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It focuses on FNT(q = 2^p + 1).
Given a block of n symbols of FNT(q), a metadata part should indicate indices of all symbols whose value = 2^p.
If these symbols are uniformly distributed, we can use the frequency of 2^p to determine the size of metadata.
For general cases, the naive approach needs n indicators for n elements, i.e. metadata size = n.
For FNT(65537), the following approach results to metadata of only one indicator.
The indicator contains the index, e.g. i, of the first symbol = 2^p. Then the ith symbol will be used to contain index of the next symbol 2^p, and so on. If there is no more symbol 2^p, this symbol will be zero.
Example,
data = [1, 2, 65536, 3, 4, 65536, 5, 6, 7]
=> metadata = [2]
data = [1, 2, 5, 3, 4, 0, 5, 6, 7]
The approach can be used for FNT(257) for n < 257.
The text was updated successfully, but these errors were encountered:
It focuses on
FNT(q = 2^p + 1)
.Given a block of
n
symbols ofFNT(q)
, a metadata part should indicate indices of all symbols whose value =2^p
.If these symbols are uniformly distributed, we can use the frequency of
2^p
to determine the size of metadata.For general cases, the naive approach needs
n
indicators forn
elements, i.e. metadata size =n
.For FNT(65537), the following approach results to metadata of only one indicator.
The indicator contains the index, e.g.
i
, of the first symbol =2^p
. Then theith
symbol will be used to contain index of the next symbol2^p
, and so on. If there is no more symbol2^p
, this symbol will be zero.Example,
data = [1, 2, 65536, 3, 4, 65536, 5, 6, 7]
=> metadata = [2]
data = [1, 2, 5, 3, 4, 0, 5, 6, 7]
The approach can be used for FNT(257) for
n
< 257.The text was updated successfully, but these errors were encountered: