07c backtracking

8/13/2019 07c BackTracking

1/28

Dr. Ahmad R. Hadaegh

A.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 1

More on

Stacks and Queues

and

Backtracking


2/28



As we mentioned before, two common introductory

Abstract Data Type (ADT) that worth studying are Stack

and Queue

Many problems (or parts of problems) can be thought of as

stack problems, queue problems, or other problems that can

be solved with the help of another (perhaps user-defined)ADT

You need to decide what ADTs you need, build them, testthem, then tackle your larger problem


3/28



Stack ADT

A stack is an ADT with specific properties and operations

Stack properties are: Only one item can be accessed at a time

Always the last item inserted is the first to come out (LIFO)

New items must be inserted at the top

The size of the stack is dynamic

The stack must be able to grow and shrink

Normally, items stored on the stack are homogeneous (i. e.they are all integers or all are string, etc


4/28



Stack Operations

Stack operations are:

Push:Puts a new item on the top of the stack. Cant pushinto a full stack

Pop:Removes the top item from stack. Cant pop an

empty stack

Top:Examine top item without modifying stack. Cant topan empty stack

Init:Set existing stack to empty

Full:Returns true if the stack is full

Empty:Returns true if the stack is empty


5/28



Problem Solving Using Stack:

Bracket Checking

Many problems can be solved using stacks

Given a mathematical expression with 3 types of brackets ( ) [ ]

{ }, we want to determine if the brackets are properly matched

For example, bracketing in the following expression is CORRECT

{ 5 * (8-2) } / { [4 * (3+2)] + [5 * 6] }

However, the bracketing in the following two expressions areINCORRECT

{ (A+C) * D

[A + {B+C} )


6/28



This problem is easily solved with a stack

Algorithm:

Scan expression from left to right

Each time a left bracket is encountered, push it onto thestack

When a right bracket is encountered, compare it to the topitem on the stack

If its a match, pop the stack

If not, report illegal bracketing

If the stack is prematurely empty, report illegal bracketing

If there are items left on the stack after the expression hasbeen scanned, report illegal bracketing


7/28Dr. Ahmad R. HadaeghA.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 7

bool BracketMaching (string expression)

{

char curr, temp;

Stack bracket_stack;

for (int i=0; i< stg.size(); i++){

curr = stg[i];

if (curr == ( || curr == [ || curr == {)

bracket_stack.push (curr);

else if (curr == ) || curr == ] || curr == })if (bracket_stack.empty() ) then

return false;

else if (bracket_stack.pop() does not match with curr)

return false;

end if

end if

} // for loop

return true;

}

This program assumes

that we have the

following routines

available

bool empty ();

push (item);

char top ();




Evaluation of An Expression

Consider the problem of evaluating an arithmetic expression usinga stack

Traditionally, when we write an arithmetic expression as follows:

2 + 3 * (6 - 5)

This is known as infix notation

Operators are in-between the operands

When computing arithmetic expressions, its sometimes useful to

represent them using postfix (operator after) or prefix (operator

before) notation



Prefix Infix Postfix

- 6 1

* + 4 3 2

/ + 2 3 - 9 4

6 - 1

(4 + 3) * 2

(2 + 3) / (9 - 4)

6 1 -

4 3 + 2 *

2 3 + 9 4 - /

Its easy to evaluate postfix and prefix expressions using astack in part because no brackets are necessary



prefix postfix

* + 4 3 2

4

3

2

4 3 + 2 *

3

4

evaluate

147

2+

*+

2

714*

Push 2

Push 3Push 4Pop 4Pop 3Apply + and get 7Push 7Pop 7Pop 2Apply * and get 14

Push 14

push 4

push 3

Pop 3

Pop 4

Apply addition and get 7

Push 7

Push 2Pop 2

Pop 7

Apply * and get 14

Push 14



Algorithm for evaluating postfix:

Parse expression from left to right

When an operand is encountered, push it onto the stack

When an operator is encountered, pop the top twooperands, apply the operator, and push the result onto thestack

When the expression is completely scanned, the finalresult will be on the stack

What if we want to convert an expression from infix to

postfix?

For example:2 + 3 * [(5 - 6) / 2] becomes

2 3 5 6 - 2 / * +



Use the following algorithm to change infix to postfix:

Scan infix string from left to right

Each time an operand is encountered, copy it to the output

When a bracket is encountered, check its orientation. Push left brackets onto the stack If its a right bracket, pop all operators out of the stack and

copy them to output until a matching left bracket isencountered. Discard the left bracket.

When an operator is encountered, check the top item of thestack.

If the priority is >= the current operator, pop the topoperator and copy it to output

Continue until an operator of lesser priority is encountered

or until stack is empty Assume left brackets have lowest priority Finally, push current operator onto the stack

When the end of the expression is reached, copy the remainingstack contents to output in the order popped



Consider the following example that is scanned from left to right

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

2 is placed on the output

line because it is an

operand

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

+ is an operator and it is placedin the stack. Note there is no

operator on the top of the stack

with priority >= the + operator+

2

2



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

3 is placed on the output

line because it is an

operand+

2 3

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

* is an operator and it is placedin the stack without taking

anything out of the stack. Note

that * has higher priority than +

*

2 3

+



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

Left brackets, no matter whattype of bracket it is, it is placed

in the stack*

2 3

+

[

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

Again another left encountered

and it is placed in the stack*

2 3

+

[

(



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

5 is placed on the output line

because it is an operand*

2 3 5

+

[(

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

Minus sign (-) is placed in the

stack. Note that the top of the

stack is not an operator to

compare minus sign with it

*

2 3 5

+

[

(

-



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:



2 3 5 6

+

[

(

-

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

( is a closing roundbracket. Keep onpopping from the stack till we you find

the matching opening roundbracket.

DO NOT COPY THE OPEN ROUND

BRACKET TO THE OUTPUT

*

2 3 5 6 -

+

[



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

Division (/) is placed in the stack.Note that the top of the stack is

not an operator to compare

division sign with it

*

2 3 5 6 -

+

[

/

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:



2 3 5 6 - 2

+

[

/



Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

] is a closingsquarebracket. Keep on

popping from the stack till we you find

the matching openingsquarebracket.

DO NOT COPY THE OPEN SQUARE

BRACKET TO THE OUTPUT

*

2 3 5 6 - 2 /

+

Expression:

2 + 3 * [ ( 56 ) / 2 ]

StackOutput:

We are done. So keep popping

from the stack and place thecharacters to the output line

2 3 5 62 / * +

What we have on the output line is the corresponding postfixexpression




Adding large numbers using stack

Another example of stack application is adding two large numberstogether.

Note that the largest magnitude of integers are limited so we are

not able to normally add numbers like:

243678489505654679857465758598595950949463535 +

462668435748955758056548577946

Since the integer variables cannot hold such a big numbers we can

use other tools to add these numbers together

The next slide shows the algorithm of adding two numbers

together using stack



Algorithm for adding two numbers

Read the numerals of the first number and store the numbers

corresponding to them on the stack;

Read the numerals of the second number and store them in anotherstack;

Let R = 0;

While at least one stack is not empty Pop a number from each stack and add them to R; Push the unitpart of R on the result stack;

Store carrypart of R in R overwriting the old value;

Push R on the result stack if it is not zero

Pop numbers from the result stack and display them

2 1



592

+ 3784

4376

29

5

4

8

7

3

2

+ 4

6

9

+ 8

17

15

+ 713

9

5

8

7

3

6

5

7

3

7

6

3

3

7

6

Stack 1

Stack 2

Result

Stack

4

3

7

6

+ =6 + =17 + =13 + =4

1

+ 3

4



Problem Solving Using Stack:Exiting A Maze Consider the problem of a trapped mouse that tries to find its way

to an exit in a maze

The mouse systematically moves and if it hits a wall, it backtracksand tries another path.

The maze is implemented as a two dimensional character array inwhich passages are marked with oand walls are marked with

1. When a cell is visited, it is marked with .so that the mouse does

not try that again.

In general, the mouse always takes right cell first, if it cannot

succeed, it takes left and if it still fails, it takes down and finally ittries the upper cell to reach its destination.

As long as it finds a space it moves in that path and tries its best tofind the exit door.

We represent the mouse with Mand exit door with E



In order to ensure that the mouse does not fall off the maze, wecreate a wall around the maze.

Thus the mouse will be trapped in the maze until it finds the door.

If the mouse tries all the possible solutions and still cannot find theexit door, we report a failure; otherwise, we report a success.

The algorithm is:

Initialize stack, exitCell, entryCell, currentCell = entryCell;Create a wall of 1s around the mazeWhile currentCell is not exitCell

Mark currentCell as visited;Push onto the stack the unvisited neighbors of current Cell in theorder of up, down left and rightIf stack is empty

Report failureElse

Pop off a cell from the stack and make it currentCellsuccess



1 1 1 1 1 1

1 1 1 o o 1

1 o o o E 1

1 o o M 1 1

1 1 1 1 1 1

0

1

2

3

4

0 1 2 3 4 5

(3, 2)

(2, 3)

Push the Up and

Left cell into stack

Start with the following maze

where the mouse is at position

(3, 3) and the exit door is at(2, 4)

1 1 1 1 1 1

1 1 1 o o 1

1 o o o E 1

1 o o M 1 1

1 1 1 1 1 1

0

1

2

34

0 1 2 3 4 5Make the current

position visited (.)

Pop the first element

of the stack

Move the mouse to

that position

0 1 2 3 4 5


26/28

Dr. Ahmad R. HadaeghA.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 26

Push the Up cell

into stack1 1 1 1 1 1

1 1 1 o o 1

1 o o o E 1

1 M . . 1 1

1 1 1 1 1 1

0

12

3

4




of the stack

Move the mouse to

that position

(3, 1)(2, 2)

(2, 3)

Push the Up and

Left cells into stack1 1 1 1 1 1

1 1 1 o o 1

1 o o o E 1

1 o M . 1 1

1 1 1 1 1 1

0

1

2

3

4



Pop the first elementof the stack

Move the mouse to

that position

(2, 1)

(2, 2)

(2, 3)

0 1 2 3 4 5


27/28

Dr. Ahmad R. HadaeghA.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 27

Push the Right

cell into stack1 1 1 1 1 1

1 1 1 o o 1

1 . M o E 1

1 . . . 1 1

1 1 1 1 1 1

0

12

3

4




of the stack

Move the mouse to

that position

(2, 2)(2, 2)

(2, 3)

Push the Right


1 1 1 o o 1

1 M o o E 1

1 . . . 1 1

1 1 1 1 1 1

0

1

2

3

4




Move the mouse to

that position

(2, 3)

(2, 2)

(2, 3)

0 1 2 3 4 5


28/28

28

Push the UP


1 1 1 o o 1

1 . . . E/M 1

1 . . . 1 1

1 1 1 1 1 1

0

12

3

4

0 1 2 3 4 5

The exit door is found

(2, 4)(1, 3)

(2, 2)

(2, 3)

Push the UP and

Right cell into

stack

1 1 1 1 1 1

1 1 1 o o 1

1 . . M E 1

1 . . . 1 1

1 1 1 1 1 1

0

1

2

3

4




Move the mouse to

that position

(1, 3)

(2, 2)

(2, 3)

07c backtracking

Documents