No komi. White to play and win.
Enjoy!
Edited to add coordinates. I hope I didn't mess up your comment, jts.
| Life In 19x19 http://prod.lifein19x19.com/ |
|
| Mid-summer endgame problem http://prod.lifein19x19.com/viewtopic.php?f=15&t=8815 |
Page 1 of 1 |
| Author: | Bill Spight [ Sun Jul 28, 2013 11:22 am ] |
| Post subject: | Mid-summer endgame problem |
No komi. White to play and win. Enjoy! Edited to add coordinates. I hope I didn't mess up your comment, jts.
|
|
| Author: | jts [ Sun Jul 28, 2013 3:52 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
| Author: | Kirby [ Sun Jul 28, 2013 8:50 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
This is not a cute problem! As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible. After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example? |
|
| Author: | lightvector [ Sun Jul 28, 2013 9:45 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
Ooooh, I think I got it. I think there's a good chance this is the answer, though I might have overlooked a detail. |
|
| Author: | Bill Spight [ Mon Jul 29, 2013 6:28 am ] |
| Post subject: | Re: Mid-summer endgame problem |
@ lightvector |
|
| Author: | daal [ Mon Jul 29, 2013 7:16 am ] |
| Post subject: | Re: Mid-summer endgame problem |
Kirby wrote: This is not a cute problem! As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible. After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example? Yes, and please show us the easy way to do it. 
|
|
| Author: | Kirby [ Mon Jul 29, 2013 2:14 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
daal wrote: Kirby wrote: This is not a cute problem! As I see it, there are something like 8 areas to play here. In a real game, I could try evaluating the score by iterating through each of the permutations of moves to the end of the game, but that's not feasible. After everyone's answered, instead of getting a solution, could it be possible to explain the process of deducing the precise optimal solution from this example? Yes, and please show us the easy way to do it. Well, easy or hard, I want to know the correct way to do it. I see people talking about some move being worth X points or this stuff being sente or gote, but it's confusing to me whether these methods give you the precise answer all of the time, or if it's really just an approximation. I can understand reading out permutations, but I don't understand how this fits up with assigning point values, because as far as I know, this can lead to error... maybe? Maybe this example is too hard, but as a general rule for these type of problems, I don't really know how to do them unless I just try to estimate the point values in each location, and maybe pick the biggest from each spot. With the issue of sente/gote, I just try to pick sente moves whenever possible, defining sente as saying that the point value that I gain by playing again is greater than what he'd get if he took a different spot. It all just seems imprecise, I feel, so I wonder if it's correct. Based on Bill's responses to lightvector's answer, is lightvector's approach the correct way to do this kind of a problem? Is it exact? At what point do you start reading out variations instead of calculating some sort of points for various areas? All in all, I'm confused. |
|
| Author: | Kirby [ Mon Jul 29, 2013 2:41 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
I'll give a simpler example to explain what I mean. Linked on the SL page for tedomari, I see the following problem: The problem is small enough that *maybe* I can read out each combination of play to see who will win in the end. But the solution shows this approach: 1.) Calculate miai values of each location - seems easy enough. Maybe this is simple to do following the way of calculating miai values. 2.) Get overall count - kind of makes sense how this is done, but I don't see what the point is in terms of finding optimal play. 3.) Get tedomari and win. The solution shows how using the greedy approach (the one I would usually use unless I can read it out exhaustively) fails, and how getting the last play gives you a score that wins the game. ---- OK, easy example, and I can see what is said is true, for the most part. But this still leaves me confused: 1.) Why do you need overall count to play optimally? 2.) In this example, trying to "get" tedomari seemed to work to give you the right answer. But: a. Is tedomari always optimal? b. In general, is this the approach: calculate miai values, try to get tedomari? Let's say there are 10 spots on the board. So I calculate the miai value of each of those 10 spots and then I try to play to get the last play? Basically, I wonder if this is imprecise, and when it fails, condition the method fails, and so on. My current understanding: * You can use miai values as a heuristic to estimate the order of play. Naïve way to go about this is to then take greedy approach, taking largest miai value as you get it. * Tedomari is another heuristic that (usually?) works that lets you say, "OK. I'm not going to take the biggest moves in order, but I'm going to give up a big move so I can get the last play, and probably have more points..." If my current understanding is correct, it would seem that my endgame approach should be as follows: 1.) If possible, read out entire game tree with all combinations to see best sequence (many search possibilities, but straightforward). 2.) If #1 is not feasible, calculate miai values for each spot. Then iterate all combinations of miai value plays until I see which gives greatest result. Is this approach optimal? 3.) If #1 and #2 are not feasible, try to figure out ordering that gives me tedomari. This approach is not always optimal...? 4.) If #1, #2, and #3 are all not feasible, order miai values largest to smallest, and play in that order. 5.) If #1, #2, #3, and #4 are not feasible... Try to keep sente. 6.) If #1, #2, #3, #4, and #5 are not feasible... Try to play big spots. 7.) If #1, #2, #3, #4, #5, and #6 are not feasible... Resign...? --- Is this the best approach? Basically, to me it sounds like we have a bunch of heuristics and stuff, and I want to define precisely when an approach is optimal, and if we don't have time for the optimal approach, the next best approach to getting the solution. Am I making any sense, or am I just typing? |
|
| Author: | daal [ Mon Jul 29, 2013 2:50 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
Kirby wrote: All in all, I'm confused. There's at least a bookfull of things to understand, and each of the 8 areas in the problem could probably fill a chapter. edit: link to the quite useful example Kirby quoted above: http://senseis.xmp.net/?EndgameProblem24 |
|
| Author: | Kirby [ Mon Jul 29, 2013 3:01 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
daal wrote: There's at least a bookfull of things to understand, and each of the 8 areas in the problem could probably fill a chapter. ... I suppose that makes sense. Given that, are each of these techniques (tedomari, lightvector's mention of asymmetry, etc.) just estimates that help us refine our endgame skills to be more widely applicable to more and more problems, or are some techniques superior to others? For example, if tedomari and this asymmetry stuff are both heuristics (are they?), then can I get a different candidate move by trying to get tedomari and by trying to avoid some sort of asymmetrical position? If so, is one usually superior? I haven't really gotten the knack for when these techniques are precise and when they are guesses, so it's hard for me to tell the utility of using one method over another. That's probably why I usually just try to play the moves that have the biggest miai value (if I can calculate it), in order. |
|
| Author: | xed_over [ Tue Jul 30, 2013 1:14 am ] |
| Post subject: | Re: Mid-summer endgame problem |
Kirby wrote: That's probably why I usually just try to play the moves that have the biggest miai value (if I can calculate it), in order. As I read lightvector's post, he's doing just that.. calculating the largest play(s) and playing those in order. |
|
| Author: | Hanmanchu [ Sun Aug 11, 2013 9:25 am ] |
| Post subject: | Re: Mid-summer endgame problem |
Quote: g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows: @ Lightvector, Bill Spight: Where do you find the information on SL on how to calculate the value of "g"? I am asking because I got it wrong 
|
|
| Author: | Bill Spight [ Sun Aug 11, 2013 2:27 pm ] |
| Post subject: | Re: Mid-summer endgame problem |
Hanmanchu wrote: Quote: g: 7/8 point. I think "g" is the correct move for both white and black, based finding this position on sensei's library and by working out and verifying the variations as follows: @ Lightvector, Bill Spight: Where do you find the information on SL on how to calculate the value of "g"? I am asking because I got it wrong http://senseis.xmp.net/?OneMoreNumber
|
|
| Page 1 of 1 | All times are UTC - 8 hours [ DST ] |
| Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group http://www.phpbb.com/ |
|
rather than the one on the left edge, because it's better to leave a plain gote move rather than an asymmetric move that favors black.
gets rid of the asymmetric black-favoring position at the top, transforming it into a 1 point gote. Black tries to do the same to white with 
.
to convert it to an ordinary 1 point gote. So when the remaining 1 point gote moves trade off, white gets the last one with 
and wins tedomari.
. I tried a few variations and they all seem to come out the same. Either way, white takes the other one, and we end up with the same result.
does not.