Timeline for Maximal score for the 2048 game [duplicate]
Current License: CC BY-SA 3.0
35 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Aug 29, 2014 at 14:38 | history | edited | Willie Wong |
edited tags
|
|
| Aug 24, 2014 at 3:08 | review | Reopen votes | |||
| Aug 24, 2014 at 12:19 | |||||
| Aug 21, 2014 at 3:37 | comment | added | Włodzimierz Holsztyński | @Pietro, thank you (no problem :-). You and others are right that the problem was easy to solve. On the other hand the generalizations of the game 2048 are mathematically appealing (to me). I am even thinking about other games which are based on sliding and merging. This time they would have a strong number theory flavor. (Would they be still fun to play by the general public, like 2048?--I doubt it :-) | |
| Aug 20, 2014 at 15:07 | comment | added | Pietro Majer | however, I agree that making everything rigorous leads to considerations similar to those in your proof. Apologies. | |
| Aug 19, 2014 at 19:36 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typo
|
| Aug 19, 2014 at 18:38 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typo
|
| Aug 19, 2014 at 18:07 | comment | added | Pietro Majer | Maybe just proving by induction: "in the process to create $2^n$, at some moment, at least $n-1$ cells must be occupied". Thus $2^{18}$ is not possible. Since two $2^n$'s are needed to create $2^{n+1}$ the induction seems to work. | |
| Aug 19, 2014 at 17:47 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
PS. A question about a quote, and a clarification of the additional issue of the TOTAL sum.
|
| Aug 19, 2014 at 17:22 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
PS. Q. and a clarification of the additional issue of the TOTAL sum.
|
| Aug 19, 2014 at 16:47 | comment | added | Włodzimierz Holsztyński | @Pietro and some others: the bound on the total sum of tiles as given by the consecutive powers of 2: $\ 4+8+\ldots+2^17 =2^18-4\ $ can naturally obtained AFTER one proves the upper bound on the maximal tile. Can you prove the bound on the total sum at the same time and before you get a bound on the maximal tile? | |
| Aug 19, 2014 at 16:13 | comment | added | David E Speyer | math.stackexchange.com/a/902535/448 | |
| Aug 19, 2014 at 15:59 | comment | added | Włodzimierz Holsztyński | @David: could you provide a link to your upper bound proof? | |
| Aug 19, 2014 at 15:49 | comment | added | David E Speyer | I put up a quick proof that $2^{16}$ is an upper bound on the math.SE with only $2$'s on the math.SE thread. Similarly, $2^{17}$ is an upper bound with $2$'s and $4$'s. I agree that the state of that thread is a little embarrassing; I'm not sure that anyone has given a proof that $2^{17}$ is achievable. | |
| Aug 19, 2014 at 15:33 | comment | added | Włodzimierz Holsztyński | @S.Carnahan: there is a difference between magic and illusion, as well as between a proof and an impression. | |
| Aug 19, 2014 at 15:21 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
italics
|
| Aug 19, 2014 at 15:09 | comment | added | Włodzimierz Holsztyński | @Pietro: if it's quite obvious then prove it. | |
| Aug 19, 2014 at 15:04 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A comment about an earlier post on 2048
|
| Aug 19, 2014 at 10:40 | comment | added | S. Carnahan♦ | @EricWofsey To be more precise, Douglas Zare mentioned the magic words "binary expansion" which, when invoked, make the question rather trivial. | |
| Aug 19, 2014 at 10:28 | review | Reopen votes | |||
| Aug 19, 2014 at 13:35 | |||||
| Aug 19, 2014 at 7:43 | comment | added | Eric Wofsey | Never mind, I see that the trivial upper bound was shown to be sharp in the MSE post linked in the second comment. Strictly speaking I would say this post should be closed as a duplicate of that MSE question, but I assume that is not possible. | |
| Aug 19, 2014 at 7:43 | comment | added | Pietro Majer | I think it is quite obvious that no power higher than $2^{17}$ can appear, because to create it, all lower powers of 2 must be there together at some moment, starting by a 4, plus an additional 4 as a last entry to start the doubling process. On the other hand, a bustrophedon sequence of increasing powers may actually be done, if one is lucky enough, allowing to reach $2^{17}$. | |
| Aug 19, 2014 at 7:37 | comment | added | Eric Wofsey | @S.Carnahan: That answer merely gives the trivial upper bound and is certainly not a complete solution. This question is neither an actual duplicate of the older question, nor does Douglas Zare's answer completely answer this question. | |
| Aug 19, 2014 at 7:31 | comment | added | S. Carnahan♦ | Douglas Zare's answer at mathoverflow.net/questions/160703/… makes this question superfluous. | |
| Aug 19, 2014 at 7:30 | history | closed | S. Carnahan♦ | Duplicate of Expected halting time for "The 2^n Game" (aka 2048) -- with random moves | |
| Aug 19, 2014 at 5:54 | answer | added | Włodzimierz Holsztyński | timeline score: 1 | |
| Aug 19, 2014 at 3:31 | comment | added | Włodzimierz Holsztyński | TYPO: read above upper case may... as upper bound may.... | |
| Aug 19, 2014 at 1:36 | comment | added | Gerry Myerson | See also mathoverflow.net/questions/160703/… | |
| Aug 19, 2014 at 1:07 | comment | added | Włodzimierz Holsztyński | CAPS and Daniel, thank you for the links. I have certain difficulties to read them with full understanding (my fault). I'll present a clean proof for the upper bound $X$, while the construction of the maximal game perhaps must be at least a bit messy (I'll not plan to do it at this time). | |
| Aug 19, 2014 at 0:59 | comment | added | Daniel Soltész | reddit.com/r/2048/comments/214njx/… | |
| Aug 19, 2014 at 0:55 | comment | added | Włodzimierz Holsztyński | I used to teach Finite Mathematics, a section about Markov chains. I prepared a special game which would illustrate the topic--and, why not, would be entertaining. Instantly a student objected. He wanted serious things... I could present an $n\times x\ $ game. (and I will). Would it make it research? Do I have to say suchn things explicitly that a generalization should be expected? A lowering the upper case may present obstacles (don't confuse them with my Archimedes obstructions in dimension 6 :-) which are worthwhile to address; both in general and in the popular case, as an illustration. | |
| Aug 19, 2014 at 0:51 | comment | added | NAME_IN_CAPS | math.stackexchange.com/questions/716469/… | |
| Aug 19, 2014 at 0:13 | review | Low quality posts | |||
| Aug 19, 2014 at 0:23 | |||||
| Aug 19, 2014 at 0:08 | comment | added | Włodzimierz Holsztyński | I have obtained many down-votes but this time under a thirty seconds. I deserve a special badge! | |
| Aug 18, 2014 at 23:58 | history | asked | Włodzimierz Holsztyński | CC BY-SA 3.0 |