Previously pre-posted:

• She-Shell — *Perseus Rescuing Andromeda* (1611) by Wtewael

Previously pre-posted:

• She-Shell — *Perseus Rescuing Andromeda* (1611) by Wtewael

In this work,

Edition of the “Way”[Tahrir al-Wasila], Khomeini gives opinions on such innovations as insurance, banking, lotteries, artificial insemination, anatomical dissection, sex-change operations, artificial insemination, space travel and marriage with extra-terrestrials. — James Buchan,Days of God: The Revolution in Iran and Its Consequences(2012), ch. 3, “The Iranian Religion”, pg. 135 of the 2013 John Murray paperback

Pre-previously on Overlord-In-Terms-of-Issues-Around-Engagement-with-the-Über-Feral, I’ve looked at various ways of creating fractals by restricting the moves of a point jumping towards the vertices of a polygon. For example, the point can be banned from jumping towards the same vertex twice in a row. This time, I want to look at fractals created not by restriction, but by compulsion. If the point jumps towards vertex *v* and then tries to jump towards vertex *v* again, it will be forced to jump towards vertex *v*+1 instead, and so on.

You could call *v* → *v*+1 a forced increment or finc. So these are finc fractals. In some cases, restriction and compulsion create the same fractals, but I’ve found some new fractals using compulsion. Consider the fractal created by the rule *v*_{[-2]}+1, *v*_{[-1]} → +0,+1, where the subscripts refer to the history of jumps: *v*_{[-2]} is the jump-before-last, *v*_{[-1]} is the last jump. If the new vertex, *v*_{[0]}, chosen is the same as *v*_{[-2]}+1 (e.g., *v*_{[0]} = 2 = *v*_{[-2]}+1 = 1+1), then the forced increment is 0, i.e., the point is allowed to choose that jump. However, if *v*_{[0]} = *v*_{[-1]}, then the forced increment is 1 and the point must jump towards *v*_{[-1]}+1.

Here is the fractal in question:

*v*_{[-2]}+1, *v*_{[-1]} → +0,+1 (black-and-white)

*v*_{[-2]}+1, *v*_{[-1]} → +0,+1 (colour)

1,0 → +0,+1 (animated)

1,0 → +1,+0 (bw)

1,0 → +1,+0 (col)

1,0 → +1,+0 (anim)

1,0 → +1,+1 (bw)

1,0 → +1,+1 (col)

1,0 → +1,+1 (animated)

0,1 → +2,+1 (anim)

0,1 → +3,+1

1,0 → +0,+1

1,0 → +1,+0

1,1 → +0,+1

1,1 → +1,+2

1,1 → +1,+3

1,1 → +2,+1

1,2 → +0,+3

1,3 → +0,+1

2,2 → +0,+1

But suppose the history of jumps records not actual jumps, but the jumps the point wanted to make instead. In some cases, the jump made will be the same as the jump originally chosen, but in other cases it won’t. Here are some fractals using this method:

0 → +2

0 → +3

2 → +1

2 → +2

I didn’t feel the need to read this. Just knowing it’s there is enough.

• Coming Out as a Gay Orthodox Talmud Teacher

It would have been even better if it had been in *The Guardian*, but this is an imperfect world.

Papyrocentric Performativity Presents:

• Queen and Not Heard – *The African Queen*, C.S. Forester (1935)

• Ley Ho, Let’s Go! – *Britannia Obscura: Mapping Britain’s Hidden Landscapes*, Joanne Parker (Vintage 2014)

• Hymn Pickings – *The Church Hymnary*, Various (Oxford University Press 1927)

• Clock, Stock and Biswell – *A Clockwork Orange: The Restored Edition*, Anthony Burgess, edited by Andrew Biswell (Heinemann 2012)

• Vid Kid – *Violated by Video: The Eldritch Chills of an ’Eighties Childhood… from Video Nasties to Nazi Xploitation…*, Paolo Nanderson (TransVisceral Books 2018)

Or Read a Review at Random: RaRaR

Cats are of divers colours, but for the most part griseld, like to congealed ise, which cometh from the condition of her meat: her head is like unto the head of a Lion, except in her sharp ears: her flesh is soft and smooth: her eyes glister above measure, especially when a man cometh to see them on the suddain, and in the night they can hardly be endured, for their flaming aspect. Wherefore Democritus describing the Persian Smaragde saith that it is not transparent, but filleth the eye with pleasant brightness, such as is in the eyes of Panthers and Cats, for they cast forth beams in the shadow and darkness, but in sunshine they have no such clearness, and thereof Alexander Aphrodise giveth this reason, both for the sight of Cats and Bats, that they have by nature a most sharpe spirit of seeing. — Edward Topsell,

Historie of Foure-Footed Beastes(1658).

Imagine a game with six players, numbered #1 to #6, and one six-sided die. Someone rolls the die and the player who matches the number wins the game. That is, if the die rolls 1, player #1 wins; if the die rolls 2, player #2 wins; and so on. With a fair die, this is a fair game, because each player has exactly a 1/6 chance of winning. You could call it a simultaneous game, because all players are playing at once. It has one rule:

• If the die rolls *n*, then player #*n* wins.

Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:

• If player #*n* rolls *n* on the die, then he wins.

• If player #*n* doesn’t roll *n*, then player *n*+1 rolls the die.

Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.

To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:

• Player #1 wins 7776 games

• Player #2 wins 6480 games

• Player #3 wins 5400 games

• Player #4 wins 4500 games

• Player #5 wins 3750 games

• Player #6 wins 3125 games

• 15625 games end without a winner.

In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.

But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.

To make the sequential game fair, you add an extra rule:

1. If player #*n* rolls *n* on the die, he wins the game.

2. If player #*n* rolls a number greater than *n*, he loses and the die passes to player *n*+1.

3. If player #*n* rolls a number less than *n*, then he rolls again.

Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.

But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: *at that stage*. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.

However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.

A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.

However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.

And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.

It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:

1. If Player #1 rolls a 1, he wins the game. Otherwise…

2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…

3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…

4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…

5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…

6. Player #6 wins the game.

Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:

• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.

• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.

• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.

This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:

01. (1,1)

02. (1,2)*

03. (2,1)*

04. (1,3)*

05. (3,1)*

06. (1,4)*

07. (4,1)*

08. (1,5)

09. (5,1)

10. (1,6)

11. (6,1)

12. (2,2)*

13. (2,3)*

14. (3,2)*

15. (2,4)

16. (4,2)

17. (2,5)

18. (5,2)

19. (2,6)

20. (6,2)

21. (3,3)

22. (3,4)

23. (4,3)

24. (3,5)

25. (5,3)

26. (3,6)

27. (6,3)

28. (4,4)

29. (4,5)

30. (5,4)

31. (4,6)

32. (6,4)

33. (5,5)

34. (5,6)

35. (6,5)

36. (6,6)