# Extension to the Tower of Hanoi

Discussion in 'Off-Topic Discussion' started by qqwref, Jan 24, 2012.

1. ### qqwrefMember

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I didn't know where else to put this, but I'm sure some people here will find this interesting. Hopefully this is actually new and not an obvious thing that I somehow couldn't find on Google.

So, you're probably all familiar with the Tower of Hanoi problem. You have three pegs, and $n$ disks of different sizes on one of the pegs, arranged with the largest disks on the bottom. The goal is to move all the disks to a different peg; the rules are that (a) you can only move one disk at a time (from one peg to another peg), and (b) you can't place a disk on top of a smaller disk. The problem is not too difficult to solve if you are just trying to finish it, but the mathematically interesting thing is that it requires $O(2^n)$ moves - roughly doubling for every new disk you add. The reason for this is that, to move $n$ disks, you must first move the smallest $n-1$ disks to another peg, then move the biggest disk, and finally move the rest of the disks a second time.

For a while I've been casually trying to think of a way to extend this to create puzzles which require $O(k^n)$ moves (for $n$ disks, and for a $k$ as large as we want). I think I've got an idea which will let us do this. Here is a rough picture of the idea, for a small case:

So, here's what's going on here. Choose some number $j$. Instead of 3 pegs you have $2j+1$ pegs, and instead of disks that cover one peg, they cover $j$ adjacent pegs. The trick is that the only way to move a disk is to move all of the smaller disks to $j$ of the other $j+1$ pegs - and, even then, this will only allow you to move the remaining disk clockwise or counterclockwise by one peg.

Figuring out the number of moves is a little trickier than in the normal 3-peg version, but at least we know that we can only move the biggest disk if all of the other disks are in a pile. To move a pile of $n$ disks by one space, we need to (1) move $n-1$ disks by $j$ spaces, (2) move the biggest disk, and then (3) move $n-1$ disks by $j$ spaces again. To move a pile of $n$ disks by $k$ spaces, though, we need to (1) move $n-1$ disks by $j$ spaces, (2) repeat (move the biggest disk, then move $n-1$ disks one space) $k-1$ times, (3) move the biggest disk one last time, and then (4) move $n-1$ disks by $j$ spaces again to finish. Only moves of 1 and $j$ spaces seem to be important here, so if we write those as $m_1(n)$ and $m_j(n)$ the recursion looks like this:
$m_1(n) = 2 m_j(n-1) + 1$
$m_j(n) = 2 m_j(n-1) + (j-1) m_1(n-1) + j$
Or, just in terms of the 1-move numbers:
$m_1(n) = 2 m_1(n-1) + (2j-2) m_1(n-2) + 2j-1$

Anyway, it looks like the $k$ constant (the number for which $m_1(n)$ grows like $k^n$) works out to $1 + \sqrt{2j-1}$. So that's pretty interesting, and we can definitely get $k$ as large as we like by choosing a sufficiently high $j$. In fact, we can set $k$ to be any even integer: for instance, let $j=41$, and then the sequence $m_1(n)$ grows as fast as $10^n$. Of course, boards with 83 pegs and disks with 41 holes are not particularly nice to play with, but seeing as the 5-disk puzzle takes 14,751 moves to just move the disks one space over, I don't think a lot of people would complain.

Incidentally, if you don't like the idea of disks spanning multiple pegs, you can get the same effect by using 1-peg disks and adding the rule that you cannot put a disk onto a peg if a smaller disk is on that peg or one of the next $j-1$ pegs in either direction.

2. ### MaeLSTRoMMember

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Sounds pretty interesting. Are you planning to make a virtual version?

3. ### qqwrefMember

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Hm, not really. I guess I could, but I'm not really sure how - I imagine the controls would be pretty finicky, and I'm not really sure how I'd display it...

4. ### OwenMember

Nov 2, 2009
You are a genius and deserve the Noble prize in Awesome.

5. ### qqwrefMember

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Alright guys, I made a sim :tu Try it here:
http://mzrg.com/js/hanoi/

I basically was thinking about how to expand the problem while keeping the restricted feel of the original one (only really one way to move stuff around) and I randomly thought of the idea of disks that would cover more than one peg, and of just barely having enough pegs to move those disks around. I initially had a wrong idea of how the math would turn out (since I didn't realize you could take shortcuts when moving a group of disks more than one space at a time) but the idea turned out to be what I wanted anyway.

PS: Wolfram Alpha gives the following closed form for $m_1$:
$m_1(n) = -1 + \frac{(1+\sqrt{2j-1})^n - (1-\sqrt{2j-1})^n}{\sqrt{2j-1}}$