#acl Björn Buchhold:read,write Elmar Haussmann:read,write Claudius Korzen:read,write Florian Bäurle:read,write Jonas Sternisko:read,write

== TASK 1 (Compression + entropy) ==

1, x, x+1, 2x, 2x+1, 3x, 3x+1, 4x

=== 1.1 For x = 10, gap-encode using Golomb with M = 8 ===

{{{
Gaps are: +1, +9, +1, +9, +1, +9, +1, +9 (eight gaps, not seven)
Golomb code for +1 : 1001
Golomb code for +9 : 01001
Encoding of the whole sequence: 1001 01001 1001 01001 1001 01001 1001 01001
}}}

=== 1.2 No larger M gives shorter sequence ===

For M > 8, writing the residual (r) part alone costs >= 4 Bits. Together with the 1 separating it from the q part, this is >= 5 Bits per code. That is >= 40 Bits for the sequence above. With M = 8 we need only 36 Bits.

=== 1.3 Function that computes Golomb code for M = 8 ===

{{{
ArrayList<Bool> golombCode(int x) {
  result = new ArrayList<Bool>; 
  int q = x / 8;
  int r = x % 8;
  for (int i = 0; i < q; i++) { result.add(0); }
  result.add(1);
  result.add(r & 4 > 0);
  result.add(r & 2 > 0);
  result.add(r & 1 > 0);
  return result;
}}}
 
=== 1.4 Entropy-optimal code for sequence above, for abitrary x ===

We have two symbols, each occuring the same number of times. The entropy of the corresponding random variable (X taking on either value with probability 1/2) is 1/2 * log_2 1/2 + 1/2 * log_2 1/2 = 1. And there is indeed a (prefix-free) encoding with just 1 Bit per value:

{{{
1 : 0
x-1 : 1
}}}