stack overload

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 x_iare 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.

This entry was posted on March 3, 2021 at 12:21 am and is filed under Books, Kids, R with tags FiveThirtyEight, gravity, riddle, Stack Overflow, The Riddler, tio.run. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “stack overload”

cellocgw Says:
March 8, 2021 at 6:10 pm

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.

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- xi'an Says:
  March 9, 2021 at 3:44 pm
  
  Thanks : I think this variant is more “classical”, unless I misunderstand the constraint.
  
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stack overload | R-bloggers Says:
March 4, 2021 at 8:11 am

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