Feb 01 Challenge Volume Number: 17 (2001)
Issue Number: 2
Column Tag: Programmer's Challenge

# Programmer's Challenge

By Bob Boonstra, Westford, MA

### Trilite

Tic-Tac-Toe is a trivial game. There are less than 9! possible games, far fewer if symmetry is taken into account, and certainly few enough for the outcome to be calculated in advance. But there is a variant of Tic-Tac-Toe that allows many more possible move sequences, and for which there may or may not be a guaranteed winning solution. This month you are going to have an opportunity to compete in the game of Trilite against your Challenge peers.

Trilite is like Tic-Tac-Toe in the sense that it is played on a 3x3 board, where two players alternate occupying squares with the objective of occupying three positions in a row. It differs from Tic-Tac-Toe in that a player may occupy only three positions at a time. When a player occupies a fourth position, one of the three previously occupied positions, the one that has been occupied the longest, becomes vacant. So after any move, there are always three vacant positions on the board, and one more that is about to become unoccupied when the current player occupies one of the three vacant positions. Sounds simple, right?

The prototype for the code you should write is:

```typedef enum {                        /* numbering system for Board positions */
kNoPosition=-1,
kTopLeft=0, kTopCenter, kTopRight,
kCenterLeft, kCenter, kCenterRight,
kBottomLeft, kBottomCenter, kBottomRight
} BoardPosition;

typedef enum {                        /* possible values for a Board Position */
kEmpty=-1,
kPlayer1Staying=0, kPlayer1Disappearing,
kPlayer2Staying, kPlayer2Disappearing
} PositionValue;

typedef PositionValue Board[9];   /* state of the Board */

BoardPosition PlayTrilite(
const Board triliteBoard,   /* current state of the Board */
BoardPosition opponentPreviousPlay,
/* the BoardPosition your opponent last played */
int playerNumber,      /* 1 if you are player 1, 2 if you are player 2 */
Boolean newGame         /* true the first time you are called for a new game */
);
```

For each game of Trilite, your PlayTrilite routine and that of your opponent will be called alternately until one of you wins by occupying three positions in a row, horizontally, vertically, or diagonally. The first time PlayTrilite is called for a new game, newGame will be set to TRUE. When newGame is TRUE, playerNumber will indicate whether you are the first (playerNumber==1) or second (playerNumber==2) player. Each time PlayTrilite is called, the BoardPosition last occupied by your opponent will be provided as opponentPreviousPlay. Finally, the current state of the Board will be provided to you as triliteBoard.

Trilite board positions have five possible values. Unoccupied positions have the value kEmpty. Positions occupied by player 1 have the value kPlayer1Staying or kPlayer1Disappearing, with the latter value distinguishing positions that will become empty following player 1's next move. Similarly, positions occupied by player 2 have the value kPlayer2Staying or kPlayer2Disappearing.

A sequence of moves works like this. Suppose the game has been going on for at least three pairs of turns, and it is player 1's turn to play. The Board will have six occupied positions, three by player 1 and three by player 2. One position for each player will be marked as "disappearing" on the next move. Player 1 will occupy one of the three remaining unoccupied positions, and - at the same time - the kPlayer1Disappearing position will become kEmpty. If player 1 now occupies three positions in a row, s/he is the winner. Otherwise, player 2 then occupies one of the three empty positions and the kPlayer2Disappearing position becomes kEmpty. Note that a player may not reoccupy the position about to disappear - the opponent is the first player with a chance to occupy that position. The astute reader might detect one element of a potential game strategy here.

Entries will compete against one another in a tournament structured so that each entry plays each other entry an even number of times, half playing first, and half playing second. If the number of entries is large, some other fair tournament scheme will be used. A game will be considered drawn when a time limit and a move count limit, not specified as part of the problem statement, are exceeded.

The winner will be the entry that scores the most points, where each game won is worth 1000 points, each game drawn is worth 500 points, and 1 point is deducted for each millisecond of execution time. The Challenge prize will be divided between the overall winner and the best scoring entry from a contestant that has not won the Challenge recently.

Your code and data must live in an application heap of 40MB. Any nontrivial tables used by your solution must be calculated at run time. Any entry that precalculates any significant part of the solution will be disqualified.

Those of you interested in experimenting with Trilite might want to check out the shareware game by John Mauro, at <http://screech.cs.alfred.edu/~maurojc/software/software.html#Trilite>.

This will be a native PowerPC Challenge, using the CodeWarrior Pro 6 environment. Solutions may be coded in C, C++, or Pascal. You can also provide a solution in Java, provided you also provide a test driver equivalent to the C code provided on the web for this problem.

### Three Months Ago Winner

Three people entered the November FreeCell Challenge, where contestants had to write code to solve the FreeCell solitaire puzzle. FreeCell requires players to move cards from eight tableaus to four home piles in ascending order based on suit, but it also provides four "free cells" where cards may be stored temporarily. Congratulations to Ernst Munter (Kanata, Ontario) for his victory in this Programmer's Challenge.

Ernst's entry performs a depth-first search of possible moves, enumerated by the GenerateMoveList routine. Moves are assigned a value that combines a heuristic weight assigned a priori to the type of move (e.g., kFreeToHome), a measure of the degree to which the cards in a tableau are in the correct order, and the presence in a tableau of cards that could be moved home. The code (IsNotRedundant) avoids moves that return a card to the position it occupied previously when no intervening move would have made the return nonredundant. A key to the speed of Ernst's entry is the way it avoids looping back into a previously encountered configuration. The Execute routine computes a hash value for the game state resulting from a prospective move and compares that hash value to that of previously encountered game states. If the prospective move results in a previously encountered state, the move is rejected. Assuming a move is not redundant, the move is made and a new set of possible moves is generated. The move search gives up and restarts if it is forced to backtrack too many times, using the list of previously encountered states to ensure that a different search path results.

As the top-placing contestant without a previous Challenge win, Greg Sadetsky wins a share of this month's Developer Depot merchandise credit prize. His second place solution also keeps track of past game states, but in a very large array instead of in a hash value. Greg employs a number of devices to reduce the storage required, but the resulting logic for detecting a repeat game state is more complex and time consuming. Greg's entry generates move sequences that are about 50% longer on average than those generated by Ernst's entry. It cuts off the search after 10 seconds, the point at which the time penalty exceeded the point value of solving the hand. As a result, his solution gave up on about 6% of the test cases.

The third entry I received this month was a recursive solution only slightly slower than the winning entry, but it crashed for 9 of the test cases. Even after I increased the heap and stack sizes significantly, the code crashed with heap corruption after apparently entering a recursion loop. To measure performance on the remaining cases, I needed to modify the test code to bypass the problematic hands and, for that reason, the entry was disqualified.

I tested the entries to this Challenge with more than 20,000 deals, including roughly one third of the 32,000 deals included in the Linux xfreecell package, 10,000 random deals, and a few manually constructed deals. Ernst's solution solved all but two of the test cases, both of which were a single deal that is known to be unsolvable. His solution required just over three minutes to run the entire set of tests, and generated an average of 156 moves to solve each deal. As you can see in the table below, a small number of test cases required more than 1500 moves to solve - the most complicated deal, excluding the ones that could not be solved, required 1863 moves.

 >100 Moves (# of cases) >500 Moves (# of cases) >1000 Moves (# of cases) >1500 Moves (# of cases) No Solution (# of cases) Ernst Munter 18190 444 24 7 2 Greg Sadetsky 19410 784 49 11 1274 C. W. 20303 0 0 0 278

The table below lists, for each of the solutions submitted, the number of test cases solved by each entry, the total execution time, the number of points earned, and the number of moves generated to solve the entire test suite. It also provides the code size, data size, and programming language used for each entry. 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. The solution marked with an asterisk was disqualified for reasons explained above.

 Name Test Cases Solved Test Cases Unsolved Time (secs) Points x100000 Moves x1000 Code Size Data Size Lang Ernst Munter (681) 20694 2 181.1 206.8 3220 9800 1793 C++ Greg Sadetsky (2) 19422 1274 24399.2 169.8 4705 8156 18.31M C C. W. (*) 20409 278 198.2 203.9 4103 7276 1858 C

### Top Contestants...

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.

 Rank Name Points 1. Munter, Ernst 271 2. Saxton, Tom 76 3. Maurer, Sebastian 68 4. Rieken, Willeke 65 5. Boring, Randy 52 6. Shearer, Rob 48 7. Taylor, Jonathan 36 8. Wihlborg, Charles 29

### ... and the Top Contestants Looking For a Recent Win

Starting this month, in order to give some recognition to other participants in the Challenge, we are also going to list the high scores for contestants who have accumulated points without taking first place in a Challenge. Listed here are all of those contestants who have accumulated 6 or more points during the past two years.

 9 Downs, Andrew 12 10 Jones, Dennis 12 11 Day, Mark 10 12 Duga, Brady 10 13 Fazekas, Miklos 10 14 Flowers, Sue 10 15 Sadetsky, Gregory 10 16 Selengut, Jared 10 17 Strout, Joe 10 18 Hala, Ladislav 7 19 Miller, Mike 7 20 Nicolle, Ludovic 7 21 Schotsman, Jan 7 22 Widyyatama, Yudhi 7 23 Heithcock, JG 6

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

Here is Ernst's winning FreeCell solution:

```FreeCell.cp
Ernst Munter, Kanata, ON, Canaca

/*
Solves FreeCell games by a guided trial and error search.

At each stage, all possible moves are listed, ranked according to a fixed heuristic which
prefers moves towards home, and towards aggregating strings of alternating colors on the
tableau.

All reached states are recorded in a database to avoid loops.  The hash method to compress
states takes care of some redundancies;  for example it does not care which column a
particular set of cards is in, and it distinguishes cards only bycolor, not suit.

If a search is making little progress, it is cut off after a specific number of undo
steps, and a fresh search started.  The same happens when the maximum number of moves has
been reached.  The new search still respects the accumulated database of previously seen
states, and so is forced to take a different path, improving its chances.

The resulting move sequences are not optimal, and certainly not elegant.  The search also
does not include macro moves (moving columns of several cards).  I tried this but it was
counter-productive:  by listing the macro moves, the move lists became longer, and more
false paths ended up being explored.

Version 2 changes
---------
- replaced the STL set<> with a simpler, faster custom set;
- replaced qsort (of move lists) with an integrated custom heap sort;
- policy constants tuned.

Version 3 change
--------
Reduced the amount of redundant moves by scanning back through the move stack to avoid any
move that would simple put a card back where it was earlier.  Such moves are truly
redundant if the to- and from- card positions were not used by intermediate moves of other
cards.  This strategy improved both time, and average number of moves to solve, by about
18%.
*/

#include "FreeCell.h"
#define NDEBUG
#include <assert.h>
#include <string.h>   // for memset()

#define VERSION 3

// I need to have the suits in alternating red-black order.
enum {
myRed=16,
myNull=0,myA=1,my2,my3,my4,my5,my6,my7,my8,my9,myT,myJ,myQ,myK,
mySpots=15,   kSignificant=myRed|mySpots
};

typedef unsigned char MyCard;// 2 bits suit + 4 bits spot
typedef unsigned char uchar;
typedef unsigned long ulong;
typedef unsigned short ushort;

enum {
kFreeCell=0,   // a single set of 16 card stacks defines the tableau
kTableau=4,      // card stack offsets
kHome=12,      // home must be last group

kAvgMoveListLength=16,// just an estimate

// Policy constants affect the order in which moves are tried:
kFreeToHome=10000,
kTableauToHome=10000,
kTableauToTableau=2000,
kFreeToTableau=500,
kFreeToEmptyTableau=500,
kTableauToEmptyTableau=50,
kTableauToFree=24,
kSrcPriority=2000,
kBlockedOnly=0,

kLongestPossibleMoveList=63,// actually no more than 31 have been observed
kUndoLimitMul=16,
kMaxRestartsDiv=65536
};

inline MyCard MakeCard(int spot,int suit)
{return spot | (suit<<4);}
inline int MySuit(MyCard c) {return c>>4;}

struct CRC
// Standard CRC based hash method.
static struct CRC {
enum {POLYNOMIAL=0x04c11db7L};
ulong table[256];
CRC()
{
long i,j,x;
for (i=0;i<256;i++) {
x=i<<24;
for (j=0;j<8;j++) {
if (x<0) x=(x<<1) ^ POLYNOMIAL;
else x=(x<<1);
}
table[i]=x;
}
}
ulong HashFunction(const uchar* ufrg,int frgLen,int type) const
{
// Uses CRC on length type, and all chars of a fragment
ulong accum=table[frgLen];
for (int i=0;i<frgLen;i++)
accum=(accum<<8) ^ table[(accum>>24) ^
(kSignificant & ufrg[i])];
accum=(accum<<8) ^ table[(accum>>24) ^ type];
return type + accum;
}
} crc;

struct Legal
// A pair of lookup tables to indicate legality of placng one card upon another.
static struct Legal {
bool   redBlack[64][64];         // legal to put second card on (first) in tableau
bool   inSequence[64][64];      // legal to send second card home (first)
Legal()
// setup  red-black and inSequence card lookup tables
{
first<=(myK|myDiamond);first++)
{
second<=(myK|myDiamond);second++)
{
if ( ((mySpots & (first - second))==1) &&
((myRed & (first ^ second))==myRed) )
redBlack[first][second]=true;
// else =0;
if ( ((mySpots & (second - first))==1) &&
((mySuits & (first ^ second))==0) )
inSequence[first][second]=true;
// else =0;
}
}
}
} gLegal;

inline MyCard Convert2MyCard(const Card c)
// converts a "Card" defined in "FreeCell.h" to an instance of "MyCard"
{
switch (c.suit)
{
case kHeart:   return myHeart   | c.spot;
case kDiamond:   return myDiamond | c.spot;
case kClub:      return myClub    | c.spot;
}
return 0;
}

struct CardStack
struct CardStack {
// Generic card stack, serving for tableau, freecell, and home columns
MyCard*   SP;
uchar   stackType;
MyCard   cards[27];// only 19 needed, struct is padded out to 32 bytes
void Init(const Tableau * theTableau,int num,int type)
{
stackType=type;
SP=cards;
if (theTableau)
for (int i=0;i<num;i++)
*SP++=Convert2MyCard(theTableau->theCard[i]);
}
void InitHome(int suit)
{
stackType=kHome;
SP=cards+1;
cards[0]=MakeCard(myNull,suit);// null card of correct suit to build upon
}
MyCard TopCard() const {return SP[-1];}
ulong Hash() const
{
return crc.HashFunction(cards,NumCards(),stackType);
}
bool IsEmpty() const {return SP==cards;}
{
assert(NumCards()<19);
*SP++=c;
}
MyCard Remove()
{
assert(SP>cards);
return *-SP;
}
int AllInOrder()
// If the entire tableau stack is in order, returns numCards.
// If not, this function returns 0.
{
int num=0;
if (SP>cards)
{
num++;
MyCard* c1=SP-1;
while (c1>cards)
{
MyCard* c2=c1-1;
if (!gLegal.redBlack[*c2][*c1])
return 0;
num++;
c1=c2;
}
}
return num;
}
int NumInOrder()
// Returns the number of cards at the top of the stack which are in order.
{
int num=0;
if (SP>cards)
{
num++;
MyCard* c1=SP-1;
while (c1>cards)
{
MyCard* c2=c1-1;
if (!gLegal.redBlack[*c2][*c1])
break;
num++;
c1=c2;
}
}
return num;
}
int SourcePriority(MyCard home[])
// Scans the stack including (or excluding) the top card, to set a priority value
// for the stack if it contains cards that could go home right away.
// kBlockedOnly=1 limits priority to blocked cards.
// Returns the priority value
{
int srcPriority=0;
MyCard* cp=cards;
for (;cp<SP-kBlockedOnly;cp++)
{
MyCard c=*cp;
for (int k=0;k<4;k++)
{
if (c==home[k])
srcPriority+=kSrcPriority;
}
}
return srcPriority;
}
int NumCards() const {return SP-cards;}
};

struct MyMove
struct MyMove {
// My move is represented in a  32-bit ulong
ulong   gameValue:16;   // value of this move or cardToMove
ulong   toPile:8;
ulong   fromPile:8;
void Init(int from,int to,int val)
{
gameValue=val;
toPile=to;
fromPile=from;
}
void Clear() {fromPile=toPile=gameValue=0;}
ulong IsValid() const {return Int();}// Null-move indicated by all-0 fields
ulong FromPile() const {return fromPile;}
void SetValue(MyCard c) {gameValue=c;}
bool IsInverseOf(MyMove m) const {
return ((fromPile == m.toPile) && (toPile == m.fromPile));
}
bool ToHome() const {return (toPile>=kHome);}
void MoveCard(CardStack* stacks)
{
assert(stacks[fromPile].NumCards());
assert(stacks[toPile].NumCards()<19);
MyCard c=stacks[fromPile].Remove();
}
void UndoMove(CardStack* stacks)
{
assert(stacks[toPile].NumCards());
assert(stacks[fromPile].NumCards()<19);
MyCard c=stacks[toPile].Remove();
}
void Convert(Move* m)
// Converts this instance of "MyMove" to a "Move" as defined in "FreeCell.h"
{
m->theSource = Source(fromPile-kFreeCell+dFreeCellA);
m->theDestination = (toPile>=kHome) ? dHome:
Destination(toPile-kFreeCell+dFreeCellA);
}
int Int() const {return *((int*)this);}// cast all three fields as single int
};
typedef MyMove* MyMovePtr;

inline bool operator > (const MyMove & a,const MyMove & b) {return a.Int() > b.Int();}

struct MoveHeap
// The custom heap for sorting moves.
struct MoveHeap {
int      heapSize;
MyMove   heapBase[kLongestPossibleMoveList];
MoveHeap() : heapSize(0) {}
int Size() const {return heapSize;}

void Insert(MyMove k)
{
int i=++heapSize;
int j=i>>1;
MyMove z;
while (j && ((z=heapBase[j]) > k) )
{
heapBase[i]=z;
i=j;
j=i>>1;
}
heapBase[i]=k;
}

MyMove Pop()
{
MyMove rc=heapBase[1];
MyMove k=heapBase[heapSize-];
if (heapSize<=1) {
heapBase[1]=k;
return rc;
}
int i=1,j=2;
while (j<=heapSize)
{
if ((j<heapSize)
&& (heapBase[j] > heapBase[j+1]))
j++;
if (heapBase[j] > k)
break;
heapBase[i]=heapBase[j];
i=j;j+=j;
}
heapBase[i]=k;
return rc;
}
};

struct Bucket
// The set (MySet below) is implemented as a hash table of buckets.
// Each bucket can hold kBucketSize values, and can be extended indefinetely
enum {kBucketSize=17,kNumBuckets=1024};
struct Bucket {
int      numEntries;
ulong   entry[kBucketSize];
// bucket size of 9 or 17 makes full use of allocated memory (CW 6)
Bucket(ulong firstEntry) :
void Insert(ulong x)
// Insert x only if x is not in the set already
{
Bucket* b=Find(x);
if (b==0) return;
}
Bucket* Find(ulong x)
// Scans this and linked buckets looking for x
{
ulong* ep=entry+numEntries;
do {
if (*-ep == x) return 0;
} while (ep>entry);
return this;
}
{
if (numEntries < kBucketSize)
entry[numEntries++]=x;
else
}
};

struct MySet
struct MySet {
// A set to record all states (represented by their hash value) which have occurred.
Bucket*   buckets[kNumBuckets];
MySet(){memset(buckets,0,sizeof(buckets));}
~MySet(){
for (int i=0;i<kNumBuckets;i++)
{
Bucket* b=buckets[i];
if (b) delete b;
}
}
void Insert(ulong x)
{
Bucket* b=buckets[x % kNumBuckets];
if (b==0)
{
b=new Bucket(x);
buckets[x % kNumBuckets]=b;
} else   b->Insert(x);
}
bool Find(ulong x)
{
Bucket* b=buckets[x % kNumBuckets];
return (b && (0==b->Find(x)));
}
};

struct MyGame
struct MyGame {
// MyGame is the top level struct which holds all local data
CardStack   stacks[16];   //    my version of the tableau, the current state
ulong      hashedState;      //   current state, compressed
long      numCardsOutstanding;
MyMove*      movePool;         //   single pool allocated for movelists
MyMove*     endMovePool;
MyMovePtr*   moveStack;      //   move stack tracks the history of executed moves
MyMovePtr*   moveStackPointer;
MyMovePtr*   lastMoveStack;
MyCard       nextHome[4];   //    next cards (1 per suit) to go home
MySet      stateSet;            //   all visited states are recorded in this set, as hash values
MyGame(long maxMoves) :
movePool(new MyMove[kLongestPossibleMoveList+
maxMoves*kAvgMoveListLength]),
endMovePool(movePool+kLongestPossibleMoveList+
maxMoves*kAvgMoveListLength),
moveStack(new MyMovePtr[maxMoves]),
moveStackPointer(moveStack),
lastMoveStack(moveStack+maxMoves-1)
{}

~MyGame(){
delete [] moveStack;
delete [] movePool;
}

void InitTableau(const Tableau theTableau[8])
// Copies the initial tableau to the local representation
{
for (int tid=0;tid<8;tid++)
stacks[tid+kTableau].Init(&theTableau[tid],
7-tid/4,kTableau);
numCardsOutstanding=52;
for (int i=0;i<4;i++)
{
stacks[i+kFreeCell].Init(0,0,kFreeCell);
stacks[i+kHome].InitHome(i);
nextHome[i]=MakeCard(myA,i);
}
hashedState=Hash();
}

MyGame::Hash
ulong Hash() const
// Hashes the game state into a single 32-bit integer
{
const CardStack* cs=stacks;
ulong h=cs->Hash();
for (int i=1;i<16;i++,cs++)
h ^= cs->Hash();
return h;
}

MyGame::GenerateMoveList
MyMove* GenerateMoveList(MyMove* mp)
{
//   Lists all legal moves in a list, starting with a null-move;
//   sorts the moves and returns the highest value move on the list
//   Each move is given a "value" reflecting its relative merit.
if (mp+kLongestPossibleMoveList >= endMovePool)
return 0; // no room for movelist, should not really happen
// but if it does, we just have to backtrack
MyMove m;
MoveHeap heap;
int src,dest;
CardStack* srcPtr;
CardStack* destPtr;
int cardToMove,topCardDest,value,srcPriority;

for (src=kFreeCell,srcPtr=stacks+src;
src<kFreeCell+4;src++,srcPtr++)
// from any freecell to: home, or tableau
{
if (srcPtr->IsEmpty()) continue;
cardToMove=srcPtr->cards[0];
srcPriority=srcPtr->SourcePriority(nextHome);

topCardDest=stacks[kHome+MySuit(cardToMove)].TopCard();
if (gLegal.inSequence[topCardDest][cardToMove])
// to correct home
{
value = kFreeToHome +
srcPriority;
m.Init(src,MySuit(topCardDest)+kHome,value);
heap.Insert(m);
}

bool toEmptyFlag=true;
for (dest=kTableau,destPtr=stacks+dest;
dest<kTableau+8;dest++,destPtr++)
// to every matching tableau
{
if (destPtr->IsEmpty())
{
if (toEmptyFlag)
{
value = kFreeToEmptyTableau +
(2<<(cardToMove&mySpots)) +
srcPriority;
m.Init(src,dest,value);
heap.Insert(m);
toEmptyFlag=false;
}
continue;
}
topCardDest=destPtr->TopCard();
if (gLegal.redBlack[topCardDest][cardToMove])
{
value = kFreeToTableau +
destPtr->AllInOrder() +
srcPriority;
m.Init(src,dest,value);
heap.Insert(m);
}
}
}

for (src=kTableau,srcPtr=stacks+src;
src<kTableau+8;src++,srcPtr++)
// from any tableau to: freecell, home or tableau
{
if (srcPtr->IsEmpty()) continue;
int srcInOrder=srcPtr->AllInOrder();
int longestInOrder=srcPtr->NumInOrder();
srcPriority=srcPtr->SourcePriority(nextHome);
int maxBlock=0;

cardToMove=srcPtr->TopCard();// single card moves
topCardDest=stacks[kHome+MySuit(cardToMove)].TopCard();
if (gLegal.inSequence[topCardDest][cardToMove])
// to matching home
{
value = kTableauToHome +
srcPriority;
m.Init(src,MySuit(topCardDest)+kHome,value);
heap.Insert(m);
}

for (dest=kFreeCell,destPtr=stacks+dest;
dest<kFreeCell+4;dest++,destPtr++)
// to first available freecell
{
if (destPtr->IsEmpty())
{
value = kTableauToFree -
srcInOrder -
4*longestInOrder +
srcPriority;
m.Init(src,dest,value);
heap.Insert(m);
break;
}
}

bool toEmptyFlag=true;
for (dest=kTableau,destPtr=stacks+dest;
dest<kTableau+8;dest++,destPtr++)
// to every matching tableau
{
if (src==dest) continue;
if (destPtr->IsEmpty()) // to empty tableau
{
if (toEmptyFlag)
{
value = kTableauToEmptyTableau +
srcInOrder +
(2<<(mySuits & cardToMove)) +
srcPriority;
m.Init(src,dest,value);
heap.Insert(m);
toEmptyFlag=false;
}
continue;
}

topCardDest=destPtr->TopCard();
if (gLegal.redBlack[topCardDest][cardToMove])
{
value = kTableauToTableau +
destPtr->AllInOrder() -
4*srcInOrder +
srcPriority;
m.Init(src,dest,value);
heap.Insert(m);
}
}
}

mp->Clear();               // puts a sentinel 0-move at the start of the movelist

while (heap.Size())   // sorts moves from heap into the movelist space
*++mp = heap.Pop();

return mp;
}

void PushMove(MyMove* m){
*moveStackPointer++=m;
}

MyMove* PopMove()
{
assert(moveStackPointer>moveStack);
return *-moveStackPointer;
}

MyGame::Execute
int Execute(MyMove* mp)
{
// Attempts to execute one move.
// Return codes:
//      -2:  failed, cannot push the last move because the move stack is full
//      -1:    failed, would have reached a previous state
//       0:  success, final move and game solved
//       >0:  normal execution succeeded
MyMove m=*mp;
stateSet.Insert(hashedState);   // save last state in hashed state set

if (moveStackPointer >= lastMoveStack)
return -2;

if (m.ToHome() && (numCardsOutstanding==1)) // The game is solved.
{
PushMove(mp);
return 0;
}

// do the move and compute a new hashed state
ulong newHash=hashedState ^
stacks[m.FromPile()].Hash() ^
stacks[m.ToPile()].Hash();

MyCard cardToMove=stacks[m.FromPile()].TopCard();
m.MoveCard(stacks);

newHash ^=
stacks[m.FromPile()].Hash() ^
stacks[m.ToPile()].Hash();

if (stateSet.Find(newHash))
{
m.UndoMove(stacks);
return -1;
} else
{
hashedState=newHash;// record new hash value
mp->SetValue(cardToMove);
PushMove(mp);
if (m.ToHome())
{
nextHome[m.ToPile()-kHome]++;
numCardsOutstanding-;
}
}
return 1;
}

MyMove* Undo()
// Undoes the last stacked move, returns this move, or 0 if no move found
{
MyMove* mp=PopMove();
if (mp==0) return mp;
MyMove m=*mp;
ulong newHash=hashedState ^
stacks[m.FromPile()].Hash() ^
stacks[m.ToPile()].Hash();

m.UndoMove(stacks);
if (m.ToHome())
{
nextHome[m.ToPile()-kHome]-;
numCardsOutstanding++;
}

hashedState=newHash ^
stacks[m.FromPile()].Hash() ^
stacks[m.ToPile()].Hash();

return mp;
}

long CopyMovesBack(Move theMoves[])
// Scans movestack, converts MyMoves to Moves, and returns the number of moves
{
int numMoves=0;
MyMovePtr* endMoveStack=moveStackPointer;
for (MyMovePtr* index=moveStack+1;index<endMoveStack;index++)
{
MyMove* mp=*index;
mp->Convert(theMoves+numMoves);
numMoves++;
}
return numMoves;
}

int IsNotRedundant(MyMove m)
{
int from=m.FromPile();
int to=m.ToPile();
MyCard cardToMove=stacks[from].TopCard();
MyMovePtr* mps=moveStackPointer;
while (mps>moveStack)
{
MyMove* oldMove=*-mps;
int oldFrom=oldMove->FromPile();
int oldTo=oldMove->ToPile();
MyCard oldCard=oldMove->gameValue;
if (oldCard==cardToMove)
{
return ((oldTo^from) | (oldFrom^to));
} else
{
if ((oldFrom==to)||(oldTo==to)||(oldFrom==from))
break;
}
}
return 1;
}

long Solve(const Tableau theTableau[8],Move theMoves[],long maxMoves)
{
// Solves the game by systematic depth-first exploration of the move tree
// Several fresh starts are possible if the move stack is exhausted
// or if the search seems to be stuck with a large number of backtracks
// In any case, all visited states are recorded in the hashed state set,
// and never entered twice.  The hash is not perfect, and some states might
// be accidentally excluded.  It is hoped that there is always enough redundancy in
// the possible solution sequences to allow an alternative solution to be found.
int cycle=kMaxRestartsDiv/maxMoves,rc;
do {
int undoLimit=kUndoLimitMul*maxMoves;
InitTableau(theTableau);
moveStackPointer=moveStack;
// Put a sentinel null move at start of move stack
PushMove(0);
MyMove* moveList=movePool;
#if VERSION<3
MyMove previousMove;
previousMove.Clear();
#endif

get_new_movelist:
MyMove* nextMove=GenerateMoveList(moveList);
// moveList to nextMove defines a movelist which always starts with a 0-move
// and is processed in order nextMove, nextMove-1, ... until 0-move is found
for (;;)
{
while (nextMove && nextMove->IsValid())
{
#if VERSION>=3
if (!IsNotRedundant(*nextMove))
#else
if (nextMove->IsInverseOf(previousMove))
#endif
{
nextMove-;
continue; // while
}
rc=Execute(nextMove);
if (rc==-1) // would have reached a previous state
{
nextMove-;   // use next best move in list
undoLimit-;
if (undoLimit<=0) // enough! let's restart
goto restart_search;
} else

if (rc>0)// move was executed, get next movelist
{
moveList=1+nextMove;
#if VERSION<3
previousMove=*nextMove;
#endif
goto get_new_movelist;
} else

if (rc==0) // copy moves back for the caller and return
return CopyMovesBack(theMoves);

else // else rc<=-2: move stack is full
{
goto restart_search;
}

} // end while

// no move is possible, try to backtrack
do {
MyMove* prevMove=Undo();
if (!prevMove)  // no solution!, stack is completely unwound
return 0;

// try to use the last move:
nextMove = prevMove-1;
assert(nextMove>=movePool);
assert(nextMove<moveList);
} while (!nextMove->IsValid());

moveList=nextMove;
while ((moveList>=movePool) && (moveList->IsValid()))
moveList-;
assert(moveList>=movePool);
}
restart_search:;
} while (-cycle > 0);    // restart only so many times
return 0;            // then give up and return 0
}
};

FreeCell
long  FreeCell(   // returns the number of moves in theMoves[]
const Tableau   theTableau[8],
Move    theMoves[],
long   maxMoves
) {
MyGame* G=new MyGame(maxMoves);
long numMoves=G->Solve(theTableau,theMoves,maxMoves);
delete G;
return numMoves;
}
```

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