Double counting individual points is a valid method and I use it for certain areas. But when I have to pair too many points that are on a diagonal and/or separated by stones or if I’m counting long stretches of points in the top corners, I often lose track of what has been counted and what has not. Counting squares seems easier. If I count the total ‘complete’ blocks within a territory then add the two longest rows of blocks that represent the width and the height of the area I am counting, then I can get an accurate number by adding one to this number and the number of any stray points. This method can get tricky and takes some practice but for me it's better than losing my place when counting intersections. A simple example would be 6 complete blocks = 6 Total + 3 in Width + 2 in Height + 1 = 12 Points.
The six blocks have the following values
|4| |2| |2|
|2| |1| |1|
If a block is represented in either the X-axis or Y-axis it is actually counted twice or has a value of 2. If a block represents a point within the matrix of these two lines it is counted once or has a value of 1. The block where the X and Y-axis intersect is counted four times or has a value of 4. A block that is not aligned with both the Y and X-axis has a value of 2. A block that is not aligned with either axis has a value of 4 points. Ignore interior stones that do not represent what you assume to be the outside border when counting. You will also need to add in any stray points within the border. If there are a number of these, I recommend counting them in pairs to save time. Obviously you will need to subtract or add the difference between friendly stones and enemy stones within the territory as well. This has worked well for medium to large territories with contingent but irregular shapes. Territories do not have to be uniform in shape.
The following is a simple but more realistic example.
Here is how the blocks in this corner might be counted:
|4| |2| |2| |2| |2| |2| |0|
|2| |1| |1| |0| |0| |2| |0|
|2| |1| |0| |0| |0| |0| |0|
Try to ignore the |0|s and visualize the numbered blocks in the diagram below.
The Block Counting equals 31.
Add the marked stray point.
The dead white stone and the black stone in the interior offset each other.
Territory = 32
The blocks above and adjacent to the marked black stone are not counted because the marked stone covers one corner of each so they are not 'complete'.
The complete block that is to the right of the marked stone is given a value of 2 because it has been separated from the axis group running down the left side of the board by the marked stone but is still aligned with axis across the top.
The last block at the bottom on the left is given a value of 4 because it is separated from the matrix above it. However, it is easier in most cases to extend the count of 2 down to this block and then subtract the two black stones that separate it from the matrix to arrive at the same number.
With this method, the first thing I do is look for the longest line of ‘blocks’.
Near the edge of the board, I move up or along the line, one row or column at a time. I count points by assigning the appropriate value as above (This does take some practice, but not that much and can be done on-the-fly) to each block in the row or column and adding them up as I go. I also add in any stray points on the border. Since I have ignored interior stones, I compare the number of friendly verses enemy stones; if there are more enemy stones, then I add the difference to my total. If not, I subtract the difference. Here is how I might count the example given.
“First row…Ignore the black stone…4 plus 2 is 6, 8, 10, 12, 14 and one stray point…
Next row…14 plus 2 is 16, 17, 18, and skip these, and this one is worth 2…2 plus 18 is 20…
Next row…20 plus 2 is 22, and this one's okay with the white stone for 23…and the stray point makes 24…
Next row…24 and 2 is 26…moving down 28…" and...
Either "the isolated block is 4 for 32." OR "30, 32 and the last block is offset by the two stones for 32".
The black and white stones offset each other so the total is 32.”
Square or rectangular territories are normally counted by multiplying the number of intersections on the X-axis by those on the Y-axis. If you want to use ‘blocks’ simply add 1 to the X and Y-axis before multiplying.
In the middle of the board, count as above realizing that the X and Y-axis may meet in the middle of the territory, should extend to the edge of the territory being counted, do not need to be in a straight line and that blocks considered within the matrix could be in four different directions. In large complicated areas I keep it simpler by counting all the blocks and stray points, add the X and Y-axis blocks and add 1.
I call this Block Counting. Comments are welcome. --BostonGo
DrStraw: The problem with this is that it is hard to see the blocks. The interesections are readily visible on the board but seeing the blocks squares required a definite shift of perception.
AJP: A case could be made along the lines (heh) that the whole idea of playing stones on the intersections drawn on the board is screwy and that go should be played by putting the stones in the middle of squares, just like in every other game (chess, checkers, hex, othello, tic-tac-toe). In particular, the edge of the go board is a confusing realm... is that somewhere you can play pieces, or is it just a border? Sure, the lines indicate the connectivity of the stones, but that's pretty obvious anyway. Playing on the intersections is immediately distressing to most western people I try to explain the game to. That being said, I've never actually tried to play go on the squares, still keeping with an overall 19 x 19 playing matrix, of course. Does anyone have experience with this kind of thing?
Phelan: I must admit I haven't quite understood it, but from what I read it seems to be easier to count areas with random stones and strange shapes. But, as has been mentioned, it seems strange to me to shift from thinking about intersections while playing, to squares, when counting.
BostonGo After contemplating the comments about the shifting of perception it has occurred to me that this shift of perception may be helpful rather than hurtful in that it seems to use different brain cells and allows a brief respite for the others; that in my case are always working so hard. AJP, I want to thank you for your comments. I assure you that I have no intention of advocating that the game be fundamentally changed to playing on the squares rather than the intersections. It’s hard enough for me to manage a maximum of four liberties per stone, I can’t imagine dealing with nine. In that case I might be compelled to figure out a way to count points by using the intersections instead of the squares just to complicate it further. ;)
Phelan: After rereading your first explanation and the clarification, I think I almost understood it.
Counting for black:
|4| |2| |2| |2| |2| |2| |2| |2| |0| = 18 |2| |1| |1| |1| |1| |1| |1| |1| |0| = 9 |2| |1| |1| |1| |1| |1| |1| |0| = 8 |2| |1| |1| |1| |1| |0| |0| |0| = 6 |2| |1| |1| |1| |0| |?| |?| |0| = 5? |2| |1| |1| |1| |0| |?| |0| = 5? |2| |1| |1| |1| |0| |0| =5 |2| |1| |0| |0| |0| = 3 |0| |0| |0| |0| |0| = 0
2 stray stones for white
5 stray stones for black
Total count : 59+2-5+1=57
Counting for white:
|0| |4| |2| |2| |2| |0| = 10 |0| |2| |1| |1| |1| |0| = 5 |0| |2| |1| |1| |1| |0| = 5 |0| |2| |1| |1| |1| |0| = 5 |0| |?| |0| |0| |1| |1| |0| = 2 |0| |0| |0| |0| |0| |0| |0| = 0
total count: 30+1=31
The blocks marked with ? are the ones I don't know what value to give.
BostonGo: I think you’re getting it Phelan. Thank you so much for your feedback. In fact, you have taught me something about my own method; and for that, I promise not to mention your adding error ; ). Great counting example that you put together!
Counting Example 1 for Black:
|4| |2| |2| |2| |2| |2| |2| |2| |0| = 18 |2| |1| |1| |1| |1| |1| |1| |1| |0| = 9 |2| |1| |1| |1| |1| |1| |1| |0| = 8 |2| |1| |1| |1| |1| |0| |0| |0| = 6 |2| |1| |1| |1| |0| |A| |A| |0| = 5+2=7 |2| |1| |1| |1| |0| |A| |0| = 5+1=6 |2| |1| |1| |1| |0| |0| =5 |2| |1| |0| |B| |0| = 3+1=4 |0| |0| |0| |0| |0| = 0
Counting Example 1: The blocks marked with ‘A’ are not ‘complete’. You would simply add these ‘stray border points’ to your total. The block marked with ‘B’ is not complete either. You counted this block as 0, which is correct. However, there is a ‘stray border point’ here in the lower right corner of the block. You added a 1 at the end of your addition to get a the total of 59. It’s not clear to me if this point represents this stray point, as it should, or if you got confused by part of my earlier explanations where I say, “…and add one.” There is no need to add an extra point to your total if you count the point where the two axis cross, which you did by giving the upper left corner block a value of 4.
Correct numbers: 18+9+8+6+7+6+5+4=63
2 stray stones for white
5 stray stones for black
Total count : 59+2-5+1=57
Correct count: 63 + ((2-5)=-3) = 60
Counting Example 2 for Black:
|4| |2| |2| |2| |2| |2| |2| |2| |0| = 18 |2| |1| |1| |1| |1| |1| |1| |1| |0| = 9 |2| |1| |1| |1| |1| |1| |1| |0| = 8 |2| |1| |1| |1| |1| |1| |1| |0| = 8 |2| |1| |1| |1| |1| |1| |1| |0| = 8 |2| |1| |1| |1| |1| |1| |0| = 7 |2| |1| |1| |1| |0| |0| = 5 |2| |1| |1| |1| |0| = 5 |0| |0| |0| |0| |0| = 0
Counting Example 2: This is how I would count this area. This is simpler for me. No stray points or isolated blocks. I just need to add a few more friendly stones to my stray stone count. The key is recognizing the outside border and complete blocks.
2 stray white stones
10 stray black stones
Correct count: 68 + ((2-10)=-8) = 60
Counting for white:
|0| |4| |2| |2| |2| |0| = 10 |0| |2| |1| |1| |1| |0| = 5 |0| |2| |1| |1| |1| |0| = 5 |0| |2| |1| |1| |1| |0| = 5 |0| |A| |0| |0| |B| |B| |0|= 3+2+1=6
Here is where you taught me something. The block marked ‘A’ is a ‘complete’ block. You were not sure how to count it. According to my previous rules, it should be given a value of 4 because it is not fully aligned with both the X and Y-axis. In fact, the upper right corner of this block is also the lower left corner of a block within our matrix. This means that this ‘complete’ block, which is not fully within the matrix of our X and Y-axis, but is partially within it, should be given a value of 3. Let’s look at another error in your counting that may help to make this clearer. You gave both the blocks marked ‘B’ a value of 1. If you look closely, you will see that both of these blocks are aligned with the axis running across the top edge but neither of them is aligned with the axis running north and south. A block that is aligned with only one axis is assigned a value of 2. However, only one of these blocks (either one) would be given a value of 2 and would then represent the north-south axis for the other stone which would be given a value of 1 because it would fall within the overall matrix. This is not as difficult in practice as it may sound in writing. The concept I just explained (however poorly) is rather important when it comes to counting territories in the middle of the board. It is important to realize that the blocks that represent the X and Y-axis do not have to be side by side or run in a straight line, as long as they represent the dimensions of the overall matrix.
Correct numbers: 10+5+5+5+6=31
total count: 30+1=31 (No need to add 1 as explained above.)
As shown in Counting Example 2 for black, One might count the last row of White's territory this way:
|0| |2| |2| |1| |1| |1| |0|= 7
This gets you to the same number once you subtract the white stone that is now counted as a stay. I prefer this method.
I’d like to make the point that double counting is still in my repertoire, especially for long meandering territories. This method is for when it’s easier to see the holes than the net.
Hicham: Though I am sure that this method might work for some people, it definetely wouldn't for me. The change in perspective is a bit confusing. The intersections are what I look for, not just when counting points, but also where to play my next move or to count liberties. If I would have to check different things for each of these, then the advantage of this method are outweighed by the extra work they bring into the thought process. The intersections are points because you can play a stone there, not because they are in your territory. Especially when counting unfinshed area you have to see future stones on the intersections to be able to count, block counting seems to be a bit counterintuitive here.
BostonGo: As you can read above, a lot of people agree with you. Either way you count you have to envision the future borders. My goal was to count more quickly and more accurately. To figure out if I was winning or losing. The biggest thing that slowed me down and reduced my accuracy, was losing my place while counting large to medium territories that were irregular in shape. I don't think there is much of a time difference in counting in pairs or block counting, if you don't lose your place, but I am less likely to lose my place if I'm counting those big holes instead of those little knots in the net. It has been worth it for me to figure out how to value the squares on the fly. Now if I can figure out an easy way to calculate the value of end game moves.
Bill: It strikes me that block counting is similar to counting by twos, which, I have heard, is how many pros count.
Counting by twos you get 2 + 2 + 2 + 2 + 2 + 2 = 12. Counting by blocks you get 4 + 2 + 2 + 1 + 2 + 1 = 12 ?