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2018-12-08 20:16

ChristopherJam

Registered: Aug 2004
Posts: 1409

Tape loaders using more than two pulse widths for data

I’ve been thinking a bit about a novel tape encoding, but it would rely on (among other things) having more than two pulse widths. So far as I can see, none of the old turbo loaders used a long pulse for anything beyond end of byte/end of block - the bitstream itself was just a sequence of short and medium pulses (usually around 200 cycles for short, 300 to 450 for long).

Is there any particular reason none of the popular loaders used (eg) four pulse widths for each bitpair? Even using quite widely separated durations of 210/320/450/600 would then lead to an average time per bit of just 197 cycles, a massive improvement on the 265 cycles per bit you’d get for 210/320.

(The idea I’ve been working on is a bit more complex than that, but if a simple bitpair scheme wouldn’t work for some reason, then the idea I have would need to be scaled back somewhat. Promising that long pulses were used for framing, mind..)

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2018-12-17 21:18

Hoogo

Registered: Jun 2002
Posts: 105

Quoting ChristopherJam

...But can I ask what pulse lengths you were using? 00/01/10/11 will always minimise the count of pulses you output, but I wouldn’t expect it to minimise the total recording length unless your four pulse lengths are quite similar to each other.

I have used 1 pulse + 1 to 4 gaps of equal size, so 2...5 overall length.

For the random test data:

00 207
01 232
10 239
11 202
=>1800 Bits, length 3156 pulses+gaps.
If all appeared 225 times, length would be 3150.

1 365
00 297
01 311
111 73
=>1800 Buts, length 3230.

The arithmetic distribution would be best at 39%, 26%, 19%, 16%, but I don't get the idea how to find a nice Huffmann for that.

2018-12-17 22:17

ChristopherJam

Registered: Aug 2004
Posts: 1409

oh!

Um, that's not a huffman code. (you can't tell from the bits read so far whether you've finished the current codeword or not)
Try
0
10
110
111

Algorithm is to combine the two smallest probability items into a tree node that has the two items as children, and replace the children in your list of things to process with the parent. (Have a look at the function huffman_encode in compare_encodings.py linked above - I lifted it from rosettacode,org )

Naming the codes for the four pulse lengths A B C & D
codes C and D combine first, so whether you get a flat code (two bits each) or not depends on whether the next combine step joins code B with {parent of C & D} or with code A

If the gap is more than around 28% of the minimum pulse length, the likelyhood of C or D is less than the likelyhood of A, so an unbalanced tree is generated.

spacing of (100+ 100*n)
+----------------+-----------------+--------------+
| Pulse duration | arithmetic code | huffman code |
+----------------+-----------------+--------------+
| 100 cycles     | 0.51879         | 1            |
| 200 cycles     | 0.26914         | 01           |
| 300 cycles     | 0.13963         | 001          |
| 400 cycles     | 0.07244         | 000          |
+----------------+-----------------+--------------+
mean cycles per bit, arithmetic code = 105.6
mean cycles per bit,    huffman code = 107.1

spacing of (100+  28*n)
+----------------+-----------------+--------------+
| Pulse duration | arithmetic code | huffman code |
+----------------+-----------------+--------------+
| 100 cycles     | 0.36380         | 0            |
| 128 cycles     | 0.27410         | 10           |
| 156 cycles     | 0.20651         | 111          |
| 184 cycles     | 0.15559         | 110          |
+----------------+-----------------+--------------+
mean cycles per bit, arithmetic code =  68.6
mean cycles per bit,    huffman code =  71.1

spacing of (100+  27*n)
+----------------+-----------------+--------------+
| Pulse duration | arithmetic code | huffman code |
+----------------+-----------------+--------------+
| 100 cycles     | 0.36057         | 11           |
| 127 cycles     | 0.27376         | 10           |
| 154 cycles     | 0.20785         | 01           |
| 181 cycles     | 0.15781         | 00           |
+----------------+-----------------+--------------+
mean cycles per bit, arithmetic code =  68.0
mean cycles per bit,    huffman code =  70.2

2018-12-17 23:07

Hoogo

Registered: Jun 2002
Posts: 105

Got it! Soo obvious now :)

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