-
Notifications
You must be signed in to change notification settings - Fork 64
/
elevation.js
157 lines (141 loc) · 6.03 KB
/
elevation.js
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
/*
* From http://www.redblobgames.com/maps/mapgen2/
* Copyright 2017 Red Blob Games <[email protected]>
* License: Apache v2.0 <http://www.apache.org/licenses/LICENSE-2.0.html>
*/
'use strict';
/**
* Coast corners are connected to coast sides, which have
* ocean on one side and land on the other
*/
function find_coasts_t(mesh, r_ocean) {
let coasts_t = [];
for (let s = 0; s < mesh.numSides; s++) {
let r0 = mesh.s_begin_r(s);
let r1 = mesh.s_end_r(s);
let t = mesh.s_inner_t(s);
if (r_ocean[r0] && !r_ocean[r1]) {
// It might seem that we also need to check !r_ocean[r0] && r_ocean[r1]
// and it might seem that we have to add both t and its opposite but
// each t vertex shows up in *four* directed sides, so we only have to test
// one fourth of those conditions to get the vertex in the list once.
coasts_t.push(t);
}
}
return coasts_t;
}
/**
* Elevation is based on breadth first search from the seed points,
* which are the coastal graph nodes. Since breadth first search also
* calculates the 'parent' pointers, return those for use as the downslope
* graph. To handle lakes, which should have all corners at the same elevation,
* there are two deviations from breadth first search:
* 1. Instead of pushing to the end of the queue, push to the beginning.
* 2. Like uniform cost search, check if the new distance is better than
* previously calculated distances. It is possible that one lake corner
* was reached with distance 2 and another with distance 3, and we need
* to revisit that node and make sure it's set to 2.
*/
exports.assign_t_elevation = function(
t_elevation, t_coastdistance, t_downslope_s,
mesh,
r_ocean, r_water, randInt
) {
t_coastdistance.length = mesh.numTriangles;
t_downslope_s.length = mesh.numTriangles;
t_elevation.length = mesh.numTriangles;
t_coastdistance.fill(null);
t_downslope_s.fill(-1);
const t_ocean = (t) => r_ocean[mesh.s_begin_r(3*t)];
const r_lake = (r) => r_water[r] && !r_ocean[r];
const s_lake = (s) => r_lake(mesh.s_begin_r(s)) || r_lake(mesh.s_end_r(s));
let out_s = [];
let queue_t = find_coasts_t(mesh, r_ocean);
queue_t.forEach((t) => { t_coastdistance[t] = 0; });
let minDistance = 1, maxDistance = 1;
while (queue_t.length > 0) {
let current_t = queue_t.shift();
mesh.t_circulate_s(out_s, current_t);
let iOffset = randInt(out_s.length);
for (let i = 0; i < out_s.length; i++) {
let s = out_s[(i + iOffset) % out_s.length];
let lake = s_lake(s);
let neighbor_t = mesh.s_outer_t(s);
let newDistance = (lake? 0 : 1) + t_coastdistance[current_t];
if (t_coastdistance[neighbor_t] === null || newDistance < t_coastdistance[neighbor_t]) {
t_downslope_s[neighbor_t] = mesh.s_opposite_s(s);
t_coastdistance[neighbor_t] = newDistance;
if (t_ocean(neighbor_t) && newDistance > minDistance) { minDistance = newDistance; }
if (!t_ocean(neighbor_t) && newDistance > maxDistance) { maxDistance = newDistance; }
if (lake) {
queue_t.unshift(neighbor_t);
} else {
queue_t.push(neighbor_t);
}
}
}
}
t_coastdistance.forEach((d, t) => {
t_elevation[t] = t_ocean(t) ? (-d / minDistance) : (d / maxDistance);
});
};
/**
* Set r elevation to the average of the t elevations. There's a
* corner case though: it is possible for an ocean region (r) to be
* surrounded by coastline corners (t), and coastlines are set to 0
* elevation. This means the region elevation would be 0. To avoid
* this, I subtract a small amount for ocean regions. */
exports.assign_r_elevation = function(r_elevation, mesh, t_elevation, r_ocean) {
const max_ocean_elevation = -0.01;
r_elevation.length = mesh.numRegions;
let out_t = [];
for (let r = 0; r < mesh.numRegions; r++) {
mesh.r_circulate_t(out_t, r);
let elevation = 0.0;
for (let t of out_t) {
elevation += t_elevation[t];
}
r_elevation[r] = elevation/out_t.length;
if (r_ocean[r] && r_elevation[r] > max_ocean_elevation) {
r_elevation[r] = max_ocean_elevation;
}
}
return r_elevation;
};
/**
* Redistribute elevation values so that lower elevations are more common
* than higher elevations. Specifically, we want elevation Z to have frequency
* (1-Z), for all the non-ocean regions.
*/
// TODO: this messes up lakes, as they will no longer all be at the same elevation
exports.redistribute_t_elevation = function(t_elevation, mesh) {
// NOTE: This is the same algorithm I used in 2010, because I'm
// trying to recreate that map generator to some extent. I don't
// think it's a great approach for other games but it worked well
// enough for that one.
// SCALE_FACTOR increases the mountain area. At 1.0 the maximum
// elevation barely shows up on the map, so we set it to 1.1.
const SCALE_FACTOR = 1.1;
let nonocean_t = [];
for (let t = 0; t < mesh.numSolidTriangles; t++) {
if (t_elevation[t] > 0.0) {
nonocean_t.push(t);
}
}
nonocean_t.sort((t1, t2) => t_elevation[t1] - t_elevation[t2]);
for (let i = 0; i < nonocean_t.length; i++) {
// Let y(x) be the total area that we want at elevation <= x.
// We want the higher elevations to occur less than lower
// ones, and set the area to be y(x) = 1 - (1-x)^2.
let y = i / (nonocean_t.length-1);
// Now we have to solve for x, given the known y.
// * y = 1 - (1-x)^2
// * y = 1 - (1 - 2x + x^2)
// * y = 2x - x^2
// * x^2 - 2x + y = 0
// From this we can use the quadratic equation to get:
let x = Math.sqrt(SCALE_FACTOR) - Math.sqrt(SCALE_FACTOR*(1-y));
if (x > 1.0) x = 1.0;
t_elevation[nonocean_t[i]] = x;
}
};