Binary digits system - PAO users?

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Hi gang,

I started doing binary sequences and I used my PAO to do so. Encoding 18 binary bits per image.

Example:

101011110001111010

Gets broken down to:

101-011-110-001-111-010

53-61-72

Resulting PAO scene:

Eric Clapton / SQUEEZING WATER / out of an AMERICAN FLAG

Can someone tell me if a lot of PAO users are doing this method for binary as well?

Thanks in advance.

Best,

Rado

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Yes this is a pretty standard way to go. 3 binary digits per base10 digit, so 6 binary per 2-digit number mnemonic element, for 18 binary digits per scene when using a PAO.

Binary is a discipline that gets a huge jump in data compression when using a 3-digit number system. 9 digits per mnemonic element (or even 10 digits if using a 2-block technique.)

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Hi Tim,

I hope you’re doing well.

Could you clarify what are the potential benefits of 3-digit system in the binary disciplines? I got pretty fast with PAO for binary but what a 3-digit system has to offer there? What are the max amount of digits per locus that one can encode?

Are the digits translated via 4-bit sequences or with 3-bit sequences - same way PAO does?

Also I heard that the Ben-type of systems can encode 30 binary bits per locus. How is that possible?

Thanks in advance.

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Using a 3-digit system is just like the 2-digit approach. Each decimal digit encodes 3 binary digits. If you have a system where 3 decimal digits are represented by a single mnemonic element, you can compress 9 binary digits into that one element.

So with a single digit system “1” = 001, maybe represented by a Toe. With a two-digit system “12” = 001010, maybe represented by ToNy. “127” = 001010111, maybe represented by a TaNK.

The number of digits per locus depends just on how many mnemonic elements you use in a scene. I’ve talked before about more complex scenes giving you more information per loci but with the trade of making it take longer and be more complicated to encode and decode.

I usually recommend 2 or 3 elements per scene as the sweet spot between “easy to build and remember” and “fewest practical number of scenes without being too complex. This means you’d encode 18 or 27 binary digits per scene. If you used a 2-block technique, you’d encode 10 digits per element and however many elements you used in each location would be 10x binary digits.

This is just someone using 3 elements per loci, encoding 10 binary digits per element.

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