Write a generator function named count_seq that doesn't take any parameters and generates a sequence that starts like this: 2, 1

Question

1 month ago

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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, 1

2, 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.

Your generator function won't just go up to some limit - it will keep going indefinitely. It may need to treat the first one or two values as special cases, which is fine.

Your file must be named: count_seq.py

Computers and Technology

1 answer:

zubka84 [1K] · Answer 1 · 2026-03-29T00:49:19+03:00

Response:

#code (count_seq.py)

def count_seq():

n='2'

while True:

yield int(n)

next_value=''

while len(n)>0:

first=n[0]

count=0

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

count+=1

n=n[1:]

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

n=next_value

if __name__ == '__main__':

gen=count_seq()

for i in range(10):

print(next(gen))

Clarification:

Begin with the number 2. Utilize a string for easier manipulation rather than integers.
Engage in an infinite loop.
Yield the current integer value of n.
Continue looping until n becomes an empty string.
Repeat as long as n has content and the first digit matches the leading digit.
Concatenate the count and the first digit to form next_value.