All natural solutions to x^2-y^z=1
http://uva.onlinejudge.org/external/127/12739.pdf
Minmax like Mastermind? FFT?
If A-B>C-D, A-C>B-D
http://uva.onlinejudge.org/external/127/12740.pdf
You have a graph of information: one node for every known tribe with the amount of known members in it, one edge between nodes if you know they are not the same tribe. You must make sure that at all times you either have one node with number bigger than the sum of the others or no independent set with sum bigger than the sum of the rest. Probably the proof of no big independent set comes from having cliques in your graph (you can only take at most 1 node from each clique for an independent set)
https://dl.dropboxusercontent.com/u/28563282/graph.html (wip)
Idea: a "good set" has a candidate tribe, a number of unknown tribes and a smaller good set completely connected to the candidate. When adding a node it goes to the unknown tribes, strategy is to request an edge between the candidate and an unknown iff (calculate condition from size of candidate, size of good set and number of unknowns), else give the edge to the smaller good set. Adding an edge between an unknown and the candidate pushes the unknown into the smaller good set (as unknown)
Conjecture: one strategy is having for each x a k where you queried all pairs (x,x-y) with ylast move and use that to do the searches better
quadrilateral thinking (on kong, could possibly use rubiks cube techniques too)
sat (run state is current variable to be set + clauses to satisfy)
pokemon speedruns (really hard)
hard kong badges: sonny 2, ebf2, max mesiria, ...
puzzle games: bloody fun day, digital upgrade, lock-n-roll, ...
protocol design for two team information games without public randomness/cryptography:
resistance
montainous mafia, one round
montainous mafia, full game
reverse mafia single round (trying to choose town instead of mafia)
spyfall
two player competitive:
super tic-tac-toe (combinatorial)
jaipur (almost combinatorial, but a lot of randomness involved)
truco
bfjoust (really hard)
mtg, pokemon, castle wars, ...
Try to apply polynomial methods on zero-sum games