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Four color theorem

Please don't add your map supposedly requiring 5 colors to the Four-color theorem article again. First, the image you replaced is an illustration of a map requiring four colors, and it's a well-drawn picture -- yours isn't. Second, your image only requires four colors anyway, please see the image on the right. -- ArglebargleIV 12:41, 22 October 2007 (UTC)

I guess, you forgot the white surrounding. --Fsswsb 13:49, 22 October 2007 (UTC)

Didn't care to put THAT much work into it. -- ArglebargleIV 19:11, 22 October 2007 (UTC)

Please do NOT replace the diagram on Four-color theorem with any version of your diagram. The diagram currently used is nicely and aesthetically drawn; your diagram's boxes don't even touch each other most of the time. If you think your diagram is superior, I suggest you discuss it on the talk page first, because perpetually reverting to your version without consensus and discussion is likely to earn you a block. -- ArglebargleIV 19:11, 22 October 2007 (UTC)

Please stop changing the image on Four color theorem; it is incorrect, redundant, and violates Wikipedia:No original research. Specifically:

1. Please do not introduce what appears to be original research into Wikipedia; it is against Wikipedia rules: see Wikipedia:No original research. Even if you had an example that disproved the four color theorem (which you don't — see below), it would be unacceptable to use until it had been published in a proper third-party source.
2. The type of images you are producing are already discussed in Four color theorem#False disproofs, with an existing example and how to solve it. Please read the article before removing other people's work.
3. As discussed in Four color theorem#False disproofs, your own examples are wrong. Just because you can't do it in 3 minutes, that doesn't mean mapmakers (or anyone else) can't; indeed, your "impossible" examples have been very easy to solve.
4. Line art shouldn't be in JPEG. Please use PNG or SVG, per Wikipedia standards.

For your convenience, I have colored both of the examples you have uploaded. I have colored each of them both ways: with the border lines ignored, and with the border lines considered their own area, in case you're "clever" enough to claim that someone forgot the "black part".

• 
  4col2X with black considered just border: solved in four
• 
  4col2X with black considered separate area: solved in three
• 
  Color5 with black considered just border: solved in four
• 
  Color5 with black considered separate area: solved in two!

Further attempts to change the article with incorrectly colored images of any kind will probably be interpreted as introducing intentional factual errors and dealt with accordingly. Please do not introduce original or unverified research into Wikipedia. --Closeapple 22:36, 22 October 2007 (UTC)

Thank you for providing the nice gallery. It shows, that it is indeed fairly easy to construct a figure, that requires at least five colors taking into account the surrounding. In the first example that still doesn't work with a single surrounding. But with a surrounding splitted into three parts it will work. In this case the surrounding requires three colors. Therefore, where is no solution, if all three parts have borders to areas having the same two colors. --Fsswsb 07:41, 23 October 2007 (UTC)

I think that, perhaps, you are very confused. There isn't any surrounding area on the first picture (4col2X). And if there was, not only could it be done, but it's already colored for it: Just use yellow for the new outside area (which doesn't even exist right now) and you won't even have to re-color anything else. --Closeapple 11:33, 23 October 2007 (UTC)

Yes, as I mentioned already, in this example a single surrounding is not sufficient. If the surrounding area is splitted in three parts of about equal size, five colors are required, since three colors are used for that splitted surrounding. --Fsswsb 11:57, 23 October 2007 (UTC)

If the areas are separated, the gaps between the separated areas may also be regarded as colored areas and hence a one color solution never exists. A common point can always be expanded to a small area, like the yellow spots in the first example. One may also combine the various figures, so that they have common borders. The four color theorem is therefore clearly disproved. --Fsswsb 08:41, 24 October 2007 (UTC)

It is impossible for 4 areas to all be against each other unless one of them is in the "middle" or wraps around the outside; therefore, you cannot add anything that touches all four old things at once unless the new thing separates 2 of the old things that were touching, in which case you can just recolor the map. Try it on a piece of paper: draw 4 areas that all touch each other, then try to add a 5th one that touches everything - you can't touch all 4 old areas without separating at least one old border. And once they're separated, they can both the same color when the map is re-colored. --Closeapple 10:22, 24 October 2007 (UTC)

You are right, if you choose only five areas, four old things plus one surrounding or something inside, it is always possible to color them with four colors. But, that doesn't mean that this will work for any map having more than five areas. If you have one area surrounded by five, you need at least three colors for the surrounding and one additional for the center. Nevertheless, if you choose five of the six areas, you never will need more than three colors. That shows, that your proof for the four color theorem is wrong. Not really surprising, since the four color theorem is just wrong. --Fsswsb 10:44, 24 October 2007 (UTC)

I emphasize that your examples are incorrect, and all maps can be coloured with 4 colours. This is a proven formal result and there is no counterexample. Closeapple has been unusually accomodating in demonstrating your error to you, and I strongly suggest you stop attempting to introduce these incorrect and inconsistent facts into the article. Dcoetzee 11:54, 31 October 2007 (UTC)

It's easy to construct a huge labyrinth of regions having borders to two, three or four other regions and to one large "background". In such a situation there exists one unique — besides trivial permutations of colors — solution with four colors. In order to get a counterexample it is necessary to split the large background area. It is quite easy to find some counterexample in this way, but it's difficult to prove formaly, that really no solution with four colors exists. --Fsswsb 19:34, 31 October 2007 (UTC)

You're wrong. There is no counterexample. Your proposed counterexample is not one, and the regions in it can be coloured with 4 colours, and in fact there is a fast algorithm to do so, even if it's not obvious to you. It's good that you're exploring the problem, but you're not going to so easily disprove a formally proven result that was attacked thoroughly for 100 years by amateurs and professionals alike. I suggest you move on to learning about something else. Dcoetzee 14:25, 1 November 2007 (UTC)

I noticed that, after the "ultimate disproof of the four color theorem" (ColorXXX.jpg) was solved in 4 colors, you created a "fixed" version. Is this one the ultimate disproof? It's hard for me to tell since I solved it in 4 colors again— or 3, depending on how you look at it:

• Fsswsb's improvement of the "ultimate disproof of the four color theorem"
  Fsswsb's improvement of the "ultimate disproof of the four color theorem"
• 
  "ColorXXX fixed" with black lines as borders: solved in four
• 
  "ColorXXX fixed" with all black as area: solved in three

And yes, there are other areas between the big purple rectangle and the purple ones above it; those little areas were there in your version, so go click on it and look.

You can split the so-called "background" (which is no different than any other area) as much as you like. It will still be solvable in 4 colors. Your counterexamples only appear valid to you because you are coloring them incorrectly and "painting yourself into a corner". Your error, which you've repeated 4 times now, is not even unique; this kind of error you're making is already addressed in the article itself. The only person who seems unable to solve your diagrams in 4 colors is yourself. --Closeapple 21:44, 1 November 2007 (UTC)

• Aw, you beat me to it. (My four-coloring is at Talk:Four color theorem. -- ArglebargleIV 22:43, 1 November 2007 (UTC)

If you start drawing some areas on a piece of paper with a certain background color you only have three colors left for the rest. After coloring two of them, the color of the third is already defined, since only one color is remaining. That means, after defining the color of the background and the color of two other regions, the colors for all the rest is already defined. Therefore, it seems, that a disproof is fairly simple. This is not the case, since it is necessary to split the background or some regions will not have borders to the background any more. Therefore, it's allowed to give the background different colors or use the background color for some regions. Nevertheless, if you draw a huge labyrinth, it's quit unlikely, that a solution with four colors will exist, although it's quit hard to prove that. --Fsswsb 11:34, 2 November 2007 (UTC)

Four color theorem article changes

Please do not add content without citing reliable sources, as you did to Four color theorem. Before making potentially controversial edits, it is recommended that you discuss them first on the article's talk page. If you are familiar with Wikipedia:Citing sources please take this opportunity to add your original reference to the article. Contact me if you need assistance adding references. Thank you. --Closeapple 17:16, 24 October 2007 (UTC)

Please do not add unsourced or original content, as you did to Four color theorem. Doing so violates Wikipedia's verifiability policy. If you would like to experiment, please use the sandbox. Thank you. --Closeapple 19:30, 24 October 2007 (UTC)

You continue to ignore our core policies like Wikipedia:no original research and add material to the article against the Wikipedia:consensus. Additionally, your claims are wrong. This is the last warning you'll receive. If you don't stop this, you will be blocked from editing Wikipedia. Yours, Jitse Niesen (talk) 12:49, 31 October 2007 (UTC)

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Blocked

You have been blocked from editing for a period of three days in accordance with Wikipedia's blocking policy for because you continue to change four color theorem against consensus and our no-original-research policy. Once the block has expired, you are welcome to make constructive contributions. If you believe this block is unjustified, you may contest the block by adding the text {{unblock|your reason here}} below. Jitse Niesen (talk) 12:48, 2 November 2007 (UTC)

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Thanks for uploading Image:ColorXXX.png. However, there is a concern that the rationale you have provided for using this image under "fair use" may be invalid. Please read the instructions at Wikipedia:Non-free content carefully, then go to the image description page and clarify why you think the image qualifies for fair use. Using one of the templates at Wikipedia:Fair use rationale guideline is an easy way to ensure that your image is in compliance with Wikipedia policy, but remember that you must complete the template. Do not simply insert a blank template on an image page.

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Continued disruption of Four color theorem

If you continue to disrupt the Four color theorem article with edits such as this, it is likely that you will be blocked again. -- ArglebargleIV 15:58, 5 November 2007 (UTC)

I should note that I am not an administrator, but if needed, I'll go find one. -- ArglebargleIV 19:55, 5 November 2007 (UTC)

I will support this block action, and will block Fsswsb myself if I notice the disruption first. Do not use Wikipedia as a platform for pushing false counterexamples to established true results. Dcoetzee 19:51, 5 November 2007 (UTC)

Sorry, the conditions e==k || e==l || e==m were missing. Nevertheless, even with these three conditions the computer still finds many four color solutions. I have changed now the surrounding, so that region '2' reaches region 'a', which have borders to four other regions now. So far the computer doesn't find a solution, but it is still calculating. --Fsswsb 09:35, 6 November 2007 (UTC)

My computer is still busy, but may be someone has a faster one and can tell me how to color the map --Fsswsb 10:11, 6 November 2007 (UTC).

main(){

int a,b,c,d,e,f,g,h,i,j,k,p; int l,m,n,o,r,s,t,u; int q,v,w; int cnt=0;

for (a=1;a<4;a++) for (b=1;b<4;b++) for (c=1;c<4;c++) for (d=2;d<4;d++)

for (e=0;e<4;e++) for (f=0;f<4;f++) for (g=0;g<4;g++) for (h=0;h<4;h++) for (i=0;i<4;i++) for (j=0;j<4;j++) for (k=0;k<4;k++) for (l=0;l<4;l++) for (m=0;m<4;m++) for (n=0;n<4;n++) for (o=0;o<4;o++) for (p=0;p<4;p++) for (q=0;q<4;q++) for (r=0;r<4;r++) for (s=0;s<4;s++) for (t=0;t<4;t++) for (u=0;u<4;u++) for (v=0;v<4;v++) for (w=0;w<4;w++) {

if (a==b || b==c || c==d || e==f || f==g || g==h || h==i || e==2 || a==2 || 
    e==k || e==l || e==m) continue;   
if (a==e || b==e || b==f || c==f || c==g || d==g || d==h ) continue;
if (j==e || j==f || j==g || j==h || j==i || i==1 || h==1 ) continue;
if (j==m || j==n || j==o || k==l || l==m || m==n || n==o) continue;
if (p==q || q==r || r==s || s==t) continue;
if (k==2 || k==p || l==p || l==q || m==q || m==r || n==r || n==s || o==s
   || o==t || o==i || t==i) continue;
if (u==v || v==w || u==w || v==t || w==t) continue;
if (i==2 || t==2 || u==2 || w==2) continue;
if (u==p || u==q || u==r || v==t || v==s ) continue;
cnt++;
printf("solution found \n");

}

printf("%d solutions found\n",cnt);

}

Ok, my Computer found some solutions now. It is quite obvious, that for no map with more than 30 areas the computer ever will find a solution. --Fsswsb 14:36, 6 November 2007 (UTC)

I have a (much) faster program, with file input describing the map, and it's had no problem with 100+ region maps. Actually implementing advanced techniques such as reductions would speed it up even more. BTW, there were more problems with your program (in the comments of the image you uploaded) than just the e-connections. -- ArglebargleIV 15:37, 6 November 2007 (UTC)

Yes, you are right. Some other conditions are also missing. I corrected for that and restarted the calculation. Since my program is somewhat slow, I have to wait some hours until the computer may find some correct solutions. --Fsswsb 09:06, 7 November 2007 (UTC)

Fsswsb: Your naive algorithms are just really slow. The fastest known algorithms (like this one) ought to scale easily to graphs with tens of thousands of regions. I'd like to write more about this algorithm in the article but don't understand it myself yet. Dcoetzee 19:29, 6 November 2007 (UTC)

I've taken a look at that -- I think I'd need at least a weekend with reference materials to figure out the outline of that one. Maybe. -- ArglebargleIV 20:29, 6 November 2007 (UTC)

So, I just read the abstract:

Abstract: The four-colour theorem, that every loopless planar graph admits a vertex-colouring with at most four different colours, was proved in 1976 by Appel and Haken, using a computer. Here we announce another proof, still using a computer, but simpler than Appel and Haken's in several respects.

Already at this point I don't understand several things. First, I don't understand what loopless and planar exactly means. Therefore, I'm not sure, if really the four-colour theorem is exactly the same as described here. Then, I don't understand, why to prove something that is already proven. Obviously, there must be some doubt about the proof. Today computers are much faster than in 1976, about one million times according to Moore's law. Hence, it shouldn't be any problem to reproduce the proof of Appel and Haken on many computers, implementing the algorithm using different programming languages and techniques. To develop a completely new technique, still requiring a computer, is not a real advantage.

"Loopless" and "planar" are basic graph theory terms relevent to the problem - you should study more graph theory if you're interested in this problem. The paper describes a different proof because that proof forms the constructive basis of their algorithm; an algorithm based on the old proof yields a less efficient algorithm. Dcoetzee 11:33, 7 November 2007 (UTC)

Loopless obviously just means, that no region has a border to itself. Moreover, I can't see a essential difference wether the map is drawn in a plane or on a surface somehow curved. At least it seams that indeed the four-colour theorem is just the same as the four color theorem. If already in 1976 the algorithm was fast enough, there is no need for a more efficient one. --Fsswsb 12:36, 7 November 2007 (UTC)

This is not an algorithm for proving the four colour theorem - it's an algorithm for four-colouring the regions of a given map. There are always new and larger maps to colour, so this is certainly a worthwhile problem to solve. Dcoetzee 12:40, 7 November 2007 (UTC)

Not an algorithm for proving ? "Here we announce another proof, still using a computer, but simpler than Appel and Haken's in several respects." Why should it be an advantage to use four rather than five colors for a map ? A real map will use anyway more colors for rivers, streets, railways and so on. In most cases it's possible to find a four color solution – if it exists – even without computer. --Fsswsb 13:14, 7 November 2007 (UTC)

Because many important practical problems (some seemingly unrelated to maps) can be encoded in the form of maps such that colouring them helps solve the original problem. For example, factoring integers can be restated in terms of determining whether a particular map can be colored with 3 colors. And for very large and complex maps, no, they are not at all easy to colour by hand - I could construct a relatively small one that a human would spend weeks trying to solve but that a good algorithm could complete almost instantly. Dcoetzee 23:39, 7 November 2007 (UTC)

Corrected program

Still, the condition p==2 was missing. Note, that region '2' reaches region 'a' now. My program is still busy again, but didn't found a solution so far. --Fsswsb 10:02, 7 November 2007 (UTC)

My program is still busy. --Fsswsb 17:06, 7 November 2007 (UTC)

#include <stdio.h>

main(){

int a,b,c,d,e,f,g,h,i,j,k,p;
int l,m,n,o,r,s,t,u;
int q,v,w;
int cnt=0;


for (a=1;a<4;a++) for (b=1;b<4;b++) for (c=1;c<4;c++) for (d=2;d<4;d++)
for (e=0;e<4;e++) for (f=0;f<4;f++) for (g=0;g<4;g++) for (h=0;h<4;h++) 
for (i=0;i<4;i++) for (j=0;j<4;j++) for (k=0;k<4;k++) for (l=0;l<4;l++) 
for (m=0;m<4;m++) for (n=0;n<4;n++) for (o=0;o<4;o++) for (p=0;p<4;p++) 
for (q=0;q<4;q++) for (r=0;r<4;r++) for (s=0;s<4;s++) for (t=0;t<4;t++)
for (u=0;u<4;u++) for (v=0;v<4;v++) for (w=0;w<4;w++) {

 if (a==2 || e==2 || k==2 || p==2 || u==2 || w==2 || t==2 || i==2) continue;
 if (a==b || b==c || c==d || e==f || f==g || g==h || h==i) continue;   
 if (a==e || b==e || b==f || c==f || c==g || d==g || d==h ) continue;
 if (e==k || e==l || e==m || 
 j==e || j==f || j==g || j==h || j==i || i==1 || h==1 ) continue;
 if (j==m || j==n || j==o || k==l || l==m || m==n || n==o) continue;
 if (p==q || q==r || r==s || s==t) continue;
 if (k==p || l==p || l==q || m==q || m==r || n==r || n==s || o==s
    || o==t || o==i || t==i) continue;
 if (u==v || u==w || u==p || u==q || u==r ) continue;
 if (v==s || v==t || v==r || v==w) continue;
 if (t==w) continue;
 cnt++;
 printf("solution found %d %d %d %d %d %d %d %d %d %d %d\n",
                         a,b ,c,d,e,f,g,h,i,j,h);
 }

 printf("solutions found\n",cnt);

}

— Preceding unsigned comment added by Fsswsb (talk • contribs)

• Here's a better program specific for your one map. Note that you have two regions marked "i" -- I have decided to call the thin white region near the middle 'j'. The three outer regions are colored 0, 1, and 2 -- program will assign 'colors' 0, 1, 2, and 3 to the rest of the regions. This program is more efficient in that it does tests as early as possible, and also it has the correct region touching information (as far as I can determine from your image). Note that on your image, region 2 does not touch region a. If region 2 touches region a, please explain what happens to region e. (Also note that a general program with input is quite doable.)

#include <stdio.h>
// calling the thin white region "j", 
// and auto-coloring the three outside regions 0, 1, and 2 as marked on the map
// the four colors are 0, 1, 2, and 3
void main(int argc, char* argv[])
{
int count = 0;
for (int a=0; a<4; a++) { if ((a==0))  continue; 
for (int b=0; b<4; b++) { if ((b==0) || (b==a))  continue; 
for (int c=0; c<4; c++) { if ((c==0) || (c==b))  continue; 
for (int d=0; d<4; d++) { if ((d==0) || (d==1) || (d==c))  continue; 
for (int e=0; e<4; e++) { if ((e==0) || (e==2) || (e==a) || (e==b))  continue; 
for (int f=0; f<4; f++) { if ((f==b) || (f==c) || (f==e))  continue; 
for (int g=0; g<4; g++) { if ((g==c) || (g==d) || (g==f))  continue; 
for (int h=0; h<4; h++) { if ((h==1) || (h==d) || (h==g))  continue; 
for (int i=0; i<4; i++) { if ((i==1) || (i==d) || (i==h))  continue; 
for (int j=0; j<4; j++) { if ((j==e) || (j==f) || (j==g) || (j==h) || (j==i))  continue; 
for (int k=0; k<4; k++) { if ((k==2) || (k==e))  continue; 
for (int l=0; l<4; l++) { if ((l==e) || (l==k))  continue; 
for (int m=0; m<4; m++) { if ((m==e) || (m==j) || (m==l))  continue; 
for (int n=0; n<4; n++) { if ((n==i) || (n==j) || (n==m))  continue; 
for (int o=0; o<4; o++) { if ((o==i) || (o==j) || (o==n))  continue; 
for (int p=0; p<4; p++) { if ((p==2) || (p==k) || (p==l))  continue; 
for (int q=0; q<4; q++) { if ((q==l) || (q==m) || (q==p))  continue; 
for (int r=0; r<4; r++) { if ((r==m) || (r==n) || (r==q))  continue; 
for (int s=0; s<4; s++) { if ((s==n) || (s==o) || (s==r))  continue; 
for (int t=0; t<4; t++) { if ((t==2) || (t==i) || (t==o) || (t==s))  continue; 
for (int u=0; u<4; u++) { if ((u==2) || (u==p) || (u==q) || (u==r))  continue; 
for (int v=0; v<4; v++) { if ((v==r) || (v==s) || (v==t) || (v==u))  continue; 
for (int w=0; w<4; w++) { if ((w==2) || (w==u) || (w==v) || (w==t))  continue; 
count++;
printf("solution found : ");
printf("a=%d b=%d c=%d d=%d e=%d f=%d g=%d h=%d i=%d j=%d k=%d l=%d ",
       a, b, c, d, e, f, g, h, i, j, k, l);
printf("m=%d n=%d o=%d p=%d q=%d r=%d s=%d t=%d u=%d v=%d w=%d\n", 
       m, n, o, p, q, r, s, t, u, v, w);
fflush(stdout);
}}}}}}}}}}}}}}}}}}}}}}}
if (count == 0) printf("Nothing found\n");
return;
}

This program runs in 2 seconds on my laptop, returning 171 different colorings. One of the colorings is
solution found : a=3 b=2 c=3 d=2 e=1 f=0 g=1 h=3 i=0 j=2 k=3 l=2 m=0 n=1 o=3 p=0 q=3 r=2 s=0 t=1 u=1 v=3 w=0
-- ArglebargleIV 18:16, 7 November 2007 (UTC)

Your are right, your algorithm is much faster, since it stops coloring immediately, if no color is remaining for one region with the regions already colored, instead of checking all combination of colors. Actually you forgot the condition i==2, so that not all solutions are really valid. It means, that my naive algorithm could be optimized enormously just by changing the sequence of checking. This algorithm is now fast enough for any realistic map, to list all solutions and to determine the minimum number of colors required. Although the algorithm is still quite simple and corresponds to the method most humans would probably use to find a solution for the coloring problem. There is actually no need to look for a more efficient one. --Fsswsb 09:25, 8 November 2007 (UTC)

Orphaned non-free media (Image:ColorXXX.png)

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Continued disruption of Four color theorem, continued.

First, the image on the right is a properly colored version of your (new) ColorXXX.png file.

Second, you have been warned about changing the images on the Four color theorem article. The image that is there is a demonstration of a four-colored image. You continued replacement of that with your latest twiddle of an image you created as your latest fallacious disproof of the theorem is considered disruption, and it has to stop now, or it is certain that you will be blocked. -- ArglebargleIV 22:52, 7 November 2007 (UTC)

Fssweb, it is clear that you are not willing to play by the rules; specifically No original research. This policy is necessary in writing an encyclopaedia, and that's our purpose here on Wikipedia. I thus blocked you again, this time for two weeks. If you persist, the next block will be indefinite. -- Jitse Niesen (talk) 12:12, 8 November 2007 (UTC)

I saw your latest edit to four color theorem via the account User:FsswsbX and I have now blocked you indefinitely. -- Jitse Niesen (talk) 15:30, 16 November 2007 (UTC)

Notification of automated file description generation

Your upload of File:ColorXXX.jpg or contribution to its description is noted, and thanks (even if belatedly) for your contribution. In order to help make better use of the media, an attempt has been made by an automated process to identify and add certain information to the media's description page.

This notification is placed on your talk page because a bot has identified you either as the uploader of the file, or as a contributor to its metadata. It would be appreciated if you could carefully review the information the bot added. To opt out of these notifications, please follow the instructions here. Thanks! Message delivered by Theo's Little Bot (opt-out) 15:22, 19 May 2014 (UTC)

• Another one of your uploads, File:ComputerProof.jpg, has also had some information automatically added. If you get a moment, please review the bot's contributions there as well. Thanks! Message delivered by Theo's Little Bot (opt-out) 15:04, 22 May 2014 (UTC)

Your account will be renamed

Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Fsswsb. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Fsswsb~enwiki that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name. If you think you might own all of the accounts with this name and this message is in error, please visit Special:MergeAccount to check and attach all of your accounts to prevent them from being renamed.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours,
Keegan Peterzell
Community Liaison, Wikimedia Foundation

00:03, 20 March 2015 (UTC)

Renamed

This account has been renamed as part of single-user login finalisation. If you own this account you can log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: Special:GlobalRenameRequest. -- Keegan (WMF) (talk)

13:03, 22 April 2015 (UTC)

Proposed deletion of File:Color5.jpg

The file File:Color5.jpg has been proposed for deletion because of the following concern:

Unused file. No description or useful file name. Looks out of scope.

While all constructive contributions to Wikipedia are appreciated, pages may be deleted for any of several reasons.

You may prevent the proposed deletion by removing the {{proposed deletion/dated files}} notice, but please explain why in your edit summary or on the file's talk page.

Please consider addressing the issues raised. Removing {{proposed deletion/dated files}} will stop the proposed deletion process, but other deletion processes exist. In particular, the speedy deletion process can result in deletion without discussion, and files for discussion allows discussion to reach consensus for deletion. Sreejith K (talk) 23:30, 25 March 2021 (UTC)