Brent Yorgey @byorgey@mathstodon.xyz
Assistant professor of CS at Hendrix College in Conway, AR. Verify my identity: https://byorgey.keybase.pub/
Euclid's Orchard, the Euclidean algorithm, and Fibonacci numbers, with pretty pictures: https://mathlesstraveled.com/2017/07/22/a-few-words-about-pww-20/ https://mathstodon.xyz/media/JemTCvyUmis2JaQmpWo
Here's a better one with a white background instead of transparent. I wish there were a way to preview toots before posting. https://mathstodon.xyz/media/fePe1dPgQqME4_mx9MY
A Venn diagram on five sets. https://mathlesstraveled.com/2017/05/24/post-without-words-17/
Produced using the code here: https://hub.darcs.net/byorgey/mathlesstraveled/browse/bijections/6-PIE-pww/Venn5Subsets.hs
The first AlphaGo - Ke Jie match was exquisite. Definitely worth staying up until 2am my local time to watch the whole thing, though I fear doing that for all the matches is not really feasible, sleep-wise. #alphago17
Super excited for the first match of Ke Jie vs AlphaGo tonight (2:30am UTC). http://events.google.com/alphago2017/ #alphago17
https://mathlesstraveled.com/2017/05/19/post-without-words-16/ Contemplating how the existence of symmetric Venn diagrams relates to the existence of symmetric planar embeddings of hypercube graphs...
@jeffgerickson You probably hear this a lot, but thanks so much for making all your algorithms notes and assignments available. I just finished teaching my second iteration of an undergraduate algorithms course, and being able to draw on your collection of homework and exam problems made my job a lot easier!
Let \(G\) be a graph with \(|V|=n\). Any two of the following imply the third: 1. \(G\) is connected; 2. \(G\) is acyclic; 3. \(G\) has \(n-1\) edges.
1,2 => 3: by induction. Any walk must reach a leaf. Delete it and apply the IH.
1,3 => 2: by induction. Sum of degrees is \(2(n-1)\), so there are at least two leaves. Delete one and apply the IH.
2,3 => 1: Let \(G\) have \(c\) connected components. Since 1,2 => 3 for each, the total number of edges is \(n-c\), hence \(c=1\).
