1
$\begingroup$

There are two stacks A and B.

A :  a,b,c,d   ('a' is on top and 'd' is at the bottom of the stack)
B :  (empty)

There are two rules.

If an element of A is popped, it must be printed immediately or pushed into B.
If an element of B is popped, it can only be printed.

So, how many permutations of a,b,c,d are possible? (continue reading)

P.S. Well, I did calculations manually(didn't use any formula) and got 14 as the answer. However, it took around 10 minutes to do the lengthy steps. So, is there an easy way to do this?

$\endgroup$
2
$\begingroup$

This problem looks like the problem of single stack sorting. "Single" because onle your stack B is used to sort, the stack A is only for input. Donald Knuth has analysed this, and characterized it as the 231-avoiding permutations, and has shown their number equals the Catalan numbers.

Now if my observation is correct, you must have counted wrong (original number you gave was 28). The Catalan numbers are 1, 2, 5, 14, 42, 132,

(added, after you recomputed) Thus, we get a formula, the number of "single-stack sortable permutations of length n" is the $n$-th Catalan number $C_n = \frac 1{n+1}{2n \choose n}$. That does not yet explain the connection. There is a lot to learn about Catalan numbers, see the wiki-page for a lot of situations where they arise! Also the page to the online encyclopedia of integer sequence I referred to has more than enough links to keep you busy for a long time.

$\endgroup$
  • $\begingroup$ yeah.. I did check again and it was 14.. (i'll change it now) and wow.. amazed at your brilliance.. so, is there any way to calculate it more easily?? like a formula? thanks in advance :) $\endgroup$ – Vishnu Vivek Dec 1 '12 at 12:39
  • 1
    $\begingroup$ Formula added to Answer. You will enjoy yourself with the many examples in the wikipedia page I linked to, I am certain. $\endgroup$ – Hendrik Jan Dec 1 '12 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.