Sequence of ones a zeroes

Question

This question was found in a CSAT entrance exam for Cambridge computer science course, and I had a look at it but I'm not really sure where to start with it:

Given a sequence of zeros and ones of length $n$, let $L_n$ be the number of sequences that have no adjacent zeros. Give a recursive formula for $L_n$

I suggest computing the first few $L_n$ by hand: something should jump out at you! — Angina Seng, Jul 01 '17 at 07:26
@LordSharktheUnknown i should start writing down sequences without adjacent zeroes? — RhysD, Jul 01 '17 at 07:30
Well 0 and 1 say $L_1=2$ and 01, 10 and 11 say $L_2=3$, what about $L_3$? $L_4$? $L_5\ldots$? — Angina Seng, Jul 01 '17 at 07:33
@LordSharkTheUnknown Fibonacci? I did It to $L_5$ and the numbers of sequences went $1,2,3,5,8,18$ so would o claim that the recursive formula is the Fibonacci sequence formula? — RhysD, Jul 01 '17 at 08:27

W.R.P.S · Accepted Answer · 2017-07-01T08:51:21.173

I hope it can help you

let $L_{n}$ be the number of bit strings of length n that do not have two consecutive zeros.

To obtain a recurrence relation for $L_{n}$ , note that the number of bit strings of length n that do not have two consecutive $0$s equal the number of such bit strings ending in $0$ plus the number of bit strings ending in $1$.

We can assume that n is greater than or equal to three, so that only bit strings of length three or more need be considered.

The valid bit strings of length n ending with $1$ are precisely the bits strings of length $n - 1$ with no consecutive zeros with a $1$ appended onto the end. Consequently, there are $L_{n−1}$ such bit strings.

Valid bit strings of length n ending in a zero must have $1$ as their second to last bit; otherwise, they would end with a pair of $0$s, and that just can’t be.

It follows that the legal bit strings of length n ending in zero are the valid bit strings of length $n - 2$ with '$10$' tacked on to the end. Consequently, there are $L_{n−2}$ such bit strings.

We conclude that, for n greater than or equal to $3$, that: $L_{n} =L_{n−1}+L_{n−2}$

The initial conditions are $L_{1} = 2$ , since both bit strings of length one, $0$ and $1$, do not have two consecutive $0$s, and $L_{2} = 3$ .

To obtain $L_{5}$ , we simply use the recurrence relation three more times to find that:

$L_{3} =L_{2}+L_{1} =3+2=5 $

$L_{4} =L_{3} +L_{2} =5+3=8$

$L_{5} =L_{4}+L_{3} =8+5=13$

Recurrence Relations examples

score 0 · Answer 2 · answered Jul 01 '17 at 09:00

Let $Z_{n}$ be the number of sequences length $n$ with no adjacent zeros that end in zero. Let $O_n$ be the number with no adjacent zeros that end in one. Now, consider taking length $n-1$ strings and adding on one more digit to get a new string with no adjacent zeros. You can add a one to either type, so $$ O_n = O_{n-1} + Z_{n-1}$$ but you can only add a zero to the strings that end in one (or you'd be making an adjacent zero) so $$ Z_n = O_{n-1}$$ Combining gives the fibonacci reccurance $$ O_{n} = O_{n-1} + O_{n-2}.$$

We have $O_1 = 1$, $O_2 = 2$ so $O_n = F_{n+1}$ and $Z_n = O_{n-1} = F_{n},$ so the total number is $$L_n = O_n + Z_n = F_{n+1} + F_{n} = F_{n+2} $$

Sequence of ones a zeroes

2 Answers2