Beware the ideas of Mach!

Just some totally random connections here, documented in case they develop into anything later.

In a comment on my Mini T. Rex post, Wandering Gondola posted a link to this video of a T. rex race at a racetrack called Emerald Downs.

The combination of emerald and down made me think of the Emerald Tablet of Hermes Trismegistus, the source of the maxim "As above, so below." This in turn led me to Kevin McCall's blog, since I remembered several of his early posts had been about the Emerald Tablet. While there, I read the comments on his latest post. One comment by an anonymous person included the line "The ideas of Mach (and Leibniz) appeal to me," which I at first misread as "the Ides of March."

I knew nothing about Mach beyond the use of Mach number to express airspeed, so I looked up some of his major works. I unexpectedly discovered that two different paperback editions of his book The Science of Mechanics were published -- in different years, by different publishers -- on the Ides of March.

Since this came up in connection with a T. rex race, I thought about how Mach's name suggests the Greek mache, "battle," which sometimes has the broader meaning of "contest, game." I wondered if it could be used to describe the T. rex race, with the modern use of Mach as a measure of speed adding a felicitous connotation. Then, thinking of such terms as Titanomachy, I had the thought that tyrannomachy -- tyranno- as in Tyrannosaurus -- was probably an English word, too, so I googled it. The very first result was one with a rather obvious connection to the Ides of March!

I'm pretty sure tyrannomachy is an error in that article -- surely the proper term is tyrannicide -- but then reading "ideas of Mach" as "Ides of March" was an error, too.

Since I'm talking about the Ides of March and reposting videos from the comments, enjoy this insanely good cover of "Vehicle," courtesy of regular commenter Debbie.

Update: How could I have written about T. rexes and airspeed and forgotten to include this?

Synchronicity: Diversity in forests

Last Friday, I posted "Calculating beta diversity," in which I explored different types of diversity by considering various hypothetical forests and the tree species in them.

The next day, Saturday, a student had some questions about an article on an English reading comprehension test he had taken. The article was called "What Is a Community?" and began thus:

The Black Hills forest, the prairie riparian forest, and other forests of the western United States can be separated by the distinctly different combinations of species they comprise. It is easy to distinguish between prairie riparian forest and Black Hills forest -- one is a broad-leaved forest of ash and cottonwood trees, the other is a coniferous forest of ponderosa pine and spruce trees.

Not only is that an example of diversity in forests, it is specifically the beta diversity I focused on in my post -- that is, one forest differing from another in terms of its species composition (as opposed to alpha diversity among trees within a single forest).

Incidentally, Kevin McCall (who, unlike me, is a trained mathematician) has taken up the quest for a formula interrelating the various types of diversity. Check out his "Summary and discussion of ecological formulas" if you're interested.

My sister and the Maid on adjacent lines of text

Kevin McCall recently posted "Some thoughts on psychics," which begins by citing my own old post "The influence of adjacent lines of text."

In Kevin’s post, I found this:

The reference is to Tycho Brahe, the astronomer, but it's also an abbreviation of my own surname. I have recently received several email messages about my sister's portrait of Joan of Arc -- whom my correspondent usually refers to by her title "the Maid."

Note added: Kevin's post also mentions' Swedenborg's clairvoyant vision of "a fire in Stockholm." The stock in Stockholm means "stake, pole," so this is another nod to St. Joan.

That syncing feeling

Kevin McCall jumps into the sync hole. He is generous enough to say that it is "by far the craziest" thing he has ever posted "but, inspired by William James Tychonievich" -- but, mind you, not because!

I trust it will come as no surprise that Saint Joan puts in an appearance. Maid of Heaven, pray for us!

The distribution of mixed polyhedral dice rolls

In a recent post, Kevin McCall wrote,

We might compare [a particular hypothetical distribution of IQs] to rolling all 5 Platonic solids: one 4 sided die, one 6 sided, one 8 sided, one 12 sided, and one 20 sided, which would have a completely different distribution from rolling five 6-sided dice.

And I commented,

Would rolling a mix of polyhedral dice really result in anything significantly different from a normal bell curve. I haven’t done the calculations, but my assumption is that it would not.

Then I almost immediately retracted this statement ("Never mind. I've checked it, and my assumption was totally wrong!") because I'd put the possible rolls of three dissimilar dice (a d4, a d6, and a d8) into a spreadsheet and it had given me a histogram that looked nothing like a normal distribution.

But now I have to retract that retraction and reaffirm my original assumption. The weird-looking histogram is an artifact. Rolling the three dice mentioned yields one of 16 possible values, from 3 to 18, but the spreadsheet software (Google Sheets) for some reason made a histogram with only 13 bars. Most of the bars represent a single value, but 3-4, 9-10, and 15-16 are grouped together, which is why those three bars are abnormally tall. Making a 16-bar histogram by hand, I find that rolling mixed dice does after all yield a normal bell curve.

The moral of the story: If it comes down to trusting either your own instincts or the basic competence of Google programmers, go with your own instincts every time!

But you already knew that.

Kevin McCall is now blogging

Kevin McCall, whose name some of you will recognize for his mathematical and Tarot-related contributions here, has finally started his own blog, No Longer Reading. His most recent post there is Synchronicity, prophecy, and the Magi, and he's already posted on a wide variety of other topics, from Goethe's IQ to the Emerald Tablet of Hermes Trismegistus. Go take a look.

Applying Kevin McCall's logic to squares and other non-centered figurate numbers

Note: this post uses special terminology and notation introduced in the last post. You should read that first in order to understand what follows.

⁂

Kevin McCall's proof of the RPS theorem for reduced triangular numbers is based on the following observation:

The series of triangular numbers is generated by starting with 0, then adding 1, then adding 2, then 3, and so on through the succession of natural numbers. Due to the fact that k - n ≡ -n (mod k), one you've added numbers up to a certain point, you start doing the modular equivalent of subtracting those same numbers in reverse order, creating a palindrome.

Thus, if we consider the sequence of triangular numbers reduced modulo 10:

• 0
• +1
• +2
• +3
• +4
• +5
• +6 ≡ -4 (mod 10)
• +7 ≡ -3 (mod 10)
• +8 ≡ -2 (mod 10)
• +9 ≡ -1 (mod 10)
• +10 ≡ 0 (mod 10)
• +11 ≡ +1 (mod 10)
• +12 ≡ +2 (mod 10)
• +13 ≡ +3 (mod 10)
• +14 ≡ +4 (mod 10)
• +15 ≡ -5 (mod 10)
• +16 ≡ -4 (mod 10)
• +17 ≡ -3 (mod 10)
• +18 ≡ -2 (mod 10)
• +19 ≡ -1 (mod 10)
• +20 ≡ +0 (mod 10)
• etc.

At the end of this 20-step cycle, we are back where we started, with a number that is congruent to 0 (mod 10), and the cycle starts again.

Note that we have to go through two cycles of adding and subtracting the numbers because +5 ≡ -5 (mod 10). We count it as +5 the first time around and -5 the second time, so that it cancels out. If the modulus is odd, only one cycle is necessary.

⁂

Now let's consider the sequence of square numbers. We generate this series by starting with 0, then adding 1, then adding 3, then 5, and so on through the succession of odd natural numbers. This results in an RPS for much the same reason that the triangular series does: adding successive numbers is the modular equivalent of adding up to a certain point and then subtracting the same numbers in reverse order. Here's how it works modulo 10.

• 0
• +1
• +3
• +5
• +7 ≡ -3 (mod 10)
• +9 ≡ -1 (mod 10)
• +11 ≡ +1 (mod 10)
• +13 ≡ +3 (mod 10)
• +15 ≡ -5 (mod 10)
• +17 ≡ -3 (mod 10)
• +19 ≡ -1 (mod 10)
• etc.

As with the triangular numbers, we have to go through two cycles so that the two 5s cancel each other out. Notice that, unlike the triangular numbers, this sequence never returns to adding 0 (mod 10). That is why the triangular numbers reduced mod 10 = RPS (0136051865), while the squares are RPS (0)1496(5) -- the extra parentheses indicating that there are not two 5s in a row in the middle of the cycle, nor two 0s in a row at the end of one cycle and the beginning of the next.

⁂

Moving on to the pentagonal numbers, they are generated by starting with 0, then adding 1, then 4, then 7, then 10, and so on -- every third natural number. The pattern should be obvious by now: The sequence of n-gonal numbers is generated by starting with 0 and adding, successively, every (n - 2)th natural number, beginning with 1. Here's the generation of the pentagonal sequence modulo 10.

• 0
• +1
• +4
• +7
• +10 ≡ +0 (mod 10)
• +13 ≡ -7 (mod 10)
• +16 ≡ -4 (mod 10)
• +19 ≡ -1 (mod 10)
• +22 ≡ +2 (mod 10)
• +25 ≡ +5 (mod 10)
• +28 ≡ +8 (mod 10)
• +31 ≡ +1 (mod 10)
• +34 ≡ +4 (mod 10)
• +37 ≡ +7 (mod 10)
• +40 ≡ +0 (mod 10)
• +43 ≡ -7 (mod 10)
• +46 ≡ -4 (mod 10)
• +49 ≡ -1 (mod 10)
• +52 ≡ -8 (mod 10)
• +55 ≡ -5 (mod 10)
• +58 ≡ -2 (mod 10)
• +61 ≡ +1 (mod 10)
• etc.

The cycle here is more involved because we are adding every third natural number, which means that after we reach the -1 which cancels out the original +1, we do not go on to either - or +1 and the cycle does not yet begin anew.

⁂

Skipping hexagonal numbers for the time being, let's jump straight to what we're really interested in: the heptagonal numbers -- the only figurate numbers yet examined which do not yield an RPS when reduced modulo 10. In keeping with the pattern, the heptagonal numbers are generated by adding, successively, every 5th natural number -- yielding, modulo 10:

• 0
• +1
• +6
• +11 ≡ +1 (mod 10)
• +16 ≡ +6 (mod 10)
• +21 ≡ +1 (mod 10)
• +26 ≡ +6 (mod 10)
• +31 ≡ +1 (mod 10)
• +36 ≡ +6 (mod 10)
• etc.

As can be seen, we just continue adding 1 and 6 (or subtracting 9 and 4) forever. This gives us a repeating cycle with a period of 20 -- because 10(1 + 6) ≡ 0 (mod 10) -- but no palindrome is created because we never reach -1/+9 or -6/+4.

⁂

My tentative conclusion is that the sequence of (non-centered) n-gonal numbers reduced modulo k will always be an RPS if n - 2 and k are relatively prime. When that condition holds, adding every (n - 2)th natural number in succession will (I think) mean in hitting all possible modular values, resulting in an RPS. The triangular numbers are a special case because for that sequence n - 2 = 1, which is coprime to every integer.

Where n - 2 and k are not coprime, an RPS may result, but not necessarily. I need to think a little more about what exactly determines which such sequences are RPSs and which are not.

The difference between proof and understanding

The mathematical proofs laid out in my previous post (which I am sure very few of you have bothered to read) left me both satisfied and disappointed. Having had only the patchiest of mathematical educations (basic algebra and statistics, plus such rudiments of set theory and symbolic logic as linguists require; no trigonometry or calculus), I took a certain satisfaction in having been able to do it at all -- but it was disappointing to realize that I didn't seem much closer to understanding the patterns than I had been before. Why are they always palindromic, for instance? Saying that their palindromicity can be expressed algebraically as n(n + 1) ÷ 2 ≡ (2k - (n + 1))(2k - n) ÷ 2 (mod k), and that that equation turns out to be true for all natural number values of n and k, just doesn't count as an answer to that question. I can follow each step of the algebra, but in the end I do not feel enlightened; I do not think, "Oh, now I get it!" It is possible to prove something without really understanding it.

Then I read Kevin McCall's much better proofs of the same postulates. What a difference! Where I had hammered out my proofs by algebraic brute force, McCall had understood. -- and left me thinking, in T. H. Huxley's much-quoted words, "How extremely stupid not to have thought of that!"

⁂

The heart of McCall's proof is the observation that, in modular arithmetic with modulus k, adding k - n is equivalent to subtracting n. You can easily see this in the most familiar everyday use of modular arithmetic, which is our 12-hour clock, with modulus 12. If you want to get from 11:00 to 7:00, for example, you can either subtract 4 hours or add 12 - 4 = 8 hours.

The series of triangular numbers is generated by starting with 0, then adding 1, then adding 2, then 3, and so on through the succession of natural numbers. Due to the fact that k - n ≡ -n (mod k), one you've added numbers up to a certain point, you start doing the modular equivalent of subtracting those same numbers in reverse order, creating a palindrome. For example, if the modulus is 7:

• 0
• +1
• +2
• +3
• +4 ≡ -3 (mod 7)
• +5 ≡ -2 (mod 7)
• +6 ≡ -1 (mod 7) 
• etc.

Obviously, this will create a repeating palindromic pattern with a period of 7.

Why is the period twice as long for even moduli? Consider the case when the modulus is 6.

• 0
• +1
• +2
• +3
• +4 ≡ -2 (mod 6)
• +5 ≡ -1 (mod 6)
• +6 ≡ 0 (mod 6)
• +7 ≡ +1 (mod 6)
• +8 ≡ +2 (mod 6)
• +9 ≡ -3 (mod 6)
• +10 ≡ -2 (mod 6)
• +11 ≡ -1 (mod 6)
• etc.

Because the modulus k is even, k ÷ 2 ≡ -k ÷ 2 (mod k). In this case 3 ≡ -3 (mod 6). That means that, when we go through the first cycle of 6 integers, we add 1, add 2, add/subtract 3, subtract 2, and subtract 1. We can think of this as adding or subtracting 3, but whichever it is, we only perform the operation once in a single cycle, so at the end of the cycle we still have something congruent to 3 (mod 6); we have not yet returned to the 0 with which we began. However, if we go through two cycles, we can think of ourselves as adding 3 the first time around and then subtracting it the second time, bringing us back to our starting point. Thus, the period is twice as long for even moduli, and the palindrome is centered on k ÷ 2, a number which is special because it alone is congruent to its own negation.

⁂

Now I understand -- and just about anyone else can understand, too, without any need to pore over complicated algebraic operations. Congratulations, Kevin McCall; you've really solved this, whereas I was just crunching numbers.

Proving triangular number congruence patterns

For the patterns proven in this post, see here and here.

1. The formula for triangular numbers is T_n = n(n + 1) ÷ 2

This is common knowledge, but I include it for the benefit of any readers who may be even less mathematically inclined than myself.

The n^th triangular number (written T_n) is the sum of the natural numbers from 0 to n. Such numbers are called "triangular" because that number of points can be arranged in a triangular configuration as shown below.

Deriving the formula for triangular numbers is fairly straightforward. Take a given number, n, and write out 1 + 2 + ... + n. The sum of those numbers will be T_n. Now, below that, write the same sum in the opposite direction, n + n-1 + ... + 1. The example below shows what this looks like for n = 5.

Each of the two rows adds up to T_n, so the total of the two rows is 2T_n. If we look at the vertical columns, each of them adds up to n + 1. The first column is 1 + n; the second is 2 + (n - 1); the third is 3 + (n - 2); and so on. There are n columns in all, so 2T_n = n(n + 1). The formula for T_n, therefore, is n(n + 1) ÷ 2.

(Sorry if using an obelus instead of fractional notation makes these equations a little hard to read. It allows me to type them inline instead of inserting an image file for each equation.)

⁂

2. For any modulus k, the series of triangular numbers reduced modulo k repeats, with a period no greater than 2k.

This part was discovered by Kevin McCall, though I've reformulated it somewhat.

It is postulated that, for any modulus k, T_2k ≡ 0 (mod k). This means that T_2k is evenly divisible by k; in other words, that T_2k ÷ k is an integer.

Plugging 2k into our triangular number formula, we get T_2k = 2k(2k + 1) ÷ 2. Dividing by k gives us T_2k ÷ k = 2k(2k + 1) ÷ 2k = 2k + 1. Since k is itself an integer, 2k + 1 is an integer as well. Therefore, T_2k ≡ 0 (mod k).

Now consider the next triangular number in the series, T_{2k + 1}. For any given triangular number,  T_n , T_{n + 1} = T_n  + n + 1. (For example, for n = 10, the 10th triangular number is 55, and 55 + 10 + 1 = 66, which is the 11th triangular number.) Since 2k + 1 ≡ 1 (mod k), this amounts (in the modular arithmetic we are using) to adding 1, and T_{2k + 1} will be congruent to 1 (mod k).

Now the whole series of triangular numbers begins with 0. Then we add 1, then 2, then 3, and so on to infinity to generate the whole series.

• T₀ = 0
• T₁ = T₀ + 1 = 1
• T₂ = T₁ + 2 = 3
• T₃ = T₂ + 3 = 6
• T₄ = T₃ + 4 = 10
• etc.

It should be clear that, when we reach T_2k , the whole process begins again.

• T_2k ≡ 0 (mod k)
• T_{2k + 1} = T_2k + 2k + 1 ≡ 1 (mod k)
• T_{2k + 2} = T_{2k + 1} + 2k + 2 ≡ 1 + 2 ≡ 3 (mod k)
• T_{2k + 3} = T_{2k + 2} + 2k + 3 ≡ 3 + 3 ≡ 6 (mod k)
• T_{2k + 4} = T_{2k + 3} + 2k + 4 ≡ 6 + 4 ≡ 10 (mod k)
• etc.

We can see that any time T_n ≡ 0 and T_{n + 1} ≡ 1 (mod k), the series will repeat. We know that this always happens where n = 2k. Therefore, the period of the repetition must be either 2k or else a number by which 2k is evenly divisible.

⁂

3. The period of repetition is always either k (if k is odd) or 2k (if k is even).

If k is and odd number, the series of triangular numbers reduced modulo k will begin to repeat at k rather than 2k because T_k ≡ 0 (mod k), meaning that T_k ÷ k is an integer. T_k ÷ k = k(k + 1) ÷ 2k = (k + 1) ÷ 2. Since k is odd, k + 1 is even, so (k + 1) ÷ 2 is an integer. T_{k + 1} will then be equal to T_k + k + 1 ≡ 1 (mod k), and so on.

At this point my direct reliance on Kevin McCall's proof stops. Everything below is my own. (Kevin has proved it all, too, but I have not yet read his proof and will not do so until I have first proved it all myself.)

How do we know that the period is never smaller than k? Might it sometimes be k ÷ 2, for instance? No. While it sometimes happens that T_j ≡ 0 (mod k), where j < k, it is never also true that T_{j + 1} ≡ 1 (mod k), which is what is necessary for the series to repeat. Recall that T_{n + 1} = T_n  + n + 1. Therefore, where T_j ≡ 0 (mod k) and j < k, it follows that T_{j + 1} ≡ j + 1 (mod k). The only way j + 1 can be congruent to 1 (mod k) is if j ≡ 0 (mod k), which we have specified that it is not. (With the exception of 0, no number less than k can be congruent to 0 modulo k.)

Therefore, the period of repetition is always precisely k (if k is odd) or 2k (if k is even).

⁂

4. The repeating series is always palindromic.

It is postulated that for any modulus k, the series for triangular numbers from T₀ to T_{2k - 1} reduced modulo k (which series goes on to repeat itself forever, as proven above) is palindromic. This means that

• T₀ ≡ T_{2k - 1} (mod k)
• T₁ ≡ T_{2k - 2} (mod k)
• T₂ ≡ T_{2k - 3} (mod k)
• etc.

Stating this generally, we can say that T_n ≡ T_{2k - (n +1)} (mod k). Replacing T_n with the formula for triangular numbers, we arrive at the following equation: n(n + 1) ÷ 2 ≡ (2k - (n + 1))(2k - n) ÷ 2 (mod k).

Since multiplying both sides by the same number preserves congruence, we can simplify this to: n(n + 1) ≡ (2k - (n + 1))(2k - n) (mod k).

Doing the math, we find that this is equivalent to: n(n + 1) ≡ n(n + 1) + k(4k - 2(2n + 1)) (mod k).

Subtracting n(n + 1) from both sides, we get: k(4k - 2(2n + 1)) ≡ 0 (mod k). Trivially, any multiple of k is congruent to 0 modulo k, so this is true. Therefore, the series is palindromic.

⁂

5. Where k is even, the number at the center of the palindrome is k ÷ 2

The period of the repeating palindrome is 2k where k is even, so the number at the center of the palindrome is T_k = k(k + 1) ÷ 2. This number should be congruent to k ÷ 2 (mod k).

Multiplying both sides by 2, we get k(k + 1) ≡ k (mod k). Trivially, nk ≡ k ≡ 0 (mod k), for any integer value of n, so this is true.

⁂

Now that everything has been proven that I set out to prove, I will look at Kevin McCall's full proof and, if it is different from my own (as I suspect it will be), post it as well.

Triangular number congruence patterns proven

Kevin McCall, whose name you may recognize because his thoughts on dice and the Tarot (qv) have appeared on this blog in the past, has worked out a proof of the congruence patterns in the series of triangular numbers which I postulated here. He has proven that the series of triangular numbers reduced modulo k repeats itself, that the repetition has a period of k (if k is odd) or 2k (if k is even), and that the repeating series is always palindromic.

I have only seen his proof of the first of those statements; I am not going to look at the remainder of his proof until after I have proven it myself independently -- at which point I will post both his proof and mine.

Dice and the Tarot trumps: another approach

Having read my earlier posts on the subject (here and here), Kevin McCall has the following thoughts on mapping the Tarot trumps to rolls of the dice.

As far as the association of cards of the Major Arcana with dice rolls, it seems like another way into the system rather than an ordering of die rolls could be by considering the symbolism of the numbers 1 – 6 and then trying to associate each number and each pair with a card.  After reading about Pythagorean number symbolism and tarot symbolism, here are some of my thoughts (highly speculative):

I think the Air Hexactys is the best of the four, but perhaps with some modification:

We know the Magician is (1,1), Priestess is (1,2), World is (6,6) and Judgment is (5,6).  I think that 1 must mean magic or beginning, 2 female or passivity, 3 male or activity, 4 terrestrial but in a negative sense, 5 combining 2 and 3 as representing balance, and I think Opsopaus is right that 6 represents finality and the celestial.

In that case, the Magician (1,1) would be pure beginning
The Empress (2,2) would be pure feminine, and the Emperor (3,3) pure masculine.
Then, it makes sense that the Priestess is (1,2) for magic, feminine and then the Hierophant “should be” (1,3) for magic, masculine.  Then, the Lovers “should be” (2,3) for masculine and feminine coming together.

Justice “should be” (5,5) as complete balance.  The Chariot (3,5) combining activity, victory with balance and self-control.  The Hanged Man (1,6) at the apex of the pyramid makes sense because I think this card represents the midpoint and beginning (1) and ending (6) coming together.   Temperance should be (1,5) for wisdom combined with balance, i.e., good judgement.  Fortitude should be (2,5) for passive balance, resistance rather than the activity represented by the Chariot.

The Devil, Death, and the Tower should all have 4, since these are all cards with a negative connotation, i.e., “terrestrial” cards, with terrestrial in a pejorative sense.  Death as (4,4) as given in the Air Hexactys makes sense because if 4 represents change, time, corruption, then Death is of all the cards, most representative of this archetype.  I’m not as sure about the Devil and the Tower, but these could be (2,4) and (3,4) respectively, with the Devil being passive corruption, sin (in a “Luciferian” sense, as Steiner would put it) and the Tower combining a dramatic change (with 3 as activity) and 4 as change but in a destructive, harmful sense.  The Hermit as (1,4) can also make sense, in particular if Opsopaus’s version of the Hermit as Time is correct, in either case representing either beginning in an earthly sense, caused by time or a lonely search for wisdom in the world.

I think Opsopaus is right that the World, Sun, Moon, Star, Judgement, and Hanged Man are all celestial cards, so they should all have a six somewhere.  (3,6) for Sun makes sense because this would be masculine/celestial.  (2,6) for Star gives contemplative/celestial.  (4,6) for Moon would mean the moon is bridge between earthly and celestial.  And, the idea of the sublunary realm, with the sphere of the moon being the lowest heaven, the boundary between nature and the Heavens makes sense for this correspondence. 

We can view Judgement and Wheel of Fortune as a pair.  Judgement representing is perfectly fair and final judgement, while the Wheel of Fortune is arbitrary and fickle.  Fortune metes out rewards and punishments on earth, while Judgement does in Heaven.  Then, Fortune should be (4,5) corresponding to Judgement as (5,6).

*

My comments:

McCall's approach here is similar to what I did in my post on the Tarot-astrology correspondences worked out by the Golden Dawn, which resulted in reordering the Justice and Strength cards. Before considering the accepted (Marseille) ordering of the Arcana and how to align it with the accepted (Sepher Yetzirah) ordering of the planets, signs, and elements, I looked at each card and considered what astrological correspondences, if any, seemed to make sense for it, irrespective of traditional ordering schemata. I could then evaluate the Golden Dawn mappings against these "natural" correspondences. Dice-Tarot mappings like those given by McCall here, based on qualitative considerations without taking into account the order of the trumps or the ranking of rolls, could play a similar role vis-à-vis the various systems I have been examining, helping to choose among the four Hexactyses (and, for that matter, among the various historical orderings of the trumps, of which the now-standard Marseille sequence is just one; Opsopaus's original Fire and Water Hexactyses were in fact applied to a variant of the Ferrara sequence).

Regarding the specific mappings proposed by McCall, I am in broad agreement with him as to the basic symbolic meaning of 1, 2, 3, and 6. However, the traditional meaning of 4 is rest, stability, stasis -- quite far from McCall's "change, time, corruption." (Interestingly, despite our near-opposite understandings of 4, we are agreed that it is appropriately mapped to Death -- which can be seen either as the ultimate change or as final rest.) I also have trouble accepting 5 as "complete balance"; For an odd number to represent balance is, well, odd. Five more often represents disruption, breakdown, crisis -- but also creativity, novelty, transcendence. Basically, it's something completely different being added to the stable arrangement represented by 4 and shaking things up.

By the way, while McCall thinks the symbolism of 2 and 3 as feminine and masculine means that 3-3 "should be" the Emperor, but I find the Triumphal Chariot even more appropriate. The charioteer is just as much a man as the emperor is, and I find the achieved status of the conquering hero to be far more archetypically masculine than the ascribed status of a passive throne-sitter.

I'll probably revisit this idea later and consider what numerical associations seem, irrespective of "ordering" considerations, most natural and appropriate to me.