TRUE OR FALSE Finding a matching string can be effectively solved using a divide-and-conquer algorithm.

agents: An agent is trying to eat all the food in a maze that contains obstacles, but he now has the help of his friends! An agent  cannot occupy  a  square that has an obstacle. There are initially k pieces of food (represented by dots), at positions (f1,...,fk). There are also n agents at positions (p1,...,pn). Initially, all agents start at random locations in the maze. Consider a search problem in which all agents move simultaneously ;that is, in each step each agent moves into some adjacent position (N, S, E, or W, or STOP). Note that any number of agents may occupy the same position.           Figure 1: A maze with 3 agents For each of the following heuristics, indicate (yes/no) whether or not it is h1: The number of dots (representing food) remaining. [ True, False ]. h2(s)=0, where s is a state node. [ True, False ]. h3(s)=1, where s is a state node. [ True, False ].

Computer Science Q2.Write a java program to implement the concept of Horspool string matching algorithm to find the pattern in the Text Text STRING MATCHING Pattern : ING Java code i want solved very quickly

def solve_n_queens(n):    def is_safe(board, row, col):        # Verificar la columna        for i in range(row):            if board[i][col] == 1:                return False               # Verificar la diagonal izquierda        i, j = row, col        while i >= 0 and j >= 0:            if board[i][j] == 1:                return False            i -= 1            j -= 1        # Verificar la diagonal derecha        i, j = row, col        while i >= 0 and j < n:            if board[i][j] == 1:                return False            i -= 1            j += 1        return True    def solve(board, row):        if row >= n:            solution = []            for r in board:                solution.append("".join(['R' if x == 1 else '.' for x in r]))            solutions.append(solution)            return               for col in range(n):            if is_safe(board, row, col):                board[row][col] = 1                solve(board, row + 1)…