This is a nice problem. The trick is always defining the recurrence
in an artful way.
Here let E(L, e) be the number of bracket expressions of length m that
are proper _except_ that there are e extra left brackets.
So for L = 1 and 0 <= e <= n, we have
E(1, e) = (e == 1) ? 1 : 0
That is, the only unit length proper bracket expression possibly
having extra left brackets is a single left bracket. It obviously has
exactly 1 extra left bracket.
Let S = { s1, s2, ... sk}
E(L, e) = (L \in S) ? E(L - 1, e - 1) : // only exprs ending in left
bracket
E(L - 1, e - 1) + E(L - 1, e + 1) // exprs ending in left +
ending in right bracket
To make this complete, we need to add E(L, e) = 0 for any e < 0 or e >
n because in these cases no proper bracket expression exists.
Finally we get the answer: E(2n, 0)
So for fun, let's suppose the set S is empty and we want to know the
answer for the case n=3, which is L=6.
We have
e = 0 1 2 3
L = 1: 0 1 0 0
2: 1 0 1 0
3: 0 2 0 1
4: 2 0 3 0
5: 0 5 0 3
6: 5 0 8 0
The answer is E(6,0) = 5 corresponding to [[[]]], [[][]], [][[]], [[]]
[], and [][][].
I'll let you try some examples where S is non-empty.
On Sep 7, 4:22 am, hari <[email protected]> wrote:
> You are given:
>
> * a positive integer n,
> * an integer k, 1<=k<=n,
> * an increasing sequence of k integers 0 < s1 < s2 < ... < sk <=
> 2n.
>
> What is the number of proper bracket expressions of length 2n with
> opening brackets appearing in positions s1, s2,...,sk?
> plz.. explain how to solve this problem
>
> thanks in advance
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