice to a robust data connection for voice, text, medical/demographic data.... Read more
Cu Parachute 1.1 Retirement Success PLanning...
Tucson, Arizona based Indie developer Bradley McCarthy has announce the release of Cu (Copper) Parachute 1.1 for iPhone, iPad, and iPod touch devices — a tool with which users can continuously... Read more
Research and Markets Releases iPhone 6s Plus...
A new analysis report from Dublin-based Research and Markets observes that with the iPhone 6s Plus, Apple introduced a new rear camera module. The new device has similar structure and technology than... Read more
Apple refurbished Retina MacBook Pros availab...
Apple has Certified Refurbished 2015 13″ and 15″ Retina MacBook Pros available for up to $380 off the cost of new models. An Apple one-year warranty is included with each model, and shipping is free... Read more
Apple refurbished 11-inch MacBook Airs availa...
Apple has Certified Refurbished 11″ MacBook Airs (the latest models), available for up to $170 off the cost of new models. An Apple one-year warranty is included with each MacBook, and shipping is... Read more
Apple price trackers, updated continuously
Scan our Apple Price Trackers for the latest information on sales, bundles, and availability on systems from Apple’s authorized internet/catalog resellers. We update the trackers continuously: - 15″... Read more
12-inch 32GB and 128GB WiFi iPad Pros on sale...
B&H Photo has 12″ 32GB & 128GB WiFi iPad Pros on sale for up to $80 off MSRP, each including free shipping. B&H charges sales tax in NY only: - 12″ Space Gray 32GB WiFi iPad Pro: $749 $50... Read more
6-core Mac Pro available for $3799, save $200
B&H Photo has the 6-core 3.5GHz Mac Pro on sale for $200 off MSRP. Shipping is free, and B&H charges sales tax in NY only: - 3.5GHz 6-core Mac Pro (sku MD878LL/A): $3799.99, $200 off MSRP Read more
Apple refurbished Apple TVs available for up...
Apple has Certified Refurbished 32GB and 64GB Apple TVs available for up to $30 off the cost of new models. Apple’s standard one-year warranty is included with each model, and shipping is free: -... Read more

Jobs Board

*Apple* iPhone 6s and New Products Tester Ne...
…we therefore look forward to put out products to quality test for durability. Apple leads the digital music revolution with its iPods and iTunes online store, Read more
*Apple* iPhone 6s and New Products Tester Ne...
…we therefore look forward to put out products to quality test for durability. Apple leads the digital music revolution with its iPods and iTunes online store, Read more
*Apple* iPhone 6s and New Products Tester Ne...
…we therefore look forward to put out products to quality test for durability. Apple leads the digital music revolution with its iPods and iTunes online store, Read more
*Apple* Retail - Multiple Positions, Towson...
Job Description: Sales Specialist - Retail Customer Service and Sales Transform Apple Store visitors into loyal Apple customers. When customers enter the store, Read more
*Apple* iPhone 6s and New Products Tester Ne...
…we therefore look forward to put out products to quality test for durability. Apple leads the digital music revolution with its iPods and iTunes online store, Read more
All contents are Copyright 1984-2011 by Xplain Corporation. All rights reserved. Theme designed by Icreon.