![RANDOM GRAPH INITIALIZE
int **graph = (int **)malloc(n * sizeof(int *)); // Dynamic space allocation //
for (i=0; i<n; i++)
graph[i] = (int *)malloc(n * sizeof(int));
for(i=0; i<n; i++ )
for(j=i+1; j<n; j++)
graph[i][j] = graph[j][i] = rand()%2; // Random undirected graph initialize //
7](https://image.slidesharecdn.com/slide-240827051234-073475fc/85/Discrete-mathematics-introductionSlide-pptx-7-320.jpg)
![CALCULATING TOTAL EDGE & DEGREE
int total_edges = 0;
for(i = 0 ; i < n ; i++)
for(j = i + 1 ; j < n ; j++)
if (graph[i][j] == 1)
total_edges++; // Counting edges //
int total_degrees = 0;
for(i = 0 ; i < n ; i++ )
for(j = 0 ; j < n ; j++)
if (graph[i][j] == 1)
total_degrees++; // Computing degrees //
8](https://image.slidesharecdn.com/slide-240827051234-073475fc/85/Discrete-mathematics-introductionSlide-pptx-8-320.jpg)
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eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi
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erat. Proin a tellus sed risus lobortis sagittis eu quis est.
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Equation :- 6x106
X2
+ 0.029X - 23.51
f(n) = 6x106
n2
+ 0.029n - 23.51
For f(n)=Og(n)
f(n) ≤ c.g(n)
=> 6x106
n2
+ 0.029n - 23.51 ≤ 6x106
n2
+ 0.029 n2
- 23.51 n2
[n ≤ n2
,1 ≤ n2
]
Þ6x106
n2
+ 0.029n - 23.51 ≤ 5.99998x106
n2
f(n) = 6x106
n2
+ 0.029n - 23.51 O(n2
)](https://image.slidesharecdn.com/slide-240827051234-073475fc/85/Discrete-mathematics-introductionSlide-pptx-13-320.jpg)
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eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi
diam, sit amet lacinia massa sodales ac. Fusce
condimentum egestas nunc a maximus. Quisque et orci
purus. Proin dolor mi, ultrices sit amet ipsum placerat,
congue mattis turpis. Donec vestibulum eros eget mauris
dignissim, ut ultricies dolor viverra. Phasellus efficitur ante
nec sem convallis, in ornare est accumsan. Lorem ipsum
dolor sit amet, consectetur adipiscing elit. Ut gravida eros
erat. Proin a tellus sed risus lobortis sagittis eu quis est.
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Your text here
int total_edges = 0;
for(i = 0 ; i < n ; i++)
for(j = i + 1 ; j < n ; j++)
if (graph[i][j] == 1)
total_edges++; // Counting edges //
int total_degrees = 0;
for(i = 0 ; i < n ; i++ )
for(j = 0 ; j < n ; j++)
if (graph[i][j] == 1)
total_degrees++; // Computing degrees //
int flag;
if(total_degrees == 2 * total_edges) // Verifying Handshaking theorem //
flag = 1;
else
flag = 0;
For assignment k1
For nested loop n(2n-1)
For If statement k2
For Summation and assignment k3
Total = k1 + k0 n(2n-1) + k2 + k3
For assignment k4
For nested loop n(2n-1)
For If statement k k5
For Summation and assignment k6
Total = k4 + k7 n(2n-1) + k5 + k6
For If statement k8
Total = k8
For total time = k1 + k0 n(2n-1) + k2 + k4 + k7 n(2n-1) + k5 + k6 + k8
= k + k (2n2
– n)
0(n2
) [After omitting constant and taking most determine value ]](https://image.slidesharecdn.com/slide-240827051234-073475fc/85/Discrete-mathematics-introductionSlide-pptx-15-320.jpg)
Discrete mathematics introductionSlide.pptx
- 1. 1 DISCREATE MATH M i n i - P r o j e c t Supervised by Dr. Mohammad Salah Uddin
- 3. 3 WORK DETAILS C Program Undirected Graph Adjacency Matrix Computational Time F(n) F(n)
- 4. GOOD AFTERNOON 4 Good Morning Good Day Happy Morning
- 5. 5 Vertices People, Pages, Photo, Comments, Places, Group Edges User Posting a video, Your Greeting to your beloved
- 6. 6 GRAPH USED Facebook Graph API Google Knowledge Graph Flight Network Recommendation Engines. Path Optimization Algorithms. Scientific Computations.
- 7. RANDOM GRAPH INITIALIZE int **graph = (int **)malloc(n * sizeof(int *)); // Dynamic space allocation // for (i=0; i<n; i++) graph[i] = (int *)malloc(n * sizeof(int)); for(i=0; i<n; i++ ) for(j=i+1; j<n; j++) graph[i][j] = graph[j][i] = rand()%2; // Random undirected graph initialize // 7
- 8. CALCULATING TOTAL EDGE & DEGREE int total_edges = 0; for(i = 0 ; i < n ; i++) for(j = i + 1 ; j < n ; j++) if (graph[i][j] == 1) total_edges++; // Counting edges // int total_degrees = 0; for(i = 0 ; i < n ; i++ ) for(j = 0 ; j < n ; j++) if (graph[i][j] == 1) total_degrees++; // Computing degrees // 8
- 9. VERIFYING THE HANDSHAKING THEOREM int flag; if(total_degrees == 2 * total_edges) // Verifying Handshaking theorem // flag = 1; else flag = 0; if(flag == 1) // Printing The Verification of Handshaking theorem // printf("ttIt holds Handshaking theorem! n"); else printf("ttIt does not holds Handshaking theorem! n"); 9
- 10. CALCULATION OF TIME struct timespec start, end; clock_gettime(CLOCK_REALTIME, &start); // Start time for computation // clock_gettime(CLOCK_REALTIME, &end); // End time for computation // double nanos = (double)(end.tv_nsec - start.tv_nsec); double ms = nanos/1000000; printf("ttComputational time = %.2f ms.nnn",ms); // Showing Computational Time // 10
- 11. Add a Footer 11 YOUR TITLE GOES HERE Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagitti Output Values for 1000, 2000, 3000, 4000, 5000 Vertices
- 12. Add a Footer 12 TITLE YOUR TITLE GOES HERE • Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. • Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicla mi diam, • sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a YOUR TITLE GOES HERE • Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. • Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi diam, • sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a
- 13. Add a Footer 13 TITLE GOES HERE S U B T I T L E G O E S H E R E Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi diam, sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a maximus. Quisque et orci purus. Proin dolor mi, ultrices sit amet ipsum placerat, congue mattis turpis. Donec vestibulum eros eget mauris dignissim, ut ultricies dolor viverra. Phasellus efficitur ante nec sem convallis, in ornare est accumsan. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. 11% 19% 17% 28% 24% CHART TITLE Jan Feb Mar Apr May Equation :- 6x106 X2 + 0.029X - 23.51 f(n) = 6x106 n2 + 0.029n - 23.51 For f(n)=Og(n) f(n) ≤ c.g(n) => 6x106 n2 + 0.029n - 23.51 ≤ 6x106 n2 + 0.029 n2 - 23.51 n2 [n ≤ n2 ,1 ≤ n2 ] Þ6x106 n2 + 0.029n - 23.51 ≤ 5.99998x106 n2 f(n) = 6x106 n2 + 0.029n - 23.51 O(n2 )
- 14. Add a Footer 14 TITLE GOES HERE S U B T I T L E G O E S H E R E Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi diam, sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a maximus. Quisque et orci purus. Proin dolor mi, ultrices sit amet ipsum placerat, congue mattis turpis. Donec vestibulum eros eget mauris dignissim, ut ultricies dolor viverra. Phasellus efficitur ante nec sem convallis, in ornare est accumsan. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. 11% 19% 17% 28% 24% CHART TITLE Jan Feb Mar Apr May Your text here For assignment, summation, , addition, divition we take contant. Because they are not dependent on n. The for loop will be executed (n – 1) times. For each for loop, two comparisons will be done. One more comparison will be needed to exit the for loop. Therefore, exactly 2(n – 1) + 1 = 2n – 1 comparisons are used. Hence, the worst-case time complexity of the algorithm is 2n – 1 O(n). For each nested loop execution will be n*(2n-1) times. Hence, the worst-case time complexity of the algorithm is 2n2 – n O(n2 ). Because the second loop will execute n time for every first n time loop.
- 15. Add a Footer 15 TITLE GOES HERE S U B T I T L E G O E S H E R E Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi diam, sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a maximus. Quisque et orci purus. Proin dolor mi, ultrices sit amet ipsum placerat, congue mattis turpis. Donec vestibulum eros eget mauris dignissim, ut ultricies dolor viverra. Phasellus efficitur ante nec sem convallis, in ornare est accumsan. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. 11% 19% 17% 28% 24% CHART TITLE Jan Feb Mar Apr May Your text here int total_edges = 0; for(i = 0 ; i < n ; i++) for(j = i + 1 ; j < n ; j++) if (graph[i][j] == 1) total_edges++; // Counting edges // int total_degrees = 0; for(i = 0 ; i < n ; i++ ) for(j = 0 ; j < n ; j++) if (graph[i][j] == 1) total_degrees++; // Computing degrees // int flag; if(total_degrees == 2 * total_edges) // Verifying Handshaking theorem // flag = 1; else flag = 0; For assignment k1 For nested loop n(2n-1) For If statement k2 For Summation and assignment k3 Total = k1 + k0 n(2n-1) + k2 + k3 For assignment k4 For nested loop n(2n-1) For If statement k k5 For Summation and assignment k6 Total = k4 + k7 n(2n-1) + k5 + k6 For If statement k8 Total = k8 For total time = k1 + k0 n(2n-1) + k2 + k4 + k7 n(2n-1) + k5 + k6 + k8 = k + k (2n2 – n) 0(n2 ) [After omitting constant and taking most determine value ]
- 16. Add a Footer 16 TITLE GOES HERE S U B T I T L E G O E S H E R E Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. Duis ut aliquam nisi. Suspendisse vehicula mi diam, sit amet lacinia massa sodales ac. Fusce condimentum egestas nunc a maximus. Quisque et orci purus. Proin dolor mi, ultrices sit amet ipsum placerat, congue mattis turpis. Donec vestibulum eros eget mauris dignissim, ut ultricies dolor viverra. Phasellus efficitur ante nec sem convallis, in ornare est accumsan. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut gravida eros erat. Proin a tellus sed risus lobortis sagittis eu quis est. 11% 19% 17% 28% 24% CHART TITLE Jan Feb Mar Apr May Your text here Theoretically we got T(n) 0(n2 ) for the nested loop. From the graph equation we got 5.99998x106 n2 . Which is T(n) k * n2 0(n2 ). Compering this with statistical result verifies the claim.
- 17. THANK YOU 17