令 a n表示长度为 n 的包含两个连续 1 的位字符串的数量。 a n的递推关系是什么?
(A) a n–2 + a n–1 + 2 n–2
(B) a n–2 + 2a n–1 + 2 n–2
(C) 2a n–2 + a n–1 + 2 n–2
(D) 2a n–2 + 2a n–1 + 2 n–2答案：(一)
说明：简单的解决方案
解决此问题的一种方法是尝试较小的值并排除选项。
a0 = 0
a1 = 0
a2 = 1 ["11"]
a3 = 3 ["011", "110", "111"]
a4 = 8 ["0011", "0110", "0111", "1101",
"1011", "1100", "1110", "1111"]
如果我们检查3 ，我们可以看到只有 A 和 C 满足该值。在(A)和(C)中，只有(A)满足a 4 。
另一个解决方案(有证据)
A string of length n (n >= 2) can be formed by
following 4 prefixes
1) 11 followed by a string of length n-2
2) 00 followed by a string of length n-2
3) 01 followed by a string of length n-2
4) 10 followed by a string of length n-2
Number 1 has already two consecutive 1's so number
of binary strings beginning with number 3 is 2n-2
as remaining n-2 bits can have any value.
Number 2 has two 0's so remaining n-2 bits must have
two consecutive 1's. Therefore number of binary strings
that can be formed by number 2 is an-2.
Number 3 and Number 4 together form all strings of length
n-1 and two consecutive 1's.这个问题的测验