General formula for a recurrence relation?

问题

I was solving a coding question, and found out the following relation to find the number of possible arrangements:

one[1] = two[1] = three[1] = 1

one[i] = two[i-1] + three[i-1]

two[i] = one[i-1] + three[i-1]

three[i] = one[i-1] + two[i-1] + three[i-1]

I could have easily used a for loop to find out the values of the individual arrays till n, but the value of n is of the order 10^9, and I won't be able to iterate from 1 to such a huge number.

For every value of n, I need to output the value of (one[n] + two[n] + three[n]) % 10^9+7 in O(1) time.

Some results:

For n = 1, result = 3
For n = 2, result = 7
For n = 3, result = 17
For n = 4, result = 41

I was not able to find out a general formula for n for the above after spending hours on it. Can someone help me out.

Edit:

n = 1, result(1) = 3
n = 2, result(2) = 7
n = 3, result(3) = result(2)*2 + result(1) = 17
n = 4, result(4) = result(3)*2 + result(2) = 41

So, result(n) = result(n-1)*2 + result(n-2) OR T(n) = 2T(n-1) + T(n-2)

回答1:

You can use a matrix to represent the recurrence relation. (I've renamed one, two, three to a, b, c).

(a[n+1]) = ( 0 1 1 ) (a[n])
(b[n+1])   ( 1 0 1 ) (b[n])
(c[n+1])   ( 1 1 1 ) (c[n])

With this representation, it's feasible to compute values for large n, by matrix exponentation (modulo your large number), using exponentation by squaring. That'll give you the result in O(log n) time.

(a[n]) = ( 0 1 1 )^(n-1) (1)
(b[n])   ( 1 0 1 )       (1)
(c[n])   ( 1 1 1 )       (1)

Here's some Python that implements this all from scratch:

# compute a*b mod K where a and b are square matrices of the same size
def mmul(a, b, K):
    n = len(a)
    return [
        [sum(a[i][k] * b[k][j] for k in xrange(n)) % K for j in xrange(n)]
        for i in xrange(n)]

# compute a^n mod K where a is a square matrix
def mpow(a, n, K):
    if n == 0: return [[i == j for i in xrange(len(a))] for j in xrange(len(a))]
    if n % 2: return mmul(mpow(a, n-1, K), a, K)
    a2 = mpow(a, n//2, K)
    return mmul(a2, a2, K)

M = [[0, 1, 1], [1, 0, 1], [1, 1, 1]]

def f(n):
    K = 10**9+7
    return sum(sum(a) for a in mpow(M, n-1, K)) % K

print f(1), f(2), f(3), f(4)
print f(10 ** 9)

Output:

3 7 17 41
999999966

It runs effectively instantly, even for the n=10**9 case.

来源：https://stackoverflow.com/questions/40829549/general-formula-for-a-recurrence-relation

标签

algorithm

formula

recurrence