The document discusses arrays and sparse matrices as data structures. It defines array and sparse matrix abstract data types, including methods for creating, accessing, and manipulating the structures. Examples are given of representing polynomials using arrays or as a sparse matrix to illustrate different implementations of these data structures.
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Breadth First Search & Depth First SearchKevin Jadiya
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This Presentation expalins what are data structures and linked lists with animations.
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DATA STRUCUTRES AND LINKED LIST
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This is a presentation on Linked Lists, one of the most important topics on Data Structures and algorithms. Anyone who is new to DSA or wants to have a theoretical understanding of the same can refer to it :D
Breadth First Search & Depth First SearchKevin Jadiya
The slides attached here describes how Breadth first search and Depth First Search technique is used in Traversing a graph/tree with Algorithm and simple code snippet.
What is Link list? explained with animationsPratikNaik41
This Presentation expalins what are data structures and linked lists with animations.
It also contains a sample implementation of Linked List using C++language.
It uses C++ Pointers to define linked lists.
DATA STRUCUTRES AND LINKED LIST
WHAT ARE DATA STRUCTURES?
Classification Of Data Structures ?
WHAT IS LINKED LIST?
WHY USE LINKED LIST?
INSERTION AND DELETION IN LINKED LIST
LINKED LIST CODE
This is a presentation on Linked Lists, one of the most important topics on Data Structures and algorithms. Anyone who is new to DSA or wants to have a theoretical understanding of the same can refer to it :D
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Sectoral targets and attacks as well as the cost of ransom
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Major cyber events in 2024
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https://alandix.com/academic/papers/synergy2024-epistemic/
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Epistemic Interaction - tuning interfaces to provide information for AI support
Arrays
1. CSCE 3110
Data Structures &
Algorithm Analysis
Rada Mihalcea
http://www.cs.unt.edu/~rada/CSCE3110
Arrays
2. Arrays
Array: a set of pairs (index and value)
data structure
For each index, there is a value associated with
that index.
representation (possible)
implemented by using consecutive memory.
3. Objects: A set of pairs <index, value> where for each value of index
there is a value from the set item. Index is a finite ordered set of one or
more dimensions, for example, {0, … , n-1} for one dimension,
{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)} for two dimensions,
etc.
Methods:
for all A ∈ Array, i ∈ index, x ∈ item, j, size ∈ integer
Array Create(j, list) ::= return an array of j dimensions where list is a
j-tuple whose kth element is the size of the
kth dimension. Items are undefined.
Item Retrieve(A, i) ::= if (i ∈ index) return the item associated with
index value i in array A
else return error
Array Store(A, i, x) ::= if (i in index)
return an array that is identical to array
A except the new pair <i, x> has been
inserted else return error
The Array ADT
4. Arrays in C
int list[5], *plist[5];
list[5]: five integers
list[0], list[1], list[2], list[3], list[4]
*plist[5]: five pointers to integers
plist[0], plist[1], plist[2], plist[3],
plist[4]
implementation of 1-D array
list[0] base address = α
list[1] α + sizeof(int)
list[2] α + 2*sizeof(int)
list[3] α + 3*sizeof(int)
5. Arrays in C (cont’d)
Compare int *list1 and int list2[5] in C.
Same: list1 and list2 are pointers.
Difference: list2 reserves five locations.
Notations:
list2 - a pointer to list2[0]
(list2 + i) - a pointer to list2[i] (&list2[i])
*(list2 + i) - list2[i]
6. Address Contents
1228 0
1230 1
1232 2
1234 3
1236 4
Example:
int one[] = {0, 1, 2, 3, 4}; //Goal: print out
address and value
void print1(int *ptr, int rows)
{
printf(“Address Contentsn”);
for (i=0; i < rows; i++)
printf(“%8u%5dn”, ptr+i, *(ptr+i));
printf(“n”);
}
Example
7. ne
n
e
xaxaxp ++= ...)( 1
1
Polynomials A(X)=3X20
+2X5
+4, B(X)=X4
+10X3
+3X2
+1
Other Data Structures
Based on Arrays
•Arrays:
•Basic data structure
•May store any type of elements
Polynomials: defined by a list of coefficients and
exponents
- degree of polynomial = the largest exponent in a
polynomial
8. Polynomial ADT
Objects: a set of ordered pairs of <ei,ai>
where ai in Coefficients and
ei in Exponents, ei are integers >= 0
Methods:
for all poly, poly1, poly2 Polynomial, coef Coefficients, expon
Exponents
Polynomial Zero( ) ::= return the polynomial p(x) = 0
Boolean IsZero(poly) ::= if (poly) return FALSE
else return TRUE
Coefficient Coef(poly, expon) ::= if (expon poly) return its
coefficient else return Zero
Exponent Lead_Exp(poly) ::= return the largest exponent in
poly
Polynomial Attach(poly,coef, expon) ::= if (expon poly) return error
else return the polynomial poly
with the term <coef, expon>
inserted
9. Polyomial ADT (cont’d)
Polynomial Remove(poly, expon) ::= if (expon poly) return the
polynomial poly with the term
whose exponent is expon deleted
else return error
Polynomial SingleMult(poly, coef, expon)::= return the polynomial
poly • coef • xexpon
Polynomial Add(poly1, poly2) ::= return the polynomial
poly1 +poly2
Polynomial Mult(poly1, poly2) ::= return the polynomial
poly1 • poly2
10. Polynomial Addition (1)
#define MAX_DEGREE 101
typedef struct {
int degree;
float coef[MAX_DEGREE];
} polynomial;
Addition(polynomial * a, polynomial * b, polynomial* c)
{
…
}
advantage: easy implementation
disadvantage: waste space when sparse
Running time?
11. Use one global array to store all polynomials
Polynomial Addition (2)
2 1 1 10 3 1
1000 0 4 3 2 0
coef
exp
starta finisha startb finishb avail
0 1 2 3 4 5 6
A(X)=2X1000
+1
B(X)=X4
+10X3
+3X2
+1
12. Polynomial Addition (2) (cont’d)
#define MAX_DEGREE 101
typedef struct {
int exp;
float coef;
} polynomial_term;
polynomial_term terms[3*MAX_DEGREE];
Addition(int starta, int enda, int startb, int endb, int startc, int endc)
{
…
}
advantage: less space
disadvantage: longer code
Running time?
14. Sparse Matrix ADT
Objects: a set of triples, <row, column, value>, where row
and column are integers and form a unique combination, and
value comes from the set item.
Methods:
for all a, b ∈ Sparse_Matrix, x item, i, j, max_col,
max_row index
Sparse_Marix Create(max_row, max_col) ::=
return a Sparse_matrix that can hold up to
max_items = max _row max_col and
whose maximum row size is max_row and
whose maximum column size is max_col.
15. Sparse Matrix ADT (cont’d)
Sparse_Matrix Transpose(a) ::=
return the matrix produced by interchanging
the row and column value of every triple.
Sparse_Matrix Add(a, b) ::=
if the dimensions of a and b are the same
return the matrix produced by adding
corresponding items, namely those with
identical row and column values.
else return error
Sparse_Matrix Multiply(a, b) ::=
if number of columns in a equals number of rows in b
return the matrix d produced by multiplying
a by b according to the formula: d [i] [j] =
(a[i][k]•b[k][j]) where d (i, j) is the (i,j)th
element
else return error.
16. (1) Represented by a two-dimensional array.
Sparse matrix wastes space.
(2) Each element is characterized by <row, col, value>.
Sparse Matrix Representation
Sparse_matrix Create(max_row, max_col) ::=
#define MAX_TERMS 101 /* maximum number of terms +1*/
typedef struct {
int col;
int row;
int value;
} term;
term A[MAX_TERMS]
The terms in A should be ordered
based on <row, col>
17. Sparse Matrix Operations
Transpose of a sparse matrix.
What is the transpose of a matrix?
row col value row col value
a[0] 6 6 8 b[0] 6 6 8
[1] 0 0 15 [1] 0 0 15
[2] 0 3 22 [2] 0 4 91
[3] 0 5 -15 [3] 1 1 11
[4] 1 1 11 [4] 2 1 3
[5] 1 2 3 [5] 2 5 28
[6] 2 3 -6 [6] 3 0 22
[7] 4 0 91 [7] 3 2 -6
[8] 5 2 28 [8] 5 0 -15
transpose
18. (1) for each row i
take element <i, j, value> and store it
in element <j, i, value> of the transpose.
difficulty: where to put <j, i, value>?
(0, 0, 15) ====> (0, 0, 15)
(0, 3, 22) ====> (3, 0, 22)
(0, 5, -15) ====> (5, 0, -15)
(1, 1, 11) ====> (1, 1, 11)
Move elements down very often.
(2) For all elements in column j,
place element <i, j, value> in element <j, i,
Transpose a Sparse Matrix
19. Transpose of a Sparse Matrix
(cont’d)
void transpose (term a[], term b[])
/* b is set to the transpose of a */
{
int n, i, j, currentb;
n = a[0].value; /* total number of elements */
b[0].row = a[0].col; /* rows in b = columns in a */
b[0].col = a[0].row; /*columns in b = rows in a */
b[0].value = n;
if (n > 0) { /*non zero matrix */
currentb = 1;
for (i = 0; i < a[0].col; i++)
/* transpose by columns in a */
for( j = 1; j <= n; j++)
/* find elements from the current column */
if (a[j].col == i) {
/* element is in current column, add it to b */