Volume Number: 16 (2000)
Issue Number: 4
Column Tag: Programmer's Challenge

Programmer's Challenge

by Bob Boonstra, Westford, MA

This month, we've got some very important messages to send. But we're not living in a world of free bandwidth. In fact, bandwidth in our Challenge world is very expensive, so expensive that we're asking you to compress our messages and save us a few bytes.

The prototype for the code you should write is:

void * /* yourStorage */ InitCompression(void);

long /* compressedLength */ CompressText(
  char *inputText,        /* text to be compressed */
  long numInputChars,     /* length of inputText in bytes */
  char *compressedText,   /* return compressedText here */
  const void *yourStorage /* storage returned by InitCompression */
);

long /* expandedLength */ ExpandText(
  char *compressedText,   /* encoded text to be expanded */
  long compressedLength,  /* length of encoded text in bytes */
  char *expandedText,     /* return expanded text here */
  const void *yourStorage /* storage returned by InitCompression */
);

void TermCompression(
  void *yourStorage       /* storage returned by InitCompression */
);

For this Challenge, you need to provide the four routines indicated above. Your InitCompression and TermCompression routines will be called only once each, at the beginning of the test and at the end of the test, respectively. InitCompression should allocate and return a block of yourStorage where you initialize any information needed by your compression and expansion routines. That storage will be passed back to you unchanged each time you are asked to compress or decompress text. TermCompression will be called at the end of the test and should deallocate the block of yourStorage to avoid a memory leak.

In between the calls to InitCompression and TermCompression, the test code will make multiple calls to CompressText and ExpandText with different inputText values. CompressText should process the inputText, populate the compressedText, and return the number of bytes in the result. ExpandText does the opposite, processing the compressedText, converting it to expandedText, and returning the number of bytes of the original text. Multiple CompressText and ExpandText calls will occur with varying inputText and compressedText, in any order, with the obvious constraint that text must be encoded before it can be decoded.

The inputText may contain any character between 0x00 and 0x7F, inclusive. As a practical matter, the inputText will be drawn from paragraphs of English-language text, computer programs in C, C++, and Pascal, and html pages.

All text-specific information needed to decode the compressedText must be stored in the compressedText itself. Any text-independent decoding information may be stored in yourStorage or in static storage within your program. No text-specific encoding information may be stored in yourStorage or in static variables.

The winner will be the solution that correctly compresses the inputText into the least costly compressedText, where cost is a function of length and execution time. Specifically, each inputText will have a cost equal to theCompressedLength of the corresponding compressedText, plus a penalty of 10% for each 100 milliseconds required to do the encoding and decoding.

This will be a native PowerPC Challenge, using the CodeWarrior Pro 5 environment. Solutions may be coded in C, C++, or Pascal. Solutions in Java will also be accepted, but Java entries must be accompanied by a test driver that uses the interface provided in the problem statement.

Congratulations to Sebastian Maurer for winning the January, 2000, Triangle Peg Challenge. The Peg Challenge required entries to solve variously sized games of peg solitaire, a game where pegs are arranged in holes on a triangle board. The objective of the game is to repeatedly jump one peg over an adjacent one, removing the jumped peg each time, with the intent of removing as many pegs as possible. In our Challenge, scoring counted 1000 penalty points for each peg left on the board, plus one penalty point for each millisecond of execution time. Sebastian did not submit the fastest entry - in fact, he ranked third in speed - but it played the Peg game significantly better than the other solutions, resulting in a better overall score.

I evaluated the Pegs entries using 14 test cases, ranging from 5 pegs to 100 pegs on a side. Nine of the test cases were missing only one peg, one of the large puzzles was missing 50 pegs, with the remainder missing a small number of pegs. Sebastian's entry left fewer pegs on the board than the second place solution in 12 of 14 test cases, and fewer than the third place solution in 10 of 14 cases. Sebastian's entry won in all of the test cases involving larger puzzles.

Sebastian's code is rather sparsely commented. The work is done in his Search routine, which iteratively tries valid moves, backtracks when it cannot find a valid move, and saves the best solution found so far. In looking at the code for the top entries, it wasn't obvious why Sebastian's solution did so much better than the others. The entries did use different logic to prune the search and trade execution time against search depth; perhaps those differences explain the performance variation.

The table below lists, for each of the solutions submitted, the overall score, the execution time in microseconds, and the total number of pegs left on the board for all of our test cases. It also indicates the code size, data size, and programming language used by each solution. Two entries did not complete all of the test cases and are listed last. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges prior to this one.

Listed here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated 10 or more points during the past two years. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants.

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:

Here is Sebastian's winning Triangle Peg solution:

Pegs.c
Copyright © 2000 Sebastian Maurer

#include "Pegs.h"
#include "Memory.h"
#include "Timer.h"

#include <stdio.h>

const int kNumTries = 1000;
const UInt32 kStopTime = 500000;
  // in microseconds

const int kNumDirections = 6;
enum {kIllegal = 0, kFull, kEmpty};

char** gGrid;
PegJump *gCurrentJumps;
PegJump *gBestJumps;
int gBestScore;
UnsignedWide gLastImprovement;
unsigned short *gDirection;
short gDeltaRow[kNumDirections];
short gDeltaCol[kNumDirections];

static UInt32 WideDiff(UnsignedWide m1, UnsignedWide m2)
{
  UnsignedWide m;
  m.lo = m2.lo - m1.lo;
  m.hi = m2.hi - m1.hi;
  if (m1.lo > m2.lo)
    m.hi -= 1;
  return m.lo;
}

static void PrintBoard(int triangleSize) {
  for(int r = 0; r < triangleSize; r++) {
    for(int c = - triangleSize; c <= triangleSize; c++)
    {
      switch (gGrid[r][c]) {
        case kIllegal: printf(" "); break;
        case kFull: printf("X"); break;
        case kEmpty: printf("."); break;
        default: printf("?");
      }
    }
    printf("\n");
  }
  printf("\n");
}

AllocateGrid
//////
// Memory allocation for the 2D grid
static char **AllocateGrid(
  int xMin, int xMax,
  int yMin, int yMax)
{
  int i, j;
  int nx = xMax - xMin + 1;
  int ny = yMax - yMin + 1;
  char **array;
  Ptr p;
  int rowSize;

  array = (char **) NewPtr((Size)(nx * sizeof(char*)));
  if (array == 0)
    return 0;
  
  array -= xMin;
  p = NewPtr((Size)(nx * ny * sizeof(char)));
  if (p == 0)
    return 0;
  p -= yMin * sizeof(char);
  rowSize = (int)(ny * sizeof(char));
  for(i = xMin; i <= xMax; i++) {
    array[i] = (char*) p;
    p += rowSize;
  }

  for(i = xMin; i <= xMax; i++)
    for(j = yMin; j <= yMax; j++)
      array[i][j] = 0;

  return array;
}

DeallocateGrid
static void DeallocateGrid(
  char **array, long xMin, long yMin)
{
  DisposePtr((Ptr) (array[xMin] + yMin));
  DisposePtr((Ptr) (array + xMin));
}

FillGrid
//////
// Fill the grid with the initial configuration
static void FillGrid(
  short size,
  short numInitialPegs,
  TrianglePegPosition initialPegPositions[]
) {
  for(int r = -2; r < size + 2; r++)
    for(int c = - size - 2; c <= size + 2; c++)
      gGrid[r][c] = kIllegal;

  for(int r = 0; r < size; r++)
    for(int c = - r; c <= r; c += 2)
      gGrid[r][c] = kEmpty;
  
  for(int p = 0; p < numInitialPegs; p++) {
    int row = initialPegPositions[p].row;
    int col = initialPegPositions[p].col;
    gGrid[row][col] = kFull;
  }
}

SaveSolution
//////
// Store the current best solution
static void SaveSolution(int numMoves) {
  gBestScore = numMoves;

  for(int m = 0; m < numMoves; m++) {
    gBestJumps[m].from.row =
      gCurrentJumps[m].from.row;
    gBestJumps[m].from.col =
      gCurrentJumps[m].from.col;
    gBestJumps[m].to.row =
      gCurrentJumps[m].to.row;
    gBestJumps[m].to.col =
      gCurrentJumps[m].to.col;
  }
    
}

Try
//////
// Perform a move if it is legal, and return true
// If the move is illegal, return false
// I assume position (r,c) already contains a peg
static bool Try(int move, int r, int c, int dir)
{
  int dr = gDeltaRow[dir];
  int dc = gDeltaCol[dir];

  if ((gGrid[r + dr][c + dc] == kFull) &&
    (gGrid[r + 2 * dr][c + 2 * dc] == kEmpty))
  {
    gDirection[move] = dir;
      
    gCurrentJumps[move].from.row = r;
    gCurrentJumps[move].from.col = c;
    
    gCurrentJumps[move].to.row = r + 2 * dr;
    gCurrentJumps[move].to.col = c + 2 * dc;
    
    gGrid[r][c] = kEmpty;
    gGrid[r + dr][c + dc] = kEmpty;
    gGrid[r + 2 * dr][c + 2 * dc] = kFull;

    return true;
  }
  else
    return false;
}

UndoMove
//////
// Undo a move by restoring the grid
static inline void UndoMove(int row, int col, int dir) {
  int dr = gDeltaRow[dir];
  int dc = gDeltaCol[dir];

  gGrid[row][col] = kFull;
  gGrid[row + dr][col + dc] = kFull;
  gGrid[row + 2 * dr][col + 2 * dc] = kEmpty;
}

Search
//////
// This is where the search is done
static void Search(
  int numInitialPegs,
  int triangleSize
) {

  UnsignedWide now;
  int move, row, col, dir;
  bool abort = false;
  bool haveASolution = false;
  
  int numTries = 0;

  row = 0;
  col = 0;
  dir = -1;
  move = 0;

  do {
  
    // find the next valid move
    do {
      dir++;
      if (dir >= kNumDirections) {
        dir = 0;
        col++;
        if (col == row + 1) {
          row++;
          col = - row;
        }
      }
    } while ((row < triangleSize) &&
           ((gGrid[row][col] == kEmpty) ||
            !Try(move, row, col, dir)));
  
    if (row < triangleSize) {
      // we just made a valid move
      // start work on the next move
      
      move++;

      row = 0;
      col = 0;
      dir = 0;
    } else { // we reached a dead end
      
      // see if this is a better solution
      if (move > gBestScore) {
        SaveSolution(move);
        haveASolution = true;

        if (move + 1 == numInitialPegs)
          abort = true;

        Microseconds(&now);

        gLastImprovement = now;
      }
      
      // backtrack
      move-;
      
      if (move >= 0) {
        row = gCurrentJumps[move].from.row;
        col = gCurrentJumps[move].from.col;
        dir = gDirection[move];
        UndoMove(row, col, dir);        
      }
      
    }
    
    if (haveASolution) {
      // if we already reached a dead end once,
      // don't waste time searching more
    
      numTries++;
      if (numTries == kNumTries) {
        numTries = 0;
        Microseconds(&now);
        if (WideDiff(gLastImprovement, now) >
            kStopTime)
        {
          abort = true;
        }
      }
    }

  } while (!abort && (move >= 0));
}

SolvePegTriangle
short /* number of moves */ SolvePegTriangle (
  short triangleSize,
    /* number of rows in triangle to solve */
  short numInitialPegs,
    /* number of pegs in starting puzzle position */
  TrianglePegPosition initialPegPositions[],
    /* peg locations in starting puzzle position */
  PegJump pegJumps[]
    /* return peg moves that solve the puzzle here,
       in sequence */

) {
  // prepare, allocate, initialize structures
  gDeltaRow[0] = -1; gDeltaCol[0] = -1;
  gDeltaRow[1] = -1; gDeltaCol[1] = +1;
  gDeltaRow[2] = 0; gDeltaCol[2] = -2;
  gDeltaRow[3] = 0; gDeltaCol[3] = +2;
  gDeltaRow[4] = +1; gDeltaCol[4] = -1;
  gDeltaRow[5] = +1; gDeltaCol[5] = +1;
  
  gGrid =
    AllocateGrid(-2, triangleSize + 2,
      - triangleSize - 2, triangleSize + 2);
  if (gGrid == 0)
    return 0;
  
  gDirection = (unsigned short*)
    NewPtr(numInitialPegs * sizeof(unsigned short));
  if (gDirection == 0)
    return 0;
  
  gCurrentJumps = (PegJump*)
    NewPtr(numInitialPegs * sizeof(PegJump));
  if (gCurrentJumps == 0)
    return 0;

  gBestJumps = pegJumps;

  Microseconds(&gLastImprovement);
  gBestScore = 0;

  FillGrid(triangleSize, numInitialPegs,
       initialPegPositions);

  // do the work
  Search(numInitialPegs, triangleSize);

  // clean up
  DeallocateGrid(gGrid, -2, - triangleSize - 2);
  DisposePtr((Ptr)gDirection);
  DisposePtr((Ptr)gCurrentJumps);

  return gBestScore;
}