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leva [86]
3 years ago
6

Suppose we have a linearly separable dataset, and we divide the data into training and validation sets. Will a perceptron learne

d on the training dataset (assuming gradient decent works perfectly well) be guaranteed to have i) 0 error on the training dataset ii) 0 error on the validation dataset. Briefly explain.
Computers and Technology
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

i)True

ii)False

Explanation:

i. True

The data given is linearly separable. So, the subset of the data will also be linearly separable. And it will pass for all training data as the model is trained using the training set.

The output using a training set guarantees a hyper plane that perfectly fits and the data sets are separated accordingly.

ii. False

Since the data of the validation set is not known, so it may not result in 100 percent accuracy and there will be errors on the data sets.

Even though the perceptron works well with the training set, it may take a bad decision boundary as the samples are unknown.

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Answer:

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Explanation:

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Answer:

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Cual es la <br> importancia de la netiqueta cuando nos comunicamos con el internet
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Which audience characteristics should be taken into consideration when developing a presentation?
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Which audience characteristics should be taken into consideration when developing a presentation?
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Take the program written last weekend that creates a deck of cards, shuffles them, and then lists them out. Modify the program t
Mars2501 [29]

Answer: Here, you might need to make some changes to format it how you'd like, but it gives you the desired output :)

Explanation:

import random as chunky

def play():

   hand1, hand2 = [], []

   for n in range(5):

       hand1.append(chunky.choice(cardList))

       hand2.append(chunky.choice(cardList))

   print("Hand 1: ", end= "")

   for n in hand1:

       print(n, end= " ")

   print()

   print("\nHand 2: ", end= "")

   for n in hand2:

       print(n, end= " ")

   print()

   while "A" in hand1:

       hand1[hand1.index("A")] = 11

   while "A" in hand2:

       hand2[hand2.index("A")] = 11

   print(f"\nHand 1 Total: {sum(hand1)}  Hand 2 Total: {sum(hand2)}")

   print("\nHand 1 WINS!") if sum(hand1) > sum(hand2) else print("\nHand 2 WINS!")

cardList = [1,2,3,4,5,6,7,8,9,10,"A"] #Skipped King, Queen, and Jack. Did not know how to incorperate them

chunky.shuffle(cardList)

looper = True

while looper == True:

   play()

   x = input("\n\nWould you like to go again?(y/n): ")

   if x.lower() == "n":

       quit()

8 0
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