Walking city streets: Catalan Closed Form (visual proof from lattice paths)

[@robnicolaides3070]

Lovely video! Is there a proof that corresponds to an algorithm that reflects the closed form formula more directly without using the subtraction of binomial coefficients?

[@MathVisualProofs]

Thanks. Good question! There is a way to get directly there by using an idea called exceedance. You can find the argument on Wikipedia. Perhaps I will make a related video at some point :)

[@robnicolaides3070]

@@MathVisualProofs thanks for pointing me to the Wikipedia article, the proof involving exceeded e is so nice :) looking forward to the next video

[@jjbigdub]

Great explanation, Thank you!

[@MathVisualProofs]

Thanks!

[@UDHAV79]

I had a similar question in my math test once, the question was about the number of shortest paths a man can take to reach a specific intersection of a city on the north-east line in which the roads are laid out as grid lines on a plane without counting those paths that lie below the north-east. I had a similar thinking process where I thought that to reach a specific intersection (n,n) on the northeast line with the shortest path, the man must travel n north units and n east units, therefore the total number of shortest paths are the number of possible arrangements of n north units and n east units which are equal to (2n)!/(n!n!) and since we won't include those below the north-east line, we would have to remove half the approaches and therefore (2n!)/(2(n!n!)). Not sure if it's correct doe :)

[@jakobr_]

Unfortunately I think the “divide by two” step doesn’t really work here. If none of the paths on the whole square passed through the line to begin with, it would work, by pairing up each lower path with an upper path. But since some paths do cross the middle line, the real answer will be less than half. Unless n=0 or 1.

[@UDHAV79]

@@jakobr_ hmm you are right. Thanks for pointing out my mistake. Appreciate it.

[@MathVisualProofs]

@@UDHAV79 Jakob has nice reasoning. The question you ask is actually exactly the question given in this video, so the answer appears at the end. You have to take 2n!/(n!n!) and divide out by n+1 (not 2). This is a bit counterintuitive because it is not even clear from formulas that 2n!/(n!n!) will be divisible by n+1.

[@UDHAV79]

@@MathVisualProofs Oh I see! thanks for the reply!

[@Daniel-mt7xr]

not sure which one is the previous video you refer to? thanks

[@MathVisualProofs]

There are three linked in the description. Just depends on which one you want.

[@hemadhumal4036]

From,Where did u get this content?.. 😅

[@MathVisualProofs]

What do you mean?

[@hemadhumal4036]

I mean ,which book you prefer if you won't let open secret ,it's fine... 😊

[@hemadhumal4036]

Your visualizing mathematics vedios are awesome ,things get quickly Understandable.. Please, can you do face reveal ...🙏🙏🙏

[@hemadhumal4036]

Reply

[@MathVisualProofs]

@@hemadhumal4036I like the Stanley book linked in description. I also like combinatorics book by Tucker.

[@anshulmanapure1980]

Yo, I'm lost. Maybe I dumb.

[@MathVisualProofs]

Might want to start with the videos linked in the description if you aren’t familiar with lattice paths in terms of counting arguments.

[@kaaristotelancien3005]

you're lost because you overestimate it, maths are just a lego mind game