{"id":3898,"date":"2016-07-14T08:00:15","date_gmt":"2016-07-14T12:00:15","guid":{"rendered":"https:\/\/blogs.mathworks.com\/community\/?p=3898"},"modified":"2016-07-22T11:13:31","modified_gmt":"2016-07-22T15:13:31","slug":"flag-math-on-bastille-day","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/community\/2016\/07\/14\/flag-math-on-bastille-day\/","title":{"rendered":"Flag Math on Bastille Day"},"content":{"rendered":"

Let me start with a graphical riddle. In what sense is the following equivalence true?<\/p>\n

<\/p>\n

To learn the answer, read on!<\/p>\n

I recently came across Nick Higham’s fun post on one-liners in MATLAB<\/a>. That piece and the approach of Bastille Day on July 14th made me think of another one-line program.<\/p>\n

image([3 2 1]);colormap(flag)\r\n<\/pre>\n
<\/p>\n
Pretty impressive, considering that it’s only 29 characters long. Is this the shortest flag program we can write?<\/p>\n
As I thought about flags and MATLAB, I gave myself the following task. Using only the colormap “flag” and images of vectors, how many national flags can we construct? Probably more than you might guess. Let’s go on a little tour.<\/p>\n
Tidying Up the Flag<\/h3>\n
Proper flags shouldn’t have numbers running their edges, so I’d like to hide the tick marks and labels. Since I plan to do this multiple times in the script below, I’m going to use an anonymous function for compactness. It’s not necessary, but I like how it keeps the code tidy.<\/p>\n
notick = @(ax) set(ax,'XTick'<\/span>,[],'YTick'<\/span>,[])\r\n<\/pre>\n
notick = \r\n\u00a0\u00a0\u00a0\u00a0@(ax)set(ax,'XTick',[],'YTick',[])\r\n\r\n\r\n<\/pre>\n
France<\/h3>\n
In honor of Bastille Day, we start with France.<\/p>\n
france = [3 2 1]\r\n<\/pre>\n
france = \r\n\r\n     3     2     1\r\n\r\n<\/pre>\n
image(france)\r\ntitle('Vive la France!'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
France is the only red, white, and blue tricolor with vertical bars. But if we look at horizontal patterns, we find many more.<\/p>\n
Netherlands<\/h3>\n
Netherlands is a rotated<\/a> version of France.<\/p>\n
netherlands = rot90(france)\r\n<\/pre>\n
netherlands = \r\n\r\n     1\r\n     2\r\n     3\r\n\r\n<\/pre>\n
image(netherlands);\r\ntitle('The Netherlands'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Luxembourg<\/h3>\n
Luxembourg, next door to the Netherlands, uses essentially the same flag. A pedantic Luxembourger or professional vexillologist may protest that the colors are slightly lighter and the aspect ratio is slightly different. We will gloss over these details.<\/p>\n
luxembourg = netherlands\r\n<\/pre>\n
luxembourg = \r\n\r\n     1\r\n     2\r\n     3\r\n\r\n<\/pre>\n
image(luxembourg)\r\ntitle('Luxembourg'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Thailand<\/h3>\n
thailand = [netherlands; flipud(netherlands)]\r\n<\/pre>\n
thailand = \r\n\r\n     1\r\n     2\r\n     3\r\n     3\r\n     2\r\n     1\r\n\r\n<\/pre>\n
image(thailand)\r\ntitle('Thailand'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Now we see the significance of the graphical riddle that we listed at the top of this post.<\/p>\n
Russia<\/h3>\n
Russia’s flag is a variant on the three horizontal bands. We’ll use circshift<\/a> to do a circular permutation of the colors in the Netherlands flag.<\/p>\n
russia = circshift(netherlands,2)\r\n<\/pre>\n
russia = \r\n\r\n     2\r\n     3\r\n     1\r\n\r\n<\/pre>\n
image(russia)\r\ntitle('Russia'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Austria<\/h3>\n
austria = netherlands([1 2 1])\r\n<\/pre>\n
austria = \r\n\r\n     1\r\n     2\r\n     1\r\n\r\n<\/pre>\n
image(austria)\r\ntitle('Austria'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Indonesia<\/h3>\n
indonesia = netherlands(1:2)\r\n<\/pre>\n
indonesia = \r\n\r\n     1\r\n     2\r\n\r\n<\/pre>\n
image(indonesia)\r\ntitle('Indonesia'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Monaco<\/h3>\n
monaco = indonesia\r\n<\/pre>\n
monaco = \r\n\r\n     1\r\n     2\r\n\r\n<\/pre>\n
image(monaco);\r\ntitle('Monaco'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
Poland<\/h3>\n
poland = flipud(monaco)\r\n<\/pre>\n
poland = \r\n\r\n     2\r\n     1\r\n\r\n<\/pre>\n
image(poland);\r\ntitle('Poland'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
<\/p>\n
There you have it. A veritable Tour d’Image. I count nine separate vector flags using only red, white, or blue. Did I miss any?<\/p>\n
As an exercise for the reader, express all flags in terms of the vector “france”.<\/p>\n
Addendum #1: Costa Rica<\/h3>\n
In the comments, Matt Tearle points out another flag that fits the model. It’s a kind of dual to Thailand. I am now convinced that vexillology conventions should always occur in the Netherlands, its flag being the indispensable element at the center of all flag alchemy.<\/p>\n
costarica = [flipud(netherlands); netherlands]\r\n<\/pre>\n
costarica = \r\n\r\n     3\r\n     2\r\n     1\r\n     1\r\n     2\r\n     3\r\n\r\n<\/pre>\n
image(costarica);\r\ntitle('Costa Rica'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
\"flags_12\"<\/p>\n
Addendum #2<\/h3>\n
And Richard Alcock from MathWorks UK writes in to give us this timely and poignant example.<\/p>\n
reflex_blue = [0 51 153]\/255;\r\nyellow = [255 204 0]\/255;\r\n[x, y] = pol2cart(linspace(0, 2*pi, 13), 1\/3); plot(x, y, 'p'<\/span>, 'MarkerSize'<\/span>, 20, ...<\/span>\r\n    'MarkerEdgeColor'<\/span>, yellow, 'MarkerFaceColor'<\/span>, yellow); axis('equal'<\/span>); set(gca, 'Color'<\/span>, reflex_blue,  ...<\/span>\r\n    'XTick'<\/span>, [], 'YTick'<\/span>, [], ...<\/span>\r\n    'Xlim'<\/span>, [-.75 .75], 'Ylim'<\/span>, [-.5, .5])\r\n<\/pre>\n
\"flags_13\"<\/p>\n
<\/pre>\n
Addendum #3: Yemen<\/h3>\n
Sean de Wolski reminded me that black is one of the four colors of the “flag” colormap, thereby letting us add the Yemeni flag to our list. I thought the addition of black might give us a few more flags, but Yemen seems to be the end of it. <\/p>\n
yemen = [1; 2; 4];\r\nimage(yemen)\r\ntitle('Yemen'<\/span>)\r\nnotick(gca)\r\n<\/pre>\n
\"flags_14\"<\/p>\n","protected":false},"excerpt":{"rendered":"
<\/div>\n
Let me start with a graphical riddle. In what sense is the following equivalence true?<\/p>\n
To learn the answer, read on!
\nI recently came across Nick Higham’s fun post on one-liners in MATLAB. That… read more >><\/a><\/p>\n","protected":false},"author":69,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/posts\/3898"}],"collection":[{"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/users\/69"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/comments?post=3898"}],"version-history":[{"count":7,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/posts\/3898\/revisions"}],"predecessor-version":[{"id":3959,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/posts\/3898\/revisions\/3959"}],"wp:attachment":[{"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/media?parent=3898"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/categories?post=3898"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/community\/wp-json\/wp\/v2\/tags?post=3898"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}