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Subject: Carnivores: error in the Circle of Life rss

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Nick Bentley
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I realized to my horror yesterday that somehow, I'd ordered the species on the Circle of Life incorrectly when I was composing Carnivores!

The order isn't arbitrary - it's based on a polyhex construction diagram and the species are *supposed* to be ordered by how difficult they are to build (the harder to build, the higher up on the food chain). However, I noticed the order wasn't right. This should only make any real difference at high-level play, but I'm a perfectionist so I can't let it go. Here's the *correct* board:



I still need to fix this in the rules, but won't have time for that until next week.

tl;dr I'm an idiot.
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Luis Bolaños Mures
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It's funny because just yesterday I took a second (or third) look at


while musing on which types of tied groups would be more common in Carteso and noticed that you included no two-space jumps in that diagram. Is that the mistake that you fixed now?

I was also puzzled by this:

Quote:
lines segments coming from the left-hand side should be weighted more than than those coming from the right, because objects on the left of the diagram will appear more commonly on the board (and so their construction paths will be available more frequently) than objects appearing on the right

I'm sure this is right, but I'm curious as to why. Could you elaborate a little?
 
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Stephen Tavener
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Let me kno what changes you need to Ai Ai. This might be a good time to add some heuristics as well.
 
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Nick Bentley
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luigi87 wrote:
It's funny because just yesterday I took a second (or third) look at


while musing on which types of tied groups would be more common in Carteso and noticed that you included no two-space jumps in that diagram. Is that the mistake that you fixed now?


Nope. The diagram is right. Any "two-space" jumps would require making a one-space jump first, so that's included in the diagram.

The problem was, for some reason I didn't weight the lines to a couple of the 4-hexes correctly (the red lines going to them should count for about 2 and I seem not to have weighted those lines at all, so they were effectively 1), *and* there was another one out of place for reasons beyond me.

I was also puzzled by this:

Quote:
lines segments coming from the left-hand side should be weighted more than than those coming from the right, because objects on the left of the diagram will appear more commonly on the board (and so their construction paths will be available more frequently) than objects appearing on the right

I'm sure this is right, but I'm curious as to why. Could you elaborate a little?


The simple version is that the more commonly some pattern shows up on the board, the easier it is to build polyhexes that can be constructed from that pattern, simply because there's more opportunity to do so. This frequency effect needs to be taken into account.

It's even a little more complicated than that though: the "bent 3" will show up even more than expected based on this diagram, due to its utility: it can capture the 2-hex, which is the most important shape in the game, and you can construct 6 the 7 4-hexes from it (as opposed to only 2 of 7 for the other two 3-hexes). So players end up building the bent-3 A LOT, and the frequency means the red lines in the 2nd layer actually end up with the highest weights, despite being only second from farthest left.

The net effect is to devalue that 4-hexes that can be built most easily from the "bent-3"
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Nick Bentley
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mrraow wrote:
Let me kno what changes you need to Ai Ai. This might be a good time to add some heuristics as well.


Will do. I'll get to this next week

[edit] I'm also going to run an online tournament in the new year. That should help with heuristics
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Nick Bentley
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luigi87 wrote:
It's funny because just yesterday I took a second (or third) look at


while musing on which types of tied groups would be more common in Carteso and noticed that you included no two-space jumps in that diagram. Is that the mistake that you fixed now?


It occurs to me you might be talking about the 2-way stretch? That *is* in the diagram, in the upper left. I need to remake this diagram so it's more clear. At some I'm going to write about how this game was designed (probably my most involved design), and I've been sort of waiting until then to make a nice version.
 
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milomilo122 wrote:
Any "two-space" jumps would require making a one-space jump first, so that's included in the diagram.

But, by the same token, couldn't you also say that any one-space jumps require making a zero-space jump first and that therefore one-space jumps needn't be included either?

milomilo122 wrote:
the more commonly some pattern shows up on the board, the easier it is to build polyhexes that can be constructed from that pattern, simply because there's more opportunity to do so. This frequency effect needs to be taken into account.

I was actually wondering how is it that some n-stone patterns occur more frequently than others in ways other than those reflected by the line segments, but I guess you just meant that you placed the polyhexes in descending order of number of lines segments pointing at them in the diagram itself, right?
 
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milomilo122 wrote:
It occurs to me you might be talking about the 2-way stretch?

No, I was just talking about any two stones that are three single-cell steps away from eath other.
 
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Nick Bentley
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luigi87 wrote:
milomilo122 wrote:
Any "two-space" jumps would require making a one-space jump first, so that's included in the diagram.

But, by the same token, couldn't you also say that any one-space jumps require making a zero-space jump first and that therefore one-space jumps needn't be included either?


I don't follow your logic, but here's how I think of it: each layer shows all the patterns that can be made from the previous layer by the addition of a single stone. That's all I need to know to make my calculations. (noting I've left out the single space hops in the layer of 4-hexes - no need to have them because they would only be useful for calculation if 5-hexes were allowed, which they aren't)

Quote:
milomilo122 wrote:
the more commonly some pattern shows up on the board, the easier it is to build polyhexes that can be constructed from that pattern, simply because there's more opportunity to do so. This frequency effect needs to be taken into account.

I was actually wondering how is it that some n-stone patterns occur more frequently than others, but I guess you just meant that you placed the polyhexes in descending order of number of lines segments pointing at them in the diagram itself, right?


Yes, a better description than frequency is "degrees of freedom in construction" or "how many enemy stones are required to block the construction of a particular polyhex", but those are mouthfuls.
 
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milomilo122 wrote:
I don't follow your logic, but here's how I think of it: each layer shows all the patterns that can be made from the previous layer by the addition of a single stone. That's all I need to know to make my calculations.

I was specifically referring to your leaving out the patterns

. . . . .
. x . . x .
. . . . .


and

. . . .
. x . . .
. . . x .
. . . .


in the second row. I still think those need to be included to get the frequencies right, but maybe I'm missing something.
 
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Nick Bentley
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luigi87 wrote:
milomilo122 wrote:
I don't follow your logic, but here's how I think of it: each layer shows all the patterns that can be made from the previous layer by the addition of a single stone. That's all I need to know to make my calculations.

I was specifically referring to your leaving out the patterns

. . . . .
. x . . x .
. . . . .


and

. . . .
. x . . .
. . . x .
. . . .


in the second row. I still think those need to be included to get the frequencies right, but maybe I'm missing something.


hmmmm, let me think about it. So your notion here is: when you add a stone to one of these, you are jumping from the 2-stone layer to the 3-stone layer, and that this is important for calculating degrees of freedom?
 
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Nick Bentley
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by jove I think you may be right! I'm going to dwell on this for a bit more, but def leaning your way at the moment.

[edit] ok you're definitely right. equal parts fascinated and embarrassed over here

[another edit] looks like I need to do the circle again. I'll have luis check my work this time
 
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milomilo122 wrote:
by jove I think you may be right! I'm going to dwell on this for a bit more, but def leaning your way at the moment.

[edit] ok you're definitely right. equal parts fascinated and embarrassed over here

[another edit] looks like I need to do the circle again. I'll have luis check my work this time

Well, I think I'm not considering all angles of the problem. I never thought of including

milomilo122 wrote:
"how many enemy stones are required to block the construction of a particular polyhex"

in the equation, for instance, and I wouldn't know how to.

The only thing I'm reasonably sure of is that if 1-space jumps are included, leaving out 2-space jumps should be wrong.

And this is where I start brooding over how much better games I could make if I had a better understanding of mathematics.

Not to mention how much better games we could have if mathematicians were more interested in designing games.
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Nick Bentley
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luigi87 wrote:
Well, I think I'm not considering all angles of the problem. I never thought of including

milomilo122 wrote:
"how many enemy stones are required to block the construction of a particular polyhex"


This is equivalent to the number of lines going between one pattern and a pattern in the next layer down.

For example, I can place a stone in 6 different spots next to a singleton to create a 2-hex. Therefore, I'd need 6 enemy stones to make it impossible for that singleton to become a 2-hex.
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Nick Bentley
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luigi87 wrote:
Not to mention how much better games we could have if mathematicians were more interested in designing games.


Plenty of mathematicians design games. They have their own problems though: they often fail to be sensitive to human psychology, e.g. Multiple modes of thought are needed.
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milomilo122 wrote:
luigi87 wrote:
Well, I think I'm not considering all angles of the problem. I never thought of including

milomilo122 wrote:
"how many enemy stones are required to block the construction of a particular polyhex"


This is equivalent to the number of lines going between one pattern and a pattern in the next layer down.

For example, I can place a stone in 6 different spots next to a singleton to create a 2-hex. Therefore, I'd need 6 enemy stones to make it impossible for that singleton to become a 2-hex.

Oh, right. Trying to accommodate enemy stones in the system threw me off the track there.
 
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Nick Bentley
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luigi87 wrote:
milomilo122 wrote:
luigi87 wrote:
Well, I think I'm not considering all angles of the problem. I never thought of including

milomilo122 wrote:
"how many enemy stones are required to block the construction of a particular polyhex"


This is equivalent to the number of lines going between one pattern and a pattern in the next layer down.

For example, I can place a stone in 6 different spots next to a singleton to create a 2-hex. Therefore, I'd need 6 enemy stones to make it impossible for that singleton to become a 2-hex.

Oh, right. Trying to accommodate enemy stones in the system threw me off the track there.


Cool. Of course, this analysis doesn't include edge effects (degrees of freedom in construction differ at the edges), which are a bit beyond me at the moment.
 
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I've pondered on this a bit more and feel that I understand it better now. In fact, I think

milomilo122 wrote:
edge effects (degrees of freedom in construction differ at the edges)

should be easy to incorporate.

There are six ways to make a 2-stone group starting from a single stone on any cell of an infinite board. Therefore, in your diagram, you currently assign this a 6.

Instead of that, you could assign it a

37*6 + 18*4 + 6*3 = 312

based on the fact that, on the Carnivores board, there are 37 cells where a singleton can be turned into a 2-stone group in 6 ways, 18 cells where a singleton can be turned into a 2-stone group in 4 ways and 6 cells where a singleton can be turned into a 2-stone group in 3 ways.

And so on and so forth. ninja
 
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Nick Bentley
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luigi87 wrote:
I've pondered on this a bit more and feel that I understand it better now. In fact, I think

milomilo122 wrote:
edge effects (degrees of freedom in construction differ at the edges)

should be easy to incorporate.

There are six ways to make a 2-stone group starting from a single stone on any cell of an infinite board. Therefore, in your diagram, you currently assign this a 6.

Instead of that, you could assign it a

37*6 + 18*4 + 6*3 = 312

based on the fact that, on the Carnivores board, there are 37 cells where a singleton can be turned into a 2-stone group in 6 ways, 18 cells where a singleton can be turned into a 2-stone group in 4 ways and 6 cells where a singleton can be turned into a 2-stone group in 3 ways.

And so on and so forth. ninja


It gets more complicated for larger groups though, because the orientation and distance of the group with respect to the edge influences the calculation. You'd have to do the same calculation a bunch of times for multiple edge-group configurations for a single group. Not against it but it's a big job and easy to get wrong.

Also, I want one circle of life that applies to every board size. So to *really* go deep, you'd have to average over the board sizes people would realistically play on. (hexhex5 is prob smallest, but I don't know about how big it might go)
 
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milomilo122 wrote:
It gets more complicated for larger groups though, because the orientation of the group with respect to the edge influences the calculation.

Since the board is symmetrical, I think the calculation is exactly the same for each of the six rotations of a stone pattern. Rotating the stone pattern is equivalent to rotating the board, and rotating the board results in the same board.
 
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Nick Bentley
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luigi87 wrote:
milomilo122 wrote:
It gets more complicated for larger groups though, because the orientation of the group with respect to the edge influences the calculation.

Since the board is symmetrical, I think the calculation is exactly the same for each of the six rotations of a stone pattern. Rotating the stone pattern is equivalent to rotating the board, and rotating the board results in the same board.


Here's why I'm not sure that's right: consider the three-in-a-row pattern. How many ways are there to make, eg, 4-in-a-row, when the 3-in-a-row is touching an edge?

Well, if the 3-in-a-row is flush against the edge, the answer is 2. But if not, the answer is 1.

I think there are several cases like this.

Maybe another way to say this is that a multi-stone group can span multiple space-types, and how it spans them matters.
 
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milomilo122 wrote:
Also, I want one circle of life that applies to every board size. So to *really* go deep, you'd have to average over the board sizes people would realistically play on. (hexhex5 is prob smallest, but I don't know about how big it might go)

But if the order is different for different board sizes, then you wouldn't really be able to have one "correct" order for different board sizes.

And if the order is the same for different board sizes, then there's no need to worry about computing it for different board sizes and then averaging those (and making the subjective judgment about the distribution of sizes which would be played on in practice).

So it seems (at first glance to me, anyway) like it would make sense to simply ignore edge effects and (in effect) assume an arbitrarily large / infinite board (with the philosophy that larger board sizes are the "real" game anyway, analogous to 19x19 Go instead of 13x13 and 9x9 being "real" Go)

(and with the philosophy that any simple computed "objective" measure is not necessarily "correct" in any case, but is a simplifying abstraction, if it's not also taking into account a complete (hugely intractible) game tree because of higher level effects impacting difficulty of forming shapes based on whole-board stuff). (E.g. similarly I've seen games which use various poker-style hands but don't rank them merely on some simple obvious objective measure like how many ways there are to form them but involve some subjective judgment based on playtesting)

(and with the philosophy that the asymptotic limit as the board grows large is itself a hand-wavingly natural / objective and frequently used notion in math)

But I don't know.
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milomilo122 wrote:
consider the three-in-a-row pattern. How many ways are there to make, eg, 4-in-a-row, when the 3-in-a-row is touching an edge?

Well, if the 3-in-a-row is flush against the edge, the answer is 2. But if not, the answer is 1.

My point is that, in your example, you'll cover all possibilities by simply "scanning" the board with one of the six rotationally symmetrical versions of the 3-in-a-row pattern. You needn't even multiply by six afterwards because that's already accounted for in previous steps of the diagram.
 
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Nick Bentley
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luigi87 wrote:
milomilo122 wrote:
consider the three-in-a-row pattern. How many ways are there to make, eg, 4-in-a-row, when the 3-in-a-row is touching an edge?

Well, if the 3-in-a-row is flush against the edge, the answer is 2. But if not, the answer is 1.

My point is that, in your example, you'll cover all possibilities by simply "scanning" the board with one of the six rotationally symmetrical versions of the 3-in-a-row pattern. You needn't even multiply by six afterwards because that's already accounted for in previous steps of the diagram.


Ah, yes, ok, I see what you're saying. Agreed.
 
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russ wrote:
And if the order is the same for different board sizes, then there's no need to worry about computing it for different board sizes and then averaging those (and making the subjective judgment about the distribution of sizes which would be played on in practice).

So it seems (at first glance to me, anyway) like it would make sense to simply ignore edge effects and (in effect) assume an arbitrarily large / infinite board (with the philosophy that larger board sizes are the "real" game anyway, analogous to 19x19 Go instead of 13x13 and 9x9 being "real" Go)


This is why I haven't put much effort into including edge effects. In fact, if I were a betting man, I'd bet edge effects don't change the order.

Quote:
(and with the philosophy that any simple computed "objective" measure is not necessarily "correct" in any case, but is a simplifying abstraction, if it's not also taking into account a complete (hugely intractible) game tree because of higher level effects impacting difficulty of forming shapes based on whole-board stuff).


Yep. I would love to revisit the question with a group of experienced players, if any ever exist, to see if experience suggests the theory I've concocted doesn't match practice.

One great difficulty is the problem is recursive: if higher order effects change the difficulties enough to change the order on the circle, that's not the end of it, because rearranging the circle alters the higher order effects. Which means you have to start the evaluation process again.

I'd love to convince Little Golem to do the game so I could get some players.

Do you think LG might be interested in being part of a dev process, rather than a home for completely finished games?

If it turns out they would, would anybody here be willing to write a note lobbying for its inclusion?

I also wonder whether Cameron's software might be repurposed usefully to study the question. Instead of evaluating different games, it evaluates different Circle of Life arrangements. As a fairly constrained problem, it seems like something that an algorithmic approach might do well.
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