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Write a generator function named count_seq that doesn't take any parameters and generates a sequence that starts like this: 2, 12, 1112, 3112, 132112, 1113122112, 311311222112, 13211321322112, ...To get a number in the sequence, enumerate how many there are of each digit (in a row) in the previous number. For example, the first number is "one 2", which gives us the second number "12". That number is "one 1" followed by "one 2", which gives us the third number "1112". That number is "three 1" followed by "one 2", or 3112. Etc.

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abdullahfarooqi

Answer:

#required generator function

def count_seq():

   #starting with number 2. using string instead of integers for easy manipulation

   n='2'

   #looping indefinitely

   while True:

       #yielding the integer value of current n

       yield int(n)

       #initializing an empty string

       next_value=''

       #looping until n is an empty string

       while len(n)>0:

           #extracting first digit (as char)

           first=n[0]

           #consecutive count of this digit

           count=0

           #looping as long as n is non empty and first digit of n is same as first

           while len(n)>0 and n[0]==first:

               #incrementing count

               count+=1

               #removing first digit from n

               n=n[1:]

           #now appending count and first digit to next_value

           next_value+='{}{}'.format(count,first)

       #replacing n with next_value

       n=next_value

#testing, remove if you don't need this

if __name__ == '__main__':

   #creating a generator from count_seq()

   gen=count_seq()

   #looping for 10 times, printing next value

   for i in range(10):

       print(next(gen))

Explanation:

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