Write a C function that calculate the following formula ( R,n,S ( double ) are input V ( double) output and returning value of function
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Write a C function that calculate the following formula ( R,n,S ( double ) are input V ( double) output and returning value of function
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- ._4Give back an you please give me a code that runs properly this time and give a proper output not an empty one Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.A region is captured by flipping all 'O's into 'X's in that surrounded region.For example, for the following board,X X X XX O O XX X O XX O X XAfter capturing all regions, the board becomesX X X XX X X XX X X XX O X XThe algorithm should take an input file, named “input.txt”, and output the resulting board in the console.Sample input.txt6X X X X X OX O X O O XX X O O O XX X X O X XX O X X O OX X X X X XThe first line is a number indicating the size of the board; in the above example, 6 means the board is .Output for in console:X X X X X OX X X X X XX X X X X XX X X X X XX X X X O OX X X X X X.We have a vector, A, with the following integers: 22, 17, 8, 11 How does the following change those integers? A[1] <- 13 Question 2 options: no change 22, 13, 8, 11 13, 17, 8, 11 13, 13, 13, 13Test.py ## test# 3 * 3 grid : boardgrid = [' *','* ','*','*','*','*','*','*','*'] # we have two players denomted as "X" and "O" play=Truecount=0 while play:# add first playeradd_x=True# loop to add playerwhile add_x:# at which postion you would like to add the playerplayer_postion_x=int(input ("Enter postion for player X: "))## add X player into the boardif grid[player_postion_x-1]!="X" and grid[player_postion_x-1]!="O":grid[player_postion_x-1]="X"add_x=False#to count the number of added playerscount=count+1else:print ("The postion has been taken")if grid[0]=="X" and grid[1]=="X" and grid[2]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9])breakelif grid[3]=="X" and grid[4]=="X" and grid[5]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9])breakelif grid[6]=="X" and grid[7]=="X" and grid[7]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9])breakelif grid[0]=="X" and grid[3]=="X" and grid[6]=="X":print…
- briefly comment the code#include <stdio.h>#include <stdlib.h>/* ADJACENCY MATRIX int source,V,E,time,visited[20],G[20][20];void DFS(int i){int j;visited[i]=1;printf(" %d->",i+1);for(j=0;j<V;j++){if(G[i][j]==1&&visited[j]==0)DFS(j);}}int main(){int i,j,v1,v2;printf("\t\t\tGraphs\n");printf("Enter the no of edges:");scanf("%d",&E);printf("Enter the no of vertices:");scanf("%d",&V);for(i=0;i<V;i++){for(j=0;j<V;j++)G[i][j]=0;}/* creating edges :P */for(i=0;i<E;i++){printf("Enter the edges (format: V1 V2) : ");scanf("%d%d",&v1,&v2);G[v1-1][v2-1]=1; } for(i=0;i<V;i++){for(j=0;j<V;j++)printf(" %d ",G[i][j]);printf("\n");}printf("Enter the source: ");scanf("%d",&source);DFS(source-1);return 0;}Test.py ## test # 3 * 3 grid : boardgrid = [' *','* ','*','*','*','*','*','*','*'] # we have two players denomted as "X" and "O" play=True count=0 while play:# add first playeradd_x=True# loop to add player while add_x:# at which postion you would like to add the player player_postion_x=int(input ("Enter postion for player X: "))## add X player into the board if grid[player_postion_x-1]!="X" and grid[player_postion_x-1]!="O":grid[player_postion_x-1]="X"add_x=False#to count the number of added playerscount=count+1 else: print ("The postion has been taken")if grid[0]=="X" and grid[1]=="X" and grid[2]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9]) breakelif grid[3]=="X" and grid[4]=="X" and grid[5]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9]) breakelif grid[6]=="X" and grid[7]=="X" and grid[7]=="X":print ("Player X win")print(grid[0:3])print(grid[3:6])print(grid[6:9]) breakelif grid[0]=="X" and grid[3]=="X" and…what are the values of the variables when runned? int j,kboolean mj = 1k = j++m = (++k == j++) ------------------ int x = 3if (x == 2)x= 0if (x==3)x++elsex+=2 --------------- how many times will it be looped for(int x = 0; x== 2; x++){//body of the loop}
- ST_MapAlgebra has the capability to work with: One pixel at a time Neighborhoods of pixels Both a and b None of the aboveCreate 4 vectors v1 = rep(0, times=6), v2=1:4, v3=1:2, v4=1:3. Calculate v1+v2, v1+v3, v1+v4, v2*v3, v2*v4. What do you find?/* * rotate4 - Rotate x to the left by 4 * Examples: rotate4(0x87654321) = 0x76543218 * Legal ops: ~ & ^ | + << >> ! * Max ops: 10 * Rating: 2 */int rotate4(int u) { return 2;} Do not use any control constructs such as if, do, while, for, switch, etc.
- In a game of tic-tac-toe, two players take turnsmarking an available cell in a 3 x 3 grid with their respective tokens (either Xor O). When one player has placed three tokens in a horizontal, vertical, or diagonalrow on the grid, the game is over and that player has won. A draw (no winner)occurs when all the cells on the grid have been filled with tokens and neitherplayer has achieved a win. Create a program for playing a tic-tac-toe game.The program prompts two players to alternately enter an X token and O token.Whenever a token is entered, the program redisplays the board on the console anddetermines the status of the game (win, draw, or continue). Here is a sample run:int a[10] = {0,1,2,3,4,5,6,7,8,9};int *m = &a[0];int *p = &a[5];int *q = &a[1]; p = (int) m + (int) p - (int) q; what is the value of of *pThis is the code that needs to be updated: .ORIG x3000 LD R0, NUM LD R1, DEN JSR GCD ADD R4, R1, #0 ADD R1, R2, #0 JSR DIVIDE ST R2, NUM ADD R0, R4, #0 JSR DIVIDE ST R2, DEN HALT; you can try other values for NUM and DEN by replacing these values in the simulatorNUM .FILL #10 DEN .FILL #3 ; Divide R0 by R1, putting quotient in R2 and remainder in R3DIVIDE ; Euclid's algorithm for GCD of R0 and R1, result in R2GCD .END