Ch3. STACKS AND QUEUES

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CHAPTER 3 1 STACKS AND QUEUES All the programs in this file are selected from Ellis Horowitz, Sartaj Sahni, and Susan Anderson-Freed “Fundamentals of Data Structures in C”, Computer Science Press, 1992. CHAPTER 3

Stack (stack: a Last-In-First-Out (LIFO) list ):

CHAPTER 3 2 Stack (stack: a Last-In-First-Out (LIFO) list ) Stack An ordered list Insertions and deletions are made at one end, called top Illustration Inserting and deleting elements in a stack A B A C B A D C B A E D C B A D C B A top top top top top top push push pop

Some stack applications:

CHAPTER 3 3 Some stack applications Implementing recusive call Expression evaluation Infix to postfix Postfix evaluation Maze problem Breadth First Search ……

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CHAPTER 3 4 fp main fp al ( a) (b) *Figure 3.2: System stack after function call a1 (p.103) an application of stack: stack frame of function call system stack before a1 is invoked system stack after a1 is invoked fp: a pointer to current stack frame

structure Stack is objects: a finite ordered list with zero or more elements. functions: for all stack  Stack, item  element, max_stack_size  positive integer Stack CreateS(max_stack_size) ::= create an empty stack whose maximum size is max_stack_size Boolean IsFull(stack, max_stack_size) ::= if (number of elements in stack == max_stack_size) return TRUE else return FALSE Stack Add(stack, item) ::= if (IsFull(stack)) stack_full else insert item into top of stack and return :

CHAPTER 3 5 structure Stack is objects: a finite ordered list with zero or more elements. functions: for all stack  Stack , item  element , max_stack_size  positive integer Stack CreateS( max_stack_size ) ::= create an empty stack whose maximum size is max_stack_size Boolean IsFull( stack, max_stack_size ) ::= if (number of elements in stack == max_stack_size ) return TRUE else return FALSE Stack Add( stack, item ) ::= if (IsFull( stack )) stack_full else insert item into top of stack and return abstract data type for stack

Boolean IsEmpty(stack) ::= if(stack == CreateS(max_stack_size)) return TRUE else return FALSE Element Delete(stack) ::= if(IsEmpty(stack)) return else remove and return the item on the top of the stack. :

CHAPTER 3 6 Boolean IsEmpty( stack ) ::= if ( stack == CreateS( max_stack_size )) return TRUE else return FALSE Element Delete( stack ) ::= if (IsEmpty( stack )) return else remove and return the item on the top of the stack. * Structure 3.1: Abstract data type Stack (p.104)

Stack CreateS(max_stack_size) ::= #define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { int key; /* other fields */ } element; element stack[MAX_STACK_SIZE]; int top = -1; Boolean IsEmpty(Stack) ::= top< 0; Boolean IsFull(Stack) ::= top >= MAX_STACK_SIZE-1; :

CHAPTER 3 7 Stack CreateS(max_stack_size) ::= #define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { int key; /* other fields */ } element; element stack[MAX_STACK_SIZE]; int top = -1; Boolean IsEmpty(Stack) ::= top< 0; Boolean IsFull(Stack) ::= top >= MAX_STACK_SIZE-1; Implementation: using array

void add(int *top, element item) { if (*top >= MAX_STACK_SIZE-1) { stack_full( ); return; } stack[++*top] = item; } *program 3.1: Add to a stack (p.104) :

CHAPTER 3 8 void add(int *top, element item) { if (*top >= MAX_STACK_SIZE-1) { stack_full( ); return; } stack[++*top] = item; } *program 3.1: Add to a stack (p.104) Add to a stack

element delete(int *top) { if (*top == -1) return stack_empty( ); /* returns and error key */ return stack[(*top)--]; } *Program 3.2: Delete from a stack (p.105):

CHAPTER 3 9 element delete(int *top) { if (*top == -1) return stack_empty( ); /* returns and error key */ return stack[(*top)--]; } *Program 3.2: Delete from a stack (p.105) Delete from a stack

Queue (Queue: a First-In-First-Out (FIFO) list):

CHAPTER 3 10 Queue ( Queue: a First-In-First-Out (FIFO) list ) Queue An ordered list All insertions take place at one end, rear All deletions take place at the opposite end, front Illustration A B A C B A D C B A D C B rear front rear front rear front rear front rear front

Some queue applications:

CHAPTER 3 11 Some queue applications Job scheduling Event list in simulator Server and Customs ……

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CHAPTER 3 12 * Figure 3.5: Insertion and deletion from a sequential queue (p.108) Application: Job scheduling

structure Queue is objects: a finite ordered list with zero or more elements. functions: for all queue  Queue, item  element, max_ queue_ size  positive integer Queue CreateQ(max_queue_size) ::= create an empty queue whose maximum size is max_queue_size Boolean IsFullQ(queue, max_queue_size) ::= if(number of elements in queue == max_queue_size) return TRUE else return FALSE Queue AddQ(queue, item) ::= if (IsFullQ(queue)) queue_full else insert item at rear of queue and return queue :

CHAPTER 3 13 structure Queue is objects: a finite ordered list with zero or more elements. functions: for all queue  Queue , item  element , max_ queue_ size  positive integer Queue CreateQ( max_queue_size ) ::= create an empty queue whose maximum size is max_queue_size Boolean IsFullQ( queue, max_queue_size ) ::= if (number of elements in queue == max_queue_size ) return TRUE else return FALSE Queue AddQ( queue, item ) ::= if (IsFullQ( queue)) queue_full else insert item at rear of queue and return queue Queue (ADT)

Boolean IsEmptyQ(queue) ::= if (queue ==CreateQ(max_queue_size)) return TRUE else return FALSE Element DeleteQ(queue) ::= if (IsEmptyQ(queue)) return else remove and return the item at front of queue. *Structure 3.2: Abstract data type Queue (p.107):

CHAPTER 3 14 Boolean IsEmptyQ( queue ) ::= if ( queue ==CreateQ( max_queue_size )) return TRUE else return FALSE Element DeleteQ( queue ) ::= if (IsEmptyQ( queue )) return else remove and return the item at front of queue. *Structure 3.2: Abstract data type Queue (p.107)

Queue CreateQ(max_queue_size) ::= # define MAX_QUEUE_SIZE 100/* Maximum queue size */ typedef struct { int key; /* other fields */ } element; element queue[MAX_QUEUE_SIZE]; int rear = -1; int front = -1; Boolean IsEmpty(queue) ::= front == rear Boolean IsFullQ(queue) ::= rear == MAX_QUEUE_SIZE-1:

CHAPTER 3 15 Queue CreateQ( max_queue_size ) ::= # define MAX_QUEUE_SIZE 100/* Maximum queue size */ typedef struct { int key; /* other fields */ } element; element queue[MAX_QUEUE_SIZE]; int rear = -1; int front = -1; Boolean IsEmpty(queue) ::= front == rear Boolean IsFullQ(queue) ::= rear == MAX_QUEUE_SIZE-1 Implementation 1: using array

void addq(int *rear, element item) { if (*rear == MAX_QUEUE_SIZE_1) { queue_full( ); return; } queue [++*rear] = item; } *Program 3.3: Add to a queue (p.108):

CHAPTER 3 16 void addq(int *rear, element item) { if (*rear == MAX_QUEUE_SIZE_1) { queue_full( ); return; } queue [++*rear] = item; } *Program 3.3: Add to a queue (p.108) Add to a queue

element deleteq(int *front, int rear) { if ( *front == rear) return queue_empty( ); /* return an error key */ return queue [++ *front]; } *Program 3.4: Delete from a queue(p.108) :

CHAPTER 3 17 element deleteq(int *front, int rear) { if ( *front == rear) return queue_empty( ); /* return an error key */ return queue [++ *front]; } *Program 3.4: Delete from a queue(p.108) Delete from a queue

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CHAPTER 3 18 *Figure 3.6: Empty and nonempty circular queues (p.109) Implementation 2: regard an array as a circular queue front: one position counterclockwise from the first element rear: current end front = 0 rear = 0 [0] [1] [2] [3] [4] [5] Empty circular Queue front = 0 rear = 3 [0] [1] [2] [3] [4] [5] J1 J2 J3 Nonempty circular queue

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CHAPTER 3 19 * Figure 3.7: Full circular queues and then we remove the item (p.110) Problem: one space is left when queue is full front = 0 rear = 5 [0] [1] [2] [3] [4] [5] J1 J2 J3 J4 J5 front = 4 rear = 3 [0] [1] [2] [3] [4] [5] J7 J8 J9 J6 J5 Full Circular queue (waste one space )

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CHAPTER 3 20 避免出現 rear=front 而無法分辨 circular queue 是滿的 ? 還是空的 ? 所以最多存放 Maxsize -1 個空間 或是加入一個 COUNT 變數表示 queue 的個數 COUNT=0 ( 空 ) COUNT=Maxsize ( 滿 )

void addq(int front, int *rear, element item) { *rear = (*rear +1) % MAX_QUEUE_SIZE; if (front == *rear) /* reset rear and print error */ return; } queue[*rear] = item; } *Program 3.5: Add to a circular queue (p.110):

CHAPTER 3 21 void addq(int front, int *rear, element item) { *rear = (*rear +1) % MAX_QUEUE_SIZE; if (front == *rear) /* reset rear and print error */ return; } queue[*rear] = item; } *Program 3.5: Add to a circular queue (p.110) Add to a circular queue

element deleteq(int* front, int rear) { element item; if (*front == rear) return queue_empty( ); /* queue_empty returns an error key */ *front = (*front+1) % MAX_QUEUE_SIZE; return queue[*front]; } *Program 3.6: Delete from a circular queue (p.111) :

CHAPTER 3 22 element deleteq(int* front, int rear) { element item; if (*front == rear) return queue_empty( ); /* queue_empty returns an error key */ *front = (*front+1) % MAX_QUEUE_SIZE; return queue[*front]; } *Program 3.6: Delete from a circular queue (p.111) Delete from a circular queue

Evaluation of Expressions:

CHAPTER 3 23 Evaluation of Expressions Evaluating a complex expression in computer ((rear+1==front)||((rear==MaxQueueSize-1)&&!front)) x= a/b- c+ d*e- a*c Figuring out the order of operation within any expression A precedence hierarchy within any programming language See Figure 3.12

Evaluation of Expressions (Cont.):

CHAPTER 3 24 Evaluation of Expressions (Cont.) Ways to write expressions Infix (standard) Prefix Postfix compiler, a parenthesis-free notation Infix Postfix 2+3*4 2 3 4*+ a*b+5 ab*5+ (1+2)*7 1 2+7* a*b/c ab*c/ ((a/(b-c+d))*(e-a)*c abc-d+/ea-*c* a/b-c+d*e-a*c ab/c-de*+ac*-

Evaluation of Postfix Expressions:

CHAPTER 3 25 Evaluation of Postfix Expressions Left-to-right scan Postfix expression, Stack operands until find an operator, Meet operator, remove correct operands for this operator, Perform the operation, Stack the result Remove the answer from the top of stack

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CHAPTER 3 26 Token Stack Top [0] [1] [2] 6 6 0 2 6 2 1 / 6/2 0 3 6/2 3 1 - 6/2-3 0 4 6/2-3 4 1 2 6/2-3 4 2 2 * 6/2-3 4*2 1 + 6/2-3+4*2 0 Postfix evaluation of 6 2/3-4 2*+ Evaluation of Postfix Expressions

#define MAX_STACK_SIZE 100 #define MAX_EXPR_SIZE 100 /* max size of expression */ typedef enum{1paran, rparen, plus, minus, times, divide, mod, eos, operand} precedence; int stack[MAX_STACK_SIZE]; /* global stack */ char expr[MAX_EXPR_SIZE]; /* input string */:

CHAPTER 3 27 # define MAX_STACK_SIZE 100 #define MAX_EXPR_SIZE 100 /* max size of expression */ typedef enum{1paran, rparen, plus, minus, times, divide, mod, eos, operand} precedence; int stack[MAX_STACK_SIZE]; /* global stack */ char expr[MAX_EXPR_SIZE]; /* input string */ Assumptions: operators: +, -, *, /, % operands: single digit integer

int eval(void) { precedence token; char symbol; int op1, op2; int n = 0; /* counter for the expression string */ int top = -1; token = get_token(&symbol, &n); while (token != eos) { if (token == operand) add(&top, symbol-’0’); /* stack add */:

CHAPTER 3 28 int eval(void) { precedence token; char symbol; int op1, op2; int n = 0; /* counter for the expression string */ int top = -1; token = get_token(&symbol, &n); while (token != eos) { if (token == operand) add(&top, symbol-’0’); /* stack add */ exp: character array

else { /* remove two operands, perform operation, and return result to the stack */ op2 = delete(&top); /* stack delete */ op1 = delete(&top); switch(token) { case plus: add(&top, op1+op2); break; case minus: add(&top, op1-op2); break; case times: add(&top, op1*op2); break; case divide: add(&top, op1/op2); break; case mod: add(&top, op1%op2); } } token = get_token (&symbol, &n); } return delete(&top); /* return result */ } *Program 3.9: Function to evaluate a postfix expression (p.122):

CHAPTER 3 29 else { /* remove two operands, perform operation, and return result to the stack */ op2 = delete(&top); /* stack delete */ op1 = delete(&top); switch(token) { case plus: add(&top, op1+op2); break; case minus: add(&top, op1-op2); break; case times: add(&top, op1*op2); break; case divide: add(&top, op1/op2); break; case mod: add(&top, op1%op2); } } token = get_token (&symbol, &n); } return delete(&top); /* return result */ } * Program 3.9: Function to evaluate a postfix expression (p.122)

precedence get_token(char *symbol, int *n) { *symbol =expr[(*n)++]; switch (*symbol) { case ‘(‘ : return lparen; case ’)’ : return rparen; case ‘+’: return plus; case ‘-’ : return minus; case ‘/’ : return divide; case ‘*’ : return times; case ‘%’ : return mod; case ‘\0‘ : return eos; default : return operand; } } *Program 3.10: Function to get a token from the input string (p.123):

CHAPTER 3 30 precedence get_token(char *symbol, int *n) { *symbol =expr[(*n)++]; switch (*symbol) { case ‘(‘ : return lparen; case ’)’ : return rparen; case ‘+’: return plus; case ‘-’ : return minus; case ‘/’ : return divide; case ‘*’ : return times; case ‘%’ : return mod; case ‘\0‘ : return eos; default : return operand; } } *Program 3.10: Function to get a token from the input string (p.123)

Infix to Postfix:

CHAPTER 3 31 Infix to Postfix Method I Fully parenthesize the expression Move all binary operators so that they replace their corresponding right parentheses Delete all parentheses Examples: a/b-c+d*e-a*c ((((a/b)-c)+(d*e))-(a*c)), fully parentheses ab / c - de *+ ac *- , replace right parentheses and delete all parentheses Disadvantage inefficient, two passes

Infix to Postfix:

CHAPTER 3 32 Infix to Postfix Method II scan the infix expression left-to-right output operand encountered output operators depending on their precedence, i.e., higher precedence operators first Example: a+b*c , simple expression Token Stack Top Output [0] [1] [2] a -1 a + + 0 a b + 0 ab * + * 1 ab c + * 1 abc eos -1 abc*+

Infix to Postfix:

CHAPTER 3 33 Infix to Postfix Example: a*(b+c)*d , parenthesized expression Token Stack Top Output [0] [1] [2] a -1 a * * 0 a ( * ( 1 a b * ( 1 ab + * ( + 2 ab c * ( + 2 abc ) * 0 abc+ * * 0 abc+* d * 0 abc+*d eos * 0 abc+*d*

Infix to Postfix:

CHAPTER 3 34 Infix to Postfix Last two examples suggests a precedence-based scheme for stacking and unstacking operators isp (in-stack precedence) icp (in-coming precedence) precedence stack[MaxStackSize]; /* isp and icp arrays - index is value of precedence lparen , rparen, plus , minus, time divide, mod , eos */ static int isp []= { 0 , 19, 12 , 12, 13 , 13, 13 , 0}; static int icp []= { 20 , 19, 12 , 12, 13 , 13, 13 , 0}; See program 3.11-  (n)

void postfix(void) { /* output the postfix of the expression. The expression string, the stack, and top are global */ char symbol; precedence token; int n = 0; int top = 0; /* place eos on stack */ stack[0] = eos; for (token = get _token(&symbol, &n); token != eos; token = get_token(&symbol, &n)) { if (token == operand) printf (“%c”, symbol); else if (token == rparen ){ :

CHAPTER 3 35 void postfix(void) { /* output the postfix of the expression. The expression string, the stack, and top are global */ char symbol; precedence token; int n = 0; int top = 0; /* place eos on stack */ stack[0] = eos; for (token = get _token(&symbol, &n); token != eos; token = get_token(&symbol, &n)) { if (token == operand) printf (“%c”, symbol); else if (token == rparen ){

/*unstack tokens until left parenthesis */ while (stack[top] != lparen) print_token(delete(&top)); delete(&top); /*discard the left parenthesis */ } else{ /* remove and print symbols whose isp is greater than or equal to the current token’s icp */ while(isp[stack[top]] >= icp[token] ) print_token(delete(&top)); add(&top, token); } } while ((token = delete(&top)) != eos) print_token(token); print(“\n”); } *Program 3.11: Function to convert from infix to postfix (p.126) :

CHAPTER 3 36 /* unstack tokens until left parenthesis */ while (stack[top] != lparen) print_token(delete(&top)); delete(&top); /*discard the left parenthesis */ } else{ /* remove and print symbols whose isp is greater than or equal to the current token’s icp */ while(isp[stack[top]] >= icp[token] ) print_token(delete(&top)); add(&top, token); } } while ((token = delete(&top)) != eos) print_token(token); print(“\n”); } *Program 3.11 : Function to convert from infix to postfix (p.126) ( n) f(n)= (g(n)) iff there exist positive constants c 1 , c 2 , and n 0 such that c 1 g(n)f(n)c 2 g(n) for all n, nn 0 . f(n)= (g(n)) iff g(n) is both an upper and lower bound on f(n).

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CHAPTER 3 37 * Figure 3.17: Infix and postfix expressions (p.127) (1) evaluation (2) transformation 後序優於中序 : 去除運算子優先權 , 結合性和括號 方便 complier 計算運算子的值 , 掃描一次便可求結果 Infix Prefix a*b/c a/b- c+d*e- a*c a*( b+c)/d-g /* abc -+-/abc*de*ac -/ * a+bcdg

便當(全民,三天前):

CHAPTER 3 38 便當 ( 全民 , 三天前 ) 招牌 50 椒鹽雞排 60 黃金排骨 60 滷排骨 60 香檸雞排 65 票選兩種 ? 日式豬排 65 雞腿 70 沙茶雞腿 75 鮭魚 70 鱈魚 70 有沒有吃素的 ?

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CHAPTER 3 39 Multiple stacks and queues Two stacks m[0], m[1], …, m[n-2], m[n-1] bottommost bottommost stack 1 stack 2 More than two stacks (n) memory is divided into n equal segments boundary[stack_no] 0  stack_no < MAX_STACKS top[stack_no] 0  stack_no < MAX_STACKS

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CHAPTER 3 40 boundary[ 0] boundary[1] boundary[ 2] boundary[n] top[ 0] top[ 1] top[ 2] All stacks are empty and divided into roughly equal segments. *Figure 3.18: Initial configuration for n stacks in memory [m]. (p.129) 0 1 [ m/n ] 2[ m/n ] m-1 Initially, boundary[i]=top[i].

#define MEMORY_SIZE 100 /* size of memory */ #define MAX_STACK_SIZE 100 /* max number of stacks plus 1 */ /* global memory declaration */ element memory[MEMORY_SIZE]; int top[MAX_STACKS]; int boundary[MAX_STACKS]; int n; /* number of stacks entered by the user */ p.128 top[0] = boundary[0] = -1; for (i = 1; i < n; i++) top[i] =boundary[i] =(MEMORY_SIZE/n)*i; boundary[n] = MEMORY_SIZE-1; p.129:

CHAPTER 3 41 # define MEMORY_SIZE 100 /* size of memory */ #define MAX_STACK_SIZE 100 /* max number of stacks plus 1 */ /* global memory declaration */ element memory[MEMORY_SIZE]; int top[MAX_STACKS]; int boundary[MAX_STACKS]; int n; /* number of stacks entered by the user */ p.128 top[0] = boundary[0] = -1; for (i = 1; i < n; i++) top[i] =boundary[i] =(MEMORY_SIZE/n)*i; boundary[n] = MEMORY_SIZE-1; p.129

void add(int i, element item) { /* add an item to the ith stack */ if (top[i] == boundary [i+1]) stack_full(i); may have unused storage memory[++top[i]] = item; } *Program 3.12:Add an item to the stack stack-no (p.129) element delete(int i) { /* remove top element from the ith stack */ if (top[i] == boundary[i]) return stack_empty(i); return memory[top[i]--]; } *Program 3.13:Delete an item from the stack stack-no (p.130):

CHAPTER 3 42 void add(int i, element item) { /* add an item to the ith stack */ if (top[i] == boundary [i+1]) stack_full(i); may have unused storage memory[++top[i]] = item; } * Program 3.12: Add an item to the stack stack-no (p.129) element delete(int i) { /* remove top element from the ith stack */ if (top[i] == boundary[i]) return stack_empty(i); return memory[top[i]--]; } *Program 3.13: Delete an item from the stack stack-no (p.130)

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CHAPTER 3 43 b[0] t[0] b[1] t[1] b[i] t[i] t[i+1] t[j] b[j+1] b[n] b[i+1] b[i+2] b=boundary, t=top *Figure 3.19: Configuration when stack i meets stack i+1, but the memory is not full (p.130) Find j, stack_no < j < n such that top[j] < boundary[j+1] or, 0  j < stack_no meet 往左或右找一個空間 ( 往左) ( 往右)

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CHAPTER 3 44 0 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 0 0 1 1 1 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 0 1 1 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 0 0 1 0 0 1 1 1 1 1 0 1 1 1 1 0 entrance exit * Figure 3.8: An example maze(p.113) A Mazing Problem 1: blocked path 0: through path

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CHAPTER 3 45 * Figure 3.9: Allowable moves (p.113) a possible representation

typedef struct { short int vert; short int horiz; } offsets; offsets move[8]; /*array of moves for each direction*/:

CHAPTER 3 46 typedef struct { short int vert; short int horiz; } offsets; offsets move[8]; /*array of moves for each direction*/ a possible implementation next_row = row + move[dir].vert; next_col = col + move[dir].horiz;

#define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { short int row; short int col; short int dir; } element; element stack[MAX_STACK_SIZE];:

CHAPTER 3 47 # define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { short int row; short int col; short int dir; } element; element stack[MAX_STACK_SIZE]; Use stack to keep pass history

Initialize a stack to the maze’s entrance coordinates and direction to north; while (stack is not empty){ /* move to position at top of stack */ <row, col, dir> = delete from top of stack; while (there are more moves from current position) { <next_row, next_col > = coordinates of next move; dir = direction of move; if ((next_row == EXIT_ROW)&& (next_col == EXIT_COL)) success; if (maze[next_row][next_col] == 0 && mark[next_row][next_col] == 0) { :

CHAPTER 3 48 Initialize a stack to the maze’s entrance coordinates and direction to north ; while (stack is not empty){ /* move to position at top of stack */ <row, col, dir> = delete from top of stack; while (there are more moves from current position) { <next_row, next_col > = coordinates of next move; dir = direction of move; if ((next_row == EXIT_ROW)&& (next_col == EXIT_COL)) success; if (maze[next_row][next_col] == 0 && mark[next_row][next_col] == 0) {

/* legal move and haven’t been there */ mark[next_row][next_col] = 1; /* save current position and direction */ add <row, col, dir> to the top of the stack; row = next_row; col = next_col; dir = north; } } } printf(“No path found\n”); *Program 3.7: Initial maze algorithm (p.115):

CHAPTER 3 49 /* legal move and haven’t been there */ mark[next_row][next_col] = 1; /* save current position and direction */ add <row, col, dir> to the top of the stack; row = next_row; col = next_col; dir = north; } } } printf(“No path found\n”); *Program 3.7: Initial maze algorithm (p.115)

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CHAPTER 3 50 00000 1 11111 0 1 0000 1 0 11111 1 0000 1 11111 0 1 0000 1 0 11111 1 00000 * Figure 3.11: Simple maze with a long path (p.116) The size of a stack? m*p

void path (void) { /* output a path through the maze if such a path exists */ int i, row, col, next_row, next_col, dir, found = FALSE; element position; mark[1][1] = 1; top =0; stack[0].row = 1; stack[0].col = 1; stack[0].dir = 1; while (top > -1 && !found) { position = delete(&top); row = position.row; col = position.col; dir = position.dir; while (dir < 8 && !found) { /*move in direction dir */ next_row = row + move[dir].vert; next_col = col + move[dir].horiz; :

CHAPTER 3 51 void path (void) { /* output a path through the maze if such a path exists */ int i, row, col, next_row, next_col, dir, found = FALSE; element position; mark[1][1] = 1; top =0; stack[0].row = 1; stack[0].col = 1; stack[0].dir = 1; while (top > -1 && !found) { position = delete(&top); row = position.row; col = position.col; dir = position.dir; while (dir < 8 && !found) { /*move in direction dir */ next_row = row + move[dir].vert; next_col = col + move[dir].horiz;

if (next_row==EXIT_ROW && next_col==EXIT_COL) found = TRUE; else if ( !maze[next_row][next_col] && !mark[next_row][next_col] { mark[next_row][next_col] = 1; position.row = row; position.col = col; position.dir = ++dir; add(&top, position); row = next_row; col = next_col; dir = 0; } else ++dir; } } :

CHAPTER 3 52 if (next_row==EXIT_ROW && next_col==EXIT_COL) found = TRUE; else if ( !maze[next_row][next_col] && !mark[next_row][next_col] { mark[next_row][next_col] = 1; position.row = row; position.col = col; position.dir = ++dir; add(&top, position); row = next_row; col = next_col; dir = 0; } else ++dir; } }

if (found) { printf(“The path is :\n”); printf(“row col\n”); for (i = 0; i <= top; i++) printf(“ %2d%5d”, stack[i].row, stack[i].col); printf(“%2d%5d\n”, row, col); printf(“%2d%5d\n”, EXIT_ROW, EXIT_COL); } else printf(“The maze does not have a path\n”); } *Program 3.8:Maze search function (p.117):

CHAPTER 3 53 if (found) { printf(“The path is :\n”); printf(“row col\n”); for (i = 0; i <= top; i++) printf(“ %2d%5d”, stack[i].row, stack[i].col); printf(“%2d%5d\n”, row, col); printf(“%2d%5d\n”, EXIT_ROW, EXIT_COL); } else printf(“The maze does not have a path\n”); } *Program 3.8: Maze search function (p.117)