# Stampery

## Verification algorithm

o = (i / 2^n) mod 2

i and n are equals to or more than 0. In this equation, o will be 0 if i is less than 2^n. Then, here is omicron table.

i value is in first row, n value is in left most column.

0 1 2 3 4 5 6 7 8 9 10 11 12 13
0 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
2 0 0 0 0 1 1 1 1 0 0 0 0 1 1
3 0 0 0 0 0 0 0 0 1 1 1 1 1 1

0 means merge with right hash, 1 means with left hash.

    ___o___
^       ^
/ \     / \
^   ^   ^   ^
| | | | | | | |
0 1 2 3 4 5 6 7


node 5 will be merged with left, right, left. it corresponds to 1, 0, 1.