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Archy [21]
4 years ago
15

For any two documents x and z, define k(x, z) to equal the number of unique words that occur in both x and z (i.e., the size of

the intersection of the sets of words in the two documents). Is this function a kernel?
Computers and Technology
1 answer:
mars1129 [50]4 years ago
8 0

Answer:

Yes, the function k(x,z) is a kernel

Explanation:

To show that k(x,z) is a kernel.

First, we'll need to explicitly construct two features of k(x,z).

The features are:

vector φ(x) and vector φ(z) in such a way that K(x,z) =φ(x)·φ(z).

Then, we proceed to the next step...

For any given documents, a vocabulary V can be constructed.

Vocabulary V is finite size for the words in the document set.

Given V, a feature mapping φ(x) can then be constructed for x by the following:

For the kth word wk in V, if wk appears in document x, assign φ(x) k(the kth element of φ(x)) to be 1; else assign 0 to it.

Then the number of unique words common in x and z is φ(x)·φ(z).

This gives us the kernel.

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Explanation:

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What do you enter at the command prompt to prevent the shell from using to much of the systems resources?
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Answer:

Ulimit is the correct answer for the above question.

Explanation:

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The part of the computer that provides access to the Internet is the
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3 years ago
Write a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters.
vichka [17]

Answer:

Explanation:

The python code for the question is attached in the image below.

The objective here is to write a code in python with a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.

SEE BELOW FOR THE CODE.

alphabet = "abcdefghijklmnopqrstuvwxyz"

test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):

   d = dict()

   for c in s:

       if c not in d:

           d[c]=1

       else:

           d[c] += 1

   return d

def has_duplicates():

   for string in test_dups:

       dictionary=histogram(string)

       duplicates=False

       for ch in dictionary.keys():

           if dictionary[ch] > 1:

               duplicates=True

               break

       if duplicates==True:

           print(string,"has duplicates")

       else:

           print(string,"has no duplicates")

def missing_letters(string):

   answer=""

   for ch in alphabet:

       if ch not in string:

           answer=answer+ch

   return answer

print("__________________________________________")

print(" Calling has_duplicates() function")

print("__________________________________________")

has_duplicates()

print("\n______________________________________________")

print("Calling missing_letters() function in for loop")

print("______________________________________________")

for i in test_miss:

   answer=missing_letters(i)

   if len(answer)>=1:

       print(i,"is missing letters",answer)

   else:

       print(i,"uses all the letters")

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