sorry, mistakes in boundaries, correct as below:
if (n>m) then C(m,n) = 0;
for each m in {1,2...N}, C(m,1) = m;
On Wed, Sep 8, 2010 at 11:22 AM, Adam <wangyanadam1...@gmail.com> wrote:
> There is my version of DP:
>
> Define C(m,n) as the number of proper bracket expressions
> B1,B2,B3,...,Bm+n with m left brackets '(' and n right brackets ')',
> in which 'proper' means that for each i belonging to {1,2,...,m+n} the
> number of left brackets '(' in B1,B2...Bi is larger than or equal to
> the number of right brackets ')' in it. The goal is to figure out
> C(N,N), where N is the given positive number.
>
> For example, m = 2 & n = 1, then '(()' and '()(' are all proper
> expressions.
>
> First we try to find the boundaries:
>
> if (n>m) then C(m,n) = 0;
> for each n in {1,2...N}, C(1,n) = 1;
> for each m in (1,2...N), C(m,1) = m;
>
> Then we try to find the recursive relations:
>
> if (m+n is in {S1,S2...Sk}) then C(m,n) = C(m-1,n);
> if (m+n is not in {S1,S2...Sk}) then C(m,n) = C(m-1,n) + C(m,n-1);
>
> The goal is to compute C(N,N).
>
> On Sep 7, 8:45 pm, Gene <gene.ress...@gmail.com> wrote:
>> 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 L 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.
>>
>> Say S = { s1, s2, ... sk}. If k < n, then k locations are "fixed" as
>> left brackets. All the others can potentially be either left or
>> right.
>>
>> E(L, e) = (L \in S) ? E(L - 1, e - 1) : // only "[" allowed at L'th
>> position
>> E(L - 1, e - 1) + E(L - 1, e + 1) // "[" or "]" allowed
>>
>> 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 expressions exist.
>>
>> 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 [][][].
>>
>> Now let's say S = { 5 }. There are only 2 of the 5 expressions with
>> "[" in the 5th position. And 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 2 0 3
>> 6: 2 0 5 0
>>
>> So we have E(6,0) = 2 as expected.
>>
>> On Sep 7, 4:22 am, hari <harihari1...@gmail.com> 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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