## Conkers Mathematics

Yesterday I was on a walk in the Cotsworlds and our group came across a horse-chestnut tree. One of our number was unfamiliar with the game of conkers and so we ended up explaining the strange customs of English primary school playgrounds. Rules varied between different schools, but as with many such things, the rules that you grew up with will always be The Rules.

What we ended up thinking about was the ranking system of conkers. All new conkers start as none-ers. After you beat a better conker you get all of its kills plus one; so a six-er beating a ten-er becomes an eleven-er. It is a bit like inheriting your defeated opponent's wins plus adding one for your defeat of them themself. If a better conker wins against a weaker one then it just adds one to its total.

In mathematical terms the new name for the winner of a match between an m-er vs an n-er is a (max(m,n)+1)-er. (The autocorrect on my tablet was deeply unhappy with me writing that sentence.) There is a nice symmetry here in that both competitors are fighting for the same title.

A question presents itself. For a group of 10 none-ers, after an intense series of battles that leaves only one conker remaining, what are the range of values that it can have at the end? The top end is easy to see, since you can't do better than just having the victor defeat the other 9, so it can only possibly attain the rank of nine-er. The bottom end requires a bit more thought.

We want to lose as many wins along the way as possible. Let's try out the 10 case by having a first round where five conkers destroy the other give. We now have five one-ers. In round two we can have two one-ers beat two others, leaving two two-ers and one one-er. No matter which way you go from here we end up with a four-er. It is worth noting that we can attain all end values between four and nine.

Since every round of play necessarily adds one to the total and can only eliminate half of the field we can evaluate the base 2 log of the number of conkers in the field and round up. So we can write the range of values for a field ox x conkers as roof(log_{2}(x)) ≤ result ≤ x-1.

We can extend this even further by not having all of the conkers start as none-ers. For a general field of x conkers with values n_{1}, n_{2}... n_{x} then the range of values is just determined by the biggest one in the field. So roof(log_{2}(x)) + max(n_{1},n_{2}...n_{x}) ≤ result ≤ x + max(n_{1},n_{2}...n_{x}) - 1. Sometimes things look more complicated when you stick them in a general formula. The idea here is just the same as the case above, but with the value of the best conker added on since you are unable to lose those kills.