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Basic Method
BICM: Text 1
BICM: Text 2
Recursive Method
RICM: Text
Encryption
Real world uses
Using playing cards to store hidden data:
The Implied Card Method for
Encoding Data Into Playing Cards.



Using the whole pack for text:

In the previous text examples we were encoding a short message into the string and with each improvement I showed how many unused cards we ended up with after it. Technically, only that portion of the pack would need to be kept in order to store the message. However, to best show how the Implied Encoding Method works we will now try to encode a much longer string and see how much of it we can put into one pack. This means we will be using the whole of the of the pack, and we will need to be much more careful about using strings of consecutive zeroes near to the end. If we don't pay attention then the last few characters might end up decoding as nonsense. For this reason it pays to keep an eye on how many remaining cards there are, and in many cases we might have to leave a large number of unused cards at the end of the pack.

Our message will be: "Hello I am trying to encode as many characters of text as possible into a single pack of fifty two playing cards using a method that maps the cards onto carefully chosen binary numbers." This is made up of 151 characters which is more than we will ever hope to fit in.

When using Mapping Table 1 (5-bit binary numbers in order of magnitude, mapped onto the letters of the alphabet sorted by frequency), we manage to encode 29 characters of our message into the pack. We fit in the letters, "Hello I am trying to encode as many c." There are 4 cards left over: the 2, 4 and 7 of Diamonds, and the 4 of Spades.

When using Mapping Table 2 (5-bit binary numbers sorted by the number of bits in them (low to high), mapped onto the letters of the alphabet sorted by frequency), we manage to encode 33 characters of our message into the pack. We fit in the letters, "Hello I am trying to encode as many chara". There are 4 cards left over: the 6 of Clubs, the 7 and Queen of Hearts, and the 10 of Diamonds.

When using Mapping Table 3 (6-bit binary numbers sorted by the number of bits in them, mapped onto the letters of the alphabet sorted by frequency), we manage to encode 34 characters of our message into the pack. We fit in the letters, "Hello I am trying to encode as many charac". There are 6 cards left over: the 8 of Diamonds, the 8 of Spades, the King of Spades, the 5 of Clubs, the King of Hearts and the 3 of Diamonds. We can't encode into the last 6 cards as there are too many consecutive zeroes. At this level of bit length we are starting to see diminishing returns as the quantity of consecutive zeroes prevents us using the last few cards.

When using Mapping Table 4 (7-bit binary numbers sorted by the number of bits in them, mapped onto the letters of the alphabet sorted by frequency), we manage to encode only 31 characters ("Hello I am trying to encode as many cha"). The reason for this is that near the end of the encoding we have 12 cards left, but need to encode the sequence "r", "a" which are mapped onto 1000000 and 0000010. After the first 1 that gives us 11 consecutive zeroes to encode in just 11 cards. When those 11 cards are passed to the back, they end up in exactly the same position as before, so won't show up in the decoding. The following characters would then be decoded as nonsense. Supposing we had chosen a different message to encode, then we might not come up with this problem at this point. Whether it happens or not is down to luck. It is a situation where it would be helpful to use a computer to choose the wording of the message in a way that maximises the number of characters we can fit in.





There are two limits to the number of letters we can encode:

1: There is a set number of permutations of the cards. These permutations allow differing capacities for storing bits (from 52 to 1,378) but the higher capacities require unlikely data (e.g. 50 zeroes in a row), which is not to happen when encoding text. I haven't done much research into it, but I think the average maximum for real-world text seems to be around 35 characters.

2: We can only encode fifty two 1s in a pack (when using the Basic Implied Encoding Method).

Using the Basic Method, the most 1s a pack can contain is 52. If we didn't encode the letter "e" with a zero, the highest card to letter ratio that would ever be possible is 1 card per letter, and the most letters we could encode would never be more than 52. However as long as our message only used letters that used one 1 in their binary number and used at least one "e" then we could theoretically beat 52. This would be quite difficult for numbers with few bits in them. For example, with 7-bit numbers we could only use the letters, "etaoinsr".

If we tried with 10-bit numbers we would have more letters to use in our message ("etaoinsrhld") and this gives us a better chance of encoding a message of more than 52 letters. The mapping for 10-bit numbers containing one or fewer 1s would be like this:

Mapping Table 5:
Containing no 1s:
00 0000 0000 (000 in decimal): e

Containing one 1:
00 0000 0001 (001): t
00 0000 0010 (002): a
00 0000 0100 (004): o
00 0000 1000 (008): i
00 0001 0000 (016): n
00 0010 0000 (032): s
00 0100 0000 (064): r
00 1000 0000 (128): h
01 0000 0000 (256): l
10 0000 0000 (512): d

There are quite a few words we can make from these 11 letters, however because we are using 10-bit numbers, we risk running out of cards when we have used up just 42 cards. We couldn't have an "e" (00 0000 0000) as the last letter if we only had 10 cards left, or else it won't exist when we come to decode it. Similarly, we can't have a "d" (10 0000 0000) as the last letter if we only had 9 cards left.

Another major problem occurs when we have many consecutive zeroes from neighbouring letters. The letters "d", "e", "t" ("10 0000 0000" "00 0000 0000" "00 0000 0001") in that order result in 28 consecutive zeroes. This means there need to be at least 29 cards left in the pack at the point we start encoding the first zero, otherwise it will be decoded as nonsense. Other bad card combinations aren't as bad but will still throw out the encoding long before the end. Any messages over 52 letters would need to be very carefully thought out, and it's extremely unlikely that any useful message of this length would ever be encoded.

It is however possible to use 10-bit numbers to encode messages of more than 52 letters without resorting to lengthy strings of consecutive "e"s. Without putting too much effort into it, I have reached 55 letters with an extremely contrived message: "Three released deer seen in little heeled shoes and old oil stains". This works out as 1.06 letters per card. After encoding there were 11 unused cards left so technically the whole message is held within just 41 cards which works out at 1.34 letters per card. However all 52 cards were needed to actually encode the message. The data required 20 Decoding Rows to decode. (For future reference, there were only 6 cards that could have been used for the Recursive Method so that wouldn't have helped in this situation).




Other Optimisation ideas:

With bit lengths that leave plenty of unassigned numbers, we can further develop the system of mapping other things onto the numbers with few 1s in them. We could map popular words for one thing. However we could also map numerals or even the most common pairs of letters (digraphs) such as: th, he, in, er, an, re, nd, at, on, nt, ha, es, st, en, ed, to, it, ou, ea, hi, is, or, ti, as, te, et; or the most common triple letter combinations (trigraphs) such as: and, for, the, tha, ent, ion, tio, nde, has, nce and so on. As long as the words, digraphs and trigraphs use fewer 1s than the individual letters within them then it would be worthwhile. Which digraphs, trigraphs, words and numerals we would use would require some research. There is also the problem that the mapping table would stop being intuitive and calculable, and would therefore need to be learned or kept by the decoder.

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