Easy if you know the secret

Bill Spight · #1

Here is a labyrinthine problem, adapted from Berlekamp and Wolfe's Mathematical Go.

The key, as usual in these problems, is to get the last play to win by 1 pt. One misstep and Black can get jigo (or win by area scoring).

Humans can simplify the problem by playing the sente early and by noting the miai, which can be ignored. AlphaGo and other top go programs do not do that, IIUC, and I am not at all sure that they could solve this problem without taking a very long time. OTOH, for humans who know the secret, the solution is almost obvious.

Schachus and I have started a discussion about this problem, which I have proposed to move here.

I will also take Black against all comers.

Bill Spight · #2

Continuing discussion with Schachus.

Edit: Corrected my error.

Schachus · #3

so according to my previous plan, it would have to be a5 now? And then continue to push in and once it is filled start with t13?

Bill Spight · #4

Schachus wrote:

so according to my previous plan, it would have to be a5 now? And then continue to push in and once it is filled start with t13?

If I read you right:

Schachus · #5

yes you played like I thought(for white, for black I was surprised). But I havent yet considered, what to do next, so let me see..

Bill Spight · #6

Schachus wrote:

yes you played like I thought(for white, for black I was surprised). But I havent yet considered, what to do next, so let me see..

Let me correct a mistake of mine. I got the colors reversed early this morning. E-04 and F-10 are not, repeat, not miai. They are both double up stars.

Bill Spight · #7

Schachus, if you would like, here is a hint.

Schachus · #8

Well, I'm not too surprised, but still I dont know what I did wrong/ how I would have had to play.

I would assume that you played, what you consider good for black, so I could try and prevent that. The main idea there would be to take a reverse sente first(after taking the 2 sente moves). But it doesnt really seem more promising to me at all. Also at each point there is the option of blocking some of black pushes before one really has to, wich is kind of similar to taking reverse sente points. But I have no clue, how to decide, when exactly that would be the right play.

PS: I do know "Nimbers"(https://en.wikipedia.org/wiki/Nimber) and how to add them, does something like this help here?

Bill Spight · #9

Schachus wrote:

Well, I'm not too surprised, but still I dont know what I did wrong/ how I would have had to play.

The secret is counter-intuitive.

Quote:

I would assume that you played, what you consider good for black, so I could try and prevent that. The main idea there would be to take a reverse sente first(after taking the 2 sente moves). But it doesnt really seem more promising to me at all.

Right. The only reason Black played a reverse sente, :b36:

, was that it was, in effect, the last play. And Black got to play that reverse sente because :b34:

was a sente that White had to answer. But White had a sente at that point. What if the White sente had had a threat as large as, or larger than Black's?

That sounds like a rhetorical question, but I'll let you draw the conclusion and consider how White might have played differently.

Bill Spight · **#10**

Click Here To Show Diagram Code: [go]$$Wc Miai $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | . O O O O O O X X O O X . X X X O O X | $$ | . O . O . . O O X X X . X . X X X X X | $$ | . O O . O O X X X X . . X O X O O O O | $$ | . . . . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X . O O O X X X O X O , . O . | $$ | . . X . X . X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O . X X . X X O . . O . , . X . | $$ | . . . O . X . . . . . O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O . . . . X X O . . O O X . | $$ ---------------------------------------[/go]

I claim, along with Berlekamp and Wolfe, that this position is strict miai. No matter who plays first, the result will be the same. Let me illustrate that with play.

Click Here To Show Diagram Code: [go]$$Wc White plays first $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | 4 O O O O O O X X O O X . X X X O O X | $$ | 6 O . O . . O O X X X . X . X X X X X | $$ | 8 O O . O O X X X X . . X O X O O O O | $$ | 0 . . . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X 7 O O O X X X O X O , . O . | $$ | . . X . X 9 X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O . X X . X X O . . O . , . X . | $$ | . . . O . X . . 5 3 1 O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O 2 . . . X X O . . O O X . | $$ ---------------------------------------[/go]

Click Here To Show Diagram Code: [go]$$Wcm11 $$ --------------------------------------- $$ | . X . X O O . . a a . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | X O O O O O O X X O O X . X X X O O X | $$ | X O . O . . O O X X X . X . X X X X X | $$ | X O O . O O X X X X . . X O X O O O O | $$ | X 2 4 b O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X c O O O . . | $$ | . X X X X O O O O X X X O X O , . O . | $$ | . . X . X O X X O O O O O O O O O . . | $$ | X . . X X d d . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X 3 e X X X O O . O O O O X X X | $$ | O O O O 1 X X . X X O . . O . f . X . | $$ | . . . O . X . g O O O O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . h O X . . . X X O . . O O X . | $$ ---------------------------------------[/go]

After

there are 8 simply gote left that gain 1 pt. for whoever takes it, which Black and White will share equally. Such a gote is called a star, written *.

Now let's look at the result if Black plays first.

Click Here To Show Diagram Code: [go]$$Bc Black plays first $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | 3 O O O O O O X X O O X . X X X O O X | $$ | 5 O . O . . O O X X X . X . X X X X X | $$ | 7 O O . O O X X X X . . X O X O O O O | $$ | 9 . . . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X 8 O O O X X X O X O , . O . | $$ | . . X . X 0 X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O . X X . X X O . . O . , . X . | $$ | . . . O . X . . 6 4 2 O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O 1 . . . X X O . . O O X . | $$ ---------------------------------------[/go]

Click Here To Show Diagram Code: [go]$$Bcm11 $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | X O O O O O O X X O O X . X X X O O X | $$ | X O . O . . O O X X X . X . X X X X X | $$ | X O O . O O X X X X . . X O X O O O O | $$ | X 1 3 . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X O O O O X X X O X O , . O . | $$ | . . X . X O X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X 4 . X X X O O . O O O O X X X | $$ | O O O O 2 X X . X X O . . O . , . X . | $$ | . . . O . X . . O O O O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O X . . . X X O . . O O X . | $$ ---------------------------------------[/go]

After 14 plays we reach the same position. In these sequences each play by the second player is necessary to preserve the miai.

(Although not necessarily unique.)

Except for the play on the bottom edge, each play is in a corridor that leads to a star. In the top left, Black has 6 plays before reaching the star, plus 1 play on the bottom side before doing so there. In the center there are 3 corridors, where White has 2 + 2 + 3 plays before reaching the stars. Each player has 7 plays before reaching the stars.

This problem is a fight for the last play that gains 1 pt., and in the top left corridor White has a local advantage in that fight. White can get the last local play unless Black makes 6 plays to reach the star, and then takes the star. The corridor is called a sextuple down star, written v6*. (Yes, it's a peculiar notation. Take your complaints elsewhere.

) The point is that before reaching the star White has 6 chances to take the last local play, which might be the last whole board play. By moving into the corridor Black takes away a potential last (whole board) move for White. You can see the similarity to filling outside liberties in a semeai.

Black does not necessarily take the star, because, while it is a potential last move for White on the whole board, it is also a potential last move for Black on the whole board. For why we say that the corridor is 6 v's plus *, see this page: http://senseis.xmp.net/?CorridorInfinitesimals , which includes a discussion of ^^*.

In this problem White has a v6* + v = v7*. Black has ^^* + ^^* + ^3 = ^7**. Up(^) and down(v) are opposites, so v7 + ^7 cancel out. * + ** = ***. Two stars are miai, as we know, so they cancel out, leaving *. (The similarity to Nim heaps is not accidental.) As we see, when the ups and downs are all played out, there are actually 5 stars left. But 4 of them are miai, leaving one star, as advertised.

(BTW, you don't have to do the strange arithmetic. There are four corridors plus the down on the bottom edge, so playing them out will leave five stars.

) There are three stars elsewhere on the board, so everything is miai.

Note that if there had been, in effect, an odd number of stars on the board after :w35:

,

would have been a mistake, as it would have allowed White to take the last star. Instead, Black should have taken the star, allowing White to play the sente.

Edit: To underscore the miai, the second player can make a mistake by not preserving it, as in the next two diagrams.

Click Here To Show Diagram Code: [go]$$Wc Black mistake $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | . O O O O O O X X O O X . X X X O O X | $$ | . O . O . . O O X X X . X . X X X X X | $$ | . O O . O O X X X X . . X O X O O O O | $$ | . . . . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X 1 O O O X X X O X O , . O . | $$ | . . X . X 3 X X O O O O O O O O O . . | $$ | X . . X X 4 . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O 5 X X . X X O . . O . , . X . | $$ | . . . O . X . . . . . O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O 2 . . . X X O . . O O X . | $$ ---------------------------------------[/go]

is a mistake. Maybe Black thought :w3:

was sente.

Click Here To Show Diagram Code: [go]$$Bcm6 $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | 1 O O O O O O X X O O X . X X X O O X | $$ | 3 O . O . . O O X X X . X . X X X X X | $$ | 5 O O . O O X X X X . . X O X O O O O | $$ | 7 9 0 . O X O O O X X . X . X O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X O O O O X X X O X O , . O . | $$ | . . X . X O X X O O O O O O O O O . . | $$ | X . . X X X . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X 2 . X X X O O . O O O O X X X | $$ | O O O O O X X . X X O . . O . , . X . | $$ | . . . O . X . . 8 6 4 O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O X . . . X X O . . O O X . | $$ ---------------------------------------[/go]

Black's mistake allows White to get the last play with :w15:

. There are six stars left on the board.

Schachus · **#11**

Ok. Let me try to understand why the ^^* from your example is indeed a ^^*.

By definition, we have ^^*= {^*,^^|^,^^}, right?(if black plays in one of the ups, then we get ^*, if he plays in the star, we get ^^, if white plays in one of the ups, we get ^**=^, if he plays the star, we get ^^). Now ^^ is alwas better for black then ^, no matter what is there otherwise, right? so this means, white would never play to ^^, but always to ^, so we have ^^*={^*,^^|^}. On the other hand , the position we are given is {0|^}. How do I know, this is the same?

Now in a way similar to the argument given in the link, I could try to show that {^*,^^|^}+{v|0}=0 (Showing that the inverted position together with a ^^* is miai). The left hand side should be {^*+{v,0},^^+{v,0},*,^|^+{v|0},^}. I see that ^+{v|0}={{v|0},0|*+{v|0},^}={0|*+{v|0},^}. Also I have *+{v|0}={{v|0},v*|{v|0},*}.
I cant make progress with this miai argument either, because I always fail to simplify {v|0} inside, so it doesnt seem to make things easier.

The way you obtain the miai in the link is by saying "B1 - W4 is correct play for both sides. The result is obviously miai, with a score of -2." I'm exactly not sure about the "correct play" statement, because I'm still new to which infinitesimals are really always better than others(in fact even when saying ^^ is better than ^ for black above(or similarly ^is better than 0 for black, how to prove that? I see that 0=**={*|*}, so now comparing with ^={0|*}, why is this always better for black? (ok, if this is the only play, I can see that leaving 0 is better than leaving *, but if it isnt the only play!?))), and thats why I fail to simplify the above. Can you eloborate on this?

Edit: Some of my concerns are gone, once I realized the following: if I have a game A, that one player can win, no matter, whose turn it is, then the game X+A is always better for him, then the game X, for any finite game X(meaning, if he has a win a winning startegx for X, then also for X+A), which I can simply see by induction on the maximal number of moves X can take(this dereases with every move). This exmplains, for example, why ^ is always better than 0 for black.

Edit2: This argument actually onlx needs that I have a winning startegy for A on the other players turn. This explains to me also the miai concept a little better. A game A in which the starting player loses always loses is miai, right? Cause I can apply the above to show that X is at least as good as X+A for either player, hence, they are just equal.

Bill Spight · **#12**

Truth to say, I find it easier for go infinitesimals to deal with go positions.

Like so:

¿Está claro?

Schachus · **#13**

Yes, I think that is pretty much clear to me:)

Bill Spight · **#14**

OK, back to the original problem.

Click Here To Show Diagram Code: [go]$$Bcm34 Black takes sente $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | . O O O O O O X X O O X . X X X O O X | $$ | . O . O . . O O X X X . X . X X X X X | $$ | . O O . O 2 1 X X X . . X O X O O O O | $$ | . . . . O X O O O X X . X . 3 O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X . O O O X X X O X O , . O . | $$ | . . X . X . X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O . X X . X X O . . O . , . X . | $$ | . . . O . X . . . . . O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O . . . . X X O . . O O X . | $$ ---------------------------------------[/go]

After 33 moves we reach a point where each player has a sente. :b34:

takes her sente before playing the reverse sente, which is the last play, as the rest of the board is strict miai.

But what if it were White's turn?

Click Here To Show Diagram Code: [go]$$Wc Black's Zwischenzug $$ --------------------------------------- $$ | . X . X O O . . . . . . . X X O . O . | $$ | . X . . X O O O O X X . X X O . . O . | $$ | X X X X X O . X O O X . O X O O O . O | $$ | . O O O O O O X X O O X . X X X O O X | $$ | . O . O . . O O X X X . X . X X X X X | $$ | . O O . O 3 2 X X X . . X O X O O O O | $$ | . . . . O X O O O X X . X 4 1 O X X O | $$ | O O O . O . . O . X . X X X X X X O O | $$ | X X O O O O O . O X . . X . O O O . . | $$ | . X X X X . O O O X X X O X O , . O . | $$ | . . X . X . X X O O O O O O O O O . . | $$ | X . . X X . . . X X O . O X X X O O O | $$ | O X X . . X X X . X O . . O O X X X X | $$ | O X . X X O X . X X O O . O . O O O X | $$ | O X X X . . X X X O O . O O O O X X X | $$ | O O O O . X X . X X O . . O . , . X . | $$ | . . . O . X . . . . . O . . O X X X X | $$ | . O . O O X X X X X X O O . O O X . . | $$ | . . O . . O . . . . X X O . . O O X . | $$ ---------------------------------------[/go]

is sente, but before answering it, Black plays her sente, which carries a larger threat, as an intervening move (Zwischenzug in chess parlance). That way Black still gets the last play.

Back to the original problem. If White only had a sente with a larger threat than Black's (or at least as large) when we reach the point where each player has a sente, then White could play it as a Zwischenzug, get the last play, and win the game.

How to do that? Start in the corridor with the smaller sente threat at the end, saving the larger sente threat for later, when it is needed.

See the main line.

Schachus · **#15**

Nice:)

Can you now tell me, why you think the fist non-sente play would be obvious!? I could have thought for hours, I would not have come up with the thinking that the larger threat at the end of the corridor is decisive, even if its much farther away

Bill Spight · **#16**

Schachus wrote:

Nice:)

Can you now tell me, why you think the fist non-sente play would be obvious!? I could have thought for hours, I would not have come up with the thinking that the larger threat at the end of the corridor is decisive, even if its much farther away

Yes, for those who don't know the secret, the first play is counterintuitive. I remember being surprised, myself.

After all, given a choice of sente of the same size with no kos around, it is obviously right to choose the sente with the largest threat. (At least for simple threats.

) So with a choice of corridors (and other such situations) with distant sente, it also makes sense to choose the one with the largest threat. But if you have read Mathematical Go, you know the secret that this problem illustrates, and you know that you choose the one with the smallest threat. A quick glance reveals that the T-13 corridor has the smallest threat at the end.

Quote:

I don't always play in corridors with distant sente,

but when I do,

I start in the one with the smallest threat.

Easy if you know the secret

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