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Question

So I've been studying skip lists in school and we briefly spoke about if we were to use an 'unfair coin' in a skip list rather than a fair coin, (example: probability of the unfair coin flip resulting in a value of "Heads" is set to p where 0 < p < 1 (so that the probability of "Tails" is 1 −p).

There are a few things I've wondered about this that I don't really understand as we passed over the topic so quickly.

1. What would happen to the height/size of a skip list if we did this? It would obviously change things if the probability was skewed, right? Say it contained arbitrary n elements, obviously the heights would be different than if we used a fair coin.

2. How would the expected number of promotions that would be received by an arbitrary node when adding it to the skip list change? I don't know if it would in this scenario but it was a topic of discussion.

I'm not looking for someone to just give answers without me actually understanding, but if you could actually explain why these changes are happening so that I can understand how it is being affected by a change in probability, I would appreciate it.

EDIT: I think I now understand after doing some comparisons of different probabilities with the equation provided on page 99 of Pat Morin's Open Data Structures book. I'll post my solution in the comments once I figure it out in order to help others with the same question.

Answer 1

You can look up the answers to your questions here: https://en.wikipedia.org/wiki/Skip_list#Description
I will try to explain them:
Let p be the probability for an element to be promoted to the next level, then

1. the height will be log _1/p n .
   Reason:
   In the last row 0 we have n elements. As every element has the probability of p to appear in the next higher row, we have n * p elements in row 1. With the same arguing we will get n * p ² elements in line 2. As you can see we have n * p ^x elements in row x .
   The highest row will have 1 item, so we get the equation 1 = n * p ^x and solving this for x we get x = log _1/p n ( x = log ₂ n for a fair coin).
2. each element will appear in 1/(1-p) lists on average.
   Reason:
   The chance of only appearing in one list is (1 - p) (chance of promotion is p ).
   The chance of making it to the second but not the third list is p * (1 - p) (promoted once [ p ] and not on the second coin flip [ 1-p ]).
   Second but not third p ² * (1 - p) and so on
   In general:
   p ^x-1 * (1-p) for being in x lists.
   For the expected value of lists an element appears in we have:
   exp = 1 * (1-p) + 2 * p * (1-p) + 3 * p ² * (1-p) ...
   = ∑ _i=0 ^∞ (i+1) * p ⁱ * (1-p)
   = (1-p) * ∑ _i=0 ^∞ (1+i) * p ⁱ
   = (1-p) * 1/(1-p) ²
   = 1/(1-p)
   Proof for the elimination of the sum to 1/(1-p) ² can be found here: https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series