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jeka57 [31]
3 years ago
14

Please answer the following in python: Answering the ones that have #TODO

Computers and Technology
1 answer:
zalisa [80]3 years ago
5 0

Answer:

Please answer the following in python: Answering the ones that have #TODO

from DiceQuad import DiceQuadclass PigPlayer:

WINNING_SCORE = 100

AUTO_WIN_RECOGNITION_ON = True

numPlayers = 0

def __init__(self, owner, name = "Player"):

self.owner = owner

self.name = name

self.dice = DiceQuad()

self.score = 0

self.roundScore = 0

self.isPlayerTurn = False

PigPlayer.numPlayers += 1

def reset(self):

''Resets all player values to their default value at the start of a game''

#TODO

pass

def getName(self):

'''Return the name of the player'''

return self.name

def getCurrentScore(self):

'''Return the current score of the player. If it is currently the player's turn,

include the round score'''

#TODO

pass

def hasWon(self):

'''Return boolean if the player has won'''

#TODO

pass

def __str__(self):

'''String representation of this class is the current total score, along with the

current round score (which could still be lost, in the case the turn is still

on), or the last round score, if it's the other player's turn'''

return (self.name + "\'s score: \t" + str(self.getCurrentScore()) + "\t"+

(" (this round so far: " if self.isPlayerTurn else " (last round's score: ")+ "\t" + str(self.roundScore) + ")")

def displayNum1s(self):

'''Prints a message about how many 1s were rolled. Also prints what happens as a consequence.

EX:

Player 1 rolled 3 ones

Player 1 loses all points'''

num1s = self.dice.num1s()

print(self.name + " rolled " + str(num1s) + " ones")

if(num1s == 1):

#TODO

pass

elif(num1s == 2):

#TODO

pass

elif(num1s == 3):

#TODO

pass

elif(num1s == 4):

#TODO

pass

def displayDice(self):

print(self.dice)

def displayDoRoll(self):

self.owner.displayScores()

print(self.name + " rolls... ")

def doRoll(self):

'''Rolls the dice for the player. Based on the number of 1s left, either asks the player if they want to roll again

or ends the turn with the proper consequence. If AUTO_WIN_RECOGNITION_ON is True, ends the turn automatically when

player has won including the most recent roll'''

self.displayDoRoll()

self.dice.roll()

self.displayDice()

num1s = self.dice.num1s()

if(num1s == 0):

#TODO

pass

elif(num1s == 1):

#TODO

pass

elif(num1s == 2):

#TODO

pass

elif(num1s == 3):

#TODO

pass

elif(num1s == 4):

#TODO

pass

def doTurn(self):

'''Performs a full turn for the player'''

self.roundScore = 0

self.isPlayerTurn = True

while (self.isPlayerTurn):

self.doRoll()

if(self.isPlayerTurn):

self.isPlayerTurn = not self.wantsHandOver()

self.score += self.<em><u>roundScorenxnd</u></em><em><u> </u></em><em><u>dd</u></em>

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Oksana_A [137]

Answer:

1. E: II and III only

2. A: (int)(Math.random() * (upper − lower) ) + lower

3. A: The value of answer is N

4. E: while( !(userGuess == secretNumber) && numGuesses <= 15 )

5. C: 21

Explanation:

1. Which of the following is equivalent to while(userGuess != secretNumber)?

I. while( userGuess < secretNumber && userGuess > secretNumber)  

NO - This will test until the userGuess is smaller AND greater than the secretNumber, at the same time... so that condition will never be true.

II. while( userGuess < secretNumber || userGuess > secretNumber)

YES - This will test the value of userGuess and see if it's smaller OR greater than secetNumber.  So, it will loop until the user guesses right.

III. while( !(userGuess == secretNumber) )

YES, this will negate the match with the secretNumber.  In order words, if it's not a match, it will return true... so the loop will run until it finds a false condition (a match).

As you can see, only II and III are valid.

2.  If the lower limit were inclusive and the upper limit exclusive, which expression would properly generate values for the secret number?

A: (int)(Math.random() * (upper − lower) ) + lower

Since the lower limit is INCLUSIVE, we mustn't add one to the lower limit.  Also, the Math.random() function returns a value that matches our needs; it returns a value between [0,1[ (meaning the 0 is included, but not the 1).

Assuming the (int) caster does return only the integer portion doing a round down of the result, we'll be perfect.

3. What conclusion can be made about the state of the program when the while loop terminates?

while(!answer.equals( "N"))

{.....

A: The value of answer is N

The condition in the loops reads as "While the negation of the answer being 'N', loop".  If the answer equals 'N' then the method should return true... which will be negated by the '!' operator, causing the condition to be false. Thus we know that if the loop ends, the value of answer contains 'N', any other value will keep the loop going.

4. Assuming numGuesses is initialized to 1, how would the while statement be modified to include an extra criterion limiting the number of guesses to 15?

E: while( !(userGuess == secretNumber) && numGuesses <= 15 )

This modified condition will first test to see if the user has guessed the secretNumber (if he has, the first sub-parenthesis will be true... so the left side of the && operator will be false due to the negation operator.  The right side of the && operator will check to see how many tries have been attempted. Since the counter starts at 1, it needs to go up to 15 inclusively... so the <= is the right comparison operator.

5. After execution of the following code segment, what will be displayed?

int x = 1;

while(x < 18)

{

x += 5;

}

System.out.println(x);

C: 21

The x variable is initialized with 1... then enters the loop, in which it is incremented by 5 at each passage.

So after first passage, x = 6

After second passage, x = 11

After third passage, x = 16

After fourth passage, x = 21

Cannot enter the loop again because 21 > 18.

So, it will print out the value of 21.

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