PROBLEM LINK:
DIFFICULTY
SIMPLE
PREREQUISITES
Simple Math, Repeated Squaring
PROBLEM
How many sequences of 0s and 1s of length N are there such that there are at least 3 consecutive 1s.
QUICK EXPLANATION
You can see that the original problem statement is equivalent to the the above by saying that among N houses, you invite only those houses for which you put a 1 in the sequence.
Now, the number of ways you can make a sequence of 0s and 1s of length N, such that there are no more than two consecutive 1s is
f(N) = f(N1) + f(N2) + f(N3)
This is known as the Tribonacci Numbers.
The answer to the problem statement would then be 2
^{N}  f(N)
.
Since N can be very large, 2
^{N} can be calculated by repeated squaring and f(N)
can be calculated by matrix exponentiation.
EXPLANATION
First, let us see what insights are used to derive the formula.
The number of sequences of N digits where each digit is either 0 or 1 is 2
^{N}.
We wish to find the number of sequences such that there are no more than 2 consecutive 1s.
Lemma:
The entire world of such sequences can be constructed by concatenating strings taken from the following prefixfree set
S = { 0, 10, 110 }
Proof:
Let us call a sequence that contains no more than two consecutive 1s as a valid string.
(1) ...
All strings with less than or equal to 2 letters are valid.
(2) ...
Any substring of a valid string is also a valid string.
(3) ...
From (2), any suffix of a valid string is also a valid string.
(4) ...
For a string T, constructed by taking strings from S, T should also partition into S.
(5) ...
If T cannot partition into S, it cannot be constructed by taking strings from S.
(6) ...
Since S is prefixfree, every string has a unique partitioning into S, or none at all.
For example,
0101100100010110
can be partitioned into S as
0
10
110
0
10
0
0
10
110
And this is the only valid partitioning into S.
(7) ...
For a valid string A, A always has a prefix that matches a string in S.
Let us consider all possible 3 letter prefixes of a string
000x..
matches 0
from S (remaining string 00x..
)
001x..
matches 0
from S (remaining string 01x..
)
010x..
matches 0
from S (remaining string 10x..
)
011x..
matches 0
from S (remaining string 11x..
)
100x..
matches 10
from S (remaining string 0x..
)
101x..
matches 10
from S (remaining string 1x..
)
110x..
matches 110
from S (remaining string x..
)
111x..
Does not match any string in S
But, if A contains the prefix 111 then A is not a valid string.
Hence, for any valid string A, some prefix of A will always be one of the strings in S.
(8) ...
From (3) and (7), a valid string A will always have a partitioning into S.
(9) ...
We can see that concatenating strings from S will always generate a valid string.
From (8) and (9) we can say that taking strings from S will generate all possible valid strings.
Now, the number of valid strings of length N will be f(N)
, which is equal to

f(N1)
, after choosing “0” from S as prefix, all possible valid strings of length N1 can be used 
f(N2)
, after choosing “10” from S as prefix, all possible valid strings of length N2 can be used 
f(N3)
, after choosing “110” from S as prefix, all possible valid strings of length N3 can be used
Thus,
f(N) = f(N1) + f(N2) + f(N3)
We have to find the result modulo 1000000007.
Now, since N can be quite large, we cannot even iterate for O(N) complexity and find the result. We can build the following matrix representation from the above formula
1 1 0 [f(N+1) f(N) f(N1)] = [f(N) f(N1) f(N2)] * 1 0 1 1 0 0
Now we see that multiplying the 3x3
matrix above (let us call it M) k times will give us f(N+k).
We can consider the base case [f(2) f(1) f(0)] = [4, 2, 1]
and find M^{N2}. Once we have M^{N2}, we can find the answer by multiplying M^{N2} to [4, 2, 1]
, and taking the first value.
You can find M^{N2} by repeated squaring on the matrix. Repeated Squaring (and similarly Matrix Exponentiation) is a very useful technique that you should add to your problem solving arsenal.
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