# The Choice of the Circle

Here’s an elementary mathematical problem: how many ways are there to choose three numbers from a set of six numbers? If the set is (1, 2, 3, 4, 5, 6), these are the possible choices (or combinations):

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6) (c = 20)

So 6C3 = 20 (C stands for “combination”). The general formula is nCr = (n! / (n-r)!) / r!, where n is the number to choose from, r is the number of choices and n! is factorial n, or n multiplied by all numbers less than itself. For example, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. When n = 6 and c = 3, 6C3 = (6! / (6-3)!) / 3! = (720 / 6) / 6 = 20.

There isn’t much visual appeal in the choices above, but there’s a simple way to change that. Take the ways of choosing two numbers from a set of ten. They start like this:

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6)…

Suppose each choice represents the midpoint of two points chosen from a set of ten points around a pentagon, so that (1, 2) is half-way between points 1 and 2, (3, 5) is half-way between points 3 and 5, and so on:

Now take the ways of choosing three numbers from a set of ten:

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 8), (1, 2, 9), (1, 2, 10), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 8), (1, 3, 9), (1, 3, 10)…

Now the pentagon looks like this, with (1, 2, 3) representing the point midway between 1, 2 and 3, (1, 3, 9) representing the point midway between 1, 3 and 9, and so on:

Now here are 10C4, 10C5 and 10C6 for the pentagon:

You can also generate the points 5C4 = 5, then add them to the original five points and generate 10C4:

5C4

10C4

And here are 5C5, 6C5 and 12C5:

Here are 7C7 and 8C8, adding points as for 5C4:

And here is 12C6 using a dodecagon:

And various nCr for dodecagons and other polygons:

This method can also be used to represent the partitions of n, or the number of sets whose members sum to n. The partitions of 5 are these:

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)

There are seven partitions, so p(5) = 7. Partitions start small and get very large, starting with p(1), p(2), p(3) and so on:

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, 239943, 281589, 329931, 386155, 451276, 526823, 614154, 715220, 831820, 966467, 1121505, 1300156…

Suppose the partitions of n are treated as sets of points around a polygon with n vertices. Each set is then used to generate the point midway between its members. For example, (5, 4, 4, 2) is one partition of 15 and would represent the point midway between 5, 4, 4 and 2 of a pentadecagon. Here is a graphical representation of p(30):

Here are graphical representations for the partitions 5 to 15, then 15 to 60 in increments of 5 (15, 20, 25, etc):

And here are some close-ups for the partitions of 35 and 40:

# Performativizing Papyrocentricity #42

Papyrocentric Performativity Presents:

Feats for the EyesDrawn from Paradise: The discovery, art and natural history of the birds of paradise, David Attenborough and Errol Fuller (Collins 2012)

Heart of the MatherChaotic Fishponds and Mirror Universes: the maths that governs our world, Richard Elwes (Quercus 2013)

BergblumenEnchanting Alpine Flowers, Alfred Pohler, trans. Jacqueline Schweighofer

Or Read a Review at Random: RaRaR

# Yale Straight Un!

Yale tablet, YBC 7289, c. 1700 BC.

# Power Trip

Here are the first few powers of 2:

2 = 1 * 2
4 = 2 * 2
8 = 4 * 2
16 = 8 * 2
32 = 16 * 2
64 = 32 * 2
128 = 64 * 2
256 = 128 * 2
512 = 256 * 2
1024 = 512 * 2
2048 = 1024 * 2
4096 = 2048 * 2
8192 = 4096 * 2
16384 = 8192 * 2
32768 = 16384 * 2
65536 = 32768 * 2
131072 = 65536 * 2
262144 = 131072 * 2
524288 = 262144 * 2
1048576 = 524288 * 2
2097152 = 1048576 * 2
4194304 = 2097152 * 2
8388608 = 4194304 * 2
16777216 = 8388608 * 2
33554432 = 16777216 * 2
67108864 = 33554432 * 2…

As you can see, it’s a one-way power-trip: the numbers simply get larger. But what happens if you delete the digit 0 whenever it appears in a result? For example, 512 * 2 = 1024, which becomes 124. If you apply this rule, the sequence looks like this:

2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 124
124 * 2 = 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 5794
5794 * 2 = 11588
11588 * 2 = 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 9274…

Is this a power-trip? Not quite: it’s a return trip, because the numbers can never grow beyond a certain size and the sequence falls into a loop. If the result 2n contains a zero, then zerodelete(2n) < n, so the sequence has an upper limit and a number will eventually occur twice. This happens at step 526 with 366784, which matches 366784 at step 490.

The rate at which we delete zeros can obviously be varied. Call it 1:z. The sequence above sets z = 1, so 1:z = 1:1. But what if z = 2, so that 1:z = 1:2? In other words, the procedure deletes every second zero. The first zero occurs when 1024 = 2 * 512, so 1024 is left as it is. The second zero occurs when 2 * 1024 = 2048, so 2048 becomes 248. When z = 2 and every second zero is deleted, the sequence begins like this:

1 * 2 = 2
2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 1024
1024 * 2 = 2048 → 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 50794
50794 * 2 = 101588 → 101588
101588 * 2 = 203176 → 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 92704
92704 * 2 = 185408 → 18548

This sequence also has a ceiling and repeats at step 9134 with 5458864, which matches 5458864 at step 4166. And what about the sequence in which z = 3 and every third zero is deleted? Does this have a ceiling or does the act of multiplying by 2 compensate for the slower removal of zeros?

In fact, it can’t do so. The larger 2n becomes, the more zeros it will tend to contain. If 2n is large enough to contain 3 zeros on average, the deletion of zeros will overpower multiplication by 2 and the sequence will not rise any higher. Therefore the sequence that deletes every third zero will eventually repeat, although I haven’t been able to discover the relevant number.

But this reasoning applies to any rate, 1:z, of zero-deletion. If z = 100 and every hundredth zero is deleted, numbers in the sequence will rise to the point at which 2n contains sufficient zeros on average to counteract multiplication by 2. The sequence will have a ceiling and will eventually repeat. If z = 10^100 or z = 10^(10^100) and every googolth or googolplexth zero is deleted, the same is true. For any rate, 1:z, at which zeros are deleted, the sequence n = zerodelete(2n,z) has an upper limit and will eventually repeat.

Update (30×21)

Six years later, I’ve found the answer for z = 3. And uncovered a serious error in this article. See:

# He Say, He Sigh, He Sow #30

• Cognitio nostra est adeo debilis quod nullus philosophus potuit unquam perfecte investigare naturam unius muscae: unde legitur quod unus philosophus fuit triginta annis in solitudine, ut cognosceret naturam apis. — Sancti Thomae de Aquino Expositio in Symbolum Apostolorum (1273).

• Our knowledge is so weak that no philosopher has ever perfectly discovered the nature of a single fly, whence we read that one philosopher was thirty years in the wilderness that he might know the nature of a bee. — Thomas Aquinas, The Apostles’ Creed.

# Metricizing Michael…

All right-thinking folk are agreed that the Peckham-based author and visionary Michael Moorcock is a core colossus of the counter-culture. As the Guardian put it in 2007, he’s “the incendiary keystone of the visionary vortex that crystallized around New Worlds magazine in the 1960s, sparking a transgressive tornado that has sculpted paradigm-defying narratives of mutant sexuality, psychology and politics on an almost daily basis for over fifty years.”

But how often have keyly committed components of the Moorcock-fan community wished they had some objective mode of metricizing the coreness of the colossusness of his counter-culturality?

Well, the wait is over dot dot dot

site:http://www.multiverse.org/ “in terms of”

• in terms of sci-fi recommendations, I gotta go — Moorcock’s
• They’re really rebellious in terms of gender, in terms of sex, in terms of politics, the portrayal of society and race, and I really want that to be …
• In terms of games I am rediscovering Zelda: Majora’s Mask with updated graphics and sound.
• … and to describe such elements in terms of Good and Evil seems (as I hope I demonstrate) a rather useless way of looking at our problems.
• We’ve reached a point, in this new century, that can be identified as both technologically and sociologically, futuristic, even in terms of the very recent past and …
• I’m wondering about stillborn-siblings in terms of esoterica: are they the next sibling born after the stillbirth, making a short appearance (i.e. is …
• I can say I’ve had one good experience with a press release distribution service, in terms of acquiring reviews.
• In terms of chronology, however, it would have to fit in somewhere between the novels The Fortress of the Pearl and The Sailor on the Seas of [Fate]

Elsewhere other-posted:

# Terminal Logorrhoea

An SJW with a PhD writes:

It’s probably about time to collect all the issues and discussion of the 2015 Hugo Awards into one big post that is, at least in terms of what I have to say, a definitive take on it…. Three days after unveiling his slate of nominees, Torgersen wrote an essay explaining the necessity of the slate in terms of the “unreliability” of contemporary science fiction… The easiest mistake to make when trying to understand fascists is to think that they are best described in terms of a philosophy…. As a PhD in English with no small amount of training in postmodernism[,] I feel some qualification to speak here… and he does explain his beliefs in part in terms of a religious experience… Let us view it this way, since, in terms of the Hugos, we now have no other choice…. That covers the actual response in terms of the Hugos…. Your beliefs are horrible. You’re horrible. You’re a nasty, cruel little bully, and I do not like you…. in terms of brilliant, Hugo-worthy stuff that spits in the face of everything Theodore Beale loves… Norman Spinrad’s 1972 novel The Iron Dream, which imagines an alternate history where Hitler became a hack sci-fi writer in America, is probably the most notable in terms of just how much it anticipates this mess… afrofuturism, an artistic movement that uses the imaginative possibilities of science fiction to try to conceive of the African Diaspora not in terms of its tragic past but in terms of the generative potential of the future…. As a song, “Electric Lady” is an anthem in praise of Cindi Mayweather, long on braggadocio, but framed in terms of Monáe’s carefully worked out vision of black female sexuality… — Guided by the Beauty of Their Weapons: An Analysis of Theodore Beale and his Supporters, Philip Sandifer, 21/iv/2015.

Elsewhere other-posted: