Base 3 Computing?

[chsh1ca]

I was just wondering if anyone out there has any good links for me for base 3 computing information. Specifically, if it's being researched, how it could be achieved, etc..

I recently re-read When Harlie Was One v2.0, and for some reason took a liking to the idea of non-binary decision-making logic.

If someone could point me in the right direction, it'd be greatly appreciated. Also, feel free to reply to this post and discuss the topic. It's a very intriguing idea to me.

 

[figgypower]

I had no idea what you were talking about, but it sounded real interesting, so I found this. It's a great article explaining the ternary system in American Scientist; a very good read.

 

[sash1]

base 3?



~Aunix

 

[MustPost]

As I recall, the Russians actually once made a base3 computer

 

[f95toli]

They also made a base7 computer.

But I guess what chsh1ca meant is a compuyter that uses not only 0s and 1s but also 2s to compute, a 3-state computer.

 

[Bloodstein]

Duh!

 

[AMDfreak]

This might be a dumb question. Would quantum computing be considered a base 3 system? I'm getting the assosciation between the two in my head since a qubit (sp?) can be on, off, or both at once.

 

[TerryMathews]

A Trinary computer would be possible to construct, but probably not cost effective. Anything except binary cannot be simplified by the shortcuts allowed from boolean algebra.

Some parts of the computer may start to utilize four states, and internally remap that two a two-bit binary set. But the main parts of the computer will remain binary for at least the near future.

 

[chiwawa626]

no, no, no you got it all wrong. it's the break in the circut that causes waveform modulations in the FLUX CAPACITOR causing a break in the TIME SPACE CONTINUEM.

 

[figgypower]

Originally posted by: TerryMathews
A Trinary computer would be possible to construct, but probably not cost effective. Anything except binary cannot be simplified by the shortcuts allowed from boolean algebra.

Some parts of the computer may start to utilize four states, and internally remap that two a two-bit binary set. But the main parts of the computer will remain binary for at least the near future.

Actually, if you read the article that I posted about above you'll see that theoritically it'll be more cost effective to produce. It never caught on, because of whatever reason (there are suggested theories on this, too - but no definite answer). I think quantum computing might lets us build a ternary system, but I'm just speculating. I mean like AMDfreak said, you've got on, off, and both. Why not apply the ternary model to it?

 

[Peter]

Well for one, it wasn't an arbitrary choice to begin with. Computers do binary because it is simple to compute with - every single computation can be cracked down to series of AND, OR, NOT operations.

 

[Sahakiel]

No, no, no Base 3 work. That requires having three states, which is in violation of the Fundamental Laws of Physics :

I'm Right.
You're Wrong.
There is no Grey Area.

 

[imgod2u]

There is so a grey. Well, more like both white and black at the same place and time. Quantum superposition

Oh, and there is SO a #$@#$@# spoon!

 

[Smilin]

This is actually kindof an old concept. Remember we finally settled on digital (binary) computers after we found out how difficult it was to build an analog one. I've heard of analog computers using as many as 8 voltage states.

 

[chsh1ca]

Ok ok, in the book, HARLIE (a mainframe that is basically the world's first AI) is a base 255 computer (IIRC).

Essentially, it's explained that BASE 2 = binary, which offers both YES and NO as possible answers. This fits a boolean model.
Base X computing, would require X possible answers, and doesn't fit into boolean arithmetic. The options for base 3 were described as YES, NO, MAYBE. And so on.

Base 5: YES, MAYBE YES, MAYBE, MAYBE NO, NO.
Base 7: YES, MOSTLY YES, MAYBE YES, MAYBE, MAYBE NO, MOSTLY NO, NO.

None of these unfortunately fit a boolean model, which is the simplest model we have to work with.

The problems come into play when you say something like "What is the boolean value of expression?"

In binary, you can say 1 or 0 (true or false). However, in base 3/5/7/whatever, you have more options, you would never hit them, with an equation like (1 + 1 = 2), but with something more open like (green = blue + yellow).
As I understand it, it would look like the following:
(green = 10blue + 10yellow) answers YES.
(green = 10blue + 1yellow) answers MAYBE.
(green = 10blue) answers NO.
The amounts I gave are to illustrate colour weighting.
Similarly, in base 5:
(green = 10blue + 10yellow) answers YES.
(green = 10blue + 1yellow) answers MAYBE NO.
(green = 10blue) answers NO.
In base 7:
(green = 10blue + 10yellow) answers YES.
(green = 10blue + 1yellow) answers MOSTLY NO.
(green = 10blue) answers NO.

And so on. I'm not a super-cool mathematician/engineer, but alternate systems have intrigued me. Thanks for the links.

 

[f95toli]

Originally posted by: figgypower
Actually, if you read the article that I posted about above you'll see that theoritically it'll be more cost effective to produce. It never caught on, because of whatever reason (there are suggested theories on this, too - but no definite answer). I think quantum computing might lets us build a ternary system, but I'm just speculating. I mean like AMDfreak said, you've got on, off, and both. Why not apply the ternary model to it?

A quantum computer only uses two states. The fact that it also uses superpositions of these two states does not change the fact that the information in the system is encoded as 1s and 0s; and all the quantum gates that I know of use two states.

Actually it should be possible to build a ternary quantum computer but manipulating 2 states is tricky enough.

 

[TerryMathews]

Trinary computation won't be cost effective, because there are lots of mathematical shortcuts for simplifing binary computers that depend on boolean algebra, or a number system where variables are only 0 or 1. Most of you might not be familiar with boolean algebra without taking at least an intro digital computer design course.

Trinary, obviously, cannot be covered under the laws of boolean algebra and therefore none of the engineering shortcuts developed in the last 20 years would apply, also invalidating the educations of the foremost CEGs in the country.

This is why trinary would not be cost effective. It'd be like someone developing a 60V DC TV and saying it's more cost effective. Sure, it could be, but everyone would have to convert power in their house, eliminating the cost savings. Any time you discard all of the old {tech,knowledge,people}, thing become less cost effective very quickly.

 

[figgypower]

The quantum computing - it'd just be cool. And theoretically more efficient, but in the future I guess. Yes anytime you just outright chuck old technology and go for new technology, things get very expensive quickly - in the long run the more efficient system should be better/cheaper. I think you'd reach a wall of efficiency even with shortcuts. Not to mention that you could also develop shortcuts and methodolgies to improve a trinary system. I mean with that logic, Intel shouldn't have come out with IA64 instruction set and the x86-64, because it improves on already existing technologies rather then outright being something new, should win.

With quantum computers, you're doing something totally new. Why not apply a different base system? Yeah, I can't even imagine the technical complexity, but while you're at it - you're ultimately going to be more efficient with the trinary system. Who knows what we can do with trinary computing if we develop it for 20 years, since at the very basic level it's more efficeint then the binary system?

 

[TerryMathews]

By that logic, why stop at base 3? Why not go with base 10, and then computer hardware and code would be (more) understandable to humans.

There are many reasons why binary is here to stay. Not the least of which is the fact that non-binary systems are not forgiving of a lack of 'clean' power. Binary systems set a threshold, where above is high and below is low, so that you don't have to achieve the perfect 1.65V for a high signal, nor do you have to drop clear to ground for a low signal. For every base you add, the tolerance allowable for a given state would have to decrease exponentially.

Last time I checked, it takes some of the best power hardware available to power high-end Athlons and P4s. I can't even fathom the amount of filtering necessary to operate a trinary computer chip at any appreciable clock speed.

 

[f95toli]

Originally posted by: figgypower
Why not apply a different base system? Yeah, I can't even imagine the technical complexity, but while you're at it - you're ultimately going to be more efficient with the trinary system. Who knows what we can do with trinary computing if we develop it for 20 years, since at the very basic level it's more efficeint then the binary system?

Quantum computers do not use "base system" in the way an ordinary computer would. The way the actual computation is done has nothing to do with ordinary boolean algebra, what I wrote was the the gates use boolean logic (for example CNOT).
You could use a a 3-state system in order to increase the number of possible states for the system but it would probably not be very efficient since you would be sacrificing coherence time.

 

[Smilin]

Other algebraic rules will apply to number systems other than base 10 and 2. Boolean is incredibly effective at reducing complex equations/inputs down to simple equations/circuits.

It doesn't mean this can't be done with other bases though.

 

[figgypower]

Originally posted by: TerryMathews
By that logic, why stop at base 3? Why not go with base 10, and then computer hardware and code would be (more) understandable to humans.

At this point I think I'm getting over my head. My logic is based on the article that I linked to where base 3 was said to be efficient. I quote:
"Curiously, this problem is easier to solve if r and w are treated as continuous rather than integer variables?that is, if we allow a fractional base and a fractional number of digits. Then it turns out (see Figure 1) that the optimum radix is e, the base of the natural logarithms, with a numerical value of about 2.718. Because 3 is the integer closest to e, it is almost always the most economical integer radix (see Figure 2)." That's why I would indeed stop at 3. Of course, like I said this is getting a bit over my head.

 

[m0ti]

Originally posted by: figgypower

Originally posted by: TerryMathews
By that logic, why stop at base 3? Why not go with base 10, and then computer hardware and code would be (more) understandable to humans.

At this point I think I'm getting over my head. My logic is based on the article that I linked to where base 3 was said to be efficient. I quote:
"Curiously, this problem is easier to solve if r and w are treated as continuous rather than integer variables?that is, if we allow a fractional base and a fractional number of digits. Then it turns out (see Figure 1) that the optimum radix is e, the base of the natural logarithms, with a numerical value of about 2.718. Because 3 is the integer closest to e, it is almost always the most economical integer radix (see Figure 2)." That's why I would indeed stop at 3. Of course, like I said this is getting a bit over my head.

Sure for a menuing system... there's a big difference between the number of options you give for subdividing (recursively) a particular problem space and computer hardware.