Description of graph6 and sparse6 encodings
-------------------------------------------
Brendan McKay, bdm@cs.anu.edu.au
Updated May 2005.
General principles:
All numbers in this description are in decimal unless obviously
in binary.
Apart from the header, there is one object per line. Apart from the
header and the end-of-line characters, all bytes have a value in the
range 63-126 (which are all printable ASCII characters). A file of
objects is a text file, so whatever end-of-line convention is locally
used is fine).
Bit vectors:
A bit vector x of length k can be represented as follows.
Example: 1000101100011100
(1) Pad on the right with 0 to make the length a multiple of 6.
Example: 100010110001110000
(2) Split into groups of 6 bits each.
Example: 100010 110001 110000
(3) Add 63 to each group, considering them as bigendian binary numbers.
Example: 97 112 111
These values are then stored one per byte.
So, the number of bytes is ceiling(k/6).
Let R(x) denote this representation of x as a string of bytes.
Small nonnegative integers:
Let n be an integer in the range 0-68719476735 (2^36-1).
If 0 <= n <= 62, define N(n) to be the single byte n+63.
If 63 <= n <= 258047, define N(n) to be the four bytes
126 R(x), where x is the bigendian 18-bit binary form of n.
If 258048 <= n <= 68719476735, define N(n) to be the eight bytes
126 126 R(x), where x is the bigendian 36-bit binary form of n.
Examples: N(30) = 93
N(12345) = N(000011 000000 111001) = 126 66 63 120
N(460175067) = N(000000 011011 011011 011011 011011 011011)
= 126 126 63 90 90 90 90 90
Description of graph6 format.
----------------------------
Data type:
simple undirected graphs of order 0 to 68719476735.
Optional Header:
>>graph6<< (without end of line!)
File name extension:
.g6
One graph:
Suppose G has n vertices. Write the upper triangle of the adjacency
matrix of G as a bit vector x of length n(n-1)/2, using the ordering
(0,1),(0,2),(1,2),(0,3),(1,3),(2,3),...,(n-1,n).
Then the graph is represented as N(n) R(x).
Example:
Suppose n=5 and G has edges 0-2, 0-4, 1-3 and 3-4.
x = 0 10 010 1001
Then N(n) = 68 and R(x) = R(010010 100100) = 81 99.
So, the graph is 68 81 99.
Description of sparse6 format.
------------------------------
Data type:
Undirected graphs of order 0 to 68719476735.
Loops and multiple edges are permitted.
Optional Header:
>>sparse6<< (without end of line!)
File name extension:
.s6
General structure:
Each graph occupies one text line. Except for end-of-line characters,
each byte has the form 63+x, where 0 <= x <= 63. The byte encodes
the six bits of x.
The encoded graph consists of:
(1) The character ':'. (This is present to distinguish
the code from graph6 format.)
(2) The number of vertices.
(3) A list of edges.
(4) end-of-line
Loops and multiple edges are supported, but not directed edges.
Number of vertices n:
1, 4, or 8 bytes N(n) as above.
This is the same as graph6 format.
List of edges:
Let k be the number of bits needed to represent n-1 in binary.
The remaining bytes encode a sequence
b[0] x[0] b[1] x[1] b[2] x[2] ... b[m] x[m]
Each b[i] occupies 1 bit, and each x[i] occupies k bits.
Pack them together in bigendian order, and pad up to a
multiple of 6 as follows:
1. If (n,k) = (2,1), (4,2), (8,4) or (16,5), and vertex
n-2 has an edge but n-1 doesn't have an edge, and
there are k+1 or more bits to pad, then pad with one
0-bit and enough 1-bits to complete the multiple of 6.
2. Otherwise, pad with enough 1-bits to complete the
multiple of 6.
These rules are to match the gtools procedures, and to avoid
the padding from looking like an extra loop in unusual cases.
Then represent this bit-stream 6 bits per byte as indicated above.
The vertices of the graph are 0..n-1.
The edges encoded by this sequence are determined thus:
v = 0
for i from 0 to m do
if b[i] = 1 then v = v+1 endif;
if x[i] > v then v = x[i] else output {x[i],v} endif
endfor
In decoding, an incomplete (b,x) pair at the end is discarded.
Example:
:Fa@x^
':' indicates sparse6 format.
Subtract 63 from the other bytes and write them in binary,
six bits each.
000111 100010 000001 111001 011111
The first byte is not 63, so it is n. n=7
n-1 needs 3 bits (k=3). Write the other bits in groups
of 1 and k:
1 000 1 000 0 001 1 110 0 101 1 111
This is the b/x sequence 1,0 1,0 0,1 1,6 0,5 1,7.
The 1,7 at the end is just padding.
The remaining parts give the edges 0-1 0-2 1-2 5-6.