Suppose we have n elements, a1, a2, …, an, arranged in a circle. That is, a2 is between a1 and a3, a3 is between a2 and a4, an is between an-1 and a1, and so forth.
Each element can take the value of either 1 or 0. Two arrangements are different if there are corresponding ai‘s whose values differ. For instance, when n=3, (1, 0, 0) and (0, 1, 0) are different arrangements, even though they may be isomorphic under rotation or reflection.
Because there are n elements, each of which can take two values, the total number of arrangements is 2n.
Here is the question:
How many arrangements are possible, such that no two adjacent elements both have the value 1? If it helps, only consider cases where n>3.
I ask here for several reasons:
- This arose while I was solving a programming problem
- It sounds like the problem may benefit from Boolean logic/bit arithmetic
- Maybe there is no closed solution.
Let’s first ask the question ‘how many 0-1 sequences of length n are there with no two consecutive 1s?’ Let the answer be A(n). We have A(0)=1 (the empty sequence), A(1) = 2 (‘0’ and ‘1’), and A(2)=3 (’00’, ’01’ and ’10’ but not ’11’).
To make it easier to write a recurrence, we’ll compute A(n) as the sum of two numbers:
B(n), the number of such sequences that end with a 0, and
C(n), the number of such sequences that end with a 1.
Then B(n) = A(n-1) (take any such sequence of length n-1, and append a 0)
and C(n) = B(n-1) (because if you have a 1 at position n, you must have a 0 at n-1.)
This gives A(n) = B(n) + C(n) = A(n-1) + B(n-1) = A(n-1) + A(n-2). By now it should be familiar 🙂
A(n) is simply the Fibonacci number Fn+2 where the Fibonacci sequence is defined by
F0=0, F1=1, and Fn+2= Fn+1+Fn for n ≥ 0.
Now for your question. We’ll count the number of arrangements with a1=0 and a1=1 separately. For the former, a2 … an can be any sequence at all (with no consecutive 1s), so the number is A(n-1)=Fn+1. For the latter, we must have a2=0, and then a3…an is any sequence with no consecutive 1s that ends with a 0, i.e. B(n-2)=A(n-3)=Fn-1.
So the answer is Fn+1 + Fn-1.
Actually, we can go even further than that answer. Note that if you call the answer as
G(n)=Fn+1+Fn-1, then
G(n+1)=Fn+2+Fn, and
G(n+2)=Fn+3+Fn+1, so even G(n) satisfies the same recurrence as the Fibonacci sequence! [Actually, any linear combination of Fibonacci-like sequences will satisfy the same recurrence, so it’s not all that surprising.] So another way to compute the answers would be using:
G(2)=3
G(3)=4
G(n)=G(n-1)+G(n-2) for n≥4.
And now you can also use the closed form Fn=(αn-βn)/(α-β) (where α and β are (1±√5)/2, the roots of x2-x-1=0), to get
G(n) = ((1+√5)/2)n + ((1-√5)/2)n.
[You can ignore the second term because it’s very close to 0 for large n, in fact G(n) is the closest integer to ((1+√5)/2)n for all n≥2.]