Ben

Physicist and dabbler in writing fantasy/science fiction.

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Ben20

I agree with you that, if we need to tax something to pay for our government services, then inheritance tax is arguably not a terrible choice.

But a lot of your arguments seem a bit problematic to me. First, as a point of basic practicality, why 100%? Couldn't most of your aims be achieved with a lesser percentage? That would also smooth out weird edge cases.

There is something fundamentally compelling about the idea that every generation should start fresh, free from the accumulated advantages or disadvantages of their ancestors.

 

This quote stood out to me as interesting. I know this isn't what you meant, but as a society it would be really weird to collectively decide "don't give the next generation fire, they need to start fresh and rediscover that for themselves. We shouldn't give them the accumulated advantages of their ancestors, send them to the wilderness and let them start fresh!".

 

Ben20

I think I am not understanding the question this equation is supposed to be answer, as it seems wrong to me.

I think you are considering the case were we draw arrowheads on the lines? So each line is either an "input" or an "output", and we randomly connect inputs only to outputs, never connecting two inputs together or two outputs? With those assumptions I think the probability of only one loop on a shape with N inputs and N outputs (for a total of 2N "puts") is  1/N.

The equation I had ( (N-2)!! / (N-1)!!) is for N "points", which are not pre-assigned into inputs and outputs.

 

These diagrams explain my logic. On the top row is the "N puts" problem. First panel on the left, we pick a unmatched end (doesn't matter which, by symmetry), the one we picked is the red circle, and we look at the options of what to tie it to, the purple circles. One purple circle is filled with yellow, if we pick that one then we will end up with more than one loop. The probability of picking it randomly is 1/7 (as their are 6 other options). In the next panel we assume we didn't die. By symmetry again it doesn't matter which of the others we connected to, so I just picked the next clockwise. We will follow the loop around. We are now looking to match the newly-red point to another purple. Now their are 5 purples, the yellow is again a "dead end", ensuring more than one loop. We have a 1/5 chance of picking it at random. Continuing like this, we eventually find that the probability of having only one loop is just the probability of not picking badly at any step, (6/7)x(4/5)x(2/3) = (N-2)!! / (N-1)!!.

In the second row I do the same thing for the case where the lines have arrows, instead of 8 ports we have 4 input ports and 4 output ports, and inputs can only be linked to outputs. This changes things, because now each time we make a connection we only reduce the number of options by one at the next step. (Because our new input was never an option as an output). The one-loop chance here comes out as (3/4)x(2/3)x(1/2) = (N-1)! / N! = 1/N. Neither expression seems to match the equations you shared, so either I have gone wrong with my methods or you are answering a different question.

Ben20

This is really wonderful, thank you so much for sharing. I have been playing with your code.

The probability that their is only one loop is also very interesting. I worked out something, which feels like it is probably already well known, but not to me until now, for the simplest case.

In the simplest case is one tile. The orange lines are the "edging rule". Pick one black point and connect it to another at random. This has a 1/13 chance of immediately creating a closed loop, meaning more than one loop total. Assuming it doesn't do that, the next connection we make has 1/11 chance of failure. The one after 1/9. Etc.

So the total probability of having only one loop is the product: (12/13)  (10/11) (8/9) (6/7) (4/5) (2/3), which can be written as  12!! / 13!!  (!! double factorial). For a single tile this comes out at 35% ish. (35% chance of only one loop).

If we had a single shape with N sides we would get a probability of  (N-2)!! / (N-1)!! .

The probability for a collection of tiles is, as you say, much harder. Each edge point might not uniformly couple to all other edge points because of the multi-stepping in between. Also loops can form that never go to the edge. So the overall probability is most likely less than  (N-2)!!/(N-1)!! for N edge dots.

Ben20

That is a nice idea. The "two sides at 180 degrees" only occurred to me after I had finished. I may look into that one day, but with that many connections is needs to be automated.

In the 6 entries/exits ones above you pick one entry, you have 5 options of where to connect it. Then, you pick the next unused entry clockwise, and have 3 options for where to send it, then you have only one option for how to connect the last two. So its 5x3x1 = 15 different possible tiles.

With 14 entries/exits, its 13x11x9x7x5x3x1 = 135,135 different tiles. (13!!, for !! being double factorial).

You also have (I think) 13+12+11+10+... = 91 different connection pieces.

One day, I may try and write a code to make some of those. I strongly suspect that they won't look nice, but they might be interesting anyway.

Ben40

I still find the effect weird, but something that I think makes it more clear is this phase space diagram:

We are treating the situation as 1D, and the circles in the x, p space are energy contours. Total energy is distance from the origin. An object in orbit goes in circles with a fixed distance from the origin. (IE a fixed total energy).

The green and purple points are two points on the same orbit. At purple we have maximum momentum and minimum potential energy. At green its the other way around. The arrows show impulses, if we could suddenly add momentum of a fixed amount by firing the rocket those are the arrows.

Its visually clear that the arrow from the purple point is more efficient. It gets us more than one whole solid-black energy contour higher, in contrast the same length of arrow at the green position only gets us to the dashed orbit, which is lower.

Visually we can see that if we want to get away from the origin of that x, p coordinate system we should shoot when out boost vector aligns with out existing vector.

A weird consequence. Say our spaceship didn't have a rocket, but instead it had a machine that teleported the ship a fixed distance (say 100m). (A fixed change in position, instead of a fixed change in momentum). In this diagram that is just rotating the arrows 90 degrees. This implies the most efficient time to use the teleporting machine is when you are at the maximum distance from the planet (minimum kinetic energy, maximum potential). Mathematically this is because the potential energy has the same quadratic scaling as the kinetic. Visually, its because its where you are adding the new vector to your existing vector most efficiently.

Ben20

I am not sure that is right. A very large percentage of people really don't think the rolls are independent. Have you ever met anyone who believed in fate, Karma, horoscopes , lucky objects or prayer? They don't think its (fully) random and independent. I think the majority of the human population believe in one or more of those things.

If someone spells a word wrong in a spelling test, then its possible they mistyped, but if its a word most people can't spell correctly then the hypothesis "they don't know the spelling' should dominate. Similarly, I think it is fair to say that a very large fraction of humans (over 50%?) don't actually think dice rolls or coin tosses are independent and random.

Ben40

That is a cool idea! I started writing a reply, but it got a bit long so I decided to make it its own post in the end. ( https://www.lesswrong.com/posts/AhmZBCKXAeAitqAYz/celtic-knots-on-einstein-lattice )

Ben20

I stuck to maximal density for two reaosns, (1) to continue the Celtic knot analogy (2) because it means all tiles are always compatible (you can fit two side by side at any orientation without loosing continuity). With tiles that dont use every facet this becomes an issue.

Thinking about it now, and without having checked carefully, I think this compatibilty does something topological and forces odd macrostructure. For example, if we have a line of 4-tiles in a sea of 6-tiles (4 tiles use four facets), then we cant end the line of 4 tiles without breaking continuity. So the wall has to loop, or go off the end. The 'missing lines' the 4 tiles lacked (that would have made them 6's) would have been looped through the 4-wall. So having those tiles available is kind of like being able to delete a few closed loops from a 6s structure.

I might try messing with 4s to see if you are right that they will be asthetically useful. 

Ben20

That's a very interesting idea. I tried going through the blue one at the end.

Its not possible in that case for each string to strictly alternate between going over and under, by any of the rules I have tried. In some cases two strings pass over/under one another, then those same two strings meet again when one has travelled two tiles and the other three. So they are de-synced. They both think its their turn to go over (or under).

The rules I tried to apply were (all of which I believe don't work):

  • Over for one tile, under for the next (along each string)
  • Over for one collision, under for the next (0, 1 or 2 collisions, are possible in a tile)
  • Each string follows the sequence 'highest, middle, lowest, highest, middle lowest...' for each tile it enters.

My feeling having played with it for about 30-45 mins is that there probably is a rule nearby to those above that makes things nice, but I haven't yet found it.

Ben20

I wasn't aware of that game.  Yes it is identical in terms of the tile designs. Thank you for sharing that, it was very interesting and that Tantrix wiki page lead me to this one, https://en.wikipedia.org/wiki/Serpentiles ,  which goes into some interesting related stuff with two strings per side or differently shaped tiles.

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