# A problem inspired by a bicycle lock

Let there be a code $(a_1, a_2, a_3, \dots, a_n)$; with each term a real number between $0$ and $1$.

Suppose there is a lock, with $n$ circular cogs. To unlock it, each cog must be set to the value of that term in the code. That is, the first cog must be spun round to $a_1$, the second cog to $a_2$, all the way to the $nth$ cog, which is to be spun to $a_n$.

Each cog is modulo $1$; that is, the position $0$ is the same as position $1$. When unlocking it, the cogs can be spun either forwards or backwards. For instance, to get from $0.9$ to $0.1$ requires spinning the cog $0.2$ units forward or $0.8$ units backward. Any number of adjacent cogs can be spun together.

Suppose the lock starts with each cog at position $0$. It is easy to calculate the minimum distance that the cogs must be spun to reach the code and unlock it.

For instance, if the code is $(0.3, 0.6)$ the optimal solution is to spin both cogs forward $0.3$, and then the second cog forwards by $0.3$ again, for a total distance of $0.6$. If the code were $(0.2, 0.9, 0.7)$ an optimal solution would be to spin the first and second cogs forward $0.2$, and then to spin the second and third cogs backwards $0.3$; for a total distance of $0.5$.

Now for the question: for a given n, what code will maximize this minimum distance? It is clear that when $n=2$, the maximising code will be $(1/3, 2/3)$ or $(2/3, 1/3)$; for a distance of $2/3$. When $n=3$, the maximum distance that I can achieve is $0.8$, with the code $(1/5, 4/5, 2/5)$. However, I haven't proved that this is the maximum. Can anyone find a better code? What about for $n>3$ ?

• You get a higher cost for $(1/2,0,1/2)$. – Arthur Jan 12 '17 at 6:54
• Ah, good point - don't know how I overlooked that. – Thomas Delaney Jan 12 '17 at 7:01

## 1 Answer

Hint: try to solve (what appears to be) a harder problem, and compute for every $x$ and $n$ the function $g(x,n)$ defined as the maximum cost you can force out of $n$ cogs, given that the first cog needs to be turned by exactly $x$ (with $0\leq x \leq 1$). E.g. $g(x,1)=min(x,1-x)$. Can you write $g(x,n)$ recursively, i.e. as a function of $g(y,n-1)$? Can you solve the recursion?