What is the solution of $T(n, m) = T(n, m-1) + T(n-1, m) + c$?

Question

Consider the recurrence $$ T(n,m) = T(n,m-1) + T(n-1,m) + c, $$ with base cases $T(n,0) = T(0,m) = 1$.

This is the complexity of a recursive algorithm for the longest common subsequence, I know that the complexity is exponential and equals $\Theta(2^i)$, but I'm not sure if $i=\max(n, m)$ or $i=nm$. Also, how can I demonstrate it using induction?

I assume you're trying to solve that recurrence, as opposed to determining the "complexity" of computing it? — Raphael, Nov 28 '18 at 22:50
Hint: Draw the recursion tree for small n, m. Do you see a pattern? — Raphael, Nov 28 '18 at 22:51
Possible duplicate of Solving or approximating recurrence relations for sequences of numbers — David Richerby, Nov 28 '18 at 22:53
Thank you for your answers, I couldn't find a simplified formula using the recursion tree, now I'm not even sure if the function is equal to 2^max(n, m). — Sydney.Ka, Nov 28 '18 at 23:59
Can we assume that $T(n,0)=T(0,m)=1$? Those values should have been given. Otherwise, for example, let $T(n,0)=2^{2^n}$, we may have double exponential growth. — John L., Nov 29 '18 at 07:38
@DavidRicherby, surprisingly, no answer in that reference question/answers deals with two variables. So it seems not very reasonable to consider this question as a duplicate by our norm. — John L., Nov 29 '18 at 07:44
Yes, I'm sorry I forgot to mention that T(n,0)=T(0,m)=1, and as you said there is no method that could help me solve this recurrence in that reference. — Sydney.Ka, Nov 29 '18 at 14:49

score 4 · Accepted Answer · answered Nov 30 '18 at 06:53

The function $f(n,m) = \binom{n+m}{n}$ satisfies the recurrence $$ f(n,m) = f(n-1,m) + f(n,m-1), $$ with base cases $f(n,0) = f(0,m) = 1$. This already gives you the solution when $c = 0$, assuming that the base cases are $f(n,0) = f(0,m) = 1$.

One function which satisfies the recurrence, though (in general) not the base case, is $g(n,m) = -c$: $$ g(n,m) = g(n,m-1) + g(n-1,m) + c. $$ This gives the solution when $c = -1$.

Finally, note that any function of the form $\alpha f(n,m) + g(n,m)$ is also a solution of the recurrence. Since $$ \alpha f(n,0) + g(n,0) = \alpha f(0,m) + g(0,m) = \alpha - c, $$ we see that $$ T(n,m) = (c + 1) f(n,m) + g(n,m) = (c+1) \binom{n+m}{n} - c. $$

What is the solution of $T(n, m) = T(n, m-1) + T(n-1, m) + c$?

1 Answers1