Parenthize

Vance

I have a string of N characters containing M(<N) words. Can you tell me how many way I can parenthize this string ?

So so COOL not kewl

K.J.Healey

Sum(n-i,i,1,n-1)

marcus

Do we use ordinary rules of pu(nctua)tion?

Parenth(eses are not allo)wed to break( up (words), or be) nested?

If one cannot break up words, then I think the number N of characters does not matter, only the number M of words matters.

It does seem to matter whether or not you want to allow nesting parens.

marcus

hi Healey, I dont know this notation, can you give an example
or explain it simply?
If there were just 4 words, how would your formula work to give
the number of different ways to parenthesize?

K.J.Healey

I assumed that you cannot parenthesize throughout words, like splitting them up. He states that there are M words, with M < N. So the maximum amount of words would be N-2 single characters, then a double character word. like a 4 letter max would be :
a b cd

in order to have M<N.

Then possible parentheticals are
(a) b cd
(a b) cd
(a b cd)
a (b) cd
a (b cd)
a b (cd)

cause you cant split "cd"

Then you see its just what i wrote as a sum
for the starting parenthesis in front of a you have m possibilities, which is n-1 possibilities.
For the parenthesis in front of "b' you have 2, or m-1 possibilities, or n-2 possibilities.

I guesss SUM(m-i,i,0,m-1) would work too.

TenaliRaman

is this allowed ?
(a) (b) (cd)
and this
a (b) (cd)

Also we may have
a bcd
can't we?

[edit]
on healey's notation
sum(m-i,i,0,m-1)
means
sum over m-i
where i goes from 0 to m-1
[/edit]

K.J.Healey

ahh, i forgot about multiple parenthesis. Theres a lot then. Im working on the formula right now.

PS:
Is there a mathematical operator "$"(ill call it arbitrarily) such that

6$ = 6+5+4+3+2+1+0 ?
sort of like 6! = 6*5*4.... but with addition? I hate writing out the sums everytime.

PPS : Oh, and im assuming he's looking for the maximum allowed given that you cant split words. So just ignore the character number. UNLESS (crazy hard) everytime you add a parenthesis it counts as a character, so you can have that many less characters in your string...

Oh and is this allowable?

(The (swift) red) (fox).

jcsd

[tex]n + (n-1) + .... + 2 + 1 = \sum_{i=1}^n i = \frac{n^2 + n}{2}[/tex]

Rogerio

If parenthesis cant break up words, and if parenthesis cant be nested , then the answer is:

[tex] \#ways(M) = \left(\frac{5+\sqrt{5}}{10}\right) \left(\frac{3+\sqrt{5}} {2}\right)^M + \left(\frac{5-\sqrt{5}}{10}\right) \left(\frac{3-\sqrt{5}} {2}\right)^M -1 [/tex]

I'm not counting the 'no parenthesis at all' case. So, if M=1 then #ways=1, for instance .
But, of course you may remove the '- 1' at the end.

TenaliRaman

Rogerio,
Brilliantè!!

Your formula seems to be agreeing with my calculations for a first few terms .... may/can i have a hint for the method u have taken?

-- AI
P.S : By the look of the formula, u seem to have got some recursive equation is it?

Rogerio

Wow, thanks TenaliRaman !

You are right, it was a 'recurrence' approach...:-)

A little bit more in white:

Consider all the combinations with the initial words: a b...f
When you add a new word, you get:

combinations with the new last word free (not enclosed in parentheses) :
thee old '...f' becomes '...f x'

combinations with the new the last word enclosed in parentheses and alone:
the old '...f' becomes '...f (x)'

combinations with the new last word enclosed in parentheses but not alone:
the old '...f)' becomes '...f x)'

Hmmm... is it enough ?

TenaliRaman

But Ofcourse!


F(m) = 2*F(m-1)+ [\sum_{i=1}^{m-2} F(i)] .. *
F(m-1) = 2*F(m-2) + [\sum_{i=1}^{m-3} F(i)] .. **
Simplifying with * and **, we get
F(m) = 3*F(m-1) - F(m-2)
and the rest is ofcourse mechanical.


Good Job!
-- AI