bit layout

H3 cell

name	# of bits	value(s)
reserved	1	0
mode	4	1
reserved	3	0
resolution	4	0–15
base cell #	7	0–121
digit 1	3	0–7
\(\vdots\)	\(\vdots\)	\(\vdots\)
digit 15	3	0–7

Note that \(1 + 4 + 3 + 4 + 7 + 3 \cdot 15 = 64\).

Cell parents/children

name	# of bits	value(s)	parent	child
reserved	1	0	0	0
mode	4	1	1	1
reserved	3	0	0	0
resolution	4	0–15	1	2
base cell #	7	0–121	37	37
digit 1	3	0–7	2	2
digit 2	3	0–7	7	5
digit 3	3	0–7	7	7

child < parent

H3 directed edge

name	# of bits	value(s)
reserved	1	0
mode	4	2
edge direction	3	1–6
resolution	4	0–15
base cell #	7	0–121
digit 1	3	0–7
\(\vdots\)	\(\vdots\)	\(\vdots\)
digit 15	3	0–7

random

Cells

A hexagon’s 7 children are numbered

0,1,2,3,4,5,6

in the next resolution digit.

A pentagon’s 6 children are numbered

0,2,3,4,5,6 (skips 1)

in the next resolution digit.

Vertices

Since each vertex is incident to 3 cells, but we also want H3 index uniqueness, we have to choose just one of these cells to be the “owner” of the vertex.

A hexagon’ 6 vertices are numbered

0,1,2,3,4,5

in the “reserved/mode-dependent” bits.

A pentagons’ 5 vertices are numbered

0,1,2,3,4,5 (no skip)