 The Riddle this week is rather straightforward to explain: stacking identical objects (bars of length and mass two, say) on top of one another so that the center of each new bar is uniformly distributed along the previous bar, what is the distribution of the number of bars when the stack collapses? If I am not confused, the stack collapses the first time the centre of gravity of an upper stack leaves the interval represented by the bar just below. Namely $\left|\frac{1}{N-j} \sum_{i=j+1}^N x_i -x_j\right|>1$

when the xi are the bar centres, or equivalently $\max_{2\le j\le N-1} \left|\frac{1}{N-j} \sum_{i=j+1}^N \sum_{k=j+1}^i\epsilon_i \right|>1$

where the ε_i‘s are U(-1,1). Which is straightforward to code in R by looking at means of cumulated sums.

### 3 Responses to “stack overload”

1. cellocgw Says:

How about a variant: what’s the maximum allowable offset to guarantee the tower never collapses? Consider the related game where you stack blocks (or playing cards) until the top one lies entirely outside the location of the first one. In that game, the cards’ positions follow a log curve when the number of cards required is a minimum.

• xi'an Says:

Thanks : I think this variant is more “classical”, unless I misunderstand the constraint.

2. stack overload | R-bloggers Says:

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