Counting Count the number of paths from S (start) to G (goal) that stays along the white roads. Paths cannot use the same road twice but may visit the same intersection twice. As an example, the pink path shows one valid path.
Expected difficulty Insane • Answer • Comment/E-mail if you want a solution to be published
Puzzle 80: Hardcore Mode
Counting
Erm. You may use a calculator or a program. (If you managed to program the solution, then you deserve the answer. I’m not responsible if you miscalculate when you’re multiplying large numbers by hand.)
This is a rejected puzzle of a set born out of a stupid idea. Why did I even think of this? The good news is I have a free Brilliant.org problem idea, and I can practice my programming skills.
Chaos at the Sky 
Quite a diverting little puzzle, but not very hard.
I think the answer is x + 6x^2 + 2x^3 + 20x^4 + 6x^5, where x= 1225 (=35^2).
Cheers,
Andrew.
The thing is that in a competition setting, you’ll be all shaky to check whether you have obtained all the paths. (Not to mention 35^10 is a pain to compute.) That’s why the puzzle is rated insane. In a more casual setting I’d call it hard.
Insane would be if you have to compute paths for the next level of the fractal 😀
16,596,325,314,442,475 nice puzzle thanks
Spoilers ahead!
Let a_i be the number of paths from A to B in this graph (hopefully, the spacing will come out right in the actual comment as well as in the editor)
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of length 2i which do not repeat arcs, for i = 1,2,3,4,5. Then the answer is… hidden by chao 😉
Does this count?
Ahh, my previous comment has become unreadable without the LaTeX. I thought that it worked in the comments, too Please delete it (and this comment as well), the first commenter has done the job already.
I’ve edited your comment. I guess you can use code tags in place of LaTeX?