Locked in a friendly rivalry, they work hard towards their goal. Thanks for watching guys!! Smiling a little better a big laugh.! Have a nice day :))Īsta and Yuno are two orphans who want the same thing: to become the Wizard King.
#OVERFLOW EP 3 FREE#
Thanks for joining, have fun, and check out and let me know what you guys think! Feel free to leave a clap and Follow This is still 15 moves, the 3 card solution has gone up by one.īut MOST interestingly, there’s now a clear pattern where the solution is always the start and end of the previous pattern with the new highest card as the move in the middle.The life of Osman Bey, the son of Ertugrul Gazi and the founder of the Ottoman Empire.
#OVERFLOW EP 3 FULL#
Vote for it in my poll if you like it!ĮDIT: It wasn’t right, it doesn’t account for the starting case of all cards starting face up! Adding this, I got a new full list of starting cases:Ĥ cards: (1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1) I’ll likely post more of these approximations by Python as a mini-series. Thanks to Matt Parker for posting this question! Perhaps because it took so long and I can’t be sure it’s right! Getting the answer by brute force was interesting and educational but not as exciting as spotting the logic that would get you there. I’m looking forward to finding out the reasoning for this.
![overflow ep 3 overflow ep 3](https://simkl.in/episodes/90/9015810344283b778_w.jpg)
I’ve left some performance tweaks out that I had to make to get there faster too! Things like removing permutations that don’t start with 1, since they would just be different starting points and removing any patterns that have 3 consecutive numbers as that’s clearly sub-optimal.īut I got an answer of 15 with a pattern of: We have 4 objects and so far, proven, at least 14 samples. Since the number of possible permutations allowing replacement when choosing r samples from n objects is n^r this could take a while. So it’s just about possible to see that this solution does work for all starting positions so it must be good. The possible starting positions with 3 cards are (1,), (2,), (3,), (1, 2), (1, 3), (2, 3) where I’m just showing the indices of cards that are face down to start with. So let’s run it on the scenario with 3 cards. That definitely sounds like the optimal solution. This comes back quickly with a solution of flipping the first card twice and then the second card once. We can test this on a smaller scenario, where there are only two cards. def find_best():įor switch_list in product(slots,repeat = size): We can start from 4 moves as we know it will take at least this long and then we can try all ways of making moves until a solution is found. Now for the biggie, we can search through all the ways of making moves until we find one that succeeds. We’ll make use of the NOT operator for flipping states.įor performance, if it does turn them all face down for that starting position then we can stop going through the list of moves, and if it doesn’t flip one of the starting positions face down at all we can stop checking that list of moves altogether. Check every possible starting position and if it will turn them all face down at any point.
![overflow ep 3 overflow ep 3](https://cdn.field59.com/KWTV/5918a5b5bf826383f49fd83d4f134815e9dfcf5a.jpg)
Then we need a function to check a list of moves. Here I’ve just listed the positions of the cards that start face down. Now we can initialise our scenario with 4 slots and calculate all the combinations for these slots: slots = Īll_combinations = list_combinations(slots) We start with a function to find all the possible starting arrays: def list_combinations(arr):Īrray = array + list(combinations(arr,r))