Choose the person responsible for each innovation.

Assume that getPlayer2Move works as specified, regardless of what you wrote in part (a) . You must use getPlayer1Move and getPla

kakasveta [241]

Answer:

(1)

public int getPlayer2Move(int round)

{

int result = 0;

//If round is divided by 3

if(round%3 == 0) {

result= 3;

}

//if round is not divided by 3 and is divided by 2

else if(round%3 != 0 && round%2 == 0) {

result = 2;

}

//if round is not divided by 3 or 2

else {

result = 1;

}

return result;

}

(2)

public void playGame()

{

//Initializing player 1 coins

int player1Coins = startingCoins;

//Initializing player 2 coins

int player2Coins = startingCoins;

for ( int round = 1 ; round <= maxRounds ; round++) {

//if the player 1 or player 2 coins are less than 3

if(player1Coins < 3 || player2Coins < 3) {

break;

}

//The number of coins player 1 spends

int player1Spends = getPlayer1Move();

//The number of coins player 2 spends

int player2Spends = getPlayer2Move(round);

//Remaining coins of player 1

player1Coins -= player1Spends;

//Remaining coins of player 2

player2Coins -= player2Spends;

//If player 2 spends the same number of coins as player 2 spends

if ( player1Spends == player2Spends) {

player2Coins += 1;

continue;

}

//positive difference between the number of coins spent by the two players

int difference = Math.abs(player1Spends - player2Spends) ;

//if difference is 1

if( difference == 1) {

player2Coins += 1;

continue;

}

//If difference is 2

if(difference == 2) {

player1Coins += 2;

continue;

}

// At the end of the game

//If player 1 coins is equal to player two coins

if(player1Coins == player2Coins) {

System.out.println("tie game");

}

//If player 1 coins are greater than player 2 coins

else if(player1Coins > player2Coins) {

System.out.println("player 1 wins");

}

//If player 2 coins is grater than player 2 coins

else if(player1Coins < player2Coins) {

System.out.println("player 2 wins");

}

3 0

2 years ago

1- Design a brute-force algorithm for solving the problem below (provide pseudocode): You have a large container with storage si

sladkih [1.3K]

Answer:

Pseudocode is as follows:

// below is a function that takes two parameters:1. An array of items 2. An integer for weight W

// it returns an array of selected items which satisfy the given condition of sum <= max sum.

function findSubset( array items[], integer W)

{

initialize:

maxSum = 0;

ansArray = [];

// take each "item" from array to create all possible combinations of arrays by comparing with "W" and // "maxSum"

start the loop:

// include item in the ansArray[]

ansArray.push(item);

// remove the item from the items[]

items.pop(item);

ansArray.push(item1);

start the while loop(sum(ansArray[]) <= W):

// exclude the element already included and start including till

if (sum(ansArray[]) > maxSum)

// if true then include item in ansArray[]

ansArray.push(item);

// update the maxSum

maxSum = sum(ansArray[items]);

else

// move to next element

continue;

end the loop;

// again make the item[] same by pushing the popped element

items.push(item);

end the loop;

return the ansArray[]

}

Explanation:

You can find example to implement the algorithm.

3 0

2 years ago

Write a public static method named printArray, that takes two arguments. The first argument is an Array of int and the second ar

Serjik [45]

Answer:

Written in Java

public static void printArray(int myarr[], String s){

for(int i = 0; i<myarr.length;i++){

System.out.print(myarr[i]+s);

}

Explanation:

This defines the static method alongside the array and the string variable

public static void printArray(int myarr[], String s){

The following iteration iterates through the elements of the array

for(int i = 0; i<myarr.length;i++){

This line prints each element of the array followed by the string literal

System.out.print(myarr[i]+s);

}

The method can be called from main using:

<em>printArray(myarr,s);</em>

Where myarr and s are local variables of the main

5 0

2 years ago

In the simulation, player 2 will always play according to the same strategy. The number of coins player 2 spends is based on wha

baherus [9]

The simulation, player 2 will always play according to the same strategy.

Method getPlayer2Move below is completed by assigning the correct value to result to be returned.

Explanation:

You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.

#include <bits/stdc++.h>

using namespace std;

bool getplayer2move(int x, int y, int n)

{

int dp[n + 1];

dp[0] = false;

dp[1] = true;

for (int i = 2; i <= n; i++) {

if (i - 1 >= 0 and !dp[i - 1])

dp[i] = true;

else if (i - x >= 0 and !dp[i - x])

dp[i] = true;

else if (i - y >= 0 and !dp[i - y])

dp[i] = true;

else

dp[i] = false;

}

return dp[n];

}

int main()

{

int x = 3, y = 4, n = 5;

if (findWinner(x, y, n))

cout << 'A';

else

cout << 'B';

return 0;

}

8 0

2 years ago

5. Assume a computer has a physical memory organized into 64-bit words. Give the word address and offset within the word for eac

spin [16.1K]

Answer:

see explaination and attachment

Explanation:

5.

To convert any byte address, By, to word address, Wo, first divide By by No, the no. of bytes/word, and ignores the remainder. To calculate a byte offset, O, in word, calculate the remainder of By divided by No.

i) 0 : word address = 0/8 = 0 and offset, O = 0 mod 8 = 0

ii) 9 : word address = 9/8 = 1 and offset, O = 9 mod 8 = 1

iii) 27 : word address = 27/8 = 3 and offset, O = 27 mod 8 = 3

iv) 31 : word address = 31/8 = 3 and offset, O = 31 mod 8 = 7

v) 120 : word address = 120/8 = 15 and offset, O = 120 mod 8 = 0

vi) 256 :word address = 256/8 = 32 and offset, O = 256 mod 8 = 0

6. see attachment for the python programming screen shot and output

4 0

2 years ago