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curated data - Building graph based on the cities connection?


I built a Graph based on the permutations of city's connections from :


largUSCities = 
Select[CityData[{All, "USA"}], CityData[#, "Population"] > 600000 &];
uScityCoords = CityData[#, "Coordinates"] & /@ largUSCities;
Graph[#[[1]] -> #[[2]] & /@ Permutations[largUSCities, {2}] ,
VertexCoordinates -> Reverse[uScityCoords, 2], VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},ImageSize -> 650]

It looks like this: graph



My question, is there any way to have the Graph like this? enter image description here



Answer



Revised answer


This uses the connectivity between states to create the graph, and uses the coordinates of the center of each state rather than the cities. I couldn't find a way to get these easily from Mathematica or from WolframAlpha (I'm no Harry Potter, and failed to discover the correct incantation for the latter). But I found a table somewhere:


stateConnections = {{"NV", "CA", "AZ", "UT", "ID", "OR"}, {"OR", "CA",
"NV", "ID", "WA"}, {"TX", "OK", "LA", "NM", "AR"}, {"DC", "VA",
"MD"}, {"FL", "GA", "AL"}, {"RI", "MA", "CT"}, {"SC", "GA",
"NC"}, {"WA", "OR", "ID"}, {"CA", "NV", "OR", "AZ"}, {"CT", "RI",
"MA", "NY"}, {"DE", "MD", "PA", "NJ"}, {"LA", "TX", "MS",
"AR"}, {"MI", "IN", "OH", "WI"}, {"ND", "SD", "MN", "MT"}, {"NH",

"ME", "VT", "MA"}, {"NJ", "NY", "PA", "DE"}, {"VT", "NH", "MA",
"NY"}, {"AL", "GA", "MS", "TN", "FL"}, {"AZ", "CA", "NM", "UT",
"NV"}, {"IN", "OH", "MI", "IL", "KY"}, {"KS", "OK", "CO", "MO",
"NE"}, {"MD", "DE", "PA", "VA", "WV"}, {"MN", "WI", "IA", "SD",
"ND"}, {"MS", "AL", "LA", "AR", "TN"}, {"MT", "ID", "WY", "SD",
"ND"}, {"NC", "SC", "VA", "TN", "GA"}, {"NM", "TX", "AZ", "CO",
"OK"}, {"WI", "IL", "MI", "IA", "MN"}, {"GA", "FL", "SC", "NC",
"AL", "TN"}, {"IL", "IA", "WI", "IN", "KY", "MO"}, {"MA", "VT",
"NH", "NY", "RI", "CT"}, {"NV", "CA", "AZ", "UT", "ID",
"OR"}, {"NY", "NJ", "VT", "PA", "MA", "CT"}, {"OH", "IN", "WV",

"PA", "KY", "MI"}, {"UT", "CO", "WY", "ID", "NV", "AZ"}, {"VA",
"WV", "MD", "NC", "TN", "KY"}, {"WV", "VA", "OH", "PA", "MD",
"KY"}, {"AR", "TX", "LA", "OK", "MO", "TN", "MS"}, {"CO", "UT",
"WY", "NM", "NE", "KS", "OK"}, {"IA", "IL", "WI", "MN", "SD",
"NE", "MO"}, {"ID", "WA", "OR", "NV", "UT", "WY", "MT"}, {"NE",
"KS", "CO", "WY", "SD", "IA", "MO"}, {"OK", "TX", "CO", "KS",
"NM", "AR", "MO"}, {"PA", "WV", "DE", "MD", "NJ", "NY",
"OH"}, {"SD", "ND", "MT", "WY", "NE", "IA", "MN"}, {"WY", "MT",
"ID", "UT", "CO", "NE", "SD"}, {"KY", "IL", "MO", "TN", "VA",
"WV", "OH", "IN"}, {"MO", "IA", "NE", "KS", "OK", "AR", "TN",

"KY", "IL"}, {"TN", "KY", "MO", "AR", "MS", "AL", "GA", "NC",
"VA"}, {"ME", "NH"}};

stateData = {"AK,61.3850,-152.2683", "AL,32.7990,-86.8073",
"AR,34.9513,-92.3809", "AZ,33.7712,-111.3877",
"CA,36.1700,-119.7462", "CO,39.0646,-105.3272",
"CT,41.5834,-72.7622", "DC,38.8964,-77.0262",
"DE,39.3498,-75.5148", "FL,27.8333,-81.7170",
"GA,32.9866,-83.6487", "HI,21.1098,-157.5311",
"IA,42.0046,-93.2140", "ID,44.2394,-114.5103",

"IL,40.3363,-89.0022", "IN,39.8647,-86.2604",
"KS,38.5111,-96.8005", "KY,37.6690,-84.6514",
"LA,31.1801,-91.8749", "MA,42.2373,-71.5314",
"MD,39.0724,-76.7902", "ME,44.6074,-69.3977",
"MI,43.3504,-84.5603", "MN,45.7326,-93.9196",
"MO,38.4623,-92.3020", "MS,32.7673,-89.6812",
"MT,46.9048,-110.3261", "NC,35.6411,-79.8431",
"ND,47.5362,-99.7930", "NE,41.1289,-98.2883",
"NH,43.4108,-71.5653", "NJ,40.3140,-74.5089",
"NM,34.8375,-106.2371", "NV,38.4199,-117.1219",

"NY,42.1497,-74.9384", "OH,40.3736,-82.7755",
"OK,35.5376,-96.9247", "OR,44.5672,-122.1269",
"PA,40.5773,-77.2640", "RI,41.6772,-71.5101",
"SC,33.8191,-80.9066", "SD,44.2853,-99.4632",
"TN,35.7449,-86.7489", "TX,31.1060,-97.6475",
"UT,40.1135,-111.8535", "VA,37.7680,-78.2057",
"VT,44.0407,-72.7093", "WA,47.3917,-121.5708",
"WI,44.2563,-89.6385", "WV,38.4680,-80.9696",
"WY,42.7475,-107.2085"} ;


stateAbbreviations = Union[Flatten[stateConnections]];
stateToNumber =
MapThread[
Rule, {stateAbbreviations, Range[Length[stateAbbreviations]]}];
numberToState =
MapThread[
Rule, {Range[Length[stateAbbreviations]], stateAbbreviations}];
allConnections =
Flatten[Function[e, Map[UndirectedEdge[First[e], #] &, Rest[e]]] /@
stateConnections];

connections = Union[Sort /@ allConnections];
stateCenters =
First[StringSplit[#, ","]] ->
ToExpression /@ RotateLeft @ Rest[StringSplit[#, ","]] & /@
stateData;
stateCoords = (# & /@ stateAbbreviations) /. stateCenters;
temp = Graph[connections /. stateToNumber];
vertexCoordinates = stateCoords[[VertexList[temp]]];
g = Graph[connections /. stateToNumber,
VertexCoordinates -> vertexCoordinates,

VertexLabels -> numberToState,
VertexShapeFunction -> "Square",
VertexSize -> 3,
VertexLabelStyle -> Directive[Black, 12]];

Show[Graphics[{LightGray, CountryData["USA", "Polygon"]}], g,
ImageSize -> 700]

graph of the usa


Apparently the order of the vertices is required from the graph before you can draw the vertices at the right coordinates on the graph - hence the weird use of temp = Graph[connections /. stateToNumber] before creating the graph again for real.



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