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Subject: New game: Side Stitch rss

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Craig Duncan
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Over the summer of 2016 I played a fair bit of Poly-Y, which I enjoyed enough that I was inspired to create a number of boards, which I uploaded to BGG:













One "issue" with Poly-Y is that on larger boards play tends to stay close to the edges, with the center used relatively less. This led the designer Ea Ea to continue evolving the rules to the game, eventually coming to create *Star -- a quite different game, in which the center is more important.

Thinking about other directions in which Poly-Y might have evolved, I came up with the following rules for new (but related) game, which can be played on any Poly-Y board.


Side Stitch

Side stitch is played on board tessellated with hexagonal cells and possessing an odd numbered of sides. The goal is to create a group that connects more sides than your opponent.

On each turn, a player plays a single stone of his/her color to any empty cell; once placed, stones do not move. Play continues until the board is full or until both players pass in succession. The pie rule applies.

A group of same-color stones has a score equal to the number of sides it connects. (Note that a stone in the cell where two sides meet connects to both of those sides). At the end of the game, each player identifies his/her highest scoring group. The player whose highest-scoring group scores the highest is the winner. In case of tie, each player identifies his/her highest scoring group among the remaining groups; highest score wins. This process is repeated until the tie is broken.


A few notes:

** This seems a rather simple idea, so I wonder if Ea Ea (or anyone else) has already invented it. If so, please let me know!

** Since a player aims at connecting the most sides, play in the center of the board is significant. At least, it has been in my play-testing.

** Note that on a three-sided board, Side Stitch is equivalent to The Game of Y. (Poly-Y on a three-sided board is likewise equivalent to Y.)

** Most interestingly to me, I feel confident that draws are impossible in Side Stich, for reasons explained below.

On draws:


In the evolution from Poly-Y to *star, Ea Ea discovered an interesting theorem, namely, he discovered that if you call any group which touches at least 3 sides a "star", and give that star two less points than the number of sides it touches, then the sum of the points for all of the stars (the "star sum") is two less than the number of sides on the board. So a board with an odd number of sides will always produce a odd value for the "star sum."

In practice this theorem is relevant for Side Stich as follows. Call a group that touches exactly three sides (no more, no less) a "3-group," call a group that touches exactly 4 sides a "4-group", etc. This means that two players of Side Stich can't have all their groups 3 or higher tie. For in that case the star sum would be even, which according to the theorem can't happen on a board with an odd number of sides.

As a possible draw, then, that leaves only a board in which all groups are 1-groups and 2-groups. This will be exceedingly rare in practice but let's suppose such a board obtains. Now, since each stone in a corner cell produces a 2-group, and there is an odd number of corners, one player will have more corner-2-groups than the other. Thus, he will win unless the second player has the same advantage in non-corner-2-groups that the first player has in corner-2-groups (so that each has the same total number of 2-groups).

I've been unable to discover a board configuration in which this is possible. To have more non-corner-2 groups than one's opponent, one must arch one's stones over the opponent's corner-2-groups (call this an "over-arch"). But since two of one's opponent's corner-2-groups must lie on adjacent corners (given the odd number), one must oneself have at least two adjacent "over-arches" -- but if these adjacent over-arches eventually join, then one has a 3-group, contrary to the supposition that all groups are 1-groups or 2-groups. The only way to prevent the adjacent over-arches from joining together, as far as I can see, is for an opponent wall to form between them, and then for this wall to stretch to connect to another opponent group. But then that makes a 3-group for one's opponent -- again, contrary to the supposition of "only 1- and 2-groups".

Alright, that reasoning was perhaps a bit opaque. Suffice it to say I've spent a long time looking for a drawn board, to no avail. If you find one, let me know!
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Luis Bolaños Mures
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Re: New game: Side Stich
Let a group's value be the number of sides it touches.

Let a player's value be the sum of the values of their groups on a full board.

Since

a) a tie would require that the values of both players be the same,

b) the sum of two equal values is always even, and,

c) as per the *Star theorem, the sum of the values of both players will always be odd,

draws are impossible.

Am I missing something?
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Craig Duncan
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Re: New game: Side Stich
luigi87 wrote:
Let a group's value be the number of sides it touches.

Let a player's value be the sum of the values of their groups on a full board.

Since

a) a tie would require that the values of both players be the same,

b) the sum of two equal values is always even, and,

c) as per the *Star theorem, the sum of the values of both players will always be odd,

draws are impossible.

Am I missing something?


The *Star Theorem -- at least as it appeared on Ea Ea's (now defunct) website -- was phrased exclusively in terms of groups with value 3 or higher. In Side Stitch, groups can have values 1 or 2, not just 3 or more.

Perhaps the theorem is generalizable to all values of groups, in which case your elegant proof works just fine!
 
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Luis Bolaños Mures
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Re: New game: Side Stich
Hmm, yes, they seem somewhat different problems.
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Craig Duncan
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I've been thinking more about this. I'm now quite convinced that every possible arrangement of stones on any Side Stitch board (odd number of sides, corner cells that touch both intersecting sides, local hexagonal tessellation) will always contain at least one 3-group (that is, group that touches 3 sides). In other words, a full board with just 1-groups and 2-groups is impossible. If so, then Ea Ea's Star Theorem implies that draws are impossible in Side Stitch.

Here's the reasoning. Consider a pentagonal board. There will be at least five 2-groups since there are 5 corner cells.



Note though that because the number of corners is odd, at least two adjacent corners will be occupied by the same color (in this case, blue). If they connect, then a 3-group will form. So, a board with only 1-groups and 2-groups must prevent these corners from connecting. The only way to do that is have a red wall between these corners. Here is an example:



But now if the lower right red corner stone connects to this wall, a red 3-group is formed. So we need a blue wall to prevent this connection:



However, this in turn creates the potential for a blue 3-group, if the blue wall connects to the blue corner stone in the upper right. A new red wall is needed to block this group from forming, but any blocking red wall would connect with the red corner stone below, and make a red 3-group.

So necessarily, this board must contain a 3-group when full. No board arrangement = 1- and 2-groups only. But there is nothing special about this board. The reasoning will generalize to all Side Stitch boards.

Thus, a game of Side Stitch will be decided at scoring levels above 1-groups and 2-groups. But since the Star Theorem entails that opponents can’t have matching sets of groups value 3 or higher, no draws are possible.
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Craig Duncan
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OK, I'm not sure this will be of interest to anyone other than me, but I've decided at least to post my thoughts here in a case in the future I want to revisit them myself.

I want in this post and the next post to explore the overlaps and divergences between several games: Poly-Y, Side Stitch, and a game call Odd-Y.

First, let me describe Odd-Y, which isn't in the BGG database at the time of writing this post (I've submitted it to BGG, but it remains in the moderators' queue).

Odd-Y was designed by Bill Taylor. It is a generalization of The Game of Y. It is played on a hexagonal grid shaped into an equilateral polygon with an odd number of sides (triangle, pentagon, heptagon, etc.). A player wins by connecting a set of three sides of which the following is true: straight lines that are drawn from midpoint to midpoint of all three sides form a triangle that contains the gameboard's midpoint.

Here is an example of a winning set of three sides on a heptagon (I've erased the interior hexagonal cell grip for clarity), followed by an example of a set of sides that is NOT a win:


An Odd-Y winning set




NOT an Odd-Y winning set


Note that Odd-Y on a triangular board just IS the Game of Y.

Odd-Y on a pentagonal board is a very good game. The winning condition can be easily restated as follows: you win by connecting any three sides that are not all adjacent. Simple to grasp!

There are five winning sets-of-three-sides, which creates some nice strategic potential without being overwhelming. The image below shows the winning sets.



By contrast, Odd-Y on a heptagonal board gets a little unwieldy for my tastes. You CAN restate the winning condition, but it's not so intuitive as in the pentagonal case. You win by connecting any three sides of which the following is true: the smallest set of contiguous sides containing all three sides has a least five sides. Basically, the idea is that the three sides have at least two extra sides interspersed between them, no matter which way you travel between sides. Below is an image showing all winning sets.



Yikes! Not for the faint of heart. I've played a few games of this. It's a bit less imposing than it appears, but it is still quite the brain strain. The designer Bill Taylor is a mathematician so I think it comes naturally to him.

Bill assures me that every arrangement of stones on any Odd-Y board contains one and only one winning path, which is a neat property. (It makes draws impossible, for instance.) I haven't seen a proof of this claim, but I trust Bill on this!

OK, that is Odd-Y. In my follow-up post, I will explore the overlaps and divergences between Odd-Y, Poly-Y and Side Stitch.
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Craig Duncan
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OK, here is my promised discussion regarding the overlaps and divergences of Odd-Y, Poly-Y, and Side Stitch.

First, a brief summary of the Poly-Y rules. In Poly-Y a player "captures" a corner by connecting any third side to the two sides that join at the corner in question. The winner is the player who captures a majority of corners. No draws are possible since there is an odd number of corners.

OK, here are my claims regarding the games' relations to each other:

--------------------------

On a triangular board, all games are equivalent (and all are equivalent to The game of Y), in the sense that a winning group in any one game will also a winning group in any other game.

On a pentagonal board, Poly-Y and Side Stitch are equivalent games, in the sense that a winning group in one game will also be a winning group in the other game. However, Odd-Y is distinct. A winning group in Odd-Y is not necessarily a winning group in Poly-Y/Side Stitch.

On a heptagonal board and above, all three games are distinct.

--------------------------

Discussion:

The first claim is easy. Any filling of a triangular board must connect all three sides, and this is a win in all three games.

Regarding the second claim: Below is a board in which Red wins in the Odd-Y rule set, but Blue wins in the Poly-Y and Side Stitch rule sets. That shows that Odd-Y is a distinct game from the other two games.



(By the way, this shows that center play is more important in Odd-Y than the other games, which I think makes Odd-Y the best game of the three on a pentagonal board.)

What about my claim that Poly-Y and Side Stitch are equivalent on a pentagonal board? I don't have a tidy proof of this. However, the Star Theorem entails that all Side Stitch wins will fall into just three types: 5 to 0, 4 to 3, or two 3s to one 3. I've checked each of these possible wins out and each is also a win in Poly-Y.

Finally, there remains the third claim that all three games are distinct on a heptagonal board and above. The previous point showed Odd-Y to diverge from Poly-Y and Side Stitch at the pentagonal level; that will remain true of course at the heptagonal level. What about Poly-Y versus Side Stitch? They are distinct. Below is a heptagonal board that shows the divergence between these two games. On this board Red wins Side Stitch by a score of 5 to 4. However, Blue wins Poly-Y since Blue captures 4 corners compared to Red's 3 captured corners.



This also shows that the center is more important in Side Stitch than in Poly-Y. So I prefer Side Stitch to Poly-Y on heptagonal boards and up. And as noted above, I prefer Odd-Y to both games on a pentagonal board. So there is no board on which I prefer Poly-Y. For me, Poly-Y turned out to be a stepping stone to other games.

And in any case, I find that overlap/divergence between the games to be interesting.
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Luis Bolaños Mures
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Side Stitch seems a nice idea, sort of *Star-meets-Catchup. I'm surprised no one thought of it before!
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Craig Duncan
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luigi87 wrote:
Side Stitch seems a nice idea, sort of *Star-meets-Catchup. I'm surprised no one thought of it before!


Thanks. I'm surprised too. And maybe someone has thought of it before, and we're just not aware of that fact.

I suspect one obstacle standing in the way of "discovering" the game was the fact that opponents could end up with equal-valued highest-scoring groups, e.g. in a pentagonal game, both players could have 3-groups as their highest scoring group. In hindsight that is easily fixable via "recursive scoring" (in case of a tie for highest, compared the highest among remaining groups, etc.). We are now familiar with such scoring from Catchup and some other games (e.g. Minimize). But as natural as the idea of recursive scoring now seems, it might not have occurred to a long ago designer exploring the design space of Poly-Y, Star, *Star, etc.

One more thing. I've uploaded to the Poly-Y BGG site a cool new 9-sided board, which could also be used for Side Stitch. (I've submitted Side Stitch to the BGG database. Once it's approved I'll upload board images there too.)



Thanks for reading, Luis!
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Recursive scoring FTW! Its inclusion in Catchup greatly deepened Catchup without making it any harder for new players to start playing, which made it one of my favorite mechanics. I got the idea for using it in Catchup from Ingenious, another game that benefits greatly from it. It looks like it will have the same effect here.
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milomilo122 wrote:
Recursive scoring FTW! Its inclusion in Catchup greatly deepened Catchup without making it any harder for new players to start playing, which made it one of my favorite mechanics. I got the idea for using it in Catchup from Ingenious, another game that benefits greatly from it. It looks like it will have the same effect here.


Hi Nick! Thanks for chiming in, and I'm glad to learn a bit of the history of recursive scoring.
 
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cdunc123 wrote:
milomilo122 wrote:
Recursive scoring FTW! Its inclusion in Catchup greatly deepened Catchup without making it any harder for new players to start playing, which made it one of my favorite mechanics. I got the idea for using it in Catchup from Ingenious, another game that benefits greatly from it. It looks like it will have the same effect here.


Hi Nick! Thanks for chiming in, and I'm glad to learn a bit of the history of recursive scoring.


I should also mention: I think this is a brilliant idea. Even Star and *Star are too "edgy" for me, and this is a wonderful solution. My early leader for combinatorial game of 2017!
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milomilo122 wrote:
I should also mention: I think this is a brilliant idea. Even Star and *Star are too "edgy" for me, and this is a wonderful solution. My early leader for combinatorial game of 2017!


Wow, that makes me insanely happy to hear that from a designer of your caliber, Nick.

I confess that one concern I have about the game has to do with a possible first player advantage. It hasn't been evident in the few playtests, but perhaps that is just because we are beginners at the game.

The pie rule is meant to correct for first player advantage, of course, but given the radial symmetry of the board, it might turn out that there aren't many non-swappable options -- or worse yet, it might turn out that all first moves are swappable, and the game requires a 3-move pie rule (which I hate on grounds of kludginess). Time will tell.

The first player advantage problem -- if it is such -- would probably be less of a problem on a larger board, so maybe the following board will turn out to be preferable to the board from a couple of posts ago (169 cells vs 127 cells). We will see...

 
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Florent Becker
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It's a kind of points based system, so you might use komi instead of the pie rule, in the form of the score of a virtual extra group for the second player (maybe a "half-group").
 
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I could be wrong, but I don't think the radial symmetry matters much in regards to whether the pie rule is useful here. What does seem to matter is if there's a steep gradient of placement value. And there is, because the center is so obviously important, with value declining steeply away from it. My sense is that the difference in value between center and edges is so big that the pie rule will work on most boards.

If something does complicate that conclusion, I'll bet it'll be the recursive win condition. If high level play resolves frequently via that recursion, then the game will often be decided by small groups where a single extra stone can make a bigger relative difference. One reason the recursion works well in Catchup is that the players decide who has the stone advantage as a part of gameplay throughout the game, so you're always trading material advantage off against positional factors. Here the players don't exert a choice as to the number of stones they place, so the recursive win condition carries a bit more risk. But it might work out just fine.
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galbolle wrote:
It's a kind of points based system, so you might use komi instead of the pie rule, in the form of the score of a virtual extra group for the second player (maybe a "half-group").


Interesting idea. That would eliminate the need for recursive scoring, since there could never be a tie with a 0.5 komi. That said, since scores are fairly low, one concern I have is that such a device might shift the game from a first player advantage to a second player advantage. Lowering komi from 0.5 to, say, 0.1 wouldn't help things, moreover, since (for example) 3.1 would beat 3.0 just as surely as 3.5 would beat 3.0.

Still, it's an option to keep in mind as play testing continues. Thanks for the suggestion!



milomilo122 wrote:

I could be wrong, but I don't think the radial symmetry matters much in regards to whether the pie rule is useful here. What does seem to matter is if there's a steep gradient of placement value. And there is, because the center is so obviously important, with value declining steeply away from it. My sense is that the difference in value between center and edges is so big that the pie rule will work on most boards.

If something does complicate that conclusion, I'll bet it'll be the recursive win condition. If high level play resolves frequently via that recursion, then the game will often be decided by small groups where a single extra stone can make a bigger relative difference. One reason the recursion works well in Catchup is that the players decide who has the stone advantage as a part of gameplay throughout the game, so you're always trading material advantage off against positional factors. Here the players don't exert a choice as to the number of stones they place, so the recursive win condition carries a bit more risk. But it might work out just fine.



I think you're right that the center is more powerful than the edges in Side Stitch, and I think you're right that differential in power will make the pie rule an effective balancing device.

Still, the following image gives me some pause.



In Side Stitch, Blue wins this game (3,3 to Red's single 3) -- and he wins it without any play in the center. So maybe the center is not as powerful as I hope it is.

However, several further thoughts: (a) such a pattern is unlikely to arise organically; (b) there may be good defenses against it; and (c) in any case I think that on boards with more sides (e.g. 7 sides or 9 sides, say) the middle grows more powerful, so this worry might be an artifact of the pentagonal board, rather than a more general issue with the game. (And anyway, I've already said that Odd-Y is a better game than Side Stitch on the pentagonal board.)

Moving on: Your point about a potential hitch with the recursive scoring element is a good one. I'll be curious to see how often this is needed to resolve the game.

For instance, on a 7-sided board, the Star Theorem can be used to deduce the possible permutations of groups valued 3 or higher on such a board. (I'll pass on explaining the reasoning behind this, but if anyone is curious, I'd be happy to explain in a follow-up post. Note that 1-groups and 2-groups can be ignored since they will never decide the game, given that there will always be at least one group of value 3 or higher.)

The possible n-groups (where n>2) are:

7
6,3
5,4
5,3,3
4,4,3
4,3,3,3
3,3,3,3,3

The only configurations on this list requiring recursive resolution are 3,3,3,3,3 and 4,4,3 (assuming it's 4,3 vs 4 and not 4,4 vs 3!).

That's just two configurations out of seven total possible configurations, so maybe recursion will need invoking in a minority of games. That said, not all configurations are equally likely (7 should be very rare indeed!), and I wouldn't be surprised if these two configurations are the most common ones that arise between high-level players of similar skill. So it may still be that recursive scoring is frequently needed. Stay tuned.
 
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cdunc123 wrote:


Still, the following image gives me some pause.



(b) there may be good defenses against it;


I believe this is the case. Hugging the sides like that puts you in a very inflexible position compared to playing in the center.

Quote:
The possible n-groups (where n>2) are:

7
6,3
5,4
5,3,3
4,4,3
4,3,3,3
3,3,3,3,3

The only configurations on this list requiring recursive resolution are 3,3,3,3,3 and 4,4,3 (assuming it's 4,3 vs 4 and not 4,4 vs 3!).

That's just two configurations out of seven total possible configurations, so maybe recursion will need invoking in a minority of games. That said, not all configurations are equally likely (7 should be very rare indeed!), and I wouldn't be surprised if these two configurations are the most common ones that arise between high-level players of similar skill. So it may still be that recursive scoring is frequently needed. Stay tuned.


Very cool. I hadn't realized the distribution of grouping broke down like that! Promising!
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cdunc123 wrote:
galbolle wrote:
It's a kind of points based system, so you might use komi instead of the pie rule, in the form of the score of a virtual extra group for the second player (maybe a "half-group").


Interesting idea. That would eliminate the need for recursive scoring, since there could never be a tie with a 0.5 komi. That said, since scores are fairly low, one concern I have is that such a device might shift the game from a first player advantage to a second player advantage. Lowering komi from 0.5 to, say, 0.1 wouldn't help things, moreover, since (for example) 3.1 would beat 3.0 just as surely as 3.5 would beat 3.0.

Still, it's an option to keep in mind as play testing continues. Thanks for the suggestion!



You can keep the recursive scoring and lower the value of the komi by making it either "your third largest group is worth 0.5 more" (ie, you win the third round of recursive scoring) or "one of your score-2 groups is worth 2.5"
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cdunc123 wrote:
The possible n-groups (where n>2) are:

7
6,3
5,4
5,3,3
4,4,3
4,3,3,3
3,3,3,3,3

The only configurations on this list requiring recursive resolution are 3,3,3,3,3 and 4,4,3 (assuming it's 4,3 vs 4 and not 4,4 vs 3!).

That's just two configurations out of seven total possible configurations, so maybe recursion will need invoking in a minority of games. That said, not all configurations are equally likely (7 should be very rare indeed!), and I wouldn't be surprised if these two configurations are the most common ones that arise between high-level players of similar skill. So it may still be that recursive scoring is frequently needed. Stay tuned.


A further thought on this. Suppose that on a 7-sided board between two skilled players, it turns out that the all 3s configuration is rare, and suppose it also turns out that 6-groups and 7-groups are rare. That leaves 4 possible configurations, of which only 1 possibly requires recursion. Maybe that suggests 25% as a reasonable baseline for percentage of 7-sided games to require recursion.

For what it is worth, I computed the possible group values on a nine-sided board. There are 15 possible distributions in total. Of those 15, just three distributions could trigger recursive scoring:

3,3,3,3,3,3,3
4,4,3,3,3 (if each opponent has a 4)
5,5,3 (if each opponent has a 5)

Again, a caveat: not all distributions are equally alike to arise in play. I should think 4,4,3,3,3 and 5,5,3 would be quite common. Supposing it is rare to have a 9-sided game in which no 4-group is made, and rare to have a 9-sided game in which a group worth 6 or more is made, this leaves 7 possible non-rare distributions, of which two are possibly recursive. That ratio (28.6%) is fairly close to the 7-sided board ratio just computed (25%). If, though, 6-sided groups turn out to not-so-very-rare on a 9-sided board, then the non-rare possible distributions jump to 10 in total, with still just two distributions possibly requiring recursion, pushing the ratio down from 28.6% to 20%.

That makes sense to me; in general, I think the following is true: more sides to the board --> more variety of possible scoring configurations --> less need for resolution via recursion.

All things considered, my hunch is that in skilled play on 7- and 9-sided boards, somewhere between 1 in 5 and 1 in 3 games will get resolved via recursion.

That leaves me hopeful that recursive scoring won't mess up too much the balancing potential of the pie rule.
 
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christian freeling
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What about the 12* protocol for balance?
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christianF wrote:
What about the 12* protocol for balance?

For what it's worth, an odd sided board is made up of one central cell and an odd number of triangular areas that have a triangular number of cells.

Triangular numbers (1, 3, 6, 10, 15, 21, 28 ...) switch from odd to even in pairs. If an odd number is used, the resulting board will have an even number of cells (odd x odd + 1). Likewise, if an even number is used, the resulting board will have an odd number of cells (odd x even + 1).

In a 12* protocol this would matter.

For instance, if the board has an odd number of cells, then after the first placement there's an even number of cells left. This number should be divisible by 4 to guarantee an equal number of (double) moves for both players. This implies a first player advantage.
If the board has an even number of cells, then the second player should have the last (single) placement. That means that the number of cells of the board should not be divisible by 4. Any turn order advantage in this case should be insignificant enough to be leveled out by the protocol itself.
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Craig Duncan
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christianF wrote:
christianF wrote:
What about the 12* protocol for balance?

For what it's worth, an odd sided board is made up of one central cell and an odd number of triangular areas that have a triangular number of cells.

Triangular numbers (1, 3, 6, 10, 15, 21, 28 ...) switch from odd to even in pairs. If an odd number is used, the resulting board will have an even number of cells (odd x odd + 1). Likewise, if an even number is used, the resulting board will have an odd number of cells (odd x even + 1).

In a 12* protocol this would matter.

For instance, if the board has an odd number of cells, then after the first placement there's an even number of cells left. This number should be divisible by 4 to guarantee an equal number of (double) moves for both players. This implies a first player advantage.
If the board has an even number of cells, then the second player should have the last (single) placement. That means that the number of cells of the board should not be divisible by 4. Any turn order advantage in this case should be insignificant enough to be leveled out by the protocol itself.


Interesting thoughts -- thanks.

I've never seen a board fill up entirely in a connection game, so I wonder how operative these concerns would turn out to be in practice. But then again I've never tried 12* in a connection game. That would have a quite different feel, as diamond stretches (aka "2 bridges") would no longer be unbreakable. Maybe the board would fill up in that case!

One way to use 12* and preserve some familiar connection game tactics such as diamond stretches would be to stipulate that a player cannot place his two stones in a single turn such that they are part of a single group.

(It would need to be decided how to handle cases where the only possible placements remaining all entail placing stones in the same group. I could imagine a rule that says "Only 1 stone is permitted in such a turn; the second stone is forfeited," or a rule that says "stones can be placed in the same group only if no other placements are possible.")

It'd be fun to try 12* in Side Stitch, and I hope to get a chance to do so soon. Still, I'm hopeful that Pie can balance the game and I won't need to resort to 12*, so that 12* could exist as a variant rather than the main form.
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cdunc123 wrote:
Interesting thoughts -- thanks.

I've never seen a board fill up entirely in a connection game, so I wonder how operative these concerns would turn out to be in practice. But then again I've never tried 12* in a connection game. That would have a quite different feel, as diamond stretches (aka "2 bridges") would no longer be unbreakable. Maybe the board would fill up in that case!

One way to use 12* and preserve some familiar connection game tactics such as diamond stretches would be to stipulate that a player cannot place his two stones in a single turn such that they are part of a single group.

It may be worthwhile to look in more detail at the 'unfamiliar' consequences first, even if it may lead the game in a direction you didn't envision (I usually tried to follow a direction that the game seemed to prefer). If bridges are no longer safe, then ... well, for one both suffer the consequences to the same degree.

cdunc123 wrote:
(It would need to be decided how to handle cases where the only possible placements remaining all entail placing stones in the same group. I could imagine a rule that says "Only 1 stone is permitted in such a turn; the second stone is forfeited," or a rule that says "stones can be placed in the same group only if no other placements are possible.")

All possible indeed, if not always inviting.

cdunc123 wrote:
It'd be fun to try 12* in Side Stitch, and I hope to get a chance to do so soon. Still, I'm hopeful that Pie can balance the game and I won't need to resort to 12*, so that 12* could exist as a variant rather than the main form.

My thoughts on the use of a pie are not entirely clear yet. I hope there'll be more clarity after further experiments on your part. Good luck with your efforts, the game seems worth it!
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David Molnar
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cdunc123 wrote:
I've been thinking more about this. I'm now quite convinced that every possible arrangement of stones on any Side Stitch board (odd number of sides, corner cells that touch both intersecting sides, local hexagonal tessellation) will always contain at least one 3-group (that is, group that touches 3 sides). In other words, a full board with just 1-groups and 2-groups is impossible.


This is correct. The reasoning is the same as for the fact that every corner on a Poly-Y board must be claimed: pick any two consecutive sides, and consider the rest of the perimeter to be one “side”. So you have a triangle. Since Y cannot end in a draw, one player must connect any pair of consecutive side to some third side. I believe this was discussed in the Mudcrack Y book.

cdunc123 wrote:
Odd-Y on a pentagonal board is a very good game. The winning condition can be easily restated as follows: you win by connecting any three sides that are not all adjacent.


Odd-Y on the pentagonal board is then the same as Ea Ea’s Star Y. Odd-Y is thus another generalization of Star Y, along with Begird.


cdunc123 wrote:
Bill assures me that every arrangement of stones on any Odd-Y board contains one and only one winning path, which is a neat property. (It makes draws impossible, for instance.) I haven't seen a proof of this claim, but I trust Bill on this!


In the n=5 case, the proof would coincide with my proof that someone must win at Star Y, which is part of my paper on Atoll appearing in The Mathematics of Various Entertaining Subjects.

cdunc123 wrote:
And in any case, I find that overlap/divergence between the games to be interesting.


Well then, your homework is to include Begird in your classification. Admittedly, I don’t recall if I have described on this site how Begird extends to all odd-sided boards, not just odd multiples of three, as in Mark Steere’s original formulation.
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molnar wrote:
In the n=5 case, the proof would coincide with my proof that someone must win at Star Y, which is part of my paper on Atoll appearing in The Mathematics of Various Entertaining Subjects.


(Tangent) Wow, that title almost made me think you were kidding, but I see that it is a real book, and cool looking. With a forward by Raymond Smullyan! Neat.
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