Algorithms 2 CTPS 2018 · 2019. 5. 20. · •Programming Languages,2nd edition,Tucker and Noonan ... (Fall 2004) •Computing Without Computers,A Gentle Introduction to Computer

Department of CSE,Coimbatore

Algorithms 2

CTPS 2018

LN #11

(2 Hrs)

Fibonacci Sequence

The sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

The first two terms 0 and 1 are often chosen by default.

New term (c) = preceding term (a) + term before preceding term

(b).

a = 1; b = 0 then the third number is c = a + b = 0 + 1 = 1.

Similarly consecutive terms can be generated iteratively but making

necessary assignments to the preceding and term before preceding

terms!!


Given the first two numbers

a = 0

b = 1

the third number is c = a + b = 0 + 1 = 1.

Now to generate the fourth number the following reassignments

have to be made so that we can slide the preceding terms forward.

a = b

b = c

the fourth number is c = a + b = 1 + 1 = 2.

Let us work through


0, 1, 1, 2, 3, 5, 8, 13, 21, …

So, we can generate consecutive Fibonacci numbers in pairs using

a = a+b and b = a+b.

a b

Thinking in terms of alternating a's

and b's our new a is a+b.

Our new b is our old b +

new a which is a+b again!!

Room for Improvement


Given the first two numbers

a = 0

b = 1

third and fourth numbers are

a = a + b = 0 + 1 = 1

b = a + b = 1(remember a is now 1) + 1 = 2

fifth and sixth numbers are

a = a + b = 1 + 2(remember b is now 2) = 3


seventh and eighth numbers are

a = a + b = 3 + 5(remember b is now 5) = 8


Let us check...


1.Get n the number of Fibonacci numbers to be generated.

2.Initialize the first two Fibonacci numbers as a = 0 and b = 1 .

3.Initialize count of Fibonacci number generated .

4.While less than n Fibonacci numbers generated do

a) generate the next number as a = a+b.

b) generate the next to next number as b = a+b.

c) update count of number of Fibonacci numbers generated.

5. If n is even display all numbers else display all numbers except the

last.

Algorithm to Fibonacci Sequence



Set i = 1

While (i <= n) {

Display t1

nextTerm = t1 + t2

t1 = t2

t2 = nextTerm= i +1

i

}

Number pattern


12345

1234

123

12

1

Set i = 5

While( i>=1)

{

Set j = 1

While(j<=i)

{

Display j

j = j +1

}

Goto Next Line

i = i -1

}

for(i=5;i>=1;i=i-1) {

for(j=1;j<=i;j = j+1){

Display j

}

Goto Next Line

}


Number pattern


Set i =5

While(i>=1) {

k = i

Set j = 1

While (j<=5) {

if(k <= 5) then Display k

else Display 5

k= k +1

j = j +1

}

i = i -1

Goto Next Line

}

55555

45555

34555

23455

12345


for(i=5;i>=1;i-1) {

k = i;

for(j=1;j<=5;j=j+1) {

if(k <= 5)

then Display k

else Display 5

k = k +1

}

Goto Next Line

}

Number pattern


for(i=4;i>=1;i =i-1) {

for(j=1;j<=4;j=j+1) {

if(j<=i)

then Print j

else Print “*”

}

for(j=4;j>=1;j=j-1) {

if(j<=i)

then Print j

else Print “*”

}

Goto Next Line

}

12344321

123**321

12****21

1******1

Fibonacci triangle


Enter the limit:9

1

1 1

1 1 2

1 1 2 3

1 1 2 3 5

1 1 2 3 5 8

1 1 2 3 5 8 13

1 1 2 3 5 8 13 21

1 1 2 3 5 8 13 21 34

Enter the limit:5

1

1 1

1 1 2

1 1 2 3

1 1 2 3 5

for(i=1;i<=n; i= i+1) {

a=0;

b=1;

Display b

for(j=1;j<i;j=j+1) {

c=a+b;

Display c

a=b;

b=c;

}

Goto Next Line

} Department of CSE,Coimbatore

Matrix : Formatted o/p


for(i=0; i<r; i = i+1){

for(j=0; j<c; j = j +1){

Display a[i][j])

if (j == c-1)

then Goto Nextline

}

}

Matrix Transpose


for(i=0; i<r; i = i +1){

for(j=0; j<c; j = j +1 ){

transpose[j][i] = a[i][j]

}

}

Symmetric Matrix


for(i=0;i<row;i = i+1) {

for(j=0;j<col;j = j+1){

(q[j][i]!=a[i][j])

}

}


symm_flag=1

for(i=0;i<row;i = i+1) {

for(j=0;j<col;j = j+1){

if(q[i][j]!=p[i][j])

then symm_flag= symm_flag +1

}

}

if (symm_flag is equal to 1)

then Display “Symmetric”

else Display “ Not Symmetric”

Reverse an array


Declare i, temp

for(i=0;i < size/2; i++)

{

temp=a[i]

a[i]=a[((size-1)-i)]

a[((size-1)-i)]=temp

}


First run (when number of elements are EVEN)

Array elements before reverse : 11 22 33 44 55 66

Array elements After reverse : 66 55 44 33 22 11

Second run (when number of elements are

ODD)

Array elements before reverse : 11 22 33 44 55

Array elements After reverse : 55 44 33 22 11

Given the following array

The array after duplicate removal becomes

2 1362323231582

2 1323158 6

✓What has happened to the given ordered array?

✓What, do you think, is required to achieve this?

Duplicates in ordered array


Except the first element, all unique elements have shifted their

positions to as far to left as possible in the array.

A pair wise comparison to identify unique elements must be there

and it leads to two situations.

• A pair of duplicates has been encountered.

• Two elements are different.


13623232315 136231513623232315

1 2 6543 41 2 3

j i

Consider the following case.

The position in the array where each most recently encountered

unique element must be located is determined at each instance by the

number of unique elements met so far!

Can be modeled by the assignment a[j] = a[i].

While all adjacent pairs have not been compared do

if they are not equal, shift the rightmost element to the array

position determined by the current unique element count.


Initialization choices: first element should remain untouched; first

two elements may be identical too!

• i = 1 and is a[i] equal to a[i+1]?

• i = 2 and is a[i-1] equal to a[i]? (direct termination)

The assignment a[j] = a[i] should be invoked only when duplicates

are encountered. This can be realized using a loop that compares

pair of elements until a duplicate is encountered.

while a[i-1] is not equal to a[i] do i = i +1

Loop can be terminated with a condition i<n to ensure that

termination works well even if there are no duplicates .


1.Get an array a[1,...n] of n elements where n ≥ 1.

2. Set loop index i to 2 to allow correct termination.

3.Compare successive pairs of elements until a duplicate is encountered

then set unique element count j .

4.While all pairs have not been examined do

a)If next not duplicates then

⚫ Add one to unique element count j.

⚫ Move later element of pair to array position determined by

the unique element count j.

5. Return the array.

Algorithm to find duplicates


Figure: Initial configuration of the puzzle

A playing board

• A playing board consists of 7 squares in a row numbered 0 to 6. Initial

positions of three white and three black pieces are as shown below. Pieces can

either move by sliding into an adjacent empty square or by jumping a single

adjacent piece into the empty square beyond. Aim: Swap the positions of the

black and white pieces


✓ How many instructions your algorithm has?

✓ Did you observe that there may be varying number of

instructions still all solving the puzzle?

✓ Needless to say, do you realize that lesser of number of

instructions solving the puzzle is often preferred?Department of CSE,Coimbatore

HCF and LCM of two numbers

Step 1 : Start

Step 2 : Read A, B

Step 3 : Assign A to X and B to Y

Step 4 : Divide A by B.

Step 5 : Is remainder zero? If yes, go to step 7.

Step 6 : Assign A to B and B to remainder. And go

to step 4

Step 7 : Assign B to HCF

Step 8 : Multiply X and Y and divide by HCF. And

assign result to LCM.

Step 9 : Display HCF and LCM

Step 10 : Stop

What has been described?

• Characteristics of algorithms

• Algorithms for various problems using combinations of

control structures

Credits

•Programming Languages,2nd edition,Tucker and Noonan

•www.cse.msu.edu/ Organization of Programming Languages-Cheng (Fall 2004)

•Computing Without Computers,A Gentle Introduction to Computer

• Programming,Data Structures and Algorithms,Version 0.15,Paul Curzon

•Google imagesDepartment of CSE,Coimbatore

Documents

Algorithms 2 CTPS 2018 · 2019. 5. 20. · •Programming Languages,2nd edition,Tucker and Noonan ... (Fall 2004) •Computing Without Computers,A Gentle Introduction to Computer