## Subitizing/Visualizing large numbers using Signed Bases

Obsessive poster
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SenaryThe12th wrote:
Kodegadulo wrote:No, I think I have to insist that any unmarked, isolated digit must default to its plain vanilla face value.
how do we interepret any unmarked, isolated digit pair?  Like, say, "10"?
(Cue the banjo music: It's "Duelin' Nosists" time.)

The answer is: We _don't_. I've been arguing for years, the moment we start considering alternatives to vanilla decimal, we have to abandon any notion that unmarked _multidigit_ numerals have any specific unambiguous meaning (unless we give up and say decimal is privileged). We've _got_ to set context somehow to disambiguate them. Hence annotations. But then what do we use for the annotations? We can get away with using _single_ digit numerals for that, if we enhance those with the Computerese set. But then we _have_ to rely on a stable bootstrap default meaning for those.
I'm running a computer program right now which can predict which stocks are going to go up today.

Oh goody. I've got a bridge in a major east coast city I'd like to sell, cheap. Send me your customer list.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)

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SenaryThe12th
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I understand the point you are trying to make here.   There is no necessary connection between the glyphs and symbols we use and their indented meanings.  Which symbol gets paired up with which meaning is arbitrary.

But then infinite regress looms.  In order to indicate the meaning of Symbols A, I have to use symbol B.  But wait!  B is just as arbitrary as A was, so in order to tell what I *really* mean, I have to indicate that with symbol C..  But wait! ....

You think you've ended the infinite regress by giving symbols which are not arbitrary.  But no matter what symbol you use, that symbol too will be arbitrary.  What ends the infinite regress isn't another symbol---its the fact that we all have agreed on the meaning of that symbol.

But why do we have to go three levels deep for this?  If eventually we'll all have to agree on the meaning of C, why don't we just all agree on the meaning of A?   Why do we have to disambiguate A wiith B, and disambiguate B with C?  I'm getting a mental image of your Greek Grandmother reaching around behind her head again.

Obsessive poster
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SenaryThe12th wrote: When somebody starts pulling out the ad homs, that's when I know I've persuaded them that I might have a good point.
Well here's another hypothesis: Your debate opponent has gotten frustrated, because he's got a good point, too. But despite repeated attempts to get it across to you, you're either not acknowledging it, or not getting it.

But let's face it, we (you and I) have completely different agendas here. I'm always looking for ways to synthesize the ideas being explored on this forum into some kind of seamless, integrated whole that covers a lot of ground with just a few basic ideas, so that people can play with any and all of these notions of "bases" whenever they want, without necessarily needing to buy into just one of them, to the exclusion of all others. In fact, I want people to be able to mix and match bases and use them compatibly. And this discussion has certainly been fruitful towards that end. DS and Arbiter have been a big help with cool suggestions. There's been some interesting new possibilities opened up.

Obviously that's not your agenda. Oh, well, this is your thread, after all. So, let's see...you've decided to immerse yourself in your alternate-carry senary as your "primary base"? Really? You use it for everyday stuff, like going to the grocery? At work? Paying your taxes? You're putting a really high value on doing fast arithmetic, and negging on algebra as not being fast enough for you? You're looking for others to join you in testing out that kind of lifestyle and willing to commit to that program? Have I gotten all that right?
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)

Obsessive poster
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SenaryThe12th wrote:But why do we have to go three levels deep for this?  If eventually we'll all have to agree on the meaning of C, why don't we just all agree on the meaning of A?   Why do we have to disambiguate A wiith B, and disambiguate B with C?  I'm getting a mental image of your Greek Grandmother reaching around behind her head again.
Well, most of the time the annotation schemes I and others here have cooked up just involve two levels, and so far at most three. But let that go. To answer the substance of your question, Senary, it sounds like you want everybody here to just give up their favorite base and use senary instead, with your alternate-carry idea, and not try to work compatibly with any other base. And I'm just not for that.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)

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SenaryThe12th
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SenaryThe12th wrote: When somebody starts pulling out the ad homs, that's when I know I've persuaded them that I might have a good point.
Well here's another hypothesis: Your debate opponent has gotten frustrated, because he's got a good point, too. But despite repeated attempts to get it across to you, you're either not acknowledging it, or not getting it.
That possibility is not mutually exclusive with the possibility which I've mentioned :-)    Lets explore your possibilities:

A.  I'm not acknowledging that you have a good point.  In which case, why bother arguing with a stone wall?
B.  I'm not getting your point.  In which case, do you really think lashing out in frustration is going to help me get it?

Like I said, I'm not the smartest person on the forum.  If you want me to get your point, explain it to me like you'd explain it to a bright first grader or a dumb fifth grader.  I can understand it might be frustrating.  But I BELIEVE IN YOU KODO!!  You have an almost infinite potential to increase your ability to keep your frustrations from spilling over into ad homs.

you've decided to immerse yourself in your alternate-carry senary as your "primary base"? Really? You use it for everyday stuff, like going to the grocery? At work? Paying your taxes? You're putting a really high value on doing fast arithmetic, and negging on algebra as not being fast enough for you?
Yes.   For several months now, I've been doing just that.  I've already mentioned that I've found it useful to be able to to on-the-fly calculations in meetings.  I hardly ever do calculations in decimal anymore. Mostly just base conversions, to communicate with decimal-normals.  Most of those, even, can be done visually and instantly using the techniques of this thread.

Think about it this way:  In signed senary, *every* digit (except zero, of course) is a regular digit.  Do you know how *EASY* it is to do long division using a notational system in which every digit is a regular digit?      Radix expansions of most fractions are very short.  Divisibility tests abound (that's another gift which ArbiterOfTruth gave, he presented an easy way to get a divisibility test in senary for any prime up to 4 or 5 digits long).

As far as agendas go, my only agenda here is to give a report of what has been useful to me.  Its just one data point. Its an anecdote.  I've found it endlessly useful.   And this agenda isn't clashing with your agenda, Kodo.  The whole point of this thread was to show how ArbiterOfTruth's visualizations techniques could be used for dozinal, or indeed any human scale base, and beyond.
Last edited by SenaryThe12th on 6:51 PM - Oct 11, 2018, edited 1 time in total.

Dozens Demigod
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SenaryThe12th wrote:You are right, it is the same.  Positional notation is essentially a compressed polynomial notation.  And overbar notation is cool in the fact that it reveals this connection to us, giving us a better understanding of how it all fits together.

Nevertheless, positional notation has been compressed from polynomial notation in a way which really facilitates computations.   That's what made them such an advance over, say Roman Numerals.  Which, come to think of it, can *also* be viewed as a kind of compressed polynomial form.  But its compressed in a way which *doesn't* facilitate computation.    I suppose a snappy way of making my point was that I found arithmetic using overbar notation to be too much like trying to calculate using Roman Numerals.
I really don't see why that should be the case. Let's compare some polynomials and numbers, where the polynomial evaluates to the number where we take x = 10:

2x^2 + 4x + 1 corresponds to 241
2x^2 - 4x + 1 corresponds to 24̅1
-2x^2 + 4x + 1 corresponds to 2̅41

It seems to me that the second and third rows are exactly the same sort of facilitating compression as the first row, which is made even clearer if you squash all the minus signs as overlines

2x^2 + 4x + 1 corresponds to 241
2x^2 + 4̅x + 1 corresponds to 24̅1
2̅x^2 + 4x + 1 corresponds to 2̅41

Writing it this way also makes it clear that the coefficient of x in the second row is not 4 but -4. Now, in your scheme (assuming that these are decimal numbers), we would have:

2x^2 + 4x + 1 corresponds to 241
2x^2 - 4x + 1 corresponds to 261
-2x^2 + 4x + 1 corresponds to 841

We now have to add an additional step whenever we encounter a negative coefficient that says "add it to the base". And, not only that, the same polynomial might need to be translated differently if the base x changes; if these were dozenal numbers instead, we would have to have:

2x^2 + 4x + 1 corresponds to 241
2x^2 - 4x + 1 corresponds to 281
-2x^2 + 4x + 1 corresponds to a41

instead. How exactly does this help? Granted, it's not Roman numerals:

2x^2 + 4x + 1 corresponds to CCXLI
2x^2 - 4x + 1 corresponds to CLXI
-2x^2 + 4x + 1 corresponds to a negative number, which can't be written at all.

But it seems to me that the overlines are a direct transcription and that anything else adds an unnecessary step to translate from the polynomial to the number and vice versa.
SenaryThe12th wrote:I found treating integers as poynomials in the calculations slowed me down too much.  YMMV.   If you enjoy overbars, bless your heart!  You must have the patience of Job :)
Not really. I really do find it just as fast to deal with overlines as it is to deal with ordinary unsigned integers, and just as fast to deal with actual polynomials. In fact if anything polynomials seem easier, as you never have carries, and so division (while tedious to write down) is mostly mechanical and you don't have to keep trying or estimating multiples. Comparing notations of numbers, overlines also seem easier for me as cancellation always works the same way (2 always cancels 2̅, whereas in your notation it might cancel 4 in senary, 6 in octal, 8 in decimal, a in dozenal, and so on) and the way negatives work is really obvious (compare signed decimal 2 * 3 = 14̅, 2 * 3̅ = 1̅4, 2̅ * 3 = 1̅4, 2̅ * 3̅ = 14̅ in overline notation with 2 * 3 = 16, 2 * 7 = 94, 8 * 3 = 94, 8 * 7 = 16 in your notation). Just because this isn't the case for you doesn't mean it can't be the case for me. ^_^

In fact, given that I find your visualisation scheme suggests the overlines much more readily than the complement-digits, I'd suggest that you try using your visualisation scheme to handle polynomials. It shouldn't work any differently; you just forget about how many small cubes make up a rod, and just call that number the indeterminate x. I think it might well make addition, subtraction, multiplication, and division of polynomials as easy as it is with numbers. (Of course, I personally favour symbols. ^_^)

Dozens Demigod
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SenaryThe12th wrote:Which means, that the only real way you have to recognize that two numbers are equal, is to reduce them to a normal form, and then compare them digit-by-digit.  Illustrating the point with balanced senary, the only way you know that all these refer to the same number:

[code block containing 303, 13̅03, 313̅, 13̅13̅]

is if you pick one of them, say 303, as a normal form, and try to reduce the others to that form, and then compare them digit-by-digit.

In Frege's terminology, sure, they all have the same reference.  But they all have a different sense.  They all may be "pointing" to the same number in some platonic realm, as it were.  But we can't access that realm directly.  It is mediated through symbols.  If you happen to be Greek Orthodox, you know what I'm talking about here.  And if the symbols are different, and multiple, it causes our brains as much work as if they were ambiguous.
{6} default senary

Actually, that's not how I think about the equality of those numbers: I don't try to convert them all to some normal form. Instead I start from the most basic double representation, 3 = 13̅, and work from there. In every one of these cases, I see that 03 and 13̅ are interchanged for each other, so without bothering to convert them all I can see that 313̅ = 303, 303 = 13̅03, 13̅03 = 13̅13̅, and since equality is transitive they are all equal. If you like, I'm chunking "13̅" as a singular object that translates to "3". ^_^ Indeed, I'm pretty much chunking just the rule that says "(digit)3" = "(digit+1)3̅", which has the following possible forms:

33 = 12̅3̅ (43̅)
23 = 33̅
13 = 23̅
3 = 13̅
1̅3 = 3̅
2̅3 = 1̅3̅
3̅3 = 2̅3̅
1̅23 (4̅3) = 3̅3̅

Indeed, what I do at the ends when the pattern appears to fail is to think of more possible duplicate representations. After all, we all know that 10 - 1 = 5, 10 - 2 = 4, and 10 - 3 = 3. That immediately gives us 3 = 13̅, 4 = 12̅, and 5 = 11̅, along with the negatives 3̅ = 1̅3, 4̅ = 1̅2, and 5̅ = 1̅1, and those are just added to my mental "rewrite rules" that are also basic subtraction problems that we understand. It's quite useful to have those fleeting out-of-range digits around, even if you rewrite them all away. Having temporary out-of-range digits is an age-old technique - the Chinese suanpan long division algorithm generates them in decimal - so I don't think there's anything very problematic with this.

I do admit that this is not quite a "native" way to think about the base, as we in decimal don't usually consider temporary transdecimals. Signed bases are not quite native for me, and hence I find it profitable to think of 3 + 1 = 4 first before turning it into 12̅. (I still find it just confusing to think of it as "14" instead.) For a hypothetical native signed senarist, the natural name for the successor of three is "two from six", and so the extra step is not really needed. In a signed vigesimal civilisation (which I find more natural because of the larger span), I'd imagine that counting works something like this:

{k} default vigesimal

0 zero
1 one
2 two
3 three
4 four
5 five
6 six
7 seven
8 eight
9 nine
a ten
19̅ score minus nine
18̅ score minus eight
17̅ score minus seven
16̅ score minus six
15̅ score minus five
14̅ score minus four
13̅ score minus three
12̅ score minus two
11̅ score minus one
10 score
11 score plus one
12 score plus two
...
a8 tenscore plus eight
a9 tenscore plus nine
aa tenscore plus ten
19̅9̅ scorent minus ninescore minus nine
19̅8̅ scorent minus ninescore minus eight
...
19̅0 scorent minus ninescore
19̅1 scorent minus ninescore plus one
...
11̅a scorent minus score plus ten
109̅ scorent minus nine
...
102̅ scorent minus two
101̅ scorent minus one
100 scorent

along with

1̅ minus one
2̅ minus two
...
1̅1 minus score plus one
1̅0 minus score
1̅1̅ minus score minus one

(Please add whatever necessary adjustments and finesses you feel you need to make these sound less like they're asking for answers.) And then there would be nothing unnatural about adding one to a and getting 19̅ and writing that down, because after all, the name of the number after ten is indeed score minus nine. ^_^

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SenaryThe12th
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Double sharp wrote:

I really don't see why that should be the case. Let's compare some polynomials and numbers, where the polynomial evaluates to the number where we take x = 10:

2x^2 + 4x + 1 corresponds to 241
2x^2 - 4x + 1 corresponds to 24̅1
-2x^2 + 4x + 1 corresponds to 2̅41
Sure.  Mapping between the overbars and the polynomials is straightforward.  For me, it was, as it were, *too* straightforward.  Its so straightforward that my mind was always just going straight to the polynomials, i.e. I just couldn't make the overbar notation "chunk".

J
ust because this isn't the case for you doesn't mean it can't be the case for me. ^_^

Very, very true.  If they chunk nicely for you, thats great!    I mean come on guys,   I'm one of the newest members of the forum, I just meeting most of you, I wouldn't remotely presume to throw my weight around or try to dictate how you do your calculations.   Indeed, I don't even have any weight to throw around.   I just  hope when you read my posts you won't be bored.

In fact, given that I find your visualisation scheme suggests the overlines much more readily than the complement-digits, I'd suggest that you try using your visualisation scheme to handle polynomials.
Well, as you can probably tell by now, this whole thing was a journey for me.  I did try to make overbars + visualization work for a long time.  That was one stop on my journey.

When I wrote up the results of my experiements and posted them on this forum, I first posted about the wacky-ass carry rule, and only later posted about the visualizations.  But they were not created in that order.  The ideas all kind of evolved together--as I got stuck in one place, I tried to fix the problem using another method, etc etc.  Eventually, it all kinda fell into place, and I found something which I thought might actually work better than the way I was doing it before.  Even then I didn't post yet; I spent another 6 weeks to drill it into my brain, then another month or so of daily use.  I wanted to be *SURE* it actually does work better for me.  This is just one man's experience, take it for what its worth.  But it is coming from a man who walks the walk.

I can well believe that there's other ways which would work better for other people   There might even be a way which would work better for me, but most of the suggestions here I've already tried, and found a better way, for me at least.

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SenaryThe12th
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Double sharp wrote:

Actually, that's not how I think about the equality of those numbers: I don't try to convert them all to some normal form.
Ok, maybe it would help if you walked me through the process, step by step.  Suppose I have this equation I'm trying to solve:

Code: Select all

  _          _   _
31303 + X = 130313
solve for X.  What steps do you take?

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arbiteroftruth
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Amen.  In that spirit, if you think that it makes more sense if the offset is negative, and should be written with an overbar, I'd be ok with that too :-)
You know, the more I think about it, the offset probably should be negative. I had initially gone with a positive offset so that the digit in the annotation would be the same as the first unusual digit in the base, but I don't think that's the most important piece of information. After all, it's not quite as simple as just shifting where carries occur. If it were, then [d+6] 6+6 would equal 2, when it should be 92. For the purposes of arithmetic, it's important to remember that the digits 6 through 9 in that system really do work differently than usual. In particular, when two digits from that range are added to each other, a 9 is carried out. If the addition also reaches at least 6 in the modular ring (which happens with 8+8, 8+9, 9+8, and 9+9), then there is also a 1 carried out as usual, cancelling out the already carried 9 to create a net carry of 0.

So that's 4 digits that are in a special class, and the carry is happening 4 digits before rolling over to 0 mod 10. So calling it an offset of -4 seems most reasonable, with 6 being the first unusual digit since it's the complement of 4.

This convention also meshes more smoothly with vanilla bases having an offset of 0. With the positive offset convention, [d+0] is vanilla decimal, [d+9] is decimal with digit 9 functioning as -1 and all others being typical, while [d+1] has digits 1 through 9 all functioning as negatives. It seems strange for the smallest offset to be the biggest departure from the vanilla base, and for the largest offset to be the smallest departure. Using a negative convention of the offset fixes this.

So counting from [d] 0 to 20 in both systems, we'd have:

Code: Select all

[d]  : 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
[d-4]: 0,  1,  2,  3,  4,  5, 16, 17, 18, 19, 10, 11, 12, 13, 14, 15, 26, 27, 28, 29, 20
In principle you should still be able to apply a positive offset and continue the pattern created by negative offsets, but the arithmetic gets confusing, since you no longer have a digit representing the value 0. (just as negative offsets cause the omega digit to represent -1, positive offsets cause digit '0' to represent the value of the base)

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SenaryThe12th
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*delete post....linenoise while I was cleaning my keyboard...***

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To answer the substance of your question, Senary, it sounds like you want everybody here to just give up their favorite base and use senary instead, with your alternate-carry idea, and not try to work compatibly with any other base.  And I'm just not for that.
Um....dude....the very title I chose for the thread is "Subitizing/Visualizing large numbers using Signed Bases".    As in, "Bases" plural.  If I really was a Senary-uber-alles type fascist, why would I spend so many hours typing in ascii cubes to make what was originally a senary-only technique applicable to all human scale bases?  In what universe would that make any sense at all?

If I wanted to force everybody to use senary, wouldn't I have kept the visualization technique just for senary?  And denied all the rest of you the benefits of it?

You don't just think I'm a senary-nazi, you must think I'm the most inept, incompetent senary-nazi on the planet.  Like the nazi's on "Hogan's Heros".  (seriously showing my age).   My master-strategy for world domination is to remove one of Senary's greatest advantages by leveling the visualization playing field.

Well, perhaps you'll help me out then....what did I say which gave you that idea, and how could I have phrased it better?

EDITED:  ooops, looks like I just exemplified another pattern, called "Godwin's Law":  Any internet conversation which goes on long enough will eventually mention the nazis.

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arbiteroftruth wrote:

You know, the more I think about it, the offset probably should be negative
Good points.  And come to think of it, having an overbar over the digit representing the offset would be a handy reminder that this feature was introduced to support signed digit arithmetic.  One point of clarification tho:
In particular, when two digits from that range are added to each other, a 9 is carried out.
Yeah, but that's really the same carry rule.....when you roll over from 5 to 6 (in signed decimal) you carry a one.  When you roll *back* from 6 to 5, you borrow (i.e. subtract) a 1, which would be 9.

Think about using a standard decimal odometer to add two numbers together, say adding 5 and 5:
1.  First, you'd rotate the first wheel 5 times.  The numbers on the odometer spin by like this 00, 01, 02, 03, 04, 05.
2.  Then you'd rotate the first wheel another 5 times.  When the first wheel rolled over from 9 to 0, it would carry a digit.  The numbers spinning by are 05, 06, 07, 08, 09, 10

Now think about using a wacky-ass carry rule odometer to add the same two numbers together:
1.  First, you'd rotate the first wheel 5 times.  The odometer counts the numbers 00, 01, 02, 03, 04, 05.
2.  Then, you'd rotate the first wheel another 5 times.  This time, when the first wheel rolled over from 5 to 6, it would carry a digit. The odometer shows 16, 17, 18, 19, 10.

Subtraction works the same way, just in reverse.   Suppose you were adding -5 and -5 on a standard odometer:
1.  First, rotate the first wheel *backwards* 5 times.  Ooops, when the odometer rolls past 00, we get 99.  Looks like we are using complement arithmetic here...  99, 98, 97, 96, 95.
2.  Then, rotate the first wheel *backwards* 5 more times.  94, 93, 92, 91, 90.   (Please, nobody tell Kodo that we're using positive digits to represent a negative number here...)

Compare with the  the wacky-ass carry rule odometer:
1.  You rotate the first wheel backwards 5 times.   The odometer counts 09, 08, 07, 06, 95.   Recall, when we roll the first wheel over from 6 to 5, we roll the second wheel backwards by 1 spot, i.e. from 0 to 9.
2.  You rotate the first wheel backwards an additional 5 times.  The odometer counts 94, 93, 92, 91, 90.

I.e. it really is just 1 change to the carry rule.  The rest of the addition works exactly the same.  Just imagine a standard odometer, which I've hacked to move the next wheel when rolling from 5 to 6, rather than 9 to 0.
Last edited by SenaryThe12th on 9:49 PM - Oct 11, 2018, edited 8 times in total.

Obsessive poster
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SenaryThe12th wrote: Ok, maybe it would help if you walked me through the process, step by step.  Suppose I have this equation I'm trying to solve:

Code: Select all

  _          _   _
31303 + X = 130313
solve for X.  What steps do you take?
By inspection, the coefficient on the right is obviously equivalent to the one on the right, just a complement form so x is zero.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)

 Posts 194
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SenaryThe12th
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SenaryThe12th wrote: Ok, maybe it would help if you walked me through the process, step by step.  Suppose I have this equation I'm trying to solve:

Code: Select all

  _          _   _
31303 + X = 130313
solve for X.  What steps do you take?
By inspection, the coefficient on the right is obviously equivalent to the one on the right, just a complement form so x is zero.
So you just do it in one step?  Impressive.  The first number has 6 symbols in it, so I guess I could learn to chunk it.  But the second one has 8 symbols in it.  That's beyond my subitization limits.  I can't just look at 8 objects and subitize the number 8.  Similarly, I cant look at the second number above and just understand it at one glance.   Anything more than 5 or 6 symbols, I have to break down into smaller chunks to understand.  For me,  its kind of like being able to look at 11111111 and 111111111 at one glance and seeing whether they are equal.   More than five or six "1"s in a row, and I have to count them up to know how many are there.

Dozens Demigod
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I do it in one step by inspection too. However, I consider there to be 5 symbols in the number on the LHS and 6 in the one on the RHS, not 6 and 8 respectively. The overline modifies an existing symbol, but it doesn't add the load of an extra symbol for me.

Obsessive poster
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SenaryThe12th wrote:
SenaryThe12th wrote: Ok, maybe it would help if you walked me through the process, step by step.  Suppose I have this equation I'm trying to solve:

Code: Select all

  _          _   _
31303 + X = 130313
solve for X.  What steps do you take?
By inspection, the coefficient on the right is obviously equivalent to the one on the right, just a complement form so x is zero.
So you just do it in one step?  Impressive.  The first number has 6 symbols in it, so I guess I could learn to chunk it.  But the second one has 8 symbols in it.  That's beyond my subitization limits.  I can't just look at 8 objects and subitize the number 8.  Similarly, I cant look at the second number above and just understand it at one glance.   Anything more than 5 or 6 symbols, I have to break down into smaller chunks to understand.  For me,  its kind of like being able to look at 11111111 and 111111111 at one glance and seeing whether they are equal.   More than five or six "1"s in a row, and I have to count them up to know how many are there.
I didn't say I did it in one step, at a mere glance. I said I did it by inspection. Isn't the name of the game here doing as much as possible in your head?  I did see quite obvious chunks: I mean, ignoring the zeroes, the only absolute values in the digits were one and three. And the ones and negated threes were always together. I had practiced a little with doing the carrying/normalizing transformation in a few bases, so something like the equivalence $$3 = 1\overline{3}_6$$, in other words $$3 = 6 - 3$$, was already primed in my mind. Once I saw that it was all just $$3$$'s and $$1\overline{3}_6$$'s, then it was instantaneous to see that both numbers were $$30303_6$$.

All of that would have been incrementally harder to see, if you applied a suggestion you made a few posts back, of repurposing the digit representing the base to play the  role of negative half-base. In this case, that would be $$6_{6\pm3} = \overline{3} = -3$$.  If the equation had been $$[31603 + x = 160316]_{6\pm3}$$, it would have been opaque to me.  You know, I can maybe hold my nose and accept that $$[5 + 1 = 0]_{6\pm3}$$ and $$[4 + 2 = 0]_{6\pm3}$$, because modular math gives the same result: $$[5 + 1 \equiv 0]_{\,\mathrm{mod}\,6}$$ and $$[4 + 2 \equiv 0]_{\,\mathrm{mod}\,6}$$. But modular math also gives $$[3 + 3 \equiv 0]_{\,\mathrm{mod}\,6}$$, so the idea that anyone could stomach $$[6 + 3 = 0]_{6\pm3}$$ instead seems quite beyond the pale.

Now, a monotonous $$11111111_6$$ is certainly hard for me to distinguish from a monotonous $$111111111_6$$.  On the other hand, I wouldn't have any trouble spotting the difference between these two numbers if they were expressed like so:

$\left[\sum_{n=0}^{11} 10^n \ne \sum_{n=0}^{12} 10^n\right]_6$

But I notice you didn't use any annotations at all, or otherwise indicate what base you were using. All you wrote was this:

$31\overline{3}03 + x = 1\overline{3}031\overline{3}$

Above, I hazarded a guess. But it could have been _any_ base. I really ought to have hassled you about that. In decimal, the result would be

$\mathrm{\left[x = 40\overline{4}04 = 39604\right]_A}$.

In  dozenal, it would be

$\mathrm{\left[x = 60\overline{6}06 = 5B606\right]_C}$.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
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Systematic Dozenal Nomenclature
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Western encoding (not by choice)
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Double sharp wrote: However, I consider there to be 5 symbols in the number on the LHS and 6 in the one on the RHS, not 6 and 8 respectively. The overline modifies an existing symbol, but it doesn't add the load of an extra symbol for me.
Yeah, it's pretty much the same for me. I don't get why that minus has to "weigh" as much as a whole digit in Sen's mind.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)

Dozens Demigod
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SenaryThe12th wrote:Ooops, when the odometer rolls past 00, we get 99.  Looks like we are using complement arithmetic here...
Well, that is because standard positional notation doesn't really support negative numbers without a negative sign. That's why you look like you have a mirror symmetry around 0 rather than a translation symmetry like you have everywhere else: counting backwards in tens from 39, you get 29, 19, and 9, but then -1, -11, and -21; only at zero does the units digit change. So we are not representing negative numbers by building them up step by step; we are representing their additive inverses and then negating those as a whole. This is the same problem as that you get when your set of digits doesn't include 0 (e.g. decimal with digits from 1 to a): you now have no way to write zero without ad hoc tricks. Similarly, if all the digits are negative except zero (e.g. decimal with digits from -9 to 0), only negative numbers have a native representation; you need a negation sign to express the positives.

If you mix negative and positive digits, then both negative and positive numbers can be natively represented, and this problem goes away. Of course, that comes at a price of decreeing symbols for single-digit negatives from the start. But since negating the additive inverse is the same as building them up one digit at a time in the special case of single-digit negatives, we might as well carry on with overlines. ^_^

Oh, and I just realised that the trouble with division remainders can be explained by first using "10" as the divisor: then everything works out with the standard digits, as the remainder is always the last digit of the number. Then you can explain that you always measure your distance, positive or negative, to the nearest multiple. Explaining this by looking at division as sharing takes little more inspiration to come up with a way to look at negative remainders the same way as we do positive remainders; I'd suggest thinking of the remainder as the minimum signed distance we need to add or subtract so that the division can be fair. I think it's better to look at the number line and think of possible negatives and vectors than to stick to approaches where only positives are visible, though.

EDIT: BTW, this odometer analogy is not a bad way to look at the 10-adic numbers, in which the metric is such that larger powers of 10 have a smaller absolute value, so that 10^2 is closer to 0 than 10^1 is, and the powers of 10 approach zero. Then -1 really does equal ...999, because when you add 1 to that infinite string, an infinite chain of odometer wheels crank each other all the way down to 0. ^_^
Last edited by Double sharp on 8:25 AM - Oct 12, 2018, edited 2 times in total.

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SenaryThe12th
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Kodegadulo wrote: I didn't say I did it in one step,
No wonder I misunderstood you--I was under the impression you actually answered my question :-)  I asked for a step-by-step explanation.  Since you gave me one line, why wouldn't I assume you did it in one step?

then it was instantaneous to see that both numbers were $$30303_6$$.
That was a little eliptical, but is it fair to say that you did some easy computations which translated both numbers into a normal form (30303) and then since they had the same normal form you could instantaniously see that they were equal?
All of that would have been incrementally harder to see, if you applied a suggestion you made a few posts back,
Well, tha't wasn't the only suggestion I made.  I also recommend using unbalanced signed senary.   If you do, you can only get the digits 30303 for that value.  The problem never even arises.

Now, a monotonous $$11111111_6$$ is certainly hard for me to distinguish from a monotonous $$111111111_6$$.  On the other hand, I wouldn't have any trouble spotting the difference between these two numbers if they were expressed like so:

$\left[\sum_{n=0}^{11} 10^n \ne \sum_{n=0}^{12} 10^n\right]_6$
Every dog has its day---that notation would make it harder to see that 35124 and 35124 were the same number.
Last edited by SenaryThe12th on 1:18 AM - Oct 12, 2018, edited 1 time in total.

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Double sharp wrote: However, I consider there to be 5 symbols in the number on the LHS and 6 in the one on the RHS, not 6 and 8 respectively. The overline modifies an existing symbol, but it doesn't add the load of an extra symbol for me.
Yeah, it's pretty much the same for me. I don't get why that minus has to "weigh" as much as a whole digit in Sen's mind.
It doesn't count as a digit--it counts as minus sign, because that's what it is.

If I don't use them but use digits 30303 instead, it doesn't count as anything, because its not there.
Last edited by SenaryThe12th on 12:57 AM - Oct 12, 2018, edited 1 time in total.

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Double sharp wrote: I do it in one step by inspection too. However, I consider there to be 5 symbols in the number on the LHS and 6 in the one on the RHS, not 6 and 8 respectively. The overline modifies an existing symbol, but it doesn't add the load of an extra symbol for me.
There are two things which are impressive to me about doing it in one step:

1.  being able to chunk both numbers

2.  and in the same step, seeing that two different digit strings, of two different lengths, and realizing that they evaluate to the same value.

I mean, there are 8 different ways to write a digit string with the same value as 30303 in balaned senary.

Code: Select all

       _
30303  130303
_   _   _
30313  130313
_     _ _
31303  131303
_ _   _ _ _
31313  131313
There are only only 7 unique entries in signed senary's multiplication table!!  Its literally more memorization to remember that all 8 of those strings represent the same value.

Yet, unless you memorize that, I really don't see how you could--in one step---determine that any two of those strings were equal.   Even if you *could* chunk both strings in one step, which I, alas, cannot.

How the heck do you do that??????

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Double sharp wrote:
SenaryThe12th wrote:Ooops, when the odometer rolls past 00, we get 99.  Looks like we are using complement arithmetic here...
Well, that is because standard positional notation doesn't really support negative numbers without a negative sign.
Oh, I have no trouble with complement arithmetic--I'm a computer programmer, and computers use complement arithmetic with positive integers (and not overbars lol) because it simplifies the hardware circuit design.

This part I didn't quite understand, can you rephrase?
Oh, and I just realised that the trouble with division remainders can be explained by first using "10" as the divisor: then everything works out with the standard digits, as the remainder is always the last digit of the number. Then you can explain that you always measure your distance, positive or negative, to the nearest multiple. Explaining this by looking at division as sharing takes little more inspiration to come up with a way to look at negative remainders the same way as we do positive remainders; I'd suggest thinking of the remainder as the minimum signed distance we need to add or subtract so that the division can be fair. I think it's better to look at the number line and think of possible negatives and vectors than to stick to approaches where only positives are visible, though.

Dozens Demigod
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Well, now that I'm typing on my phone I can think of one little problem with overlines: they are hard to type. ^_^ So for the purpose of this post I'll have to cheat and write the negative digits as [-1], [-2], and so on, using the brackets to force the minus sign in with the digit symbol.
SenaryThe12th wrote:Yet, unless you memorize that, I really don't see how you could--in one step---determine that any two of those strings were equal.   Even if you *could* chunk both strings in one step, which I, alas, cannot.

How the heck do you do that??????
{6} default senary

I only memorised that 10 - 3 = 3, so that you can exchange +3 for -3 if you increment the preceding digit by one. Using that rule, it's easy to see that 03 = 1[-3]. Since in your numbers those two pairs are always changed into each other, a quick inspection confirms that they are all equal, except for the one at the top right that has a [-1]3 instead of a 1[-3] and hence looks wrong to me immediately.
SenaryThe12th wrote:It doesn't count as a digit--it counts as minus sign, because that's what it is.

If I don't use them but use digits 30303 instead, it doesn't count as anything, because its not there.
I agree that the overline is a minus sign that we put over a single digit to negate it. What I don't agree is that an overline over a digit counts as two symbols. I think of overline-1 as a single symbol (albeit compose of two symbols) just as I think that š (s-háček) is a single letter.

As for remainders, what I mean is this (I'll just consider canonical representations to make the exposition a little shorter and more focused on the main point):

20 / 10 = 2 r 0
21 / 10 = 2 r 1
22 / 10 = 2 r 2
23 / 10 = 2 r 3
3[-2] / 10 = 3 r -2
3[-1] / 10 = 3 r -1

The remainder in this case always matches the last digit. This should hopefully make negative remainders look a little less strange in the general case where it doesn't match the last digit:

{a} default decimal

4[-4] / 3 = 12 r 0
35 / 3 = 12 r -1

And the remainder, of course, need not be a single digit.

30 / 12 = 2 r 1[-4]

I'd think of it in terms of vectors. For 4 / 3, you take the vector +3 and keep adding it to itself on the number line: we get closest to 4 after one copy, where we are +1 away from it (we haven't reached it), so the answer is 1 r 1. For 5 / 3, we are closest to 5 after two copies, where we are -1 away from it (we have overshot it), so the answer is 2 r -1. You can apply this logic to cases with negative dividends and divisors as well. ^_^