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- Q. Given a 2d grid map of '1's (land) and '0's (water),count the number of islands.An island is surrounded by water and is formed byconnecting adjacent lands horizontally or vertically.You may assume all four edges of the grid are all surrounded by water. Example 1: 11110110101100000000Answer: 1 Example 2: 11000110000010000011Answer: 3""" def num_islands(grid): count = 0 for i in range(len(grid)): for j, col in enumerate(grid[i]): if col == 1: dfs(grid, i, j) count += 1 Please code it. .I use this code but I cant get the first graph from the picture, I get second graph. Why?? Code: t=(0:0.5:20); % t is a vector which contains time values from 0 to 20s with a step size of 0.5a=sin(t); % a is a sin(t) functionx=pi/6:15:pi/2; % phase difference φ for 30,45,60,90b=sin(t+x); % b is a sin(t+x) funcion plot(t,a) % plot the sin(t) function with respect to ttitle('sin(t) versus Time') % add title to graphxlabel('time') % add x-label to the graphylabel('sin(t)') % add y-label to the graphlegend('sin(t)') % add legend to the graph figure % to plot multiple linesplot(t,a,t,b,'--') % plot the sint(t) & sin(t+x) function with respect to ttitle('Function versus Time') % add title to the graphxlabel('time') % add x-label to the graphylabel('Function') % add y-label to the graphlegend('sin(t)','sin(t+x)') % add legend to the graph % Create a figure divided into four subplotssubplot(2,2,1) % divides the current figure into an 2-by-2 grid and creates axes in the position…I need a Multidimensional matrix to be made out of this input file for a DFS. A to B are points on a graph and 10 is the distance. A B 10 C 15B C 10 D 15C A 5D A 10 B 20 E 10E A 25 X 33
- One can manually count path lengths in a graph using adjacency matrices. Using the simple example below, produces the following adjacency matrix: A B A 1 1 B 1 0 This matrix means that given two vertices A and B in the graph above, there is a connection from A back to itself, and a two-way connection from A to B. To count the number of paths of length one, or direct connections in the graph, all one must do is count the number of 1s in the graph, three in this case, represented in letter notation as AA, AB, and BA. AA means that the connection starts and ends at A, AB means it starts at A and ends at B, and so on. However, counting the number of two-hop paths is a little more involved. The possibilities are AAA, ABA, and BAB, AAB, and BAA, making a total of five 2-hop paths. The 3-hop paths starting from A would be AAAA, AAAB, AABA, ABAA, and ABAB. Starting from B, the 3-hop paths are BAAA, BAAB, and BABA. Altogether, that would be eight 3-hop paths within this graph. Write a program…I use this code but I cant get the graph from the picture. Why?? Code: t=(0:0.5:20); % t is a vector which contains time values from 0 to 20s with a step size of 0.5a=sin(t); % a is a sin(t) functionx=pi/6:15:pi/2; % phase difference φ for 30,45,60,90b=sin(t+x); % b is a sin(t+x) funcion plot(t,a) % plot the sin(t) function with respect to ttitle('sin(t) versus Time') % add title to graphxlabel('time') % add x-label to the graphylabel('sin(t)') % add y-label to the graphlegend('sin(t)') % add legend to the graph figure % to plot multiple linesplot(t,a,t,b,'--') % plot the sint(t) & sin(t+x) function with respect to ttitle('Function versus Time') % add title to the graphxlabel('time') % add x-label to the graphylabel('Function') % add y-label to the graphlegend('sin(t)','sin(t+x)') % add legend to the graph % Create a figure divided into four subplotssubplot(2,2,1) % divides the current figure into an 2-by-2 grid and creates axes in the position 1y1=cos(t+(pi/6)); %y1 is cos(t+φ)…Consider the vector: x = c(1,2,3,4). What is the value of (x+2)[(!is.na(x)) & x > 0]? What do you find?
- a. Build an adjacency matrix ? for this map. b. How many paths of length 2 from V5 to V1 exist? c. How many paths of length 3 from V5 to V1 exist?Let l be a line in the x-y plane. If l is a vertical line, its equation is x 5a for some real number a. Suppose l is not a vertical line and its slope is m. Then the equation of l is y 5mx 1b, where b is the y-intercept. If l passes through the point (x0, y0,), the equation of l can be written as y 2y0 5m(x 2x0 ). If (x1, y1) and (x2, y2) are two points in the x-y plane and x1 ≠ x2, the slope of line passing through these points is m 5(y2 2y1 )/(x2 2x1 ). Write a program that prompts the user two points in the x-y plane. The program outputs the equation of the line and uses if statements to determine and output whether the line is vertical, horizontal, increasing, or decreasing. If l is a non-vertical line, output its equation in the form y 5mx 1b.Hippity hoppity, abolish loopity def frog_collision_time(frog1, frog2): A frog hopping along on the infinite two-dimensional lattice grid of integers is represented as a 4- tuple of the form (sx, sy, dx, dy) where (sx, sy) is its starting position at time zero, and (dx, dy) is its constant direction vector for each hop. Time advances in discrete integer steps 0, 1, 2, 3, ... so that each frog makes one hop at every tick of the clock. At time t, the position of that frog is given by the formula (sx+t*dx, sy+t*dy) that can be nimbly evaluated for any t. Given two frogs frog1 and frog2 that are guaranteed to initially stand on different squares, return the time when both frogs hop into the same square. If these two frogs never simultaneously arrive at the same square, return None. This function should not contain any loops whatsoever. The result should be calculated using conditional statements and integer arithmetic. Perhaps the best way to get cracking is to first solve a simpler…
- Let y be a column vector: y = [1, 2, 3, 4, 5, 6] so that y.shape = (6,1). Reshape the vector into a matrix z using the numpy.array.reshape and (numpy.array.transpose if necessary) to form a new matrix z whose first column is [1, 2, 3], and whose second column is [4, 5, 6]. Print out the resulting array z.Let l be a line in the x-yplane. If l is a vertical line, its equation is x = a for some real number a. Suppose l is not a vertical line and its slope is m. Then the equation of l is y = mx + b, where b is the y-intercept. If l passes through the point (x₀, y₀), the equation of l can be written as y - y₀ = m(x - x₀). If (x₁, y₁) and (x₂, y₂) are two points in the x-y plane and x₁ ≠ x₂, the slope of line passing through these points is m = (y₂ - y₁)/(x₂ - x₁). Instructions Write a program that prompts the user for two points in the x-y plane. Input should be entered in the following order: Input x₁ Input y₁ Input x₂Consider a vector ('L', 'M', 'M', 'L', 'H', 'M', 'H', 'H'), where H stands for ‘high’, L stands for ‘low’, and M for ‘medium’ in rstudio. a. Convert this vector into an unordered factor. b. Convert this vector into an ordered factor with Low < Medium < High.