# Pixel perfect world map with no border lines

I'm trying to build a real world map in Minecraft and want a pixel-perfect raster image I can copy pixel for block.

How can I modify

CountryData["World", {"Shape", "Robinson"}]


which gives

such that:

• Each country is its own separate color (random colors are fine).
• The water color is distinct from the background.
• There are no black border lines (but the borders are still clear from the color difference in countries).
• There is no anti-aliasing or color blending between regions of different colors, be they countries, water, or the background.
• I can scale the map to the exact width (or height) in pixels I want.

Sorry if this is asking a bit much. Mathematica is completely new to me and I wasn't getting anywhere fumbling around with CountryData and GeoGraphics and WorldPlot for an hour.

• What version are you on? – J. M. will be back soon Jul 22 '16 at 5:02
• @J.M. The pilot release for Raspberry Pi (10.0.0.0). – Calvin's Hobbies Jul 22 '16 at 5:06
• Would Graphics[Riffle[CountryData["World", {"SchematicPolygon", "Robinson"}] /. Polygon[{p__?MatrixQ}] :> Map[Polygon, {p}], Unevaluated[RandomColor[]], {1, -2, 2}], Background -> ColorData["Legacy", "Azure"]] suit your needs? – J. M. will be back soon Jul 22 '16 at 5:20
• @J.M. That still appears to have the anti-aliased borders between countries. I can maybe deal with those in an image editor but I'd prefer not to. – Calvin's Hobbies Jul 22 '16 at 5:29
• What if you use "FullPolygon" instead of "SchematicPolygon"? FWIW: these are all generating vector graphics, which you can easily resize before saving as an image file. – J. M. will be back soon Jul 22 '16 at 5:41

With GeoGraphics:

GeoGraphics[{GeoStyling@Opacity@1, RandomColor[],
CountryData[#, "SchematicPolygon"]} & /@ Join[CountryData["Continents"], CountryData[]],
GeoBackground -> Hue[0.56, .8, .8, .5], GeoRange -> "World",
GeoProjection -> "Robinson", Background -> White]


• Just what I wanted. (Any chance of adding antarctica?) – Calvin's Hobbies Jul 22 '16 at 6:59
• Since Antarctica is not a country, it's not returned by CountryData[] (cf. this post). I've joined continents and countries to have everything - make sure you do not change the order as then continents would cover countries. – István Zachar Jul 22 '16 at 7:49

From its doc

Graphics[{Hue[
2/3 Sqrt[
1 - (CountryData[#,
"IndependenceYear"] /. {DateObject[{y_}] :> y, _Missing ->
First[DateList[]]})/First[DateList[]]]],
CountryData[#, "SchematicPolygon"]} & /@ CountryData[]]


which makes it plot countries color coded by the length of their claimed independence. You can plot by any other method.

• This is almost just what I want :) but (unless I'm very mistaken), it's not a Robinson projection. How could I fix that? – Calvin's Hobbies Jul 22 '16 at 5:28