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- What is the termination condition (i.e. value of beta) for the following While loop? while (beta > 0 && beta < 10) { cout << beta << endl; cin >> beta; } explain pleaseCHALLENGE ACTIVITY 5.10.2: Simon says. "Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares each character of the two strings. For each matching character, add one point to user_score. Upon a mismatch, end the loop.Sample output with inputs: 'RRGBRYYBGY' 'RRGBBRYBGY' User score: 4 Use Python, please.tgn’sider the Given Loop: For(int i=0;i
- Write a while loop that prints userNum divided by 4 (integer division) until reaching 2 or less. Follow each number by a space. Example output for userNum = 160:40 10 2Maintaining the Loop Invariant (LI & not exit & codeloop → LI): Suppose that LI (the statement of the loop invariant before the iteration) is true,the exit condition exit is not, and we have executed another iteration of thealgorithm."Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings starting from index 0. For each match, add one point to userScore. Upon a mismatch, exit the loop using a break statement. Assume simonPattern and userPattern are always the same length. Ex: The following patterns yield a userScore of 4:simonPattern: RRGBRYYBGY userPattern: RRGBBRYBGY #include <iostream>#include <string>using namespace std; int main() { string simonPattern; string userPattern; int userScore; int i; userScore = 0; cin >> simonPattern; cin >> userPattern; /* Your solution goes here */ cout << "userScore: " << userScore << endl; return 0;}
- "Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings starting from index 0. For each match, add one point to userScore. Upon a mismatch, exit the loop using a break statement. Assume simonPattern and userPattern are always the same length. Ex: The following patterns yield a userScore of 4: simonPattern: RRGBRYYBGY userPattern: RRGBBRYBGY import java.util.Scanner; public class SimonSays {public static void main (String [] args) {Scanner scnr = new Scanner(System.in);String simonPattern;String userPattern;int userScore;int i; userScore = 0; simonPattern = scnr.next();userPattern = scnr.next(); /* Your solution goes here */ System.out.println("userScore: " + userScore); return;}}"Simon Says" is a memory game where "Simon" outputs a sequence of 10 characters (R, G, B, Y) and the user must repeat the sequence. Create a for loop that compares the two strings starting from index 0. For each match, add one point to userScore. Upon a mismatch, exit the loop using a break statement. Assume simonPattern and userPattern are always the same length. Ex: The following patterns yield a userScore of 4: simonPattern: RRGBRYYBGY userPattern: RRGBBRYBGY #include <iostream>#include <string>using namespace std; int main() {string simonPattern;string userPattern;int userScore;int i; userScore = 0; cin >> simonPattern;cin >> userPattern; /* Your solution goes here */ cout << "userScore: " << userScore << endl; return 0;} Please help me with this problem using c++.Test.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…
- 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…Write a program that deliberately contains an endless or infinite while loop. The loop should generate multiplication questions with single digit random integers. Users can answer the questions and get immediate feedback. After each question, the user should be able to stop the questions and get an overall result.Using loops of any kind, lists, Range, Sets, etc NOT allowed Angela loves reading books. She recently started reading an AI generated series called “Harry Trotter”. Angela is collecting books from the series at her nearest bookstore. Since the series is AI generated, the publishers have produced an infinite collection of the books where each book is identified by a unique integer. The bookstore has exactly one copy of each book. Angela wants to buy the books in the range [l,r], where l ≤ r. As an example, the range [−3,3] means that Angela wants to buy the books − 3, − 2, − 1, 0, 1, 2, and 3. Dan also loves the series (or maybe annoying Angela – who knows, really), and he manages to sneak into the bookstore very early to buy all of the books in the range [d,u], where d ≤ u. When Angela later visits, sadly she will not find those books there anymore. For example, if Angela tries to buy books [−2,3] and Dan has bought books [0,2], Angela would only receive books − 2, − 1, and 3. The…