I want to find the worst-case running time of an algorithm, which follows the following recurrence equation: The worst-case running time is $\Theta(n^2) + T(n, 2, n)$, where

$T(x, i, y) = \begin{cases} 1 & \textrm{if $(x=0)$ or $(y=0)$ or $(i > n)$} \\ x+y+ T(x-a,\quad i+1, \qquad b)\\ \ \qquad+ T(x-a,\quad i+1,\ y-b)\\ \ \qquad+ T(\quad\ \ \ a,\quad i+1,\ y-b) & \textrm{otherwise} \end{cases}$

where $a \in \{0, ..., x\}, b \in \{0, ..., y\}$, and $a,b$ can be different in each recursion step, dependent on the input.

(Note: $x$ and $y$ represent the sizes of two lists $|L_1|=x, |L_2|=y$, and in each recursion step, each list is split into two, i.e. into lists of size $a$ and $x-a$ (for $L_1$), and $b$ and $y-b$ respectively (for $L_2$).)

How can I analyze such a recurrence equation, i.e. determine $f$ s.t. $T(n, 2, n)\in O(f)$? In particular, I do not know how to deal with $a, b$, especially since they are completely dependent on the input of the algorithm.

(I can solve the recurrence equation for special cases of $a,b$, e.g. $a=\frac{x}{2}, b=\frac{y}{2}$ for all recursion steps, or $a=x, b=0$ for all recursion steps etc., but not in general.)

  • 1
    $\begingroup$ Replace the free parameters $a,b$ by a maximum over all legal choices of $a,b$. $\endgroup$ May 29, 2014 at 15:48
  • $\begingroup$ What do you mean by maximum over all legal choices of $a,b$? For me, that sounds like I should set $a:=x,b:=y$, which would result in $T(x,i,y)=x+y+1+1+1$. $\endgroup$
    – user18051
    May 29, 2014 at 16:59

1 Answer 1


In order to determine an upper bound on $T(x,i,y)$, consider instead the recurrence $$ T'(x,i,y) = \begin{cases} 1 & \text{if $x = 0$ or $y = 0$ or $i > n$} \\ x + y + \max_{\substack{0 \leq a \leq x\\ 0 \leq b \leq y}} S'(x,i,y,a,b) & \text{otherwise,} \end{cases} $$ where $S'$ is given by $$ S'(x,i,y,a,b) = T'(x-a,i+1,b) + T'(x-a,i+1,y-b) + T'(a,i+1,y-b). $$ If you have more constraints on $a,b$, you should adjust the bounds in the definition of $T'$ accordingly.

  • $\begingroup$ Thank you for your answer! I don't really know how to analyze this monster now, though, since S' is dependent on T' and vice versa. $\endgroup$
    – user18051
    May 30, 2014 at 7:24
  • $\begingroup$ @user18026 Now it's a plain recurrence relation. I agree that it looks difficult to analyze. The circular reference you mention is a red herring since once you substitute the definition $S'$ in the definition of $T$', it disappears. I only wrote it this way to simplify typesetting. $\endgroup$ May 30, 2014 at 15:09
  • $\begingroup$ Can you tell me how to solve such recurrence equations, i.e. bring them into a closed form? After substituting once, I get $$ T'(n,2,n) = 2n + \max_{\substack{0 \leq a \leq n\\ 0 \leq b \leq n}} \{\\2(x+y)-(a+b)\\ + max_{\substack{0\leq a_1 \leq x-a\\0\leq b_1 \leq b}}\{...\} + max_{\substack{0\leq a_2 \leq x-a\\0\leq b_2 \leq y-b}}\{...\} + max_{\substack{0\leq a_3 \leq a\\0\leq b_3 \leq y-b}}\{...\} \}$$ But again I end up having trouble analyzing this recurrence, since all the $a_i,b_i's$ are dependent on the input. $\endgroup$
    – user18051
    May 30, 2014 at 17:40
  • $\begingroup$ @user18051 The $a_i,b_i$ are independent of the input since we're maximizing over them. Regarding how to solve this recurrence, that's a different question. $\endgroup$ May 31, 2014 at 5:15
  • $\begingroup$ Do you happen to know how to solve such a recurrence? I would very much appreciate it if you can give me some directions (or some literature) :) $\endgroup$
    – user18051
    May 31, 2014 at 8:22

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