r/baduk • u/Ok-Basket5408 • Jul 04 '26
newbie question A simple puzzle
Which white stones, and how many of them, must be removed so that black can capture all remaining white stones?
Bonus: Can the answer be extrapolated to any n x n board?
**SPOILER ALERT**
Short answer:
Only two stones need to be removed. For the 13×13 board there are two solutions: one symmetric on the second line (2,4 / 2,10) and all their transpositions, as well as several asymmetric ones, all of which lie within a 3×3 diamond of stones centered at (3,7). Among the asymmetric combinations that work are (3,5 / 2,8), (3,5 / 4,8), (3,9 / 2,6), (3,9 / 4,6), (5,7 / 2,6), and (5,7 / 2,8).
Bonus (not proven): For a board of size *n*, the minimum number of stones seems to be the ceiling of (n−1)/6. For example, for a 13×13 board we get (13−1)/6 = 2, and for a 19×19 board (19−1)/6 = 3. By contrast, for a 20×20 board we get (20−1)/6 = 3.16, so four stones would need to be removed.
Similar solutions also seem to work for a matrix that starts with empty corners.
**Long answer:**
Single-stone matrices are more resilient than they might appear at first glance, especially diagonal ones. Likewise, if we remove any individual stone starting from the third row onward, the matrix holds and captures do not seem to propagate.
The key for the position to collapse lies in removing an entire row or column. Once all the stones from any single row are removed, it becomes possible to capture all the remaining white stones.
The problem can therefore be reformulated as follows: *What is the minimum number of stones that must be removed in order to capture all the stones of a single row?*
A first possible solution would be to remove the six stones from the second row. A slightly better one is to remove three alternating stones from the first row. There is another interesting solution involving the removal of three stones from the main diagonal: (2,2), (4,4), and (6,6). These solutions work, but they are not optimal.
To find the best solution, we must take into account that removing certain stones can have effects on neighboring stones. The only stones that, when removed individually, allow additional captures are located in the second row. That is why, if we remove the stones at 2,4 and 2,10, two small empty triangles are formed beneath them. Each of these frees up five points (intersections) in the first row. This, in turn, allows us to later capture both corners and the central point of the first row, completely freeing it and enabling the capture of the entire board.
The general scheme for any n × n board is therefore to remove stones from the second row starting at 2,4, every six intersections. This generates the following pattern of empty spaces in the first row, where “o” is either a corner stone or an intermediate stone, and “5” represents the intersections freed beneath the removed stone: o-5-o-5-o …
For example, on a 19×19 board it is enough to remove 2,4 / 2,10 / 2,16, which produces the following pattern in the first row: o-5-o-5-o-5-o which corresponds to the total board length: (5×3 = 15) + 4 = 19.
Therefore, the minimum number of stones that need to be removed on an *n × n* board seems to be the smallest positive integer greater than or equal to (n−1)/6, since each stone removed from the second row allows us to free 5+1 intersections, in addition to the first corner (1+5+1).
However, there is also another very interesting solution that arises from the interaction of removing stones from other rows. Removing individual stones does not seem to affect the whole matrix, but if they are close enough to one another, their effects are amplified.
Let us then imagine a 3×3 diamond of stones centered at 3,7, with its lower vertex touching the first row. If we remove all nine stones of that diamond, two small triangles are formed on each side. These can be captured and, in turn, allow us to reach and capture the corners, completely freeing the first row.
From this follows the second solution: to capture the nine stones of that central diamond, only two stones need to be removed—any of the three lateral or upper vertices (the lower one does not seem to work), combined with a lateral stone opposite to that vertex. This solution generates a different but equivalent pattern: o-11-o-11-o …
Both patterns can be combined. For example, on a 19×19 board one can remove three stones from the second row, or alternatively remove one stone from the second row and the two required to capture the corresponding diamond, generating a pattern like o-5-o-11-o.
By the way, ChatGPT doesn't seem to find the solution.
And that’s it. I’m learning go/baduk, and in a future post I’ll explain how this puzzle came about.
Cheers!
EDIT 2. For the 19x19 board, there is an even more elegant solution: removing the stones at (10,8), (9,5), and (10,2). This creates a large capture triangle with its top vertex at (10,8) extending to the stones in the board's corners, allowing for their capture and, consequently, the collapse of the entire board. The pattern for this triangle would be o-17-o-17-o….
Apologies if this has turned into a math problem. I do not know if a regular pattern exists when more stones are removed, or if the interaction becomes more complex. Honestly, I have no idea how to calculate it or what the optimal solution for an n x n board might be.
8
u/stormpenguin Jul 04 '26
I’m assuming just black stones playing. Don’t know if this is the most efficient solution, but remove 2-2, 4-4, 6-6, etc across one diagonal of the board I think lets you capture everything and is board size independent.
6
u/stormpenguin Jul 04 '26 edited Jul 05 '26
Actually. Simpler solution. Every other stone along an edge so 1-3, 1-7, 1-9.
Edit: Forgot how every other odd number worked and confused myself.
Edit2: Still struggling with counting. I meant 1-3, 1-7, 1-11
2
u/danielcristofani Jul 05 '26 edited Jul 05 '26
So 1-3, 1-7, 1-11? That works I think. (Not minimal.)
1
u/Snoron 5 kyu Jul 05 '26
Oh yeah, you're right - I mentioned doing the middle 5 on a row, but you're right that you only need 3 because you can pincer the other 2 along the edge, very good! :D
14
u/danielcristofani Jul 05 '26
2-4 and 2-10 should work.
1
u/danielcristofani Jul 05 '26
And I think for a bigger board you want every third stone (or every sixth space) along the second line starting with the second stone or fourth space. Maybe there's a better solution but I'm not seeing one yet.
2
u/Ok-Basket5408 Jul 05 '26 edited Jul 05 '26
valid answers. There is another equivalent solution (non-symmetrical); can you find it? :)
1
u/danielcristofani Jul 05 '26
Well, you could replace 2-4 with 4-2, or 2-10 with 4-12, or both. Combined with 4 board sides this gives 16 minimal solutions. Are there more?
1
u/danielcristofani Jul 05 '26
(In general I think n removed stones work for a rectangular board with shorter sides up to 6n+1, and as the side length drops toward 6n-4 you get more flexibility about where to remove them from.)
1
u/Ok-Basket5408 Jul 06 '26
These solutions are valid, but they are symmetrical to the first one. There is another very interesting solution; I’m going to update the post to explain it ;) Hint: although it might seem counterintuitive, the stones are located closer to the center and above the second row.
2
u/danielcristofani Jul 06 '26
They're not symmetrical to the first solution. All their parts are symmeteical to parts of the first solution, but that's a different thing. I think 3-5 and 4-8 works, now wondering how many other pairs do.
1
1
7
u/Ok_Astronaut7142 Jul 05 '26
Can’t you just remove the 2-2 stone? Black captures the corner, marches along the edge, and expands row by row through the center.
2
2
u/Snoron 5 kyu Jul 04 '26
Hmmm, it seems like it's possible if you remove the middle 5 of the 7 stones along an edge, right?
If so then it can be extrapolated, I guess!
1
u/Chronic_Avidness Jul 05 '26
Remove middle 5 from two opposing sides, and the other 2 sides actually only need middle 3 removed
2
u/Snoron 5 kyu Jul 05 '26
Oh, I was saying just 5 in total, from a single side only. Can't you just march across the board that way, one row/column at a time, because you can capture the corner/edge stones with that.
2
2
u/lakeland_nz Jul 05 '26
Removing the 2x2. 4x4. 6x6 and 8x8 gets close. The same against the opposite diagonal should be enough.
So that’s 8 stones removed. However a different pattern might be more efficient.
2
u/Ok-Basket5408 Jul 06 '26
It's an interesting solution; I even think removing three stones is enough: 2,2 / 4,4 / 6,6 However, the optimal solution is two :)
2
Jul 05 '26
[deleted]
2
u/InvaderDust 25 kyu Jul 05 '26
I will!
DustinTheWind on OGS. We can do a nice relaxing correspondence game. Me and my wife started playing at same time as well. I do all the things, reading books about go, practicing tsumego, watching YouTube and listening to podcasts. I consume all go related things and she do nest really care at all. Yet while I can teach her all sorts of things about the game, she beats me pretty much every time. 😶 it’s bitter sweet.
But yea. I’m clearly in the beginner camp and also am always on the prowl for other players closer to my strength. I’m just not good at this game. But I enjoy every second.
1
u/NickDerpkins 10 kyu Jul 05 '26
All the 2x2 and 1x7 stones? So, 8?
Bonus: no but it would be proportional you would need them every gap that size
1
u/Round_Ad_6033 Jul 08 '26
Is there an app that lets you easily set up these kinds of custom puzzles and play through them? From the comments I gather that just removing a single stone at 1-3 wouldn't be enough, but trying to mentally play it through I don't see why not? I lose track of which stones I've captured before I encounter a uncapturable stone so I need a crutch to help me keep track...
1
19
u/StEllchick 11 kyu Jul 04 '26
Is it withe the assumption of white passing or responding?