Tower of Hanoi with three Colors Bodo Zinser, 2018 Rules: Like with the known 'Tower of Hanoi' you have three rods with n discs in descending sizes. You are only allowed to move one disc at a time and you are not allowed to move to a rod where the highest disc is smaller than the disc you want to move. The same rule applies here but here on the first rod are n red discs, on the 2nd are n blue discs and on the 3rd n white ones. The task is to rearrange the discs so that red ones go to rod 2, the blue ones to rod 3 and the white one to rod 1. If you don't know the normal 'Tower of Hanoi' then become acquainted with it first. You can easily simulate it with coins of different values. If you want to use minimum moves then for the normal 'Tower of Hanoi' there is a strict rule how to move the rods. E.g. if you want to move 4 discs on rod 1 to rod 2 the sequence is 13-12-32-13-21-23-13-12-32-31-21-32-13-12-32 (15 moves) which means move the highest disc on rod 1 to rod 3, then disc on rod 1 to rod 2, etc. It is nice playing the 'Tower of Hanoi' either with the normal rules or with 3 colors. And the real task is to find the minimum moves which are needed for the rearrangement. Algorithm: I had no full research and only a helpful algorithm (hoping this was the best): I was always looking where the largest disc which is not on place has to go to. Lets have an example with 3 discs: Disc 1 of rod 1 has to go to rod 2 place 1, so disc 1 of rod 2 has to go to rod 3 place 2, so disc 2 of rod 3 has to go to rod 1 place 3, so disc 3 of rod 1 has to go to rod 2 place 4, result: move disc 3 of rod 1 to rod 2 (=12) start again: disc 1 of rod 1 has to go to rod 2 place 1, so disc 1 of rod 2 has to go to rod 3 place 2, so disc 2 of rod 3 has to go to rod 1 place 3, result: move disc 3 of rod 3 to rod 2 and disc 2 of rod 3 to rod 1 (=32, 31) start again: disc 1 of rod 1 has to go to rod 2 place 1, so disc 1 of rod 2 has to go to rod 3 place 2, but not by moving the whole group disc 1 to disc 5 of rod 2 to rod 3 rather move disc 2 to disc 5 to rod 1 and only disc 1 of rod 1 to rod 3 so groupmovement (see below) of rod 2 from disc 02 to disc 5 to rod 1 (=g2021) and move disc 1 of rod 2 to rod 3 (=23) start again: disc 1 of rod 1 has to go to rod 2 place 1, so rod 2 is free, but disc 1 of rod 1 is not free, so first groupmovment of rod 1 from disc to 2 to disc 7 to rod 3 (=g1023) then move disc 1 of rod 1 to rod 2 (=12) and so on: (=g3022, 31, g2031, 23, g1033,12, g3042, 31, 23, 23, 21. 32) (for the last movements you have to be careful again) Group-movements: (excerpt from the documentation to the program simulating the problem) This is a powerful command which allows you to move an orderly group of discs to another place in one step instead of doing all the single steps. If you want for example to move a group of 6 discs it takes 63 (2^6-1) single movements but using the group command you can do it in 1 step. Of course the rules for moving are obeyed. This is only possible when the group has all descending discs and the other two rods are free. This is a way of playing in the normal 'Tower of Hanoi', too. Example: Start with level 3, then do the single movements 12-32-31. Now you want to move the discs on rod 2, from the second disc to the highest to rod 1. This would need 7 single steps 23-23-23-21-31-31-31. But instead you can use the group-statement g2021, which means 'move the group on rod 2, from place 02 to the highest to rod 1'. The syntax is 'gfaat: f=from-rod, aa=at of rod-from, t=to-rod', i.e. 5 digits. Instead of the last group-statement you could also play g2013 which obeys the rules, too, but is not optimal for getting the minimum moves.