I had one of these as a child and was practically screaming at the screen when you'd make an error or repeat backwards. While of course Grey codes are a way of thinking of it, it's even simpler: no partials exist. You must slam to the end one way or another every time. The only decision is to turn the 1st or second position when you are at the starting area. Everything else is either going backwards (undoing), or progress. It's extremely procedural. Fun to see this puzzle after many years.
It's like the Chinese ring puzzle. At any point there are only two possible moves. One will undo previous move, the other will progress toward solution. The only decision is at the beginning, by removing the 1st or 2nd ring. [Remove 1st for odd number of rings, 2nd for even number of rings].
![[186] Solving the Crazy Loco Lock Puzzle!](http://i.ytimg.com/vi/Q_CmLsFHj6w/mqdefault.jpg)
![[186] Solving the Crazy Loco Lock Puzzle!](/img/tr.png)
![How to solve Spinout [Binary Arts/ Think Fun] ?](/img/n.gif)
What a great puzzle, awesome that you could find one and try it out, thanks for sharing
Thanks for the tip!
@@LockPickingCuber your welcome, glad you enjoyed it
I love binary puzzles. thanks for the great video. cool puzzle!
Glad you enjoyed it!
I had this! loved it SO MUCH
Oh cool!
I had one of these as a child and was practically screaming at the screen when you'd make an error or repeat backwards.
While of course Grey codes are a way of thinking of it, it's even simpler: no partials exist. You must slam to the end one way or another every time. The only decision is to turn the 1st or second position when you are at the starting area. Everything else is either going backwards (undoing), or progress. It's extremely procedural.
Fun to see this puzzle after many years.
It's like the Chinese ring puzzle. At any point there are only two possible moves. One will undo previous move, the other will progress toward solution. The only decision is at the beginning, by removing the 1st or 2nd ring. [Remove 1st for odd number of rings, 2nd for even number of rings].