Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the elemen
2 answers:
You might be interested in
Answer:
I am writing a Python program.
def concordance(filename): # function that takes a file name as parameter and returns the concordance for that file
file = open(filename,"r") #opens file in read mode
unique={} #creates and empty list
for word in file.read().split(): #loops through every word in the file
if word in unique: # if words is already present in the list
unique[word] += 1 #add 1 to the count of the existing word
else: #if word is not present in the list
unique[word] = 1 #add the word to the list and add 1 to the count of word
file.close(); # close the file
for x in sorted(unique): #sort the list and displays words with frequencies
print(x, unique[x]); #prints the unique words and their frequencies in alphabetical order
#prompts user to enter the name of the file
file_name=input('Enter the input file name: ')
concordance(file_name) #calls concordance method by passing file name to it
Explanation:
The program has a function concordance that takes a text file name as parameter and returns a list of unique words and their frequencies.
The program first uses open() method to open the file in read mode. Here "r" represents the mode. Then creates an empty list named unique. Then the for loop iterates through each word in the file. Here the method read() is used to read the contents of file and split() is used to return these contents as a list of strings. The if condition inside the loop body checks each word if it is already present in the list unique. If the word is present then it adds 1 to the count of that word and if the word is not already present then it adds that unique word to the list and assign 1 as a count to that word. close() method then closes the file. The second for loop for x in sorted(unique): is used to display the list of unique words with their frequencies. sorted() method is used to sort the list and then the loop iterates through each word in the list, through x, which acts like an index variable, and the loop displays each word along with its frequency.
Answer:
Following are the algorithm to this question:
Explanation:
In the RSA algorithm can be defined as follows:
In this algorithm, we select two separate prime numbers that are the "P and Q", To protection purposes, both p and q combines are supposed to become dynamically chosen but must be similar in scale but 'unique in length' so render it easier to influence. Its value can be found by the main analysis effectively.
Computing N = PQ.
In this, N can be used for key pair, that is public and private together as the unit and the Length was its key length, normally is spoken bits. Measure,
where
is the total function of Carmichaels. It is a privately held value. Selecting the integer E to be relatively prime from
and
that is
. D was its complex number equivalent to E (modulo
); that is d was its design multiplicative equivalent of E-1.
It's more evident as a fix for d provided of DE ≡ 1 (modulo ).E with an automatic warning latitude or little mass of bigging contribute most frequently to 216 + 1 = 65,537 more qualified encrypted data.
In some situations it's was shown that far lower E values (such as 3) are less stable.
E is eligible as a supporter of the public key.
D is retained as the personal supporter of its key.
Its digital signature was its module N and the assistance for the community (or authentication). Its secret key includes that modulus N and coded (or decoding) sponsor D, that must be kept private. P, Q, and will also be confined as they can be used in measuring D. The Euler totient operates
however, could even, as mentioned throughout the initial RSA paper, have been used to compute the private exponent D rather than λ(N).
It applies because , which can always be split into
, and thus any D satisfying DE ≡ 1, it can also satisfy (mod
). It works because
, will always be divided by
,. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >
) for time - to - time). Many RSA frameworks assume notation are generated either by methodology, however, some concepts like fips, 186-4, may demand that D<
. if they use a private follower D, rather than by streamlined decoding method mostly based on a china rest theorem. Every sensitive "over-sized" exponential which does not cooperate may always be reduced to a shorter corresponding exponential by modulo
.
As there are common threads (P− 1) and (Q – 1) which are present throughout the , it's also possible, if there are any, for all the common factors
to become very small, if necessary.
Indication: Its original writers of RSA articles conduct their main age range by choosing E as a modular D-reverse (module ) multiplying. Because a low value (e.g. 65,537) is beneficial for E to improve the testing purpose, existing RSA implementation, such as PKCS#1, rather use E and compute D.