He Say, He Sigh, He Sow #26

“What he said often had double and even triple meanings so that, while the rest of us speak and think in single notes, he thought in chords.” — Robert Trivers on W.D. Hamilton, Vignettes of Famous Evolutionary Biologists, Large and Small, Unz Review, 27/iv/2015.

Playing the Double Base

Here’s some mathematical nonsense:

10 > 12
100 > 122
1000 > 1222

How can 1000 > 1222? Well, it makes perfect sense in what you might call a double base. In this base, every number is identified by a unique string of digits, but the strings don’t behave as they do in a standard base.

To see how this double base works, first look at 9 in standard base 2. To generate the binary digits from right to left, you follow the procedure x mod 2 and x = x div 2, where (x mod 2) returns the remainder when x is divided by 2 and (x div 2) divides x by 2 and discards the remainder:

9 mod 2 = 1 → ...1
9 div 2 = 4
4 mod 2 = 0 → ..01
4 div 2 = 2
2 mod 2 = 0 → .001
2 div 2 = 1
1 mod 2 = 1 → 1001

So 9[b=10] = 1001[b=2]. To adapt the procedure to base 3, simply use x mod 3 and x = x div 3:

32 mod 3 = 2 → ...2
32 div 3 = 10
10 mod 3 = 1 → ..12
10 div 3 = 3
3 mod 3 = 0 → .012
3 div 3 = 1
1 mod 3 = 1 → 1012

So 32[b=10] = 1012[b=3].

But what happens if you mix bases and use (x mod 3) and (x div 2), like this?:

2 mod 3 = 2 → .2
2 div 2 = 1
1 mod 3 = 1 → 12

3 mod 3 = 0 → .0
3 div 2 = 1
1 mod 3 = 1 → 10

So 10 > 12, i.e. 10[b=3,2] > 12[b=3,2].

5 mod 3 = 2 → ..2
5 div 2 = 2
2 mod 3 = 2 → .22
2 div 2 = 1
1 mod 3 = 1 → 122

6 mod 3 = 0 → ..0
6 div 2 = 3
3 mod 3 = 0 → .00
3 div 2 = 1
1 mod 3 = 1 → 100

So 100 > 122.

11 mod 3 = 2 → ...2
11 div 2 = 5
5 mod 3 = 2 → ..22
5 div 2 = 2
2 mod 3 = 2 → .222
2 div 2 = 1
1 mod 3 = 1 → 1222

12 mod 3 = 0 → …0
12 div 2 = 6
6 mod 3 = 0 → ..00
6 div 2 = 3
3 mod 3 = 0 → .000
3 div 2 = 1
1 mod 3 = 1 → 1000

And 1000 > 1222. Here are numbers 1 to 32 in this double base:

1 = 1
12 = 2
10 = 3
121 = 4
122 = 5
100 = 6
101 = 7
1212 = 8
1210 = 9
1221 = 10
1222 = 11
1000 = 12
1001 = 13
1012 = 14
1010 = 15
12121 = 16
12122 = 17
12100 = 18
12101 = 19
12212 = 20
12210 = 21
12221 = 22
12222 = 23
10000 = 24
10001 = 25
10012 = 26
10010 = 27
10121 = 28
10122 = 29
10100 = 30
10101 = 31
121212 = 32

Given a number represented in this mixed base, how do you extract the underlying n? Suppose the number takes the form n = (digit[1]..digit[di]), where digit[1] is the first and leftmost digit and digit[di] the final and rightmost digit. Then this algorithm will extract n:

n = 1
for i = 2 to di
..n = n * 2
..while n mod 3 ≠ digit[i]
....n = n + 1
..endwhile
next i
print n

For example, suppose n = 12212[b=3,2]. Then di = 5 and the algorithm will work like this:

n = 1
n = n * 2 = 2.
2 mod 3 = 2 = digit[2]
2 * 2 = 4
4 mod 3 = 1 ≠ digit[3]
5 mod 3 = 2 = digit[3]
5 * 2 = 10
10 mod 3 = 1 = digit[4]
10 * 2 = 20
20 mod 3 = 2 = digit[5]

Therefore 12212[b=3,2] = 20[b=10].

Now try some more mathematical nonsense:

21 > 100
111 > 1,000
1,001 > 10,000
10,001 > 100,000

How can numbers with d digits be greater than numbers with d+1 digits? Easily. In this incremental base, the base adjusts itself as the digits are generated, like this:

5 mod 2 = 1 → .1
5 div 2 = 2
2 mod (2 + 1) = 2 mod 3 = 2 → 21

The first digit generated is 1, so the base increases to (2 + 1) = 3 for the second digit. Compare the procedure when n = 4:

4 mod 2 = 0 → ..0
4 div 2 = 2
2 mod 2 = 0 → .00
2 div 2 = 1
1 mod 2 = 1 → 100

So 21 > 100, because 4 is a power of 2 and all the digits generated by (x mod 2) are 0 except the final and leftmost. 2 + 0 = 2. Now try n = 33:

33 mod 2 = 1 → ...1
33 div 2 = 16
16 mod (2+1) = 16 mod 3 = 1 → ..11
16 div 3 = 5
5 mod (3+1) = 5 mod 4 = 1 → .111
5 div 4 = 1
1 mod (4+1) = 1 mod 5 = 1.

33[b=10] = 1111[b=2,3,4,5].

Here are numbers 1 to 60 in this incremental base (note how 21 > 100, 111 > 1000, 1001 > 10000 and 10001 > 100000):

1 = 1
10 = 2
11 = 3
100 = 4*
21 = 5*
110 = 6
101 = 7
1000 = 8*
111 = 9*
210 = 10
121 = 11
1100 = 12
201 = 13
1010 = 14
211 = 15
10000 = 16*
221 = 17
1110 = 18
1001 = 19*
2100 = 20
311 = 21
1210 = 22
321 = 23
11000 = 24
1101 = 25
2010 = 26
1011 = 27
10100 = 28
421 = 29
2110 = 30
1201 = 31
100000 = 32*
1111 = 33
2210 = 34
1021 = 35
11100 = 36
2001 = 37
10010 = 38
1211 = 39
21000 = 40
1121 = 41
3110 = 42
2101 = 43
12100 = 44
1311 = 45
3210 = 46
1221 = 47
110000 = 48
2201 = 49
11010 = 50
2011 = 51
20100 = 52
1321 = 53
10110 = 54
10001 = 55*
101000 = 56
2111 = 57
4210 = 58
1421 = 59
21100 = 60

And here are numbers 256 to 270 (Note how 8,421 > 202,100 > 100,000,000):

100000000 = 256*
11221 = 257
101110 = 258
32101 = 259
202100 = 260*
13311 = 261
41210 = 262
10321 = 263
1111000 = 264
24201 = 265
131010 = 266
23011 = 267
320100 = 268
8421 = 269*
52110 = 270

Extracting n from a number represented in this incremental base is trickier than for the double base using (x mod 3) and (x div 2). To see how to do it, examine 11221[b=incremental]. The fifth and rightmost digit is 1, so the base increases to (2 + 1) = 3 for the fourth digit, which is 2. The base increases to (3 + 2) = 5 for the third digit, which is 2 again. The base increases to (5 + 2) = 7 for the second digit, 1. But the first and rightmost digit, 1, represents (x div 7) mod (7 + 1 = 8). So n can be extracted like this:

digit[1] * 7 = 1 * 7 = 7
7 mod 7 = 0 ≠ digit[2]
8 mod 7 = 1 = digit[2]
8 * 5 = 40
40 mod 5 = 0 ≠ digit[3]
41 mod 5 = 1 ≠ digit[3]
42 mod 5 = 2 = digit[3]
42 * 3 = 126
126 mod 3 = 0 ≠ digit[4]
127 mod 3 = 1 ≠ digit[4]
128 mod 3 = 2 = digit[4]
128 * 2 = 256
256 mod 2 = 0 ≠ digit[5]
257 mod 2 = 1 = digit[5]

So 11221[b=8,7,5,3,2] = 257[b=10].

Now try 8421[b=incremental]. The fourth and rightmost digit is 1, so the base increases to (2 + 1) = 3 for the third digit, which is 2. The base increases to (3 + 2) = 5 for the second digit, 4. But the first and rightmost digit, 8, represents (x div 5) mod (5 + 4 = 9). So n can be extracted like this:

digit[1] * 5 = 8 * 5 = 40
40 mod 5 = 0 ≠ digit[2]
41 mod 5 = 1 ≠ digit[2]
42 mod 5 = 2 ≠ digit[2]
43 mod 5 = 3 ≠ digit[2]
44 mod 5 = 4 = digit[2]
44 * 3 = 132
132 mod 3 = 0 ≠ digit[3]
133 mod 3 = 1 ≠ digit[3]
134 mod 3 = 2 = digit[3]
134 * 2 = 268
268 mod 2 = 0 ≠ digit[4]
269 mod 2 = 1 = digit[4]

So 8421[b=9,5,3,2] = 269[b=10].

Performativizing Papyrocentricity #37

Papyrocentric Performativity Presents:

Maths and Marmosets – The Great Mathematical Problems: Marvels and Mysteries of Mathematics, Ian Stewart (Profile Books 2013)

Be Ear Now – Sonic Wonderland: A Scientific Odyssey of Sound, Trevor Cox (Vintage 2015)

Exquisite Bulgarity – The Future of Architecture in 100 Buildings, Mark Kushner (Simon & Schuster 2015)

Stellar StoryDiscovering the Universe: The Story of Astronomy, Paul Murdin (Andre Deutsch 2014)

Terms of EndrearmentShe Literally Exploded: The Daily Telegraph Infuriating Phrasebook, Christopher Howse and Richard Preston (Constable 2007)


Or Read a Review at Random: RaRaR

Oh My Guardian

“The Naz and Matt Foundation was announced at a special service held in London for Nazim, two weeks after his funeral. The service featured contributions from a gay Muslim, gay Hindu, a gay vicar, a trainee Rabbi and a lesbian interfaith minister.” — My boyfriend killed himself because his family couldn’t accept that he was gay, The Guardian, 21/iii/2015.

Narcischism

What have bits to do with splits? A lot. Suppose you take the digits 12345, split them in all possible ways, then sum the results, like this:

12345 → (1234 + 5) + (123 + 45) + (123 + 4 + 5) + (12 + 345) + (12 + 34 + 5) + (12 + 3 + 45) + (12 + 3 + 4 + 5) + (1 + 2345) + (1 + 234 + 5) + (1 + 23 + 45) + (1 + 23 + 4 + 5) + (1 + 2 + 345) + (1 + 2 + 34 + 5) + (1 + 2 + 3 + 45) + (1 + 2 + 3 + 4 + 5) = 5175.

That’s a sum in base 10, but base 2 is at work below the surface, because each set of numbers is the answer to a series of binary questions: split or not? There are four possible places to split the digits 12345: after the 1, after the 2, after the 3 and after the 4. In (1 + 2 + 3 + 4 + 5), the binary question “Split or not?” is answered SPLIT every time. In (1234 + 5) and (1 + 2345) it’s answered SPLIT only once.

So the splits are governed by a four-digit binary number ranging from 0001 to 1111. When the binary digit is 1, split; when the binary digit is 0, don’t split. In binary, 0001 to 1111 = 01 to 15 in base 10 = 2^4-1. That’s for a five-digit number, so the four-digit 1234 will have 2^3-1 = 7 sets of sums:

1234 → (123 + 4) + (12 + 34) + (12 + 3 + 4) + (1 + 234) + (1 + 23 + 4) + 110 (1 + 2 + 34) + (1 + 2 + 3 + 4) = 502.

And the six-digit number 123456 will have 2^5-1 = 31 sets of sums. By now, an exciting question may have occurred to some readers. Does any number in base 10 equal the sum of all possible numbers formed by splitting its digits?

The exciting answer is: 0. In other words: No. To see why not, examine a quick way of summing the split-bits of 123,456,789, with nine digits. The long way is to find all possible sets of split-bits. There are 2^8-1 = 255 of them. The quick way is to sum these equations:

1 * 128 + 10 * 64 + 100 * 32 + 1000 * 16 + 10000 * 8 + 100000 * 4 + 1000000 * 2 + 10000000 * 1
2 * 128 + 20 * 64 + 200 * 32 + 2000 * 16 + 20000 * 8 + 200000 * 4 + 2000000 * 2 + 20000000 * 1
3 * 128 + 30 * 64 + 300 * 32 + 3000 * 16 + 30000 * 8 + 300000 * 4 + 3000000 * 3
4 * 128 + 40 * 64 + 400 * 32 + 4000 * 16 + 40000 * 8 + 400000 * 7
5 * 128 + 50 * 64 + 500 * 32 + 5000 * 16 + 50000 * 15
6 * 128 + 60 * 64 + 600 * 32 + 6000 * 31
7 * 128 + 70 * 64 + 700 * 63
8 * 128 + 80 * 127
9 * 255

Sum = 52,322,283.

52,322,283 has eight digits. If you use the same formula for the nine-digit number 999,999,999, the sum is 265,621,761, which has nine digits but is far smaller than 999,999,999. If you adapt the formula for the twenty-digit 19,999,999,999,999,999,999 (starting with 1), the split-bit sum is 16,562,499,999,987,400,705. In base 10, as far as I can see, numbers increase too fast and digit-lengths too slowly for the binary governing the split-sums to keep up. That’s also true in base 9 and base 8:

Num = 18,888,888,888,888,888,888 (b=9)
Sum = 16,714,201,578,038,328,760

Num = 17,777,777,777,777,777,777 (b=8)
Sum = 17,070,707,070,625,000,001

So what about base 7? Do the numbers increase slowly enough and the digit-lengths fast enough for the binary to keep up? The answer is: 1. In base 7, this twenty-digit number is actually smaller than its split-bit sum:

Num = 16,666,666,666,666,666,666 (b=7)
Sum = 20,363,036,303,404,141,363

And if you search below that, you can find a number that is equal to its split-bit sum:

166512 → (1 + 6 + 6 + 5 + 1 + 2) + (16 + 6 + 5 + 1 + 2) + (1 + 66 + 5 + 1 + 2) + (166 + 5 + 1 + 2) + (1 + 6 + 65 + 1 + 2) + (16 + 65 + 1 + 2) + (1 + 665 + 1 + 2) + (1665 + 1 + 2) + (1 + 6 + 6 + 51 + 2) + (16 + 6 + 51 + 2) + (1 + 66 + 51 + 2) + (166 + 51 + 2) + (1 + 6 + 651 + 2) + (16 + 651 + 2) + (1 + 6651 + 2) + (16651 + 2) + (1 + 6 + 6 + 5 + 12) + (16 + 6 + 5 + 12) + (1 + 66 + 5 + 12) + (166 + 5 + 12) + (1 + 6 + 65 + 12) + (16 + 65 + 12) + (1 + 665 + 12) + (1665 + 12) + (1 + 6 + 6 + 512) + (16 + 6 + 512) + (1 + 66 + 512) + (166 + 512) + (1 + 6 + 6512) + (16 + 6512) + (1 + 66512) = 166512[b=7] = 33525[b=10].

So 33525 in base 7 is what might be called a narcischist: it can gaze into the split-bits of its own digits and see itself gazing back. In base 6, 1940 is a narcischist:

12552 → (1 + 2 + 5 + 5 + 2) + (12 + 5 + 5 + 2) + (1 + 25 + 5 + 2) + (125 + 5 + 2) + (1 + 2 + 55 + 2) + (12 + 55 + 2) + (1 + 255 + 2) + (1255 + 2) + (1 + 2 + 5+ 52) + (12 + 5 + 52) + (1 + 25 + 52) + (125 + 52) + (1 + 2 + 552) + (12 + 552) + (1 + 2552) = 12552[b=6] = 1940[b=10].

In base 5, 4074 is a narcischist:

112244 → (1 + 1 + 2 + 2 + 4 + 4) + (11 + 2 + 2 + 4 + 4) + (1 + 12 + 2 + 4 + 4) + (112 + 2 + 4 + 4) + (1 + 1 + 22 + 4 + 4) + (11 + 22 + 4 + 4) + (1 + 122 + 4 + 4) + (1122 + 4 + 4) + (1 + 1 + 2 + 24 + 4) + (11 + 2 + 24 + 4) + (1 + 12 + 24 + 4) + (112 + 24 + 4) + (1 + 1 + 224 + 4) + (11 + 224 + 4) + (1 + 1224 + 4) + (11224 + 4) + (1 + 1 + 2 + 2 + 44) + (11 + 2 + 2 + 44) + (1 + 12 + 2 + 44) + (112 + 2 + 44) + (1 + 1 + 22 + 44) + (11 + 22 + 44) + (1 + 122 + 44) + (1122 + 44) + (1 + 1 + 2 + 244) + (11 + 2 + 244) + (1 + 12 + 244) + (112 + 244) + (1 + 1 + 2244) + (11 + 2244) + (1 + 12244) = 112244[b=5] = 4074.

And in base 4, 27 is:

123 → (1 + 2 + 3) + (12 + 3) + (1 + 23) = 123[b=4] = 27.

And in base 3, 13 and 26 are:

111 → (1 + 1 + 1) + (11 + 1) + (1 + 11) = 111[b=3] = 13.

222 → (2 + 2 + 2) + (22 + 2) + (2 + 22) = 222[b=3] = 26.

There are many more narcischists in all these bases, even if you exclude numbers with zeroes in them, like these in base 4:

1022 → (1 + 0 + 2 + 2) + (10 + 2 + 2) + (1 + 02 + 2) + (102 + 2) + (1 + 0 + 22) + (10 + 22) + (1 + 022) = 1022[b=4] = 74.

1030 → (1 + 0 + 3 + 0) + (10 + 3 + 0) + (1 + 03 + 0) + (103 + 0) + (1 + 0 + 30) + (10 + 30) + (1 + 030) = 1030[b=4] = 76.

1120 → (1 + 1 + 2 + 0) + (11 + 2 + 0) + (1 + 12 + 0) + (112 + 0) + (1 + 1 + 20) + (11 + 20) + (1 + 120) = 1120[b=4] = 88.

Heads and Tells

I have a natural history book that has a photograph of a mole as it emerges from a patch of bare soil, digging itself up with its forepaws. But the photograph has been posed: the mole’s obviously dead. Its colour is wrong and you can see that there’s no tension in its body.

But the photograph might fool a hasty glance. The difference between tense life and flaccid death is small and there’s no clue in the mole’s eyes, because you can’t see them. It’s a mole, after all. In this photo, on the other hand, you can’t miss the eyes:

Ruby-tailed wasp, Chrysis ignita

Chrysis ignita, Ruby-tailed wasp

But an insect’s eyes don’t generally change when it dies, so the wasp’s eyes don’t refute or confirm a suspicion I have about the photo: that it’s also posed with a dead subject. I’m much less certain than I am about the mole, but then a dead insect is harder to read than a dead mammal. Insects have chitinous exoskeletons, not skin or fur over muscles, so their bodies don’t obviously lose tonus when they’re dead.

But something about the posture of the wasp looks wrong to me. So do its antennae: trailing on the ground, not held up. I’m far less certain than I am about the mole, but I’m suspicious. And I’m interested in my suspicion. Photographs are usually harder to read than moving pictures. There’s less information in them, because they record an object in an instant of time. You might say that you have to go by the geometry, not the trajectory.

Or lack of it. Dead things don’t have trajectories, unless an external force imposes one on them. So it’s the geometry – slumped limbs, slightly twisted heads – that betrays the true status of some of the subjects in the book Living Jewels, which collects photographs of tropical beetles. They’re museum specimens and some may have been dead for years or decades. Exoskeletons don’t corrupt and decay like skin and muscle, so the beetles retain their beauty.

And they don’t look blatantly dead. Not the way mammals would. There’s less information in an insect’s exoskeleton, so the difference between life and death is harder to read. Emotions are harder to read in an insect too. What would a photo of an angry beetle look like? Insects’ faces are immobile, like masks, which is one reason they seem so eerie and alien.

People are different: there’s lots of information in our faces and postures, let alone our voices. And sometimes imposture (and im-posture) is easy to read. We’re all familiar with false smiles and fake laughs. For me, it gets interesting when information is restricted and you can’t see someone’s face. How much can you tell from the back of a head, for example? Or a hand? Sometimes a lot. That’s why I’m interested in these photos of people adopting a stereotyped attitude of despair: slumped and heads-in-hands:

posture 1

posture 2

posture 3

posture 4

posture 5

posture 6

I don’t think any of the photos are of genuine emotion: they’re too clean and carefully lit, for one thing. And you wouldn’t expect them to be real. But the difference between posed despair and the real thing is often very slight. There are subtle differences in the outline of the body and its muscle tension. There’s a term for this in poker: tells, or slight give-aways in the posture or expression of an opposing player. The heads above tell me that the despair is being acted. So do the hands. And the outline of the bodies.

This photo I’m less certain about:

Deserter by Charles Glass

Deserter by Charles Glass

That might be real despair in a real soldier — I haven’t read the book, so I don’t know what the provenance of the photo is. The way his right foot is pointing slightly inward looks convincing to me. But I’m not certain. The photo might have been posed. If it was, the soldier was a good actor. With a posture like that, there’s little scope to express emotion: with less to do, you have to act more.

And with less information, the mind has to work harder. That’s why I find this an interesting topic. How do we read falsity or veracity in very small things like the angle of a hand or outline of a body? And will computers be able to imitate us? And then surpass us? If so, there must be mines of information buried beneath the surface of old photographs. At the moment, we intuit that information or miss it altogether. One day, computers may be able to trawl archives and come up with new facts about the psychology and relationships of historical figures, simply by reading tiny tells in expressions and postures.