in reply to Re^2: Dynamically Updating Frequency Analysis in thread Dynamically Updating Frequency Analysis
We note that HE appears 3 times and will give us the largest compression (I have used # to indicate the data has already been compressed).
But that still doesn't make sense to me. Why would you encode all the "HE"'s when "THEN" is also in your dictionary? Of the choices you could make here, that seems the strangest option. For instance if you were taking that approach I would expect you to encode from longest to shortest first. In which case I would expect the THEN to match first, and then HE to match twice. Which is actually the optimal match for this string and dictionary. IE: "THE HEN THEN" would turn into "T" "HE" "HE" "N" "THEN", which at 5 symbols is shorter than the 7 that makes up "T" "HE" "HE" "N" "T" "HE" "N".
Also consider that encoding a single THEN wins you more than encoding 3 HE's with the type of scheme you have (4:1 versus 2:1), which is not a problem you would have with huffman encoding where the bit sequence for HE might be shorter than that of THEN. Perhaps you need to account for the length of the string being replaced when you do your frequency analysis.
Anyway, there are other issues here. Once you allow overlapping strings you have a problem that is much harder. A simple greedy algorithm will not necessarily find the optimal encoding of the string. For that you have to do exhaustive search. Luckily greedy solutions tend to be "close enough" most of the time.
If I may be so bold, encouraging people to try to implement algorithms without any domain research is not a bad idea as the starting point of a learning exercise. But I think you should follow up with some proper study of some important algorithms. In this case I would suggest really studying Huffman encoding and LZ. Both are straight forward to implement in perl, and Huffman encoding in particular is nearly trivial to implement in Perl. Here is the tree construction algorithm:
sub build_huffman_tree {
my ($dict_hash)= @_;
# @node constains either:
# leafs: [ $non_ref_symbol, $count ]
# nodes: [ [ $left_node, $right_node ], $left_count + $right_count ]
+;
# note we assume the counts in the hash are > 0.
my @nodes= map { [ $_, $dict_hash>{$_} ] } keys %$dict_hash;
while (@nodes > 1) {
@nodes= sort { $b>[1] <=> $a>[1] } @nodes;
my $right= pop @nodes;
my $left= pop @nodes;
push @nodes, [ [ $left, $right ], $left>[1] + $right>[1] ];
}
return $nodes[0];
}
Doing the rest is a good exercise for your people and will teach them more than struggling with their own algorithms without understanding these ones first.

$world=~s/war/peace/g
Re^4: Dynamically Updating Frequency Analysis by Limbic~Region (Chancellor) on Aug 01, 2013 at 20:45 UTC 
demerphq,
Unfortunately, life has side tracked me and I don't have time to respond fully but I did want to leave enough details here so that when/if I get a chance, I will have enough context to respond appropriately.
But that still doesn't make sense to me. Why would you encode all the "HE"'s when "THEN" is also in your dictionary?
In my example, it turns out to be a tie. Compressing the 3 instances of 'HE' as step 1 reduces the string from 12 characters to 9. If instead, as step 1 'THEN' was chosen it would still go from 12 characters down to 9. I should have chosen a nontie example but the idea was that at that point in time, compressing 3 instances of 'HE' was supposed to be better than 1 instance of 'THEN'
I think to understand how the person arrived at this decision (and why it makes sense to him), you need to follow his logic.
 Determine the number of distinct 1 byte characters that appear in the file  96
 Use only as many bits as necessary to encode the full range of characters  2^6 = 64 < 96 < 128 = 2^7
 Use the unused 7 bit sequences to encode common multibyte sequences  he intended to sample the whole file
 Compress the file by in a recursive fashion
 Choose the sequence that results in maximal compression = $bytes_prior_to_compression  ($num_of_chars_to_encode * $occurrences) + $occurrences
 Encode those bytes
 Update the frequency data (without resampling) and go back to step 1
Also consider that encoding a single THEN wins you more than encoding 3 HE's with the type of scheme you have (4:1 versus 2:1)
Actually, no. Because there is only 1 instance of THEN, encoding it has reduced the file by 3 characters. While it is true that encoding 3 instances of HE also only reduces the file by 3 characters, the example probably should have been 'THE HEN THEN HE HE HE'
Luckily greedy solutions tend to be "close enough" most of the time.
He gave up and went with a heuristic approach that didn't require recalculating the frequency distributions. I didn't see his original approach as viable but I wanted to help him explore it as fully as possible.
If I may be so bold, encouraging people to try to implement algorithms without any domain research is not a bad idea as the starting point of a learning exercise. But I think you should follow up with some proper study of some important algorithms.
That is exactly what this exercise was about. Not everyone agrees with my philosophy of doing things the wrong way to appreciate the right way later.
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