Timeline for Shortest path connecting two opposite points on a cube
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2017 at 9:03 | comment | added | Włodzimierz Holsztyński | In the past, I removed my each Answer which was late, without a word--even when I had 3 upvotes on occasions. I can remove my Answer also on this occasion. I was busy LaTeXing my answer (in the middle of it I also got distracted by writing a hm-shortcut which I have deleted). I can only say that I haven't read Ivan's solution before Arseniy's message under my present answer. (On the occasions of some earlier Questions I was unable to decipher the respective Answers or comments but only afterward). | |
| Mar 19, 2017 at 9:00 | comment | added | Arseniy Akopyan | You are right. Ivan (and me after) did not noted that we always live in dimension one less. Thank you, I've got two beautifull solutions of the problem. | |
| Mar 19, 2017 at 8:45 | comment | added | Włodzimierz Holsztyński | Now that I have read Ivan's proof in detail I see that indeed the proofs are similar. (But only one is complete and correct). Of course, I have left impatiently the (obvious!) step that a path can be replaced by one which is linear within each wall. Nevertheless, my proof is correct. Observe also that I have avoided talking about velocity--it's enough to consider the elementary comparisons for the straight portions within each wall. | |
| Mar 19, 2017 at 8:34 | comment | added | Włodzimierz Holsztyński | Ivan's proof does not work for $\ d=2\ $ nor for $\ d=3.\ $ The proof does not make any distinction between dimensions hence it should work for all dimensions $\ d\ge 2\ $ but it does not (strange?). | |
| Mar 19, 2017 at 8:22 | comment | added | Włodzimierz Holsztyński | What is the similarity? (I don't see any--I must be a poor reader?) | |
| Mar 19, 2017 at 8:13 | comment | added | Arseniy Akopyan | What is the difference with Ivan's proof? | |
| Mar 19, 2017 at 2:07 | comment | added | Włodzimierz Holsztyński | The product of morphisms (e.g. functions) and the diagonal product was (is?) often confused, where the diagonal product is sometimes called (cartesian) product. E.g. the great mathematician Schafarevich in his classical lectures on algebraic geometry used "cartesian product" (of functions) for what more precisely should be called diagonal product. | |
| Mar 19, 2017 at 2:00 | comment | added | Włodzimierz Holsztyński | "Diagonal product" because (roughly) it is the restriction of the product of morphisms to the diagonal. | |
| Mar 19, 2017 at 1:37 | comment | added | Todd Trimble | Thanks. I've never seen that notation used in the theory of categories, although the concept is of course quite familiar; I call it "tupling": mathoverflow.net/a/220213 | |
| Mar 19, 2017 at 1:12 | comment | added | Włodzimierz Holsztyński | "Triangle" notation stands for the "diagonal product" (the triangle or $\Delta$ reminds one of letter D for Diagonal). Richard Engelking used this notation in his classical textbook on general topology. It is used in the theory of categories in the context of the product of morphisms and objects: $\ f:X\rightarrow \prod_t Y_t = \triangle_t\, f_t\ $ for the respective (coordinate) morphisms $\ f_t:X\rightarrow Y_t$. | |
| Mar 19, 2017 at 0:59 | comment | added | Todd Trimble | What does the triangle notation mean? | |
| Mar 18, 2017 at 22:11 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |