The specific problem I'm working on is the puzzle presented in this video. For those who don't want to watch the video, my summary of the puzzle is:

A frog is sitting on the edge of a pond facing the opposite bank. In front of the frog, running along the pond's diameter, is a line of $N$ lilypads. This is a very athletic frog: it is able in a single leap to land on any of the $N$ lilypads in the pond or on the opposite bank (for a total of $N + 1$ possible destinations). The frog first considers all $N+1$ positions in front of it, chooses one randomly with uniform probability, and then jumps to this location. It then repeats the process of choosing a new random position from those remaining in front of it (never jumping backwards). This goes on until the frog eventually reaches the opposite bank.

The puzzle is twofold: First, if $N=10$, what is the expectation value of the number of jumps the frog will take to reach the other side of the pond? Second, find an explicit formula (NOT a recurrence relation!) for the expectation value of the number of jumps in general for $N$ lilypads.

(It's a great puzzle; if you haven't yet given it a shot, I'd highly recommend doing so before reading any possible spoilers in the rest of this post!).

To attempt this puzzle, I first calculated the first few terms by hand and then tried to generalize my result. I eventually came up with a recursive algorithm for the expectation value $f$:

def f(n, depth=1):
    if n == 0:
        return depth
        return (n * f(n - 1, depth) + f(n - 1, depth + 1)) / (n + 1)

However, the second part of the puzzle explicitly disallows recurrence relations as solutions, so I am trying to un-roll/solve the recurrence relation. There are already several questions on strategies for solving recurrence relations (see this and this), but none I can find on solving recurrence relations which are dependent on the depth of recursion. Any help on how to do this is appreciated.

  • $\begingroup$ We denote the expected number of jumps by $A_n$. Try the following approach. Express $A_{n+1}$ through $A_0, A_1, \ldots, A_{n}$. Then, using the similar expression for $A_n$ through $A_0, A_1, \ldots, A_{n-1}$, derive a formula for $A_{n+1}$ depending only on $A_n$. The rest of the problem then becomes straightforward. $\endgroup$
    – diplodocus
    Sep 25, 2019 at 21:13
  • $\begingroup$ @diplodoc That is the approach I tried, and is how I arrived at my recurrence relation. The problem I ran into is that the value of $A_n$ depends on $A_{n-1},\cdots,A_i,\cdots,A_0$ but each $A_i$ also depends on the depth $n - i$. In my attempts I was not able to eliminate the depth dependence. $\endgroup$
    – fezziks
    Sep 26, 2019 at 17:54
  • $\begingroup$ $A_{n} = 1 + \frac{1}{n+1}\sum_{i=0}^{n-1}A_i$, then $(n+1)(A_n-1) = \sum_{i=0}^{n-1}A_i$. You can now exclude the sum from the expression for $A_{n+1}$ and obtain $A_{n+1} = A_n+\frac{1}{n+2}$ $\endgroup$
    – diplodocus
    Sep 26, 2019 at 18:24

1 Answer 1


diplodoc explained in the comments one way to solve the recursion. Your recursion is $$ A_n = 1 + \frac{A_0 + \cdots + A_{n-1}}{n+1}, \quad A_0 = 1. $$ The trick is to calculate $$ (n+1)A_n - n A_{n-1} = (n+1 + A_0 + \cdots + A_{n-1}) - (n + A_0 + \cdots + A_{n-2}) = 1 + A_{n-1}, $$ and so $A_n = A_{n-1} + 1/(n+1)$. From here it follows easily that $A_n$ is the $(n+1)$th harmonic number.

Another possibility is to use generating functions. Let $$ P(x) = \sum_{n=0}^\infty A_n x^n. $$ Since $\frac{1}{1-x} = 1 + x + x^2 + \cdots$, it follows that $$ \frac{xP(x)}{1-x} = \sum_{n=0}^\infty (A_0 + \cdots + A_n) x^{n+1}. $$ Integrating, we get $$ \int_0^x \frac{yP(y)}{1-y} \, dy = \sum_{n=0}^\infty \frac{A_0 + \cdots + A_n}{n+2} x^{n+1}. $$ It follows that $$ \frac{1}{1-x} + \frac{1}{x} \int_0^x \frac{yP(y)}{1-y} \, dy = P(x). $$ In terms of $Q(x) = \int_0^x \frac{yP(y)}{1-y} \, dy$, this reads $$ \frac{1}{1-x} + \frac{Q(x)}{x} = \frac{(1-x)Q'(x)}{x}, $$ which simplifies to $$ \frac{x}{1-x} + Q(x) = (1-x)Q'(x) = [(1-x)Q(x)]' + Q(x), $$ and so $$ (1-x)Q(x) = \int^y \frac{y}{1-y} \, dy = \int^y \left(\frac{1}{1-y}-1\right) \, dy = C - x - \log(1-x). $$ It follows that $$ Q(x) = \frac{C-x-\log(1-x)}{1-x}. $$ Since $Q(0) = 0$, we deduce $C = 0$. Differentiating and multiplying by $(1-x)/x$, we find that $$ P(x) = \frac{-\log(1-x)}{x(1-x)} = \frac{\sum_{n=0}^\infty x^n/(n+1)}{1-x}, $$ from which we can extract the coefficient of $P(x)$ to be $H_{n+1}$.

Finally, we can solve this without using any recurrences. Consider a random permutation of the numbers $1,\ldots,n+1$, which we think of as the $n$ places and the goal. The frog jumps to the first number. If it is not $n+1$, it scans the list to the first larger number, and jumps there. Repeat in this way, until you get to $n+1$. You can check that this defines the same random process.

Given a permutation, the number of jumps is the number of left-to-right maxima, that is, elements which are larger than all that precedes them. The probability that the $i$th number is a left-to-right maximum is exactly $1/i$, and so linearity of expectation shows that the expected number of jumps is $1 + 1/2 + \cdots + 1/(n+1)$.

As a bonus, the last solution allows us to calculate higher moments of the distribution. The main observation is that the indicator variables $X_i$ for the $i$th number being a left-to-right maxima are independent Bernoulli random variables. This implies that the variance of the number of left-to-right maxima is $$ \sum_{i=1}^{n+1} \frac{1}{i} \left(1-\frac{1}{i}\right) = H_{n+1} - \sum_{i=1}^{n+1} \frac{1}{i^2} = H_{n+1} - \frac{\pi^2}{6} + O\left(\frac{1}{n}\right). $$

One checks that the central limit theorem applies, and so the total number of left-to-right maxima is asymptotically normal.

Why is the answer roughly $\log n$? One might expect the answer to be closer to $\log_2 n$, since at each step, $n$ is roughly cut in half. However, this argument assumes that the changes to $n$ are additive, whereas in reality they are multiplicative. It is better to consider instead the effect of each step on $\log n$. If $m$ is the new value of $n$, then the expected value of $\log m$ is $$ E[\log m] = \frac{\log 1 + \cdots + \log n}{n+1} \approx \frac{1}{n}\int_0^n \log x \, dx = \left.\frac{1}{n} (x\log x-x)\right|_0^n = \log n - 1. $$ Therefore the value of $\log n$ decreases by roughly 1 each step, and so we expect the number of steps to be around $\log n$ rather than $\log_2 n$.

(With some work, this intuition can be made rigorous.)

  • $\begingroup$ Thanks for the answer! You have presented several solutions to the puzzle; however, each one appears to hinge on an ansatz that I am unable to connect in a direct way to the puzzle description. Could you add some more explanation as to how you arrived at the recurrence relation and the generating functions you use in your solutions? $\endgroup$
    – fezziks
    Sep 30, 2019 at 14:08
  • $\begingroup$ There are algorithms for solving such recurrences, implemented in computer algebra systems. It’s a rather broad topic. You can check the books generatingfunctionology and A=B. $\endgroup$ Sep 30, 2019 at 14:10

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