Stack. A list whose end points are pointed by top and bottom

Transcription

1 4. Stck

2 Stck A list whose end points re pointed by top nd bottom Insertion nd deletion tke plce t the top (cf: Wht is the difference between Stck nd Arry?) Bottom is constnt, but top grows nd shrinks! Not possible to ccess the items within stck, except top! Which item is removed first? : LIFO (Lst-In First-Out)

3 Implementtion : Use of Arry Use one-dimensionl rry Use one vrible : top Alwys initilize the top by - Ex : #define MAX_STACK_SIZE typedef struct { top=- chr key; } element; element stck [MAX_STACK_SIZE]; int top = -; /initilize top/ stck

4 Opertors: Push nd Pop PUSH : Insert n item into stck ; Time =?? void push (int top, element item) if (top >= MAX_STACK_SIZE - ) top=k+ return stck_full( ); top=k top = top + ; /increse top by / stck [top] = item; POP : Delete n item from stck; Time =?? push pop d 3 stck element pop (int top) if (top == ) return stck_empty(); item = stck [top] top = top ; /decrese top by / return item; top=k+ top=k d 3 stck

5 Push nd Pop... Ex: Push(E), Push(F), Pop, Pop top D Push(E) top E D Push(F) top F E D bottom C B A Pop bottom C B A Pop bottom C B A Reference from Dt Structures by Gilberg nd Forouzn

6 Applictions of Stck Evlution of Arithmetic Expressions Mze Problem Prsing (Pttern Mtching) Function Clls/Returns (Ex: Recursions) Queens Problem Bcktrcking

7 Evlution of Arithmetic Expression Arithmetic expression consists of two elements s follows: Opernd:, b, c,... x, y, z Opertor: +, -,,, (, ),... How to clculte the rithmetic expression? Ex: b + c Ex2: + b c Ex 3: (b + c) Ex4: / b c + d e c Point: Which opertor is computed first?

8 Evlution Method of Arithmetic Expression Generl Method (i.e., Humn Method) () Assign priority to ech opertor (2) Prenthesize opertors (if necessry) (3) Compute from the inside to the outside of prentheses This pproch is not efficient! c + b / c 3 (c + b)/ c 2 Compiler Method () Trnsltion: Convert Infix into Postfix (2) Evlution: Compute by Postfix This is efficient, nd does not need the prenthesizing.

9 Priority of Opertors in C Postfix form Prefix form opertor priority order () [] ->. 7 left-to-right left-to-right -- ++! ~ - + & sizeof 5 right-to-left (type) 4 right-to-left / % 3 left-to-right left-to-right << >> left-to-right > >= < <= left-to-right ==!= 9 left-to-right & 8 left-to-right ^ 7 left-to-right 6 left-to-right && 5 left-to-right 4 left-to-right?: 3 right-to-left = += -= /= = %= <<= >>= &= 2 right-to-left ^= =, left-to-right

10 Three Arithmetic Expressions () Infix Opertor loctes in the middle of opernds i.e., (opnd ) (optr) (opnd 2 ) Infix + b + b c ( + b) c (2) Postfix Opertor loctes t the end of opernds i.e., (opnd ) (opnd 2 ) (optr) (3) Prefix Opertor loctes t the first of opernds i.e., (optr) (opnd ) (opnd 2 ) Postfix b + b c + b + c Pretfix + b + b c + b c Note: Postfix nd Prefix do not use Prenthesis opertor!

11 stck stck Bsic Principle (According to Postfix property) opernds lwys loctes in front of their opertor. push To store the vlues of opernds, stck is needed. d + b c d Algorithm Evlution of Postfix () Scn Postfix from left to right. (just once run!) (2) If symbol X is opernd, push it to the stck (3) Otherwise (i.e., X is opertor) ) pop opernds (for X) from the stck 2) perform the specified opertion = 3) push the result bck on the stck 3 pop pop + d d +d +d 3

12 Evlution of Postfix : Procedures Ex: 6 2 / eos (eos is chrcter tht indictes the end of eqution) token 6 2 / stck [] [] [2] /2 6/2 3 6/2-3 6/ / / / top 2 When reching eos, print out pop. Here, the result lwys exists t the top of stck

13 Trnslte : Infix to Postfix () Generl Approch : Prenthesizing () Prenthesize the opertors of Infix expression ccording to their priorities. (2) Move the opertor in ech prenthesis to the right side of the prenthesis. (3) Delete ll the prentheses. Ex: / b c + d e c () Prenthesize : ( / b) c + (d e) ( c) ((( / b) c) + (d e)) ( c) (((( / b) c) + (d e)) ( c)) (2) Move : (((( b) / c) (d e) ) + ( c) ) (3) Delete : b / c d e + c This pproch requires t lest two psses; thus, not efficient!

14 Trnslte : Infix to Postfix (2) Another Approch : Use of Stck The order of opernds is not chnged. The order of opertors is chnged by their priorities To store the priorities, stck is used b c + Algorithm () Scn left to right; (Just once!) (2) If symbol X is opernd, print(x) Otherwise, (i.e., X is opertor) ) Compre the priority of X with the priority of opertor (t the top) in stck How to decide 2) Decide if we push or pop on opertors? - push X into stck, or - pop opertors from the stck + b c + stck

15 Trnslte : Infix to Postfix (3) How to decide?: Pop or Push top stck infix expression c + b d compre current symbol X In-Stck Priority (ISP) In-Coming-Priority (ICP) Two Cses; Cse : If ISP[stck [top]] < ICP[X], then Push (X) Cse 2 : If ISP[stck [top]] ICP[X], then ) Pop(top) nd Print it 2) Push (X)

16 Trnslte : Infix to Postfix (Procedures) Ex : + b c eos symbol + b c eos stck [] [] [2] top - - output b b bc bc+ Since (input) is higher thn + (in stck), we push

17 Trnslte : Infix to Postfix (Procedures) Ex : b + c eos symbol b + c eos stck [] [] [2] + + top - - output b b bc bc+ Since + (input) is lower thn (in stck), we pop

18 Trnslte : Infix to Postfix (Procedures) Ex : b / c eos token b / c eos stck [] [] [2] / / top - - output b b bc bc/ Since / (input) is the sme s (in stck), we pop.

19 Wht if prenthesis exists? Concept of Prenthesis Prenthesis cn be employed. Prenthesis hs the highest priority. y / ( x + y) z b + Left nd right prenthesis mens the strt nd the end opertion, respectively. Computtion procedures become complex if prenthesis exists. All prentheses must be deleted when trnslting infix to postfix Ex: (b + c) d Wht is the result of postfix? Process of Prenthesis If X is left prenthesis : Push If X is right prenthesis: Pop nd Print ll opertors in Stck until mtched left prenthesis is reched! (but, the left prenthesis is not printed) ( + b ) c pop ) + ( pop stck

20 Trnslte : Infix to Postfix (Procedures) Ex: (b + c) d eos token ( b + c ) d eos stck [] [] [2] ( ( ( ( + + top output b b bc bc+ bc+ bc+d bc+d Exercise: (((( / b) c) + (d e)) ( c))

21 stck stck Performnce Anlysis Infix Trnsltion Time : O(n) (n : no. of symbols in infix) Spce : it needs n extr stck Wht is the stck size required t this time? t lest one, t most (mx) no. of opertors Exmples + b / c d... (constntly incresing priority) b / c + d... (constntly decresing priority) / b c + d e c + b / c d e + c Postfix Trnsltion O(n) Evlution O(n) Evlution Time : O(n) (n : no. of symbols in postfix) Spce: it lso needs n extr stck Wht is the required stck size? The required size is (mx) no. of opernds 3 Result

22 A Nice Appliction of Stck! Mze Problem

23 Representtion of Mze Mze Problem Use two-dimensionl rry: mze[row][col] : open pths : closed pths entrnce exit

24 Possible Moving Direction Mze Problem Let X be the current position. There re 8 possible moving directions NW N NE [row-][col-] [row-][col] [row-][col+] [row][col-] W X [row][col] [row][col+] E [row+][col-] SW [row+][col] S [row+][col+] SE

25 Mze Problem Moving Directions Every position (except border) hs 8 directions If [row, col] is on border, then it hs 3 or 5 directions. To void checking for borders, we surround the mze by s. Size of Mze m p mze needs (m + 2) (p + 2) rry Entrnce : mze[][] Exit : mze[m][p] [] [2] [3] [m] [] [2] [3] [p]

26 Mze Problem We predefine possible directions; use move rry. typedef struct { short int short int vert; horiz; } offsets; offsets move [8] ; / rry of moves for ech direction / Tble of moves NW(7) N() nme dir move[dir].vert move[dir].hori N NE E SE S SW W NW W [2][4] [3][4] X (6) [3][5] [4][4] SW(5) S(4) NE() [2][5] [2][6] [4][5] [3][6] E (2) [4][6] SE(3)

27 Mze Problem If we re t position mze [row][col], then the position for the next move mze [next_row][next_col] we set; next_row = row + move[dir].vert; next_col = col + move[dir].horiz; We exmine the possible moves, strting from the north nd moving clockwise. #define MAX_STACK_SIZE typedef struct { short int row; short int col; short int dir; } element; element stck [MAX_STACK_SIZE]; Before moving, we must sve our current position; - Why? To sfely return bck to it!! - Use stck; To sve the position in stck, we define element s the bove. We need to void returning to previously tried pth - Use rry mrk[row][col] ; initilly, ll entries re - Mrk to when the position is visited

28 Exmple Cul-de-sc Mze Problem (row, col, dir) Next direction! mze = mrk = row=; col=; dir=; dir=; nextrow=; nextcol=; No! dir=; nextrow=; nextcol=2; No! dir=2; nextrow=; nextcol=2; No! dir=3; nextrow=2; nextcol=2; Yes! row=2; col=2; dir=; dir=; nextrow=; nextcol=2; No! dir=; nextrow=; nextcol=3; Yes! row=; col=3; dir=; push (3,3,3) push (2,2,2) (2,2,4) push (,,) (,,4) stck row=2; col=2; dir=2; pop pop dir=2; nextrow=2; nextcol=3; No! dir=3; nextrow=3; nextcol=3; Yes! row=3; col=3; dir=; dir=2; nextrow=3; nextcol=4; Yes!

29 Mze Problem Algorithm pth ( ) { mrk[][]=; top=; stck[].row=; stck[].col=; stck[].dir=; while (top > - &&!found) { position = pop(); row=position.row; col=position.col; dir=position.dir; while (dir < 8 &&!found) { / move in direction dir / nextrow = row + move[dir].vert; nextcol = col + move[dir].horiz; } } if (nextrow == EXIT_ROW && nextcol == EXIT_COL) found = TRUE; else if (!mze[nextrow][nextcol] &&!mrk[nextrow][nextcol] ) { mrk[nextrow][nextcol] = ; position.row = row; position.col = col; position.dir = ++dir; push (position); row = nextrow; col = nextcol; dir = ; } else {mrk[nextrow][nextcol] = ; ++dir;}

30 Mrk Mze Problem Wht is the mximum stck size? Ech position within the mze is visited no more thn once; The stck needs only the number of s in the mze. p push Performnce Time : O(m p) Spce : rrys(mze, move, mrk) + extr spce(stck) m pop (2,2,3) (,,2) stck Mze