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Subject: Two stones at once? rss

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Hunter Clark
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On IGGamecenter there is a game called Master Y. In master Y the game begins with player one placing 1 stone (only on the first move of the game) followed by player two placing two stones on their turn. The game continues with both players placing two stones per turn. Essentially this means that after every turn the player who moved last has placed more stones on the board than their opponent. This is unlike hex where only player one gets to enjoy this advantage.

Has this been tried with hex? Are there balance issues? Would this rule eliminate any first player advantage?i
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Russ Williams
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The often-called 12* move protocol (apparently popularized by Connect6) has been tried on many games. It seems a good way to minimize advantage for first (or second) player. But it often changes the character of a game drastically. E.g. much of what you know about playing Hex becomes invalid (e.g. 2 stones with 2 mutually adjacent empty spaces are a guaranteed connection in Hex, but not if the opponent can place 2 stones).

You might enjoy the article "Game Mutators for Restricting Play" by João Pedro Neto and Bill Taylor from Game and Puzzle Design. It's a peer-reviewed journal which sells issues/subscriptions.
http://gapdjournal.com/issues/
(Disclosure: I am a copy editor for the journal.)

The article gives various examples of game mutators allowing multiple moves per turn and presents ways to restrict them to keep the game strategically interesting and to better preserve the character of the original version.

E.g. in Hex, if you make a restriction that the 2 stones cannot be placed adjacent to each other, then the change to the character of the game is less drastic, because now at least the fundamental "Hex principle" that 2 stones with 2 mutually adjacent empty spaces are still a guaranteed connection.
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Hunter Clark
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One could argue that two hexes being able to be placed adjacently might create a better character for the game. I suppose if you want to preserve the hex feel you can with those restrictions. If you want a different game you can play without the restrictions.

I have never tried hex with these rules so I am not trying to say that it is better or not. I do think with the restriction that you had mentioned would make for a even and fair game and I'm suprised that it is not as common as the pie rule.
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Edward Jackman
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It's an open question WHICH player has the win in unrestricted balanced double move (aka *12) Hex (or Y as well), but one of the two players MUST have a theoretic win since draws aren't possible.

On the smallest boards:
1x1 board, 1st player wins.
2x2 board, 2nd player wins.
3x3 board, 1st player wins.
4x4 board, 1st player wins.
5x5 board, 1st player wins.
6x6 and up, ?


Of course, this applies only to the UNRESTRICTED version, that is, where a player's two stones can be placed anywhere. Restrictions I've seen -- (a) stones must NOT be adjacent, (b) stones MUST be adjacent, (c) stones must not be a part of a the same group. If you can't legally play both stones, play just one.
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Craig Duncan
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clark94 wrote:
One could argue that two hexes being able to be placed adjacently might create a better character for the game. I suppose if you want to preserve the hex feel you can with those restrictions. If you want a different game you can play without the restrictions.

I have never tried hex with these rules so I am not trying to say that it is better or not. I do think with the restriction that you had mentioned would make for a even and fair game and I'm suprised that it is not as common as the pie rule.

I'm coming late to this thread, but let me pipe up and report that I've tried Hex with 12* and I quite enjoyed it. The rule we used was that your two stones cannot be placed in the same group.

It had a feel that was similar in some ways to standard Hex, but still different.

For instance, one "diamond bridge" (i.e. 2 own-stones with 2 mutually adjacent empty spaces between them) is still unbreakable, but in a case of owning two diamond bridges, one of them is vulnerable.

E.g. (showing just a portion of the board) suppose I have something like the following:

. . x . .
. . x . .
. x . . .
. x . . . two diamond bridges
. . . . .
. . x . .
. . x . .


Next suppose my opponent plays as follows:


. . x . .
. . x . .
. x o . .
. x . . . threats to both bridges
. . o . .
. . x . .
. . x . .


I can save ONE of these connections on my next turn, but not both connections. That's because it's illegal for me to play in the empty space in each diamond bridge, because the two stones so played would end up in the same single group, which the rules forbid.

That fact changes the strategy of the game in interesting ways. I'm not saying that 12* Hex is better than regular Hex, just different. And of course, with 12* one doesn't need the Pie Rule, which makes for a nice change.
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Edward Jackman
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Another version of the 12* mutator that works for Hex and keeps *most* of the tactics of Hex intact and nearly all of the strategy is restricting the first player to only half of a cell.

First player chooses any cell and draws a line between any two opposing sides, essentially cutting the hex into two pentagons, (which are now surrounded by 4 hexagons and 2 *heptagons*) then filling in one of the two pentagons. Thereafter, each player fills in any one cell. (Any cell, including the other half of the first player's cell.)

Though one player still has a theoretical win, I am unaware if it is known WHICH player that is on various sized boards. Also, I think this can be played on somewhat smaller boards than 12* type games.

This idea is from "Mudcrack Y & Poly-Y" page 151-52.

Here's an example board after two full turns.

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Craig Duncan
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edwardpjackman wrote:
Another version of the 12* mutator that works for Hex and keeps *most* of the tactics of Hex intact and nearly all of the strategy is restricting the first player to only half of a cell.

First player chooses any cell and draws a line between any two opposing sides, essentially cutting the hex into two pentagons, (which are now surrounded by 4 hexagons and 2 *heptagons*) then filling in one of the two pentagons. Thereafter, each player fills in any one cell. (Any cell, including the other half of the first player's cell.)


Interesting! Thanks for sharing.

One question: Where are the heptagons? It seems to me each pentagon is surrounded by four hexagons and one pentagon. What am I missing?
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Andrew Schoonmaker
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cdunc123 wrote:
edwardpjackman wrote:
Another version of the 12* mutator that works for Hex and keeps *most* of the tactics of Hex intact and nearly all of the strategy is restricting the first player to only half of a cell.

First player chooses any cell and draws a line between any two opposing sides, essentially cutting the hex into two pentagons, (which are now surrounded by 4 hexagons and 2 *heptagons*) then filling in one of the two pentagons. Thereafter, each player fills in any one cell. (Any cell, including the other half of the first player's cell.)


Interesting! Thanks for sharing.

One question: Where are the heptagons? It seems to me each pentagon is surrounded by four hexagons and one pentagon. What am I missing?

The two hexes that touch the bisector of the initial hex are adjacent to seven cells each (five normal ones and the two pentagons); I assume that's what's meant by "heptagons" (it took me a minute).
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Craig Duncan
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Quote:
The two hexes that touch the bisector of the initial hex are adjacent to seven cells each (five normal ones and the two pentagons); I assume that's what's meant by "heptagons" (it took me a minute).

Ah, yes -- that must be it. Thanks!
 
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Craig Duncan
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Another thought on 12* Hex with the rule your two stones cannot end up in the same group.

Note that the stones below are in effect connected:

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


To see this, suppose my opponent plays as follows:

. . x . .
. . x . .
. . x . .
. . o . .
. . x . .
. . x . .
. . x . .


Then I can respond:

. . x . .
. . x . .
. . x . .
. x o x .
. . x . .
. . x . .
. . x . .


The opponent can't play at both cells adjacent to the O-stone in the same turn. However, my playing at one of these cells on my next turn will cement the connection.

That said, a one-cell gap as in the first board above is still a type of vulnerability for me. For it only requires one opponent stone to create the threat (as in the second board), and two of my stones stones to respond to the threat (as in the third board). So the opponent can gain a tempo on me by creating the threat.
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