1. Design vertices of the lower case
2. What is the adjacency matrix?
3. Using Python program to draw the lower case k.
4. Rotate the lower case k by 45◦counterclockwise.
5. Draw the backward lower case k.
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Ans 1N = 200005n, m, =0,0vis=[0 for i in range(N)]gr=[ for i in range(N)]v=[ for i in range(2)]def add_edges(x, y): gr[x].append(y) gr[y].append(x)def dfs(x, state): v[state].append(x) vis[x] = 1 for i in gr[x]: if (vis[i] == 0): dfs(i, state ^ 1)def Print_vertices(): if (len(v) < len(v)): for i in v: print(i,end=" ") # If even level vertices are less else: for i in v: print(i,end=" ")n = 4m = 3add_edges(1, 2)add_edges(2, 3)add_edges(3, 4)dfs(1, 0)Print_vertices()def display(n): v = n while ( v >= 0) : c = 65 for j in range(v + 1): print( chr ( c + j ), end = " ") v = v - 1 print() for i in range(n + 1): c = 65 for j in range( i + 1): print( chr ( c + j), end =" ") print()n = 5display(n)Ans 2The adjacency matrix, also called theconnection matrix, is a matrix containing rows andcolumns which is used to represent a simple labelled graph, with 0or 1 in the position of (Vi , Vj) accordingto the condition whether Vi and Vj areadjacent or not. It is a compact way to represent the finite graphcontaining n vertices of a m x m matrix M. Sometimes adjacencymatrix is also called as vertex matrix and it isdefined in the general form asIf the simple graph has no self-loops, Then the vertex matrixshould have 0s in the diagonal. It is symmetric for the undirectedgraph. The connection matrix is considered as a square array whereeach row represents the out-nodes of a graph and each columnrepresents the in-nodes of a graph. Entry 1 represents that thereis an edge between two nodes.Ans 3def display(n): v = n while ( v >= 0) : c = 65 for j in range(v + 1): print( chr ( c + j ), end = " ") v = v - 1 print() for i in range(n + 1): c = 65 for j in range( i + 1): print( chr ( c + j), end =" ") print()n = 5display(n)Ans 4import imutilstext = cv2.imread("k")Rotated_text = imutils.rotate(text, angle=45)cv2.imshow("Rotated", Rotated_text)cv2.waitKey(0)Ans 5s=input()new_str="k"for i in range (len(s)): if s[i].isupper(): new_str+=s[i].lower() elif s[i].islower(): new_str+=s[i].upper() else: new_str+=s[i]print(new_str) ...