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Question - 7 orbs are labeled 1−7 and are linked linearly in a vertical stack from the ceiling with orb 1 being a part of the ceiling and orb 7 being closest to the floor. Each orb is attached to adjacent orbs by a chain link. At each time step, one of the remaining links is going to be uniformly at random selected and cut. As a result, all the orbs below that link will fall and shatter. What is the expected number of cuts needed until orb 1 is the only remaining orb?

Could someone help me with an elegant solution for the same, my solution is below.

Attempt -

Let's call this a Orb(7) Problem

So we need to find Expected Cuts in Orb(7) Problem - Cuts(Orb(7))

I figured the first cut can be made in 6 ways, with one of the six way leaving us with 1 orb (Target Requirement), other way in 2 orbs, other in 3 orbs and so on till 6 orbs -

Cuts(Orb(7)) = 1/6 * 1 + 1/6 * (1 + Cuts(Orb(2))) + 1/6 * (1 + Cuts(Orb(3))) .. 1/6 * (1 + Cuts(Orb(6)))

Cuts(Orb(7)) = 1 + 1/6 * [Cuts(Orb(2)) + Cuts(Orb(3))) .. Cuts(Orb(6))]

Which can be understood as -

$$Cuts(Orb(n)) = 1 + (1/(n-1)) * \sum_{k=2}^{n-1} (Cuts(Orb(k))$$ Note : $$Cuts(Orb(2)) = 1 $$

So now it becomes a backtracking problem.

Just in case someone was curious, I plotted it in Python as I was curious about the nature of the Cuts(Orb(n)) - Number of Cuts

Any clue why such a curve is observed?

Thanks

longnightofsolace
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1 Answers1

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Let $X(n)$ represent the random number of cuts until the process stops, when there are $n$ orbs in the chain.

Then $X(1) = 0$ by definition, and $X(2) = 1$ since there is only one link to cut.

For $X(3)$, we note that with probability $1/2$, the upper chain is cut between orb $1$ and $2$, in which case $X(3) = 1$, and with probability $1/2$, the lower chain is cut, and $X(3) = 2$. So the expected number of cuts is $$\operatorname{E}[X(3)] = 1(1/2) + 2(1/2) = 3/2.$$

Another way to look at $X(3)$ is to define a discrete uniform random variable $$Y(n) \sim \operatorname{Uniform}\{1,2,\ldots,n-1\}$$ that is the chain between orbs $Y(n)$ and $Y(n+1)$ that is cut, when the chain has $n$ orbs. Then note that $$\operatorname{E}[X(3)] = 1 + \operatorname{E}[X(1)]\Pr[Y(n) = 1] + \operatorname{E}[X(2)]\Pr[Y(n) = 2] = 1 + 0(1/2) + 1(1/2) = 3/2.$$ The extra $1$ on the RHS is because we had to cut one chain.

Then it is easy to see that in general,

$$\operatorname{E}[X(n)] = 1 + \sum_{y=1}^{n-1} \operatorname{E}[X(y)]\Pr[Y(n) = y] = 1 + \frac{1}{n-1} \sum_{y=1}^{n-1} \operatorname{E}[X(y)].$$

to solve this recurrence, we rewrite it as $$(n-1)(\operatorname{E}[X(n)] - 1) = \sum_{y=1}^{n-1} \operatorname{E}[X(y)]. \tag{1}$$

Consequently, we also have by incrementing $n$ $$n(\operatorname{E}[X(n+1)] - 1) = \sum_{y=1}^n \operatorname{E}[X(y)]. \tag{2}$$

Subtracting $(1)$ from $(2)$ yields

$$n\operatorname{E}[X(n+1)] - (n-1)\operatorname{E}[X(n)] - 1 = \operatorname{E}[X(n)], \tag{3}$$

which in turn implies $$\operatorname{E}[X(n+1)] = \operatorname{E}[X(n)] + \frac{1}{n}. \tag{4}$$

Therefore,

$$\operatorname{E}[X(n)] = \sum_{k=1}^{n-1} \frac{1}{k} = H_{n-1}, \tag{5}$$ the $(n-1)^{\rm th}$ harmonic number.

heropup
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