Geohash算法的实现

geohash的算法

下面以(39.92324, 116.3906)为例，介绍一下geohash的编码算法。首先将纬度范围(-90, 90)平分成两个区间(-90, 0)、(0, 90)， 如果目标纬度位于前一个区间，则编码为0，否则编码为1。由于39.92324属于(0, 90)，所以取编码为1。然后再将(0, 90)分成 (0, 45), (45, 90)两个区间，而39.92324位于(0, 45)，所以编码为0。以此类推，直到精度符合要求为止，得到纬度编码为1011 1000 1100 0111 1001。

纬度范围	划分区间0	划分区间1	39.92324所属区间
(-90, 90)	(-90, 0.0)	(0.0, 90)	1
(0.0, 90)	(0.0, 45.0)	(45.0, 90)	0
(0.0, 45.0)	(0.0, 22.5)	(22.5, 45.0)	1
(22.5, 45.0)	(22.5, 33.75)	(33.75, 45.0)	1
(33.75, 45.0)	(33.75, 39.375)	(39.375, 45.0)	1
(39.375, 45.0)	(39.375, 42.1875)	(42.1875, 45.0)	0
(39.375, 42.1875)	(39.375, 40.7812)	(40.7812, 42.1875)	0
(39.375, 40.7812)	(39.375, 40.0781)	(40.0781, 40.7812)	0
(39.375, 40.0781)	(39.375, 39.7265)	(39.7265, 40.0781)	1
(39.7265, 40.0781)	(39.7265, 39.9023)	(39.9023, 40.0781)	1
(39.9023, 40.0781)	(39.9023, 39.9902)	(39.9902, 40.0781)	0
(39.9023, 39.9902)	(39.9023, 39.9462)	(39.9462, 39.9902)	0
(39.9023, 39.9462)	(39.9023, 39.9243)	(39.9243, 39.9462)	0
(39.9023, 39.9243)	(39.9023, 39.9133)	(39.9133, 39.9243)	1
(39.9133, 39.9243)	(39.9133, 39.9188)	(39.9188, 39.9243)	1
(39.9188, 39.9243)	(39.9188, 39.9215)	(39.9215, 39.9243)	1

经度也用同样的算法，对(-180, 180)依次细分，得到116.3906的编码为1101 0010 1100 0100 0100。

经度范围	划分区间0	划分区间1	116.3906所属区间
(-180, 180)	(-180, 0.0)	(0.0, 180)	1
(0.0, 180)	(0.0, 90.0)	(90.0, 180)	1
(90.0, 180)	(90.0, 135.0)	(135.0, 180)	0
(90.0, 135.0)	(90.0, 112.5)	(112.5, 135.0)	1
(112.5, 135.0)	(112.5, 123.75)	(123.75, 135.0)	0
(112.5, 123.75)	(112.5, 118.125)	(118.125, 123.75)	0
(112.5, 118.125)	(112.5, 115.312)	(115.312, 118.125)	1
(115.312, 118.125)	(115.312, 116.718)	(116.718, 118.125)	0
(115.312, 116.718)	(115.312, 116.015)	(116.015, 116.718)	1
(116.015, 116.718)	(116.015, 116.367)	(116.367, 116.718)	1
(116.367, 116.718)	(116.367, 116.542)	(116.542, 116.718)	0
(116.367, 116.542)	(116.367, 116.455)	(116.455, 116.542)	0
(116.367, 116.455)	(116.367, 116.411)	(116.411, 116.455)	0
(116.367, 116.411)	(116.367, 116.389)	(116.389, 116.411)	1
(116.389, 116.411)	(116.389, 116.400)	(116.400, 116.411)	0
(116.389, 116.400)	(116.389, 116.394)	(116.394, 116.400)	0

接下来将经度和纬度的编码合并，奇数位是纬度，偶数位是经度，得到编码 11100 11101 00100 01111 00000 01101 01011 00001。

最后，用0-9、b-z（去掉a, i, l, o）这32个字母进行base32编码，得到(39.92324, 116.3906)的编码为wx4g0ec1。

十进制	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
base32	0	1	2	3	4	5	6	7	8	9	b	c	d	e	f	g
十进制	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30	31
base32	h	j	k	m	n	p	q	r	s	t	u	v	w	x	y	z

解码算法与编码算法相反，先进行base32解码，然后分离出经纬度，最后根据二进制编码对经纬度范围进行细分即可，这里不再赘述。 不过由于geohash表示的是区间，编码越长越精确，但不可能解码出完全一致的地址。

#define BASE32 "0123456789bcdefghjkmnpqrstuvwxyz"
      static void encode_geohash(double latitude, double longitude, int precision, char *geohash) {
      int is_even=1, i=0;
      double lat[2], lon[2], mid;
      char bits[] = {16,8,4,2,1};
      int bit=0, ch=0;
      lat[0] = -90.0; lat[1] = 90.0;
      lon[0] = -180.0; lon[1] = 180.0;
      while (i < precision) {
      if (is_even) {
      mid = (lon[0] + lon[1]) / 2;
      if (longitude > mid) {
      ch |= bits[bit];
      lon[0] = mid;
      } else
      lon[1] = mid;
      } else {
      mid = (lat[0] + lat[1]) / 2;
      if (latitude > mid) {
      ch |= bits[bit];
      lat[0] = mid;
      } else
      lat[1] = mid;
      }
      is_even = !is_even;
      if (bit < 4)
      bit++;
      else {
      geohash[i++] = BASE32[ch];
      bit = 0;
      ch = 0;
      }
      }
      geohash[i] = 0;
      }