Timeline for A continuous function for defining unique values to a 1024x1024 image with a 24 bit 3 color channel image
Current License: CC BY-SA 3.0
29 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2014 at 1:00 | vote | accept | David | ||
| Sep 25, 2014 at 2:18 | answer | added | guest | timeline score: 7 | |
| Sep 25, 2014 at 1:46 | history | edited | David | CC BY-SA 3.0 |
Added in clarifications about the values of neighboring pixels
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| Sep 24, 2014 at 20:28 | answer | added | ARupinski | timeline score: 1 | |
| Sep 24, 2014 at 18:27 | comment | added | David | @PerAlexandersson - Ok, not all combinations 2^216 may be a bit much, but I could test different starting values and positions and get a decent idea. | |
| Sep 24, 2014 at 18:24 | comment | added | David | @PerAlexandersson - Yes, I think its definitely a good option. I have been playing with 2 bit color channels on a 6x6 grid and I could definitely write a program to test all combinations for that configuration | |
| Sep 24, 2014 at 18:21 | comment | added | Per Alexandersson | @David, Yes, but my point is that perhaps a monte-Carlo inspired search can give a solution? | |
| Sep 24, 2014 at 18:02 | comment | added | David | @PerAlexandersson - The problem with this is that with most typical patters that may come of just selecting unique adjacent values, you end up hitting the edge of the color space. For instance, red for X, blue for Y will only get you 255 pixels of 1024. You can easily extend one way with the blue, but doing it for both is the root of the issue. | |
| Sep 24, 2014 at 17:54 | comment | added | David | @RicardoAndrade - For my application, the diagonals would not matter, as they could be no more than 2 different given the direct neighbor rule of a difference of 1. | |
| Sep 24, 2014 at 14:54 | comment | added | Per Alexandersson | Have you tried to start by determining the pixels row by row, and in each step, check what possible neighbors are, and select one that have not been used yet, at random. If there are a LOT of possible solutions to your problem, this might be feasible. | |
| Sep 24, 2014 at 14:18 | comment | added | Ricardo Andrade | @David, I have a question meant to ensure that Dirk's interpretation is correct. What do you mean by neighbouring pixels? Do you mean pixels which share an edge, or do you also allow pixels which are located diagonally from each other (only share a vertex)? | |
| Sep 24, 2014 at 13:48 | history | reopened |
Ramiro de la Vega Willie Wong Stefan Kohl♦ Felipe Voloch Carlo Beenakker |
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| Sep 24, 2014 at 13:46 | comment | added | David | @PerAlexandersson - I am a Computer Engineer and I would consider myself strong in math and struggling with this. I posted here because I thought that the general StackOverflow community would not be able to solve this. Mathematica is a good idea. I may try that if I or others on here are unable to come up with a solution. | |
| Sep 24, 2014 at 13:31 | history | edited | David |
edited tags
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| Sep 24, 2014 at 12:54 | comment | added | Per Alexandersson | I am also curious to a solution; technically, I consider this a CS question, and people over at say Mathematica.stackexchange would most likely cook up a solution quite quickly... | |
| Sep 24, 2014 at 6:24 | comment | added | Dirk | Put differently: You look for an injective mapping $f:2^{10}\times 2^{10} \to 2^8\times 2^8\times 2^8$ which is Lipschitz continuous with constant 1 where we put the $1$-norm on $2^{10}\times 2^{10}$ and the $\infty$-norm on $2^8\times 2^8\times 2^8$. | |
| Sep 24, 2014 at 3:04 | comment | added | David | Now that I have clarified, I would appreciate it if this question could be taken out of hold, or if someone could tell me what else needs to be clarified. Thanks! | |
| Sep 24, 2014 at 0:50 | history | edited | David | CC BY-SA 3.0 |
Clarified the problem and added an example of a "non-continuous" solution
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| Sep 23, 2014 at 23:51 | review | Reopen votes | |||
| Sep 24, 2014 at 13:49 | |||||
| Sep 23, 2014 at 22:50 | history | closed |
Dima Pasechnik Douglas Zare Stefan Kohl♦ Ricardo Andrade Nik Weaver |
Needs details or clarity | |
| Sep 23, 2014 at 19:57 | comment | added | David | I mean that each individual neighboring pixels channel value does not change by more than one. Thanks for asking for this as I realize this does not match up with the real definition of a continuous function. | |
| Sep 23, 2014 at 19:57 | history | edited | David | CC BY-SA 3.0 |
added 206 characters in body
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| Sep 23, 2014 at 19:53 | comment | added | Per Alexandersson | Define continuous in this context. | |
| Sep 23, 2014 at 19:24 | history | edited | David | CC BY-SA 3.0 |
added 47 characters in body
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| Sep 23, 2014 at 19:21 | comment | added | David | If I can find a function f(x, y) = r, g, b, then the fact that all inverses will not exist should not be a problem as I know the values will be in range. This is admittedly a computer engineering problem and in a strictly mathematical sense, you are defiantly correct that this is not possible. | |
| Sep 23, 2014 at 19:19 | review | Close votes | |||
| Sep 23, 2014 at 22:51 | |||||
| Sep 23, 2014 at 19:15 | comment | added | Carlo Beenakker | how can you hope for an invertible relation from $2^{24}$ to $2^{20}$ elements? | |
| Sep 23, 2014 at 19:04 | review | First posts | |||
| Sep 23, 2014 at 19:07 | |||||
| Sep 23, 2014 at 19:02 | history | asked | David | CC BY-SA 3.0 |