The subset sum problem can be reliably solved optimally using the dynamic programming algorithm shown below: SubsetSum(n, W) Let B(0, w) = 0 for each w E {0, ..., W} for i ← 1 to n for w← 0 to W if w< w, then B(i, w) else B(i, w) B(i 1,w) max (w; + B(i − 1, ww₁), B(i - 1,w)) - where n is the number of requests, W is the maximum weight constraint, w, i the weight associated with request i, and B is the solution space. You are given a set of requests and their corresponding weights. The maximum weight constraint W is 12.

5.1. Trace the operation of Greedy Best-First Search, applied to the problem of getting to the goal state G from state S given below, for the following heuristic function: h,(N) = sum of the (Manhattan) distances of every tile to its goal position 341 25 876 State S 432 16 578 State G

Find f(1), f(2), f(3), f(4), and f(5) if f(n) is defined re-cursively by f(0) = 3 and for n = 0, 1, 2, ... a) f(n + 1) = −2f(n).b) f(n + 1) = 3f(n) + 7.

Machine Learning Problem Perform the optimization problem of finding the minimum of J(x) = (2x-3)2 by: (i) defining theta, J(theta), h(theta) as defined in the Stanford Machine Learning videos in Coursera; (ii) plotting J(theta) vs theta by hand then use a program (iii) determining its minimum using gradient descent approach starting from a random initial value of theta = 5. Perform the search for the minimum using the gradient descent approach by hand calculations, i.e., step 1, step 2, etc. showing your work completely

PLEASE USE PYTHONGiven a jungle matrix NxM:jungle = [ [1, 0, 0, 0], [1, 1, 0, 1], [0, 1, 0, 0], [1, 1, 1, 1,]]Where 0 means the block is dead end and 1 means the block can be used in the path fromsource to destination.Task:Starting at position (0, 0), the goal is to reach position (N-1, M-1).Your program needs to build and output the solution matrix – a 4x4 matrix with 1’s inpositions used to get from the starting position (0,0) to the ending position (N-1,M-1)with the following constraints:You can only move one space at a timeYou can only in two directions: forward and down.You can only pass thru spaces on the jungle matrix marked ‘1’If you cannot reach the ending position – print a message that you’re trapped in thejungleAlgorithm:If destination is reachedprint the solution matrixElseMark current cell in the solution matrixMove forward horizontally and recursively check if this leads to a solution If there is no solution, move down and recursively check if this leads to a solution If…