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Tackling the problem of semi-realistic looking terrain generation in clojure.
I recently started looking into the diamond-square algorithm. The following is a short-ish walkthrough of how I tackled the problem in clojure and the results. You can find the leiningen repo here and follow along within that, or simply read the code below to get an idea.
(ns diamond-square.core)
; == The Goal ==
; Create a fractal terrain generator using clojure
; == The Algorithm ==
; Diamond-Square. We start with a grid of points, each with a height of 0.
;
; 1. Take each corner point of the square, average the heights, and assign that
; to be the height of the midpoint of the square. Apply some random error to
; the midpoint.
;
; 2. Creating a line from the midpoint to each corner we get four half-diamonds.
; Average the heights of the points (with some random error) and assign the
; heights to the midpoints of the diamonds.
;
; 3. We now have four square sections, start at 1 for each of them (with
; decreasing amount of error for each iteration).
;
; This picture explains it better than I can:
; mediocregopher.com/art/diamond-square/diamond-square-dsalg.png
; (http://nbickford.wordpress.com/2012/12/21/creating-fake-landscapes/dsalg/)
;
; == The Strategy ==
; We begin with a vector of vectors of numbers, and iterate over it, filling in
; spots as they become available. Our grid will have the top-left being (0,0),
; y being pointing down and x going to the right. The outermost vector
; indicating row number (y) and the inner vectors indicate the column number (x)
;
; = Utility =
; First we create some utility functions for dealing with vectors of vectors.
(defn print-m
"Prints a grid in a nice way"
[m]
(doseq [n m]
(println n)))
(defn get-m
"Gets a value at the given x,y coordinate of the grid, with [0,0] being in the
top left"
[m x y]
((m y) x))
(defn set-m
"Sets a value at the given x,y coordinat of the grid, with [0,0] being in the
top left"
[m x y v]
(assoc m y
(assoc (m y) x v)))
(defn add-m
"Like set-m, but adds the given value to the current on instead of overwriting
it"
[m x y v]
(set-m m x y
(+ (get-m m x y) v)))
(defn avg
"Returns the truncated average of all the given arguments"
[& l]
(int (/ (reduce + l) (count l))))
; = Grid size =
; Since we're starting with a blank grid we need to find out what sizes the
; grids can be. For convenience the size (height and width) should be odd, so we
; easily get a midpoint. And on each iteration we'll be halfing the grid, so
; whenever we do that the two resultrant grids should be odd and halfable as
; well, and so on.
;
; The algorithm that fits this is size = 2^n + 1, where 1 <= n. For the rest of
; this guide I'll be referring to n as the "degree" of the grid.
(def exp2-pre-compute
(vec (map #(int (Math/pow 2 %)) (range 31))))
(defn exp2
"Returns 2^n as an integer. Uses pre-computed values since we end up doing
this so much"
[n]
(exp2-pre-compute n))
(def grid-sizes
(vec (map #(inc (exp2 %)) (range 1 31))))
(defn grid-size [degree]
(inc (exp2 degree)))
; Available grid heights/widths are as follows:
;[3 5 9 17 33 65 129 257 513 1025 2049 4097 8193 16385 32769 65537 131073
;262145 524289 1048577 2097153 4194305 8388609 16777217 33554433 67108865
;134217729 268435457 536870913 1073741825])
(defn blank-grid
"Generates a grid of the given degree, filled in with zeros"
[degree]
(let [gsize (grid-size degree)]
(vec (repeat gsize
(vec (repeat gsize 0))))))
(comment
(print-m (blank-grid 3))
)
; = Coordinate Pattern (The Tricky Part) =
; We now have to figure out which coordinates need to be filled in on each pass.
; A pass is defined as a square step followed by a diamond step. The next pass
; will be the square/dimaond steps on all the smaller squares generated in the
; pass. It works out that the number of passes required to fill in the grid is
; the same as the degree of the grid, where the first pass is 1.
;
; So we can easily find patterns in the coordinates for a given degree/pass,
; I've laid out below all the coordinates for each pass for a 3rd degree grid
; (which is 9x9).
; Degree 3 Pass 1 Square
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . 1 . . . .] (4,4)
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; Degree 3 Pass 1 Diamond
; [. . . . 2 . . . .] (4,0)
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [2 . . . . . . . 2] (0,4) (8,4)
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . 2 . . . .] (4,8)
; Degree 3 Pass 2 Square
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . 3 . . . 3 . .] (2,2) (6,2)
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . . . . . . . .]
; [. . 3 . . . 3 . .] (2,6) (6,6)
; [. . . . . . . . .]
; [. . . . . . . . .]
; Degree 3 Pass 2 Diamond
; [. . 4 . . . 4 . .] (2,0) (6,0)
; [. . . . . . . . .]
; [4 . . . 4 . . . 4] (0,2) (4,2) (8,2)
; [. . . . . . . . .]
; [. . 4 . . . 4 . .] (2,4) (6,4)
; [. . . . . . . . .]
; [4 . . . 4 . . . 4] (0,6) (4,6) (8,6)
; [. . . . . . . . .]
; [. . 4 . . . 4 . .] (2,8) (6,8)
; Degree 3 Pass 3 Square
; [. . . . . . . . .]
; [. 5 . 5 . 5 . 5 .] (1,1) (3,1) (5,1) (7,1)
; [. . . . . . . . .]
; [. 5 . 5 . 5 . 5 .] (1,3) (3,3) (5,3) (7,3)
; [. . . . . . . . .]
; [. 5 . 5 . 5 . 5 .] (1,5) (3,5) (5,5) (7,5)
; [. . . . . . . . .]
; [. 5 . 5 . 5 . 5 .] (1,7) (3,7) (5,7) (7,7)
; [. . . . . . . . .]
; Degree 3 Pass 3 Square
; [. 6 . 6 . 6 . 6 .] (1,0) (3,0) (5,0) (7,0)
; [6 . 6 . 6 . 6 . 6] (0,1) (2,1) (4,1) (6,1) (8,1)
; [. 6 . 6 . 6 . 6 .] (1,2) (3,2) (5,2) (7,2)
; [6 . 6 . 6 . 6 . 6] (0,3) (2,3) (4,3) (6,3) (8,3)
; [. 6 . 6 . 6 . 6 .] (1,4) (3,4) (5,4) (7,4)
; [6 . 6 . 6 . 6 . 6] (0,5) (2,5) (4,5) (6,5) (8,5)
; [. 6 . 6 . 6 . 6 .] (1,6) (3,6) (5,6) (7,6)
; [6 . 6 . 6 . 6 . 6] (0,7) (2,7) (4,7) (6,7) (8,7)
; [. 6 . 6 . 6 . 6 .] (1,8) (3,8) (5,8) (7,8)
;
; I make two different functions, one to give the coordinates for the square
; portion of each pass and one for the diamond portion of each pass. To find the
; actual patterns it was useful to first look only at the pattern in the
; y-coordinates, and figure out how that translated into the pattern for the
; x-coordinates.
(defn grid-square-coords
"Given a grid degree and pass number, returns all the coordinates which need
to be computed for the square step of that pass"
[degree pass]
(let [gsize (grid-size degree)
start (exp2 (- degree pass))
interval (* 2 start)
coords (map #(+ start (* interval %))
(range (exp2 (dec pass))))]
(mapcat (fn [y]
(map #(vector % y) coords))
coords)))
;
; (grid-square-coords 3 2)
; => ([2 2] [6 2] [2 6] [6 6])
(defn grid-diamond-coords
"Given a grid degree and a pass number, returns all the coordinates which need
to be computed for the diamond step of that pass"
[degree pass]
(let [gsize (grid-size degree)
interval (exp2 (- degree pass))
num-coords (grid-size pass)
coords (map #(* interval %) (range 0 num-coords))]
(mapcat (fn [y]
(if (even? (/ y interval))
(map #(vector % y) (take-nth 2 (drop 1 coords)))
(map #(vector % y) (take-nth 2 coords))))
coords)))
; (grid-diamond-coords 3 2)
; => ([2 0] [6 0] [0 2] [4 2] [8 2] [2 4] [6 4] [0 6] [4 6] [8 6] [2 8] [6 8])
; = Height Generation =
; We now work on functions which, given a coordinate, will return what value
; coordinate will have.
(defn avg-points
"Given a grid and an arbitrary number of points (of the form [x y]) returns
the average of all the given points that are on the map. Any points which are
off the map are ignored"
[m & coords]
(let [grid-size (count m)]
(apply avg
(map #(apply get-m m %)
(filter
(fn [[x y]]
(and (< -1 x) (> grid-size x)
(< -1 y) (> grid-size y)))
coords)))))
(defn error
"Returns a number between -e and e, inclusive"
[e]
(- (rand-int (inc (* 2 e))) e))
; The next function is a little weird. It primarily takes in a point, then
; figures out the distance from that point to the points we'll take the average
; of. The locf (locator function) is used to return back the actual points to
; use. For the square portion it'll be the points diagonal from the given one,
; for the diamond portion it'll be the points to the top/bottom/left/right from
; the given one.
;
; Once it has those points, it finds the average and applies the error. The
; error function is nothing more than a number between -interval and +interval,
; where interval is the distance between the given point and one of the averaged
; points. It is important that the error decreases the more passes you do, which
; is why the interval is used.
;
; The error function is what should be messed with primarily if you want to
; change what kind of terrain you generate (a giant mountain instead of
; hills/valleys, for example). The one we use is uniform for all intervals, so
; it generates a uniform terrain.
(defn- grid-fill-point
[locf m degree pass x y]
(let [interval (exp2 (- degree pass))
leftx (- x interval)
rightx (+ x interval)
upy (- y interval)
downy (+ y interval)
v (apply avg-points m
(locf x y leftx rightx upy downy))]
(add-m m x y (+ v (error interval)))))
(def grid-fill-point-square
"Given a grid, the grid's degree, the current pass number, and a point on the
grid, fills in that point with the average (plus some error) of the
appropriate corner points, and returns the resultant grid"
(partial grid-fill-point
(fn [_ _ leftx rightx upy downy]
[[leftx upy]
[rightx upy]
[leftx downy]
[rightx downy]])))
(def grid-fill-point-diamond
"Given a grid, the grid's degree, the current pass number, and a point on the
grid, fills in that point with the average (plus some error) of the
appropriate edge points, and returns the resultant grid"
(partial grid-fill-point
(fn [x y leftx rightx upy downy]
[[leftx y]
[rightx y]
[x upy]
[x downy]])))
; = Filling in the Grid =
; We finally compose the functions we've been creating to fill in the entire
; grid
(defn- grid-fill-point-passes
"Given a grid, a function to fill in coordinates, and a function to generate
those coordinates, fills in all coordinates for a given pass, returning the
resultant grid"
[m fill-f coord-f degree pass]
(reduce
(fn [macc [x y]] (fill-f macc degree pass x y))
m
(coord-f degree pass)))
(defn grid-pass
"Given a grid and a pass number, does the square then the diamond portion of
the pass"
[m degree pass]
(-> m
(grid-fill-point-passes
grid-fill-point-square grid-square-coords degree pass)
(grid-fill-point-passes
grid-fill-point-diamond grid-diamond-coords degree pass)))
; The most important function in this guide, does all the work
(defn terrain
"Given a grid degree, generates a uniformly random terrain on a grid of that
degree"
([degree]
(terrain (blank-grid degree) degree))
([m degree]
(reduce
#(grid-pass %1 degree %2)
m
(range 1 (inc degree)))))
(comment
(print-m
(terrain 5))
)
; == The Results ==
; We now have a generated terrain, probably. We should check it. First we'll
; create an ASCII representation. But to do that we'll need some utility
; functions.
(defn max-terrain-height
"Returns the maximum height found in the given terrain grid"
[m]
(reduce max
(map #(reduce max %) m)))
(defn min-terrain-height
"Returns the minimum height found in the given terrain grid"
[m]
(reduce min
(map #(reduce min %) m)))
(defn norm
"Given x in the range (A,B), normalizes it into the range (0,new-height)"
[A B new-height x]
(int (/ (* (- x A) new-height) (- B A))))
(defn normalize-terrain
"Given a terrain map and a number of \"steps\", normalizes the terrain so all
heights in it are in the range (0,steps)"
[m steps]
(let [max-height (max-terrain-height m)
min-height (min-terrain-height m)
norm-f (partial norm min-height max-height steps)]
(vec (map #(vec (map norm-f %)) m))))
; We now define which ASCII characters we want to use for which heights. The
; vector starts with the character for the lowest height and ends with the
; character for the heighest height.
(def tiles
[\~ \~ \" \" \x \x \X \$ \% \# \@])
(defn tile-terrain
"Given a terrain map, converts it into an ASCII tile map"
[m]
(vec (map #(vec (map tiles %))
(normalize-terrain m (dec (count tiles))))))
(comment
(print-m
(tile-terrain
(terrain 5)))
; [~ ~ " " x x x X % $ $ $ X X X X X X $ x x x X X X x x x x " " " ~]
; [" ~ " " x x X X $ $ $ X X X X X X X X X X X X X X x x x x " " " "]
; [" " " x x x X X % $ % $ % $ $ X X X X $ $ $ X X X X x x x x " " "]
; [" " " x x X $ % % % % % $ % $ $ X X $ $ $ $ X X x x x x x x " " x]
; [" x x x x X $ $ # % % % % % % $ X $ X X % $ % X X x x x x x x x x]
; [x x x X $ $ $ % % % % % $ % $ $ $ % % $ $ $ $ X X x x x x x x x x]
; [X X X $ % $ % % # % % $ $ % % % % $ % $ $ X $ X $ X X x x x X x x]
; [$ $ X $ $ % $ % % % % $ $ $ % # % % % X X X $ $ $ X X X x x x x x]
; [% X X % % $ % % % $ % $ % % % # @ % $ $ X $ X X $ X x X X x x x x]
; [$ $ % % $ $ % % $ $ X $ $ % % % % $ $ X $ $ X X X X X X x x x x x]
; [% % % X $ $ % $ $ X X $ $ $ $ % % $ $ X X X $ X X X x x X x x X X]
; [$ $ $ X $ $ X $ X X X $ $ $ $ % $ $ $ $ $ X $ X x X X X X X x X X]
; [$ $ $ $ X X $ X X X X X $ % % % % % $ X $ $ $ X x X X X $ X X $ $]
; [X $ $ $ $ $ X X X X X X X % $ % $ $ $ X X X X X x x X X x X X $ $]
; [$ $ X X $ X X x X $ $ X X $ % X X X X X X X X X x X X x x X X X X]
; [$ $ X X X X X X X $ $ $ $ $ X $ X X X X X X X x x x x x x x X X X]
; [% % % $ $ X $ X % X X X % $ $ X X X X X X x x x x x x x x x X X $]
; [$ % % $ $ $ X X $ $ $ $ $ $ X X X X x X x x x x " x x x " x x x x]
; [$ X % $ $ $ $ $ X X X X X $ $ X X X X X X x x " " " " " " " " x x]
; [$ X $ $ % % $ X X X $ X X X x x X X x x x x x " " " " " ~ " " " "]
; [$ $ X X % $ % X X X X X X X X x x X X X x x x " " " " " " ~ " " "]
; [$ $ X $ % $ $ X X X X X X x x x x x x x x x " " " " " " " " " ~ ~]
; [$ $ $ $ $ X X $ X X X X X x x x x x x x x " " " " " " " ~ " " " ~]
; [$ % X X $ $ $ $ X X X X x x x x x x x x x x " " " " ~ " " ~ " " ~]
; [% $ $ X $ X $ X $ X $ X x x x x x x x x x x " " " " ~ ~ ~ " ~ " ~]
; [$ X X X X $ $ $ $ $ X x x x x x x x x x x " " " " ~ ~ ~ ~ ~ ~ ~ ~]
; [X x X X x X X X X X X X X x x x x x x x x x " " " ~ ~ " " ~ ~ ~ ~]
; [x x x x x x X x X X x X X X x x x x x x x " x " " " " " ~ ~ ~ ~ ~]
; [x x x x x x x x X X X X $ X X x X x x x x x x x x " ~ ~ ~ ~ ~ ~ ~]
; [" x x x x x X x X X X X X X X X X x x x x x x " " " " ~ ~ ~ ~ ~ ~]
; [" " " x x x X X X X $ $ $ X X X X X X x x x x x x x x " " ~ ~ ~ ~]
; [" " " " x x x X X X X X $ $ X X x X X x x x x x x x " " " " " ~ ~]
; [~ " " x x x x X $ X $ X $ $ X x X x x x x x x x x x x x x " " " ~]
)
; = Pictures! =
; ASCII is cool, but pictures are better. First we import some java libraries
; that we'll need, then define the colors for each level just like we did tiles
; for the ascii representation.
(import
'java.awt.image.BufferedImage
'javax.imageio.ImageIO
'java.io.File)
(def colors
[0x1437AD 0x04859D 0x007D1C 0x007D1C 0x24913C
0x00C12B 0x38E05D 0xA3A3A4 0x757575 0xFFFFFF])
; Finally we reduce over a BufferedImage instance to output every tile as a
; single pixel on it.
(defn img-terrain
"Given a terrain map and a file name, outputs a png representation of the
terrain map to that file"
[m file]
(let [img (BufferedImage. (count m) (count m) BufferedImage/TYPE_INT_RGB)]
(reduce
(fn [rown row]
(reduce
(fn [coln tile]
(.setRGB img coln rown (colors tile))
(inc coln))
0 row)
(inc rown))
0 (normalize-terrain m (dec (count colors))))
(ImageIO/write img "png" (File. file))))
(comment
(img-terrain
(terrain 10)
"resources/terrain.png")
; https://blog.mediocregopher.com/img/diamond-square/terrain.png
)
; == Conclusion ==
; There's still a lot of work to be done. The algorithm starts taking a
; non-trivial amount of time around the 10th degree, which is only a 1025x1025px
; image. I need to profile the code and find out where the bottlenecks are. It's
; possible re-organizing the code to use pmaps instead of reduces in some places
; could help.
-----
Published 2014-01-11