Question

Consider the Catalan numbers P(n) described here

1. Show a recursion-dependence tree for P(5) according to the recurrence relation.
2. Design a bottom-up dynamic programming algorithm to compute P(n) based on the recurrence relation (without using the formula P(n) = (2n−2)!/(n!(n−1)!)). Analyze its asymptotic runtime efficiency.
3. Design a top-down memoized algorithm to compute P(n) based on the recurrence relation (without using the formula P(n) = (2n−2)!/(n!(n−1)!)). Analyze its asymptotic runtime efficiency.

Answer 1

As stated, nth Catalan number is given by P(n) = (2n-2)!/(n!(n-1)!)

Some terms of the Catalan numbers sequence using above formula are:

1, 1, 2, 5, 14, 42, ..... so on

On analysing some of the terms of obtained by the above formula, we find that, P(n+1) =   and p(0) = 1 . We can use this recurrent relation in our algorithm.

(a.) Recursion-dependence tree for P(5) is shown in the image below:

note: due to lack of space i have not drawn the nodes having reverse order multiplications. for ex, i have shown node p(0)p(4) but not p(4)p(0), those nodes will also be in the tree. I have shown these nodes by multiplying the reverse of missing node by 2.

(b.) In bottom-up DP (also called tabulation method), we start from the base case and keep calculating values one by one using the previously obtained values until we reach the desired case. For ex, if we need to find p(n), and we have p(0)=1, we will calculate , p(1), p(2), p(3),.......p(n) in the given order using previous values.

Bottom-up DP algorithm pseudo code:

catalan (n): //function to find the nth catalan number declare array c[n+1] //declare array 'c' of size n+1 to store values c[0] <- 1 //base case c[1] <- 1 //base case for i <- 2 to n do //filling c array from index 0 to index n c[i] <- 0 for j <- 0 to i-1 do c[i] <- c[i] + c[j] * c[i-j-1] //calculating the sum end for end for return c[n] //return nth index of c array as it gives the value of P(n)

Time complexity of this algorithm = O()

(c.) In top-down DP (also called memoization method), we start from the problem(desired case) instead of the base case and use the value stored in the table for its subproblems to solve the problem. If the subproblem is not solved then we solve the subproblem first. For ex, if we need to find p(n), we can find it using p(n-1),p(n-2).....p(1), if we dont have value for p(n-1), we calculate it first in the similar manner.

Top-down DP algorithm pseudo code:

declare array c[n+1]and initialize it with 0 catalan (n): //function to calculate nth catalan number if n = 0 then return 1 if n = 1 then return 1 if c[n] != 0 then //Check if we have already calculated return c[n] //the value and stored in c array for i <- 0 to n-1 do //if the value is not calculated already c[n] = c[n] + catalan(i) * catalan(n-i-1) //using recursive formula to calculate it end for return c[n] 

Time complexity of this algorithm = O()