Apr 98 Challenge
**Volume Number: 14 (1998)**

Issue Number: 4

Column Tag: Programmer's Challenge

# Apr 98 - Programmer's Challenge

*by Bob Boonstra, Westford, MA*

## Mancala

A stocking stuffer from this past Christmas provided the inspiration for this month's Challenge. Santa gave me a two player travel game called Mancala which might provide some amusement the next time I travel by car or by plane, provided the ride is smooth enough to keep the small stones inside the bowls of the game board. I thought it might make for an interesting Challenge tournament.

The basic Mancala game consists of a board with 14 hollowed-out bowls arranged in an oval form, one large bowl at each end of the board, and six smaller bowls facing each of two players seated opposite one another. Each player "owns" the large bowl, or "mancala", positioned to his right, and the six small bowls closest to him. The game starts with each small bowl containing four stones. The game begins with the first player picking up all of the stones in one of his small bowls, dropping one stone in the bowl to the right, the second stone in the second bowl on the right, continuing around the board in counterclockwise fashion until the stones he picked up are gone. The second player then picks up the stones in one of his small bowls, drops them one at a time in the bowls to the right, etc. The game ends when one player cannot move (i.e., no stones remain in that player's small bowls). The winner is the player with the most stones in his mancala. There are a number of variations to the game, and the specific restrictions on our Mancala Challenge tournament are explained below.

The prototype for the code you should write is:

#if defined(__cplusplus)
extern "C" {
#endif
Boolean Mancala(
long board[], /* on entry, board[i] is number of stones in bowl i */
/* on exit, board reflects the results of your move */
const long boardSize, /* number of bowls in the board, including mancalas */
void *privStorage, /* pointer to 1MB of storage for your use */
const Boolean newGame, /* true for your first move of a game */
const Boolean playerOne,/* true when you are the first player */
long *bowlPlayed, /* return the number of the bowl you played from */
long *directionPlayed /* return 1 if you played counter-clockwise, */
/* return -1 if you played clockwise */
);
#if defined(__cplusplus)
}
#endif

Each time your Mancala routine is called, you will be provided with a board[] array that indicates the number of stones in each bowl, including the Mancalas, at the beginning of your turn. The boardSize parameter will indicate the number of bowls in the board - in the standard Mancala game, this would be 14, but in our Challenge it might be any even number between 8 and 32, inclusive. The mancala for the first player will be board[0], while the mancala for the second player will be board[boardSize/2]. You will also be provided a pointer privStorage to 1MB of storage, preinitialized to zero, for each of your moves in a single game. For your first move of a game, newGame will be TRUE, otherwise newGame will be false. If you are the first player, playerOne will be TRUE for each of your moves, otherwise playerOne will be FALSE. You should return the index of the bowl you played from in bowlPlayed, and you should return the direction you chose to move in directionPlayed. You should also update your view of the number of stones in each bowl in board[].

There are a number of rule variations for Mancala. We will play with the following additions to the standard rules:

- the board will contain between eight and 32 bowls, instead of the standard 14.
- at the beginning of the game, each small bowl will have between two and 16 stones (instead of the standard four), with the same number in each bowl.
- players do not drop stones into their opponent's mancalas.
- players may choose to move either counter-clockwise or clockwise on a given move.
- if a player drops the last stone into his mancala, he gets to move again (my test code will call your Mancala routine again).
- if a player drops the last stone into one of the empty bowls (board[i]) on his side of the board, he takes that stone, plus all the stones in his opponent's bowl directly across from his bowl (board[boardSize-i]) and places them in his mancala.
- the game ends when one player has no stones in any of his small bowls and cannot move. The other player then places all remaining stones from his small bowls into his mancala.

At the end of the game, each player will be credited with one point for each stone in his mancala, minus one point for each 100ms of cumulative execution time. The Challenge winner will be the entry that accumulates the most points in a round-robin tournament where each entry competes against each other entry twice for each set of game parameters, once playing first, and once playing second.

This will be a native PowerPC Challenge, using the latest CodeWarrior environment. Solutions may be coded in C, C++, or Pascal.

## Three Months Ago Winner

Congratulations once again to **Ernst Munter** (Kanata, Ontario) for submitting the fastest entry to the January Cell Selection Challenge. Readers were invited to implement a C++ CellSelection class, including methods to add one selection to another, remove one selection from another, invert a selection, count the number of active cells in a selection, and determine if two selections were equal. A dual-processor 8500/2x200 was used to test this Challenge, and contestants were free to take advantage of the multiple processors, either through the Apple/Daystar Multiprocessing API, or by programming for BeOS and taking advantage of the symmetric multiprocessing features of that operating system.

Or so I thought when I structured the problem. As it turned out, of the four solutions submitted, only one took advantage of the multiprocessor opportunity. I'll discuss the use of multiprocessing in that solution later in the article.
Ernst based his solution on the "vixel" data structure he used to win the Intersecting Rectangle Challenge of two years ago. His code is well-commented, so I'll let his solution speak for itself, except to point out that one other entry made reference to the "vixel" approach. It's nice to see readers put winning code from the past to good (and efficient) use.

The table below lists the total execution time in milliseconds for all test cases, as well as execution times for five individual test cases. It also lists code size, data size, and the number of processors used for each entry. The number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges to date prior to this one. The entries marked with an asterisk are those which did not complete one or more test cases. Test times in italics are extrapolated from partial test cases that were successfully completed. (See table 1).

Ulf Schröder's entry was the only one to attempt to use both of the processors available in the test machine. I tested a version of Ulf's entry modified to use only one processor, but the results were not much different, suggesting that either the CellSelection problem is not amenable to partitioning, or that Ulf's technique for doing so didn't have much affect for the particular test cases I used.

This Challenge was intended to be about multiprocessing, so it is worth spending a minute on how Ulf tried to take advantage of it. The first step was to determine how many processors were available and to create a task entry and a pair of semaphores for each processor:

if (!MPLibraryIsLoaded())
gProcessors = 1;
else
gProcessors = MPProcessors();
// Create the tasks if more than one processor
if (gProcessors > 1)
{
OSErr err;
.....
for (int32 i = 0; i < GPROCESSORS - 1; ++I)
{
ERR = MPCREATESEMAPHORE(1, 0,
>ASKDATA[I].MSTARTSEMAPHORE);
IF (ERR != NOERR) THROW ERR;
ERR = MPCREATESEMAPHORE(1, 0,
>ASKDATA[I].MFINISHEDSEMAPHORE);
IF (ERR != NOERR) THROW ERR;
ERR = MPCREATETASK(
TASK, >ASKDATA[I], 0, GTERMINATIONQUEUE,
0, 0, 0, >ASKDATA[I].MTASKID);
IF (ERR != NOERR) THROW ERR;
}

This step is done once, at initialization time. Then later, when the code determines that it has a problem worth partitioning, it divides the problem, assigns one piece to another task (or tasks, in the general case), signals the other task to begin work, solves the remaining piece of the problem, and then waits for the other task to complete it's portion.

IF (GPROCESSORS > 1 && length > kMinLengthForMP)
{
uint32 mid = length / 2;
.....
DoRemove doRemove(area, restEnd);
DoRemove doRemove1(area, rest1End);
gTaskData->mBegin = mAreas.begin() + mid;
gTaskData->mEnd = mAreas.end();
gTaskData->mNewEnd = gTaskData->mEnd;
gTaskData->mFunction = &doRemove1;
// Start the other task
MPSignalSemaphore(gTaskData->mStartSemaphore);
// Do my part of the job
AreaList::iterator newEnd =
remove_if(
mAreas.begin(),
mAreas.begin() + mid,
doRemove);
// Wait for other task
MPWaitOnSemaphore(gTaskData->mFinishedSemaphore,
kDurationForever);
.....
}

For further information, there is an introduction to multiprocessing on the Mac in the March, 1996, issue of MacTech Magazine, or you can visit Apple's web site to obtain MultiProcessing API in the SDK, including some sample code.

## Top 20 Contestants

Here are the Top Contestants for the Programmer's Challenge, including everyone who has accumulated more than 10 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.

Rank Name Points
1. Munter, Ernst 218
2. Boring, Randy 73
3. Cooper, Greg 61
4. Lewis, Peter 51
5. Mallett, Jeff 50
6. Nicolle, Ludovic 44
7. Murphy, ACC 34
8. Gregg, Xan 28
9. Antoniewicz, Andy 24
10. Day, Mark 20
11. Higgins, Charles 20
12. Hostetter, Mat 20
13. Rieken, Willeke 20
14. Studer, Thomas 20
15. Hart, Alan 14
16. O'Connor, Turlough 14
17. Picao, Miguel Cruz 14

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 5th place 2 points
2nd place 10 points finding bug 2 points
3rd place 7 points suggesting Challenge 2 points
4th place 4 points

Here is Ernst's winning solution to the CellSelection Challenge:

**Cells.h**

© 1998 Ernst Munter

/*
"Cell Selection"

This file is a C++ header file containing the CellSelection class, as well as a number of auxiliary structs and classes.

The purpose of the CellSelection class is to contain a set of cells in a 2-D world. Cells can be added and manipulated in groups represented as sets of cells, called Areas.

**Solution Strategy**

The overall idea is the same as the one I used in the "Intersecting Rectangles"
challenge (solution in MacTech of Apr 96).

Cells are similar to black-and-white pixels, and any area of all-black (off) or all-white (on) pixels can be represented by a single bit and the 4 edge coordinates. I had called such a group of pixels a "vixel".

The CellSelection class maintains two sets of coordinates, rows and columns, and a bit map of vixels.

The empty selection starts out with a single (off) vixel, covering the area defined by the range of 32-bit integers (left=-2147483648,top=-2147483648,right=2147483647, bottom=2147483647).

As areas are added or removed, missing coordinates are inserted and the total area becomes further and further divided into sub-areas.

Coordinate sets are stored in trees to facilitate easy lookup. In addition, they are linked in linear lists to make it easier to traverse a span of coordinates corresponding to the edges of a given area which may be represented by a rectangular block of many sub-areas of vixels.

Memory Management

As row and column coordinates are added, each is allocated by "new". Each row also holds a pointer to "vixels" which are allocated with each row as needed. The default amount of vixel bits per row is set to 512. If more vixels are needed as the CellSelection grows, rows are expanded in increments of 512 bits.
The bit map is not sorted right to left; rather, each column struct contains an index to the bit-position in the vixel array. The bits are allocated from the vixels arrays in the order of arrival with the method calls.

All calls to "new" are in try/catch blocks in order to allow the test program to fail gracefully if we should run out of memory.

**Optimisations**

Basically none.

Memory, rows and columns, get allocated as needed, but are only deleted when the CellSelection is destroyed. This will probably slow things down as the vixel map gets more and more fragmented in large selections.

A future improvement could be a garbage collector which detects when adjacent rows or columns are identical and can be merged.

Alternatively, CellSelections could be kept in a "normal" state during each operation by avoiding unnecessary splits of coordinates, and detection of pairs that can be merged.

**Extras**

I have added some public methods to the CellSelection class to simplify testing:

int CellSelection::Width()
int CellSelection::Height()
return the width and height of the vixel map
void CellSelection::ResetTestAreas()
bool CellSelection::NextTestArea(Area* a)
permit stepping through the vixel map and identify
each sub-area that contains a block of cells

I have also added a method to the Area struct to test for empty areas according to the definition.

**Clarifications**

Methods with return type "bool" (except EqualSelected()) return false only if we run out of memory. They return true in all other cases, including calls with empty areas.

The class makes no assumptions which could restrict the range of areas. Specifically, an area with a right or bottom coordinate of 2147483647 will be handled correctly, i.e. we do not need a value 1 greater than the highest value coordinate of an area. */

#include <STRING.H>
// needed for memcpy and memset
#if __dest_os == __be_os
#include <SUPPORTDEFS.H>
#else
typedef long int32;
typedef unsigned long uint32;
#endif
#define EXTRA_FOR_TESTING 1
static const enum ProgConsts{
kIncrementBits=512,
kNegInfinity=0x80000000L,
kPosInfinity=0x7FFFFFFFL
}
// the following line avoids a compiler warning in CW-PRO2
Consts(kNegInfinity);
Area
struct Area {
int32 left,top,right,bottom;
/* Area coordinates are inclusive. {2,2,3,4} includes 6 cells. */
const bool IsEmpty() const {
/* Any area with left>right or top>bottom is empty. */
return ((left>right) | (top>bottom));
}
};
Node
struct Node {
int32 lo;
Node* nextNode;
Node* leftNode;
Node* rightNode;
int bal;
Node(int32 x) {
lo=x;
leftNode=rightNode=nextNode=0;
bal=0;
}
int32 Limit() const {
return (nextNode)?nextNode->lo-1:kPosInfinity;
}
uint32 Slice(int32 xlo,int32 xhi) const {
int32 upperLimit=Limit();
return 1+(xhi<UPPERLIMIT?XHI:UPPERLIMIT)-(XLO>lo?xlo:lo);
}
};
Row:Node
struct Row:Node{
int size; // number of bytes allocated
char* vixels; // 1 bit per vixel
Row(int32 x):Node(x){
size=kIncrementBits/8;
try {vixels=new char[size];}
catch (...) {vixels=0;return;}
memset(vixels,0,size);
}
Row(Row* rp):Node(rp->lo){
size=rp->size;
try {vixels=new char[size];}
catch (...) {vixels=0;return;}
memcpy(vixels,rp->vixels,size);
}
~Row(){delete[] vixels;}
Row* Next() const {return (Row*)(nextNode);}
bool Stretch(uint32 index,uint32 width) {
//if necessary, we replace this row with a wider row
if (width >= size*8) {
int newSize=size+kIncrementBits/8;
char* newVixels;
try{newVixels=new char[newSize];}
catch(...){return false;}
memcpy(newVixels,vixels,size);
memset(newVixels+size,0,kIncrementBits/8);
delete vixels;
size=newSize;
vixels=newVixels;
}
if (Vixel(index)) SetVixel(width);//else already 0;
return true;
}
#define MEMBER vixels[index>>3]
#define BIT (1L<<(INDEX&7))
INT VIXEL(UINT32 INDEX) CONST {RETURN MEMBER & BIT;}
VOID CLEARVIXEL(UINT32 INDEX) {MEMBER &= ~BIT;}
VOID SETVIXEL(UINT32 INDEX) {MEMBER |= BIT;}
VOID INVERTVIXEL(UINT32 INDEX){MEMBER ^= BIT;}
#UNDEF MEMBER
#UNDEF BIT
BOOL INITFAILED() CONST {RETURN (VIXELS==0);}
};
COL:NODE
STRUCT COL:NODE {
UINT32 INDEX;// BIT NUMBER IN EVERY ROW OF VIXELS
COL(INT32 XLEFT,UINT32 XINDEX):
NODE(XLEFT){INDEX=XINDEX;}
COL* NEXT() CONST {RETURN (COL*)(NEXTNODE);}
};
TREE
CLASS TREE {
// BASED ON AVL BALANCED BINARY TREE, SEE:
// "ALGORITHMS AND DATA STRUCTURES IN C++"
// BY LEENDERT AMMERDAAL, PUBLISHED BY WILEY 1996
// HERE, THE TREE IS NEVER ALLOWED TO BE EMPTY,
// SO WE DO NOT HAVE TO CHECK FOR NULL POINTERS
PRIVATE:
NODE* ROOT;
VOID LEFTROTATE(NODE* &P) {
NODE* Q=P;
P=P->rightNode;
q->rightNode=p->leftNode;
p->leftNode=q;
q->bal-;
if (p->bal>0) q->bal-=p->bal;
p->bal-;
if (q->bal<0) P->bal+=q->bal;
}
void RightRotate(Node* &p) {
Node* q=p;
p=p->leftNode;
q->leftNode=p->rightNode;
p->rightNode=q;
q->bal++;
if (p->bal<0) Q->bal-=p->bal;
p->bal++;
if (q->bal>0) p->bal+=q->bal;
}
int Insert(Node* &p,Node* q,Node* prev);
Node* Find(Node* p,int x) const;
public:
void Clear(){root=0;}
Node* Root(){return root;}
void Insert(Node* q,Node* prev) {
Insert(root,q,prev);
}
Node* Find(int x) const {
if (root) return Find(root,x);
return root;
}
};
CellSelection
class CellSelection {
private:
Tree rowTree; // tree of sub-areas
Row* row; // top most row, never deleted
Tree colTree; // tree of column indices
Col* col; // left most column, never deleted
uint32 width; // number of columns
#if EXTRA_FOR_TESTING
Row* testRow;
Col* testCol;
#endif
bool CreateEmpty() {
// Creates a single vixel covering the 32-bit universe
rowTree.Clear();
colTree.Clear();
try {row=new Row(kNegInfinity);}
catch (...) {return false;}
if (row->InitFailed())
return false;
rowTree.Insert(row,0);
try {col=new Col(kNegInfinity,0);}
catch (...) {delete row; return false;}
colTree.Insert(col,0);
width=1;
return true;
}
void FreeAll() {
// Uses linked lists to delete all rows and columns
Row* rp=row;
while (rp) {
Row* nextRow=rp->Next();
delete rp;
rp=nextRow;
}
Col* cp=col;
while (cp) {
Col* nextCol=cp->Next();
delete cp;
cp=nextCol;
}
// and clear the trees, in preparation for re-use
rowTree.Clear();
colTree.Clear();
}
// The next three methods expand the vixel map by
// insertion of new rows and columns
Row* SplitRow(Row* rp,int32 newTop) {
// Creates a new row as a copy of rp, and inserts
// it after it, with new coordinates marking the split
Row* newRow;
try {newRow=new Row(rp);}
catch (...) {return 0;}
if (newRow->InitFailed())
return 0;
newRow->lo=newTop;
rowTree.Insert(newRow,rp);
return newRow;
}
bool StretchRows(uint32 index,uint32 width) {
// Allocates the next available vixel column
Row* rp=row;
while (rp) {
if (!rp->Stretch(index,width)) {
// If we run out out of memory before stretching all rows
// we fail. Those rows that did get stretched stay there.
// No harm done.
return false;
}
rp=rp->Next();
}
return true;
}
Col* SplitColumn(Col* cp,int32 newLeft) {
// Creates a new column as a copy of cp, and inserts
// it after it, with new coordinates marking the split
if (!StretchRows(cp->index,width))
return 0;
Col* newCol;
try {newCol=new Col(newLeft,width);}
catch (...) {return 0;}
colTree.Insert(newCol,cp);
width++;
return newCol;
}
void OutlineArea(Area a,Col** left,Row** top) const {
// locate Top-left corner closest to area.
*top=(Row*)(rowTree.Find(a.top));
*left=(Col*)(colTree.Find(a.left));
}
bool DefineArea(Area a,Col** left,Row** top) {
// locate area and create new borders if needed
Row* rp=(Row*)(rowTree.Find(a.top));
if (rp->lo == a.top) *top=rp; else
if (0==(*top=SplitRow(rp,a.top))) return false;
rp=(Row*)(rowTree.Find(a.bottom+1));
if (rp->lo != a.bottom+1)
if (!SplitRow(rp,a.bottom+1)) return false;
Col* cp=(Col*)(colTree.Find(a.left));
if (cp->lo == a.left) *left=cp; else
if (0==(*left=SplitColumn(cp,a.left))) return false;
cp=(Col*)(colTree.Find(a.right+1));
if (cp->lo != a.right+1)
if (!SplitColumn(cp,a.right+1)) return false;
return true;
}
// The next set of methods scan blocks of vixels and perform
// the indicated function on each vixel
bool SetArea(Area a) {
Col* left;
Row* top;
if (!DefineArea(a,&left,&top)) return false;
do {
Col* cp=left;
do {
top->SetVixel(cp->index);
cp=cp->Next();
} while ((cp) && (a.right>=cp->lo));
top=top->Next();
} while ((top) && (a.bottom>=top->lo));
return true;
}
bool ClearArea(Area a) {
Col* left;
Row* top;
if (!DefineArea(a,&left,&top)) return false;
do {
Col* cp=left;
do {
top->ClearVixel(cp->index);
cp=cp->Next();
} while ((cp) && (a.right>=cp->lo));
top=top->Next();
} while ((top) && (a.bottom>=top->lo));
return true;
}
bool InvertArea(Area a) {
Col* left;
Row* top;
if (!DefineArea(a,&left,&top)) return false;
do {
Col* cp=left;
do {
top->InvertVixel(cp->index);
cp=cp->Next();
} while ((cp) && (a.right>=cp->lo));
top=top->Next();
} while ((top) && (a.bottom>=top->lo));
return true;
}
bool AllSelected(Area a,bool test) const {
Col* left;
Row* top;
OutlineArea(a,&left,&top);
do {
Col* cp=left;
do {
if (test != IsSet(top,cp)) return false;
cp=cp->Next();
} while ((cp) && (a.right>=cp->lo));
top=top->Next();
} while ((top) && (a.bottom>=top->lo));
return true;
}
uint32 CountSelectedP(Area a) const {
Col* left;
Row* top;
OutlineArea(a,&&left,&top);
uint32 count=0;
do {
Col* cp=left;
uint32 strip=0;
do {
if (IsSet(top,cp)) strip+=cp->Slice(a.left,a.right);
cp=cp->Next();
} while ((cp) && (a.right>=cp->lo));
count+=strip*top->Slice(a.top,a.bottom);
top=top->Next();
} while ((top) && (a.bottom>=top->lo));
return count;
}
bool IsSet(Row* rp,Col* cp) const{
// Tests if a vixel is set, i.e. if the cells represented
// by the sub-area defined by 1 row and 1 column, are
// in the CellSelection.
return rp->Vixel(cp->index);
}
public:
CellSelection(void) {
/* create an empty selection */
CreateEmpty();
#if EXTRA_FOR_TESTING
testRow=0;
testCol=0;
#endif
}
~CellSelection(void) {
/* free any allocated memory */
FreeAll();
}
bool Clear() {
/* make the selection empty */
FreeAll();
return CreateEmpty();
}
bool Add(Area area) {
/* add the area of cells to this selection */
if (!area.IsEmpty()) return SetArea(area);
return true;
}
bool Remove(Area area) {
/* remove the area of cells from this selection */
if (!area.IsEmpty()) return ClearArea(area);
return true;
}
bool Invert(Area area) {
/* remove cells in the area that are also in this selection
and add the area cells that are not in this selection */
if (!area.IsEmpty()) return InvertArea(area);
return true;
}
bool Add(const CellSelection & otherSelection) {
/* add the otherSelection to this selection */
Row* rp=otherSelection.row;
do {
Area a;
a.top=rp->lo;a.bottom=rp->Limit();
Col* cp=otherSelection.col;
do {
if (IsSet(rp,cp)) {
a.left=cp->lo;a.right=cp->Limit();
if (!SetArea(a)) return false;
}
cp=cp->Next();
} while (cp);
rp=rp->Next();
} while (rp);
return true;
}
bool Remove(const CellSelection & otherSelection) {
/* remove the otherSelection from this selection */
Row* rp=otherSelection.row;
do {
Area a;
a.top=rp->lo;a.bottom=rp->Limit();
Col* cp=otherSelection.col;
do {
if (IsSet(rp,cp)) {
a.left=cp->lo;a.right=cp->Limit();
if (!ClearArea(a)) return false;
}
cp=cp->Next();
} while (cp);
rp=rp->Next();
} while (rp);
return true;
}
bool Invert(const CellSelection & otherSelection) {
/* remove cells in the otherSelection that are also in this selection
and add the otherSelection cells that are not in this selection */
Row* rp=otherSelection.row;
do {
Area a;
a.top=rp->lo;a.bottom=rp->Limit();
Col* cp=otherSelection.col;
do {
if (IsSet(rp,cp)) {
a.left=cp->lo;a.right=cp->Limit();
if (!InvertArea(a)) return false;
}
cp=cp->Next();
} while (cp);
rp=rp->Next();
} while (rp);
return true;
}
bool AllSelected(Area area) {
/* return TRUE if all cells in the area are selected */
if (area.IsEmpty()) return false;
return AllSelected(area,true);
}
uint32 CountSelected(Area area) {
/* count cells that are "on" */
if (area.IsEmpty()) return 0;
return CountSelectedP(area);
}
bool EqualSelected(const CellSelection & otherSelection) {
/* return TRUE if otherSelection equals this selection */
Row* rp=row;
do {
Area a;
a.top=rp->lo;a.bottom=rp->Limit();
Col* cp=col;
do {
a.left=cp->lo;a.right=cp->Limit();
if (!otherSelection.AllSelected(a,IsSet(rp,cp)))
return false;
cp=cp->Next();
} while (cp);
rp=rp->Next();
} while (rp);
return true;
}
#if EXTRA_FOR_TESTING
const int Width() const {return width;}
const int Height() const {
int h=0;
Row* rp=row;
while (rp) {h++;rp=rp->Next();}
return h;
}
void ResetTestAreas() {
testRow=row;
testCol=col;
}
bool NextTestArea(Area* a) {
while (testRow) {
Row* rp=testRow;
while (testCol) {
Col* cp=testCol;
testCol=cp->Next();
if (IsSet(rp,cp)) {
a->left=cp->lo;
a->top=rp->lo;
a->right=cp->Limit();
a->bottom=rp->Limit();
return true;
}
}
testRow=rp->Next();
testCol=col;
}
return false;
}
#endif
};
#ifndef CELLS_DEFINITIONS_INCLUDED
#define CELLS_DEFINITIONS_INCLUDED
// We avoid Inlining of recursive methods
int Tree::Insert(Node* &p,Node* q,Node* prev) {
// insert node q in (sub-)tree rooted in p
int deltaH=0;
if (p==0) {
p=q;
deltaH=1;
// also insert q in linear list, following prev
if (prev) {
p->nextNode=prev->nextNode;
prev->nextNode=p;
}
} else if (q->lo > p->lo) {
if (Insert(p->rightNode,q,prev)) {
p->bal++;
if (p->bal==1) deltaH=1;
else if (p->bal==2) {
if (p->rightNode->bal==-1)
RightRotate(p->rightNode);
LeftRotate(p);
}
}
} else if (q->lo < P->lo) {
if (Insert(p->leftNode,q,prev)) {
p->bal-;
if (p->bal==-1) deltaH=1;
else if (p->bal==-2) {
if (p->leftNode->bal==1)
LeftRotate(p->leftNode);
RightRotate(p);
}
}
}
return deltaH;
}
Node* Tree::Find(Node* p,int x) const {
// find nearest p->lo <= X
// NEVER RETURNS A NULL POINTER
IF (P->lo<X) {
IF (P->rightNode) {
Node* q=Find(p->rightNode,x);
if (q->lo<=X) RETURN Q;
}
}
IF (P->lo>x) {
if (p->leftNode) return Find(p->leftNode,x);
}
return p;
}
#endif