# Karger's min cut and tips on bounding nonlinear recurrences

I was recently working on an old qualifying exam problem asking us to generalize Karger's randomized global min cut algorithm to that of a global min $$k$$-cut. I recalled the strategy of running a recursive version of the randomized edge contraction algorithm to help with the shrinking the error probability and ultimately help with an efficient algorithm for getting the global min $$k$$-cut with probability $$1/2$$.

Most of the analysis was going well until I started to analyze the probability of success of the recursive algorithm. I found that this probability was ultimately described as $$P_i \geq P_{i-1} - \frac{1}{4} P_{i-1}^2$$ where $$P_{i}$$ is the probability of success on some call of the algorithm at level $$i$$ of the recursion tree, with a leaf having $$i = 0$$. One could show that the depth of the tree was $$d = \frac{c \log_2(n)}{(k-1)}$$ for some constant $$c$$ and ultimately I was looking to lower bound $$P_{d}$$. However, I struggled to lower bound this recurrence. I knew from studying the algorithm for a min cut that this recurrence should have the lower bound $$P_{d} = \Omega(1/\log(n))$$. I wanted to try and show this but the best I got on my own was doing \begin{align} P_{i} &\geq P_{i-1} - \frac{1}{4} P_{i-1}^2 \\ &\geq P_{i-1}\left( 1 - \frac{1}{4}\right) \tag{ P_{i-1} \leq 1 } \\ &= \frac{3}{4} P_{i-1} \\ &\geq \left(\frac{3}{4}\right)^{i} P_{0} \end{align} which, when we evaluate $$P_{d}$$, gives us that \begin{align} P_{d} &\geq (3/4)^{d} P_0 \\ &= (4/3)^{-c \log_2(n)/(k-1)} \\ &= n^{-\frac{c}{(k-1) \log_{4/3}(2)}} \\ &\geq n^{-\frac{2.41 c}{(k-1)}} \end{align}

This is a much worse lower bound than what can be found. After this point, I investigated the original paper for the recursive randomized min cut algorithm. The trick they used was to perform a nonlinear change of variables to the recursion, something of the form $$P_{k} = 4/Z_{k}$$, where $$Z_{k}$$ is the new recurrence variable. This change leads to a new recurrence of the form $$Z_{k} \leq Z_{k-1} + 1 + 1 / (Z_{k-1} - 1)$$ Since $$Z_{k} = 4 / P_{k} \geq 4$$, this implies we can upper bound the above recurrence as $$Z_{k} \leq Z_{k-1} + 2 \leq 2 k + Z_{0} = 2 k + 4/P_{0}$$ which finally implies that indeed $$P_{d} = \Omega(1/\log(n))$$. I see that this transformation worked, but I do not have intuition for why one might think to try such a transformation. Can anyone help me with that intuition? Are there any other useful tips for tackling bounds of nonlinear recurrences?

First of all, let me explain why your estimate is lossy: when $$P_i$$ becomes small, $$P_{i-1} - P_{i-1}^2/4$$ is very close to $$P_{i-1}$$, and in particular, it doesn't decrease exponentially.
Since the function $$x - x^2/4$$ is monotone in the range $$[0,1]$$, if you solve the corresponding recurrence $$Q_i = Q_{i-1} - Q_{i-1}^2/4$$ (with base case $$Q_0 = 1$$), then $$P_i \geq Q_i$$.
Let us write the recurrence relation as $$Q_i - Q_{i-1} = -Q_{i-1}^2/4$$. This is a difference equation which can be approximated by a differential equation $$q' = -q^2/4$$, with $$q(0) = 1$$. Using standard techniques, you can determine the solution, which is $$q(i) = 4/(4+i)$$. This suggests that $$Q_i \approx 4/(4+i)$$, and so hints that $$4/Q_i$$ is a useful quantity to look at.
Here is another way to estimate the recurrence. Write the recurrence as $$Q_i = (1-Q_{i-1}/4) Q_{i-1}.$$ In a range of $$i$$ where $$Q_i \approx \theta$$, it takes $$\Theta(1/\theta)$$ steps to reduce $$Q_i$$ by a factor of $$1/2$$. In particular, after $$\Theta(1+2+\cdots+2^{\ell-1}) = \Theta(2^\ell)$$ steps, we reach $$Q_i \approx 2^{-\ell}$$. This suggests that $$Q_i$$ scales like $$\Theta(1/i)$$. With a little bit of work, this argument can be made completely formal, though it is hard to get a hold on the exact constant factor in this way.
• I just want you to know that I used the differential equation technique on a new problem I had, namely to compute a lower bound to the recurrence $V(i) \geq V(i-1) - kn/V(i-1)$ where $0 \leq k \leq n$ and $V(1) = n$ is the base case. It was very convenient to use here, ultimately allowing me to prove the lower bound $V(i) \geq \sqrt{n^2 - 2kn(i-1)}$ which worked well for my needs. I really appreciate the insight from this answer! Mar 10, 2021 at 7:28