Print the square that consists of NxN cells filled with numbers from 1 to N*N in a spiral mode(see examples). Note. Use recursion for solving this problem. Input: Only one line that contains n (1<=n<=100). Output: The matrix in spiral view Samples: No Input Output 1 3 123 894 765 4 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7 3 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9

Print the square that consists of NxN cells filled with numbers from 1 to N*N in a spiral mode(see examples). Note. Use recursion for solving this problem. Input: Only one line that contains n (1<=n<=100). Output: The matrix in spiral view Samples: No Input Output 1 3 123 894 765 4 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7 3 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter8: Arrays And Strings

Section: Chapter Questions

Problem 24PE

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#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
Consider a divide-and-conquer algorithm that calculates the sum of all elements in a set of n numbers by dividing the set into two sets of n/2 numbers each, finding the sum of each of the two subsets recursively, and then adding the result. What is the recurrence relation for the number of operations required for this algorithm? Answer is f(n) = 2 f(n/2) + 1. Please show why this is the case.
#plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
Modeling the spread of a virus like COVID-19 using recursion. Let N = total population (assumed constant, disregarding deaths, births, immigration, and emigration). S n = number who are susceptible to the disease at time n (n is in weeks). I n = number who are infected (and contagious) at time n. R n = number who are recovered (and not contagiuous) at time n. The total population is divided between these three groups: N = S n + I n + R n There are several hidden assumptions here that may or may not apply to COVID-19, such as a recovered person is assumed to not be able to get the disease a second time, at least within the time window being examined. On week 0 (the start), you assume a certain small number of people have the infection (just to get things going). Everyone else is initially susceptible, and no one is recovered. There are two constants of interest: Let period = time period that it takes for an infected person to recover (recover meaning they become not infectious to…
For each, draw the recursion tree, find the height of the tree, the running time of each layer, and the sum of running times. Then use this info to find the explicit answer for T(n). a. T(n) = 2T(n/4) + √ n (n is a power of 4 (n = 4^k) for some positive integer k) b. T(n) = 9T(n/3) + n^2 (n is a power of 3 (n = 3^k) for some positive integer k) c. T(n) = T(n/2) + 1 (n is a power of 2 (n = 2^k) for some positive integer k)
A game is played by moving a marker ahead either 2 or 3 steps on a linear path. Let cn be the number of different ways a path of length n can be covered. Given, Cn =Cn-2 + Cn-3, Ci=0, c2=1, c3=1 Write a recursive algorithm to compute Cn.
with n=6 and A=(3,5,4,1,3,2). Draw the corresponding walkthrough as shown in P.173. No subsequent recursive call to Quicksort is needed.
The binomial coefficient C(N,k) can be defined recursively as follows: C(N,0) = 1, C(N,N) = 1, and for 0 < k < N, C(N,k) = C(N-1,k) + C(N - 1,k - 1). Write a function and give an analysis of the running time to compute the binomial coefficients as follows: A. The function is written recursively.
A matrix is an array of numbers of size m by n (i.e., m x n). When we multiply 2 matrices, we multiply the matching numbers, then sum them up. Multiplying a matrix of size m x n with another matrix of size n x q will result in a matrix of size m x q. An example is shown in Figure 1: [ ? ? ? ? ? ? ] x [ ? ? ? ? ] = [ ?? +?? ?? + ?? ?? + ?? ?? + ?? ?? + ?? ?? + ?? ] Matrix 3 x 2 Matrix 2 x 2 Matrix 3 x 2 Figure 1: Multiplication of Matrix 3 x 2 and 2 x 2 Write a C++ program that multiplies 2 matrices and return the sum of the transposed matrix. Your program shall include the following: a) Define two 2 dimensional arrays, M and N that each represents a matrix of size 3 x 2 and a matrix of size 2 x 2. Request the user to enter the values for the 2 arrays using advanced pointer notations (refer to your slides for the list of array pointer notations). b) Create a function named multMatrix() that passes as arguments, the values of array M and N, multiplies them and store the results in a new…
Execute the following using Array Implementation. Use diagrams in showing your final answer A ENQUEUE(A) ENQUEUE(B) ENQUEUE(C) DEQUEUE() ENQUEUE(A) ENQUEUE(B) ENQUEUE(C) DEQUEUE() DEQUEUE() ENQUEUE(A) DEQUEUE() ENQUEUE(B) ENQUEUE(C) B PUSH (x, S) PUSH (y, S) PUSH (z, S) POP (S) PUSH (x, S) PUSH (y, S) POP (S) POP (S) C INSERT(Alice, 1, ) INSERT(Bob, 2, L) INSERT(Karen, 3, L) INSERT(Ram, 4, L) DELETE( 2, L) What will happen to your array?
Define the following variant of the Fibonacci sequence called Fib3 using recursion : ● fib3(0) = 0 ● fib3(1) = 1 ● fib3(2) = 2 ● fib3(n) = fib(n-1)+fib(n-2)+fib(n-3) for all n>=3 That is, the sequence is 0,1,2,3,6,11,20,37,68,125... unsigned int fib3(unsigned int n);