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- Write a code (phyton) implementing the Trisection method, which is analogous to the Bisection method except that at each step, the interval is subdivided into 3 equal subintervals, instead of 2; then takes a subinterval where the function values at the endpoints are of opposite signs. (i) Is the Trisection method guaranteed to converge if for the initial interval [a, b], we have f(a)f(b) < 0? Why or why not?(ii) Starting with several different initial intervals, solve e^x = x^2 using the Trisection method to within 6 correct decimal places. Compare with the results obtained using the Bisection Method.Write a program in python Using swarm basic optimization with arguments to find a minimum of Rosenbrook function, To find the minimum of the test function y = x – 2sin(x). set up the bound for the value [-5.12, 5.12]. Use Swarm to find the value of X which minimizes the test function. Use the default hyperparameters. # hyperparameters options = {'c1': 0.5, 'c2': 0.3, 'w':0.9}Information is present in the screenshot and below. Based on that need help in solving the code for this problem in python. The time complexity has to be as less as possible (nlogn or n at best, no n^2). Apply divide-and-conquer algorithm in the problem. Make sure all test cases return expected outputs. Hint: Apply bisection method/modules Output FormatOutput contains a line with two space-separated integers W_a and W_b.- W_a is the maximum matchups won by Hamiltonia- W_b is the maximum matchups won by Burrgadia. Sample Input 03 554402410 Sample Output 03 0 Sample Input 15 4833485183 Sample Output 12 2 Sample Input 27 81028121912601319851 Sample Output 27 0 The actual code """Solves a test case Parameters:a : int - number of leaders in Hamiltoniab : int - number of leaders in Burrgadias_i : array-like - rap proficiencies of Hamiltonia's leadersr_j : array-like - rap proficiencies of Burrgadia's leaders Returns:win_a : int - number of…
- 1 Implement a function performing gradient descent using numerical solution (STOCHASTIC gradient descent) def myGradientDescentFun(X,y,learning_rate,epoches, batchsize): # Inputs: Training data, training label, learrning rate, number of epoches, batch size # Output: the final Weights # the loss history along batches1. def f(x):"""Evaluates function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likeInput(s) to the functionReturns-------out : ndarrayFunction `f`, evaluated at point(s) x"""# your code here 2. def grad_f(x):"""Evaluates gradient of a function `f(x) = x^2 - 15 \sin(x * \pi/3)`Parameters----------x : array_likePoint(s) at which gradient should be avalueatedReturns-------out : ndarrayGradient of the function `f` evaluaed at point(s) x"""# your code here # TEST 1. f(x)assert f(0.) == 0.assert np.allclose(f(np.array([2.5, 7.5])), np.array([-1.25, 41.25])) assert np.allclose(f(np.arange(-10, 10, 1)), np.arange(-10, 10, 1)**2 - 15*np.sin(np.pi/3*np.arange(-10, 10, 1)))x_min = 1.33668375assert np.allclose(f(x_min), -12.9944407) # TEST 2. grad_f(x)assert np.allclose(grad_f(0.), -15.7079)assert isinstance(grad_f(0.), float) assert np.allclose(grad_f(0), - 5*np.pi*np.cos(0)), 'Gradient at point 0 is wrong'assert np.allclose(grad_f(np.arange(-10, 10, 1)), 2*np.arange(-10,…Write a program Minesweeper.java that takes three integer command-line arguments m, n, and k and prints an m-by-n grid of cells with k mines, using asterisks for mines and integers for the neighboring mine counts (with two space characters between each cell). To do so, Generate an m-by-n grid of cells, with exactly k of the mn cells containing mines, uniformly at random. For each cell not containing a mine, count the number of neighboring mines (above, below, left, right, or diagonal)
- What is the Big-O Time Complexity Analysis of BubbleSort? LC Process Step 1: Please use Loop Analysis method to analyze the functionvoid bubbleSort(int arr[])Please explain your answer. class BubbleSort { void bubbleSort(int arr[]) { int n = arr.length; for (int i = 0; i < n-1; i++) for (int j = 0; j < n-i-1; j++) if (arr[j] > arr[j+1]) { // swap arr[j+1] and arr[i] int temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; } } /* Prints the array */ void printArray(int arr[]) { int n = arr.length; for (int i=0; i<n; ++i) System.out.print(arr[i] + " "); System.out.println(); } // Driver method to test above public static void main(String args[]) { BubbleSort ob = new BubbleSort(); int arr[] = {64, 34, 25, 12, 22, 11, 90}; ob.bubbleSort(arr); System.out.println("Sorted array"); ob.printArray(arr); } }1. Consider the following dynamic programming implementation of the Knapsack problem: #include int find_max(int a, int b) { if(a > b) return a; return b; } int knapsack(int W, int *wt, int *val,int n) { int ans[n + 1][W + 1]; int itm,w; for(itm = 0; itm <= n; itm++) ans[itm][0] = 0; for(w = 0;w <= W; w++) ans[0][w] = 0; for(itm = 1; itm <= n; itm++) { for(w = 1; w <= W; w++) { if(wt[itm - 1] <= w) ans[itm][w] = ______________; else ans[itm][w] = ans[itm - 1][w]; } } return ans[n][W]; } int main() { int w[] = {10,20,30}, v[] = {60, 100, 120}, W = 50; int ans = knapsack(W, w, v, 3); printf("%d",ans); return 0; } Which of the following lines completes the above code? A. find_max(ans[itm – 1][w – wt[itm – 1]] + val[itm – 1], ans[itm – 1][w]) B. find_max(ans[itm – 1][w –…Your task is to implement Depth-first search algorithm to solve the problem of placing 8 Queens on a chess board so none can “take” each other.• A code framework, supporting videos and html documentation of the provided code-base will be provided via Moodle in assignments section.• The implementation should accept one command-line argument at the time it is invoked for the starting position of the first queen.o It will be tested with a range of values for that input.o If the input is not an integer in the range of 0-7 it should report “Invalid Input” and exit. • You will be supplied with a partially completed file ‘main.c’ which contains the code need to read the command line and sets up the initial working candidate within the code framework.o Your task is to complete the code to implement depth-first search. o Your code must finish with a call to the function PrintFinalSolutionAndExit(); when a valid final solution stored in the variable “workingCandidate”. please solve full…
- Write a Python program for applying CNN with considering the following requirements: Shuffle the cifar 10 training set x_train( each input with corresponding label in y_train ) Select 2500 images and you must ensure that each class must contain at least 180 samples Apply CNN that keeps noisy examples. Overall, the pseudocode of CNN is as follows: The code must contain at least one lambda expression The code must contain at one comprehension list It is Not allowed to use the numpy library. Euclidean distance is the distance between two samples in Euclidean space. The formula can be expressed as: USE THIS CODE import tensorflowimport cv2from tensorflow import kerasfrom PIL import Imageimport numpy as p (x_train, y_train), (_, _) = tf.keras.datasets.cifar10.load_data() x_train = [cv2.cvtColor(image, cv2.COLOR_BGR2GRAY).flatten().tolist() for image in x_train]##x_train=np.asarray(x_train)print(len(x_train))print(len(x_train[0])) # 32*32 NOTE: PLEAS USE THE CODE I ATTACHED NOTE 2:…Write a Python program for applying CNN with considering the following requirements: Shuffle the cifar 10 training set x_train( each input with corresponding label in y_train ) Select 2500 images and you must ensure that each class must contain at least 180 samples Apply CNN that keeps noisy examples. Overall, the pseudocode of CNN is as follows: The code must contain at least one lambda expression The code must contain at one comprehension list It is Not allowed to use the numpy library. Euclidean distance is the distance between two samples in Euclidean space. The formula can be expressed as:Answer in C++, show example output please There are three buckets size X, Y, M (1<=X<=Y<=M). All three buckets are initiallyempty. Using these three buckets, we can perform any number of the following twotypes of operations. We can fill the smallest bucket (of size X) completely to the top with X units of water andpour it into the size-M bucket, as long as this will not cause the size-M bucket tooverflow. We can fill the medium bucket (of size Y) completely to the top with Y units of water andpour it into the size-M bucket, as long as this will not cause the size-M bucket tooverflow. Although we may not be able to completely fill the size-M bucket, but we can stilldetermine the maximum amount of milk we can possibly add to largest bucket. Sample input:17 25 77 Sample output:76 In this example, we fill the bucket of size 17 three times and then bucket of size 25once, accumulating a total of 76 units of water. You could use additional test case to test your program:Input: 52 791…