Maximum Number Of Live Groups
Independently Live groups on a Goban.
I found this in the IGS ftp archive some years ago, and the discussion in False Eye reminded me.
Cool, no?
(I've changed some of the formatting to 'suit' the wiki, but none of the contents)
-- Morten Pahle
How many live groups can there be on a go board?
Subject: Number of live groups. From: wft@math.canterbury.ac.nz Bill Taylor
Yijun Ding asks what is the maximum possible number of groups in a game.
Yes; I know he was asking for a practical maximum, for computer programming purposes, rather than a theoretical one; but I choose to interpret it the latter way. After all, it's clear that there's way too many computer scientists on rec.games.go and not nearly enough mathematicians ;-) .
Anyway, the question, the maximum possible number of groups on the board, has two interpretations (at least).
1: Maximum number independently alive. 2: Maximum number alive in seki (or similar).
Both of these are difficult to define unambiguously, without 134 densely printed pages of Japanese-style-rule bookishness. However we'll take for granted, (temporarily), that we all know what we mean.
It's already been mentioned that things like might be considered as either one or two groups, but I'll assume that it's clearly one, in the spirit of the regular rules.
Whereas something like this (on a 4x4 board), would have to be counted as four groups, two black and two white. The separate halves of the black groups are `connected' by a uni-colored eye; this doesn't hold for the two white bits.
Anyway, none of this appears to be relevant for the answers to questions 1 and 2 above; on a 19x19 board at least.
I get answers of 28, and 60, respectively.
As always, I would be delighted to see anyone write in with improvements.
Here are my solutions.
From: smith@pell.anu.edu.au (Michael Smith)
I didn't have time to fine tune this -- It might be possible to get an extra group one the edge somewhere.
Simply based on regular pattern in capital stones :-) which are a more efficient way of covering the board than the way you chose:
Instead of 15 intersections for the groups in the centre, I only use 12.
However I only get one extra fully alive group for a total of 29:
But I think with a bit more thought I can fine tune this.
Next question is how can you PROVE that a certain number is the maximum - apart from trying every combination of alive stones on a board and counting.
TapaniRaiko: One can count the points used by each group. They should add up to at most 361. In the corner a group takes up at least 8 points. In the middle, it takes up 12. In the border, it takes up 10 points consuming 4 points of the border or 11 points consuming 3 of the border. It seems that this leads to a maximum of 33 groups.
This one is not a part of the thread, but I thought it should be added here.
I could not find a way to rearrange the marked area to any shape that could contain a live white group. (circled stones would be black then)
Anyway, 30 groups fit quite nicely :-)
-- Bass
Only took one more day of iterating to find this.
-- Bass, 12.9.2002
From: wft@math.canterbury.ac.nz (Bill Taylor)
Following up my own post, I'm able to improve one of my records for the maximum number of live groups on a go board.
The formation is....
From: wolfe@vina.Berkeley.EDU (David Wolfe)
I've worked on the problem of fitting the most live "strings" on a board, where a string is a connected group of stones, and a live string includes life in seki.
My understanding of Niek van Diespen's posting is that the solution is somewhere in the 130's:
- >Sometimes I'm simply too fast: Ger Hungerink also addressed the version with seki's. I did not put too much time in that, but the maximum is somewhere in the 130's, if I remember correctly. My last posting of course only addressed the "Independently alive" case.
I have a solution which has only 108 strings, and solves Bill Taylors original posting as well with 104 groups:
From: herbison@manage.enet.dec.com (B.J.)
- >I have a solution which has only 108 strings, and solves Bill Taylors original posting as well with 104 groups:
There are two white groups in this diagram with three liberties. (I circeled them to make them stand out.) From this configuration, white can capture most of the board.
From: wft@math.canterbury.ac.nz (Bill Taylor)
I've been plugging away at this business of maximum number of LIVE groups possible on the board. (Of course without the "live" restriction, the checkerboard patterns that have been posted will easily win).
I'm now convinced that the best patterns will have a basic shape like that shown here, which must be the max possible density of live groups in an infinite board. (These groups are only alive in seki, of course.)
This density is noticably better than any other solutions seen on the net.
It's a bit tricky fitting in stuff round the edges of a finite board, though. In view of the messy nature of my solution, and the alleged Dutch publication of a 130-odd group formation, I'm far from sure this is best possible. It must be close though.
So here's my solution:
- 128 groups (counting orthogonal connections only)
- 114 groups (counting 2 bits joined by an eye as the same)
Interestingly, (in Chinese counting), the result of the game is an exact tie ! ( 151 points each, with 59 neutral).
From: maiorana%idacrd.uucp@princeton.edu (James Maiorana)
This article is in reponse to wft@math.canterbury.ac.nz (Bill Taylor)'s message dated 21 May 92 04:19:15 GMT. He writes
- >I'm now convinced that the best patterns will have a basic shape (like that shown above) which must be the max possible density of live groups in an infinite board. (These groups are only alive in seki, of course.) This density is noticably better than any other solutions seen on the net.
There is also a variation on this configuration that proves useful, namely:
Here we have shown only a single strip. These strips can be put together to cover an infinite board. This second configuration allows the boundary connectivity to change phase.
He also writes
- >It's a bit tricky fitting in stuff round the edges of a finite board, though. In view of the messy nature of my solution, and the alleged Dutch publication of a 130-odd group formation, I'm far from sure this is best possible. It must be close though.
- >So here's my solution;
- 128 groups (counting orthogonal connections only)
- 114 groups (counting 2 bits joined by an eye as the same)
- 128 groups (counting orthogonal connections only)
We were able to get to the 130-odd group level. We found the following:
This arrangement has 132 groups (in the sense of connected components).
HandOfPaper: Shouldn't the phrase you want be " monochromatic connected components "? Otherwise, you would have only 8 of them...
If we allow connections though empty intersections, then the connected components combine to form 65 areas. The three circles are empty intersections that behave as false eyes (as opposed to seki points).
The false eyes at F6 and Q16 are the centers of "windmills". These windmills are there to solve phase problems with the edges of the board. Each one forms a single area with 4 connected components. The false eye at T1 is the center of another windmill. This rearrangement of that corner makes this clear.
This rearrangement does not change the area or component counts.
All these seki examples deal with locally stable solutions. But what's about this one?
If one player moves, is there any safe way to the opponent to win the game? Or, on the contrary, is there any safe way the first moving player wins the game? Perhaps it is something you could call a "global seki"? - Although i´ve been trying to analyse it for quite a while, I still don't know anything about that. So, if you're sure that it's not stable at all, just remove it. :) --rubilia
(Sebastian:) Well, that's what it's a game for - let's play it: Global Seki Ongoing Game
PurpleHaze: This is a silly position. A seki is by definition stable. All these cut/connections are very urgent. Many stones will die rapidly. I suspect that black wins by about 40 points regardless of whose move it is.
rubilia-- Maybe I shouldn't use the word "seki" to what I think about: a whole board position, the score of wich is better to both players than what each of them could expect to gain with additional moves. This kind of position can be locally stable (like the other examples given above), but it doesn't have to. In both cases there's a negative global temperature, that's why I used the expression "global seki". Don't stick to that single word, please.
Concerning this particular position: how to get the 40 points estimation? I am curious about details. :) --2003-11-17
PurpleHaze: You still don't get it. In this position playing is better than passing and will be for a long time. There is no sense in which this position could be called stable, either locally or globally.
There are 100 black stones versus 81 white stones. If we make the assumption that black can kill white stones at the same rate that white can kill black stones, which seems reasonable to do, then the extra stones will provide an advantage. I guesstimate this as being in the same ratio as the number of stones, though the rule set used makes a big difference.
rubilia-- As mentioned above, I am not sure a bit about this position. So, if you are, i'd like you to share a little more of your knowledge. Wich moves are the best ones to start? How to continue? To be a globally unstable position, any "perfect play" sequence must favour the first moving player. Perhaps we should try to find it (first, on a smaller board?) to understand what can happen.
Appropriate komi might be 100 - 81 = 19, but it also could be much less because of the weakness of stones at the edges. (Anyway, that doesn't affect the stabel/unstabel question. About some small so-called "unstable seki" you may wanna have a look at http://www.goban.demon.co.uk/go/bestiary/rule_challenge.html. The label "unstable" seems some mistakable, since those positions are globally stable. Of course, else they couldn't be called "seki" at all.).--2003-11-18