I have the following python code. I get the following message when running it using a test script which I cannot send here:

while textstring[iterator].isspace(): # loop until we get other than space character

IndexError: string index out of range. Please find out why and correct the code.

def createWords(textstrings):
        createdWords = [] # empty list
        for textstring in textstrings: # iterate through each string in trxtstrings
                iterator = 0
                begin = iterator # new begin variable
                while (iterator < len(textstring)): # iterate through each character in the string
                        finished = iterator # new finished character
                        while textstring[iterator].isspace(): # loop until we get other than space character
                                iterator = iterator+1
                                word = textstring[begin:finished+1].strip() #  extract the word from begin position to before the space character,
                                                                                                                        #  textstring[begin,finshed] ; strip() will remove white spaces at ends
                                if (word): # if word is not empty
                                        createdWords.append(word) # append new word to createWords
                                begin = iterator # update begin value with current iterator position
                        if (textstring[iterator]).isalpha():
                                if (textstring[iterator].isupper()): # if letter is in uppercase  change it to lower case by using letter.lower() method.
                                        textstring = textstring[:iterator] + textstring[iterator].lower() + textstring[iterator+1:]
                                print(textstring[iterator],"this a letter")
                        elif (textstring[iterator]).isdigit():
                                print(textstring[iterator],"is a digit")
                        else:
                                print(textstring[iterator],"this a symbol")
                                word = textstring[begin:finished].strip()       #  extract the word from begin position to before the symbol character,
                                                                                                                        #  textstring[begin,finshed] ; strip() will remove white spaces at ends
                                createdWords.append(word) # append new word to createWords
                                begin = iterator  # update begin value with current iterator position
                        iterator= iterator+1 # increment iterator value
                word = textstring[begin:iterator].strip()  #  extract the word from begin position to the end of the string
                createdWords.append(word) # append new word to createWords
        return createdWords # return createWords

A (7, 4) cyclic code is designed with a generator polynomial, g(D) = D3 + D + 1. a) (10 points) Determine the code word for the message, 1010. b) (10 points) Determine the code word for the message, 1100. c) (9+1= 10 points) Determine the error polynomial for the received word, 1110101. Is the received word correct?

A (7, 4) cyclic code is designed with a generator polynomial, g(D) = D3 + D + 1. a) (10 points) Determine the code word for the message, 1010. b) (10 points) Determine the code word for the message, 1100. c) (9+1= 10 points) Determine the error polynomial for the received word, 1110101. Is the received word correct?

pseudo-code, please

If a word has a tag O, it means that the word has nothing to do with the named entity (it is not a part of a named entity such as a location, person name, organization, and etc.)

If a word has a tag starting with B, it means that the word is the beginning of a named entity. For instance, the tag B-per means that the associated word is the beginning of a person's name. On the other hand, if a word has a tag starting with I, it means the word is inside a named entity. Hence, the tag I-per means that the associated word is inside a person's name.

Thousands O

of O

demonstrators O

have O

marched O

through O

London B-geo

Take the following lines in the file for example.

President B-per

Mahmoud I-per

Ahmadinejad I-per

said O

We know the words “President, Mahmoud, and Ahmadinejad” cover a named entity of the type person name.

Could you write pseudo code to pull the names out of a file that are stated in the example

Question 1

1. Pain in a joint Blank 1 (Blank 2 - root word, Blank 3 - suffix)

2.00000 points

Question 2

1. With reference to something within a muscle : Blank 1 Blank 2 – prefix,(Blank 3 - root word, Blank 4 - suffix)

2.00000 points

Question 3

1. Refers to “beneath the skin : Blank 1 Blank 2 – prefix,(Blank 3 - root word, Blank 4 - suffix)

2.00000 points

Question 4

1. The process of viewing the esophagus, stomach and duodenum : Blank 1 (Blank 2 – root word,(Blank 3 - root word, Blank 4 - root word, Blank 5 - process of viewing)

2.00000 points

Question 5

1. Pertaining to (being) against seizures: Blank 1 (Blank 2 – prefix,(Blank 3 - root word, Blank 4 - suffix)

2.00000 points

Question 6

1. Inflammation of meninges : Blank 1 Blank 2 - root word, Blank 3 - suffix)

2.00000 points

Question 7

1. Inflammation of many nerves : Blank 1 ( Blank 2-prefix, Blank 3 - root word, Blank 4 - suffix)

2.00000 points

Question 8

1. Heart specialist : Blank 1 (Blank 2 - root word, Blank 3 - suffix)

2.00000 points

Question 9

1. Taking place within veins : Blank 1 Blank 2 – prefix,(Blank 3 - root word, Blank 4 - suffix)

2.00000 points

Question 10

1. Fat cell : Blank 1 Blank 2 - root word, Blank 3 - suffix)

2.00000 points

Public static boolean isPalindrome(String word)
* Check to see if a word is a palindrome. Should be case-independent.
* @param word a String without whitespace
* @return true if the word is a palindrome
  

Public static int longestWordLength(String words)

* Returns length of the longest word in the given String using recursion (no loops).
* Hint: a Scanner may be helpful for finding word boundaries. After delimiting by space,
* use the following method on your String to remove punctuation {@code .replaceAll("[^a-zA-Z]", "")}
* If you use a Scanner, you will need a helper method to do the recursion on the Scanner object.
* @param words A String containing one or more words.
* @return The length of the longest word in the String.
* @see Scanner#Scanner(String)
* @see Scanner#next()
* @see String#split(String)
* @see String#replaceAll(String, String)
* @see Math#max(int, int)

1. Word Match: From the list of words provided, enter in the table below the word that best describes each of the listed meanings

1. Word Match: From the list of words provided, enter in the table below the word that best describes each of the listed meanings