When to play a reverse sente
This page is an editorial page. It discusses when a reverse sente must be played, in comparison with a gote move. Read these definitions first.
Table of contents | Table of diagrams Diagram 1, W first Diagram 1, B first Diagram 2, W first Diagram 2,B first Diagram 3, W first Diagram 4, B first Mutual damage 7x7 endgame, Black to play B: 11, W: 12 B:13, W:12 |
Popular conception and basic endgame theory
Popular conception of the endgame (see basic endgame theory) tends to state that sente moves of a certain size are twice as big as gote plays of the same size. As a reverse sente forestalls such sente, it is also considered to be twice the value it would have if it were a double gote (or simply gote). This leads to think the timing of a reverse sente equals the timing of a same-size sente play and should come before gote plays. This is not true
The reasoning behind the conception is that, if a sente play of value x is available, then surely there will be a gote play of about the same size available to take next. While this is a useful heuristic, especially in the early endgame, in the late endgame it is easy to find counterexamples, and while it is safe to execute sente plays first, the question of when to play a reverse sente is more subtle.
Get the last big point
Another heuristic to guide your endgame is to get tedomari. If you have a choice of gote vs. reverse sente that are worth the same (or often approximately the same), you usually want to get the last such play. Doing so depends on whether the number of gote is even or odd. (None are ko, which complicates things). If the number of gote is even -- they are miai --, play the reverse sente to get tedomari. If the number of gote is odd, play one of them and answer the sente.
However, useful as it is as a heuristic, there are many situations where the available reverse sente is not equal to the remaining largest gote and still either of them could be the best play, depending on the complete picture. Of course, almost everything is a heuristic unless you can read the whole game out.
The stack of coins
One can get some feel from Charles Matthews's stacks of coins model.
Given a chance of a reverse sente move you can play of value v, added onto a stacks position with largest stack s, there are just two ways the game goes. Either you reverse the sente and your opponent starts on the stacks: or you start the stacks, your opponent plays the sente v-point play, and then the stacks play out unaffected.
Here you are clearly faced with saving v at the cost of the advantage of starting the stacks. If that is guessed at s/2 we get the recipe 'double for reverse sente'. It could be as much as s, or smaller than s/2 in the presence of much miai: in which case the reverse sente is going to stand out as tedomari.
The gote delta versus the reverse sente
The stacks of coins idea applied to Go, in the (late) endgame is
- What is higher: the delta of the gote plays or the largest reverse sente play?
(There is a strong assumption, namely that each side's sente plays are indeed sente, i.e. will be answered no matter what. Of course this is not true, because players will try to apply mutual damage. So, in the end, it always boils down to calculating the particular line of play that leads to the best result for both.)
However, if we do accept the assumption, then the above rule applies. Here is an example and a mathematical argument.
Example
is 1 point in sente. is 2 points in gote, since it is 1 point gote with a 1 point sente follow up. is a 1 point gote.
In the previous diagram, is a 1 pt.reverse sente. The difference of the remaining gote is 1 pt. So it doesn't matter for Black to start at a or b. (Black can even start at c!)
This is a modified situation:
Now D is available as sente against White. So there is the question of mutual damage. If White plays A and Black plays D, and both continue in damaging the other, White will gain.
Hence we can say that White's sente at A will be absolutely answered, but B not and we regard B as 2 pt gote, not 1 pt sente.
So the gote delta = 1 and the reverse sente = 1.
Again it doesn't matter where White starts after A: B or D. (Even A must not be necessarily played first, because it is a reverse sente for Black of the same value as the gote delta).
For Black the situation is remarkably identical. If he can start, D is sente because now White comes too late in the mutual damage exchange.^{[1]}
So, D is absolute sente, but the reverse sente at A is equal to the delta gote and it doesn't matter where he starts next, A or B. (And here too, the absolute sente at D need not be played first.)
With respect to diagram 2, B has now become a 4 pt. gote, as it is a 3 point gote move with a 1 point sente follow up.
This now biases the order.
If White starts and executes her sente at A, next B is better than D, not because B>2D but because B-C>D. Do the correct order is a, answered, then b, d/c, c/d. The sente also must be played first, to get the big gote.
For Black first, this is the correct order. (However, it is still OK for Black to start at 3, because he can get the last play at 6.)
[1] Bill: Well, Bd is sente, but White can still interpolate the sente at a, because White does more damage than Black.
White wins big.
Some Math
- Let s_1, ..., s_j be Black's sente plays in descending order and S be there sum.
- Let r_1, ..., r_k be White's sente plays, hence Black's reverse sente plays in descending order and R be their sum.
- Let g1, ..., g_l be the gote plays, G = g1 + g3 + ... and G' = g2 + g4 + ... (it really doesn't matter if l is even or odd)
First way of playing for both sides is to apply all their sente plays, then the largest gote available. The result is then S + G - R - G'
Second way of playing is to apply all their sente, then the largest reverse sente available (which can only happen once). Now the result is: S + r_1 + G' - (R - r_1) - G
The net difference between these results is 2G - 2G' - 2r_1 So, the first way of playing (gote first) is better than the second way of playing (reverse sente first) if
- G-G' > r_1
G-G' is easier to calculate than it seems, for one can strip any even number of miai moves from the calculation.
RobertJasiek The theory above and the principle below rely on implicit, hidden, essential assumptions: 1) Mutual reduction strategy does not describe perfect play. 2) The two ways of playing are the only ways one needs to consider. Usually, this is wrong if some gote plays are larger than all follow-up move values of the local sentes. 3) All local sentes occur in only one group so that their follow-up move values are at least the value of the largest gote play. 4) Kos and ko threats do not play a role.
The result applied to deiri counting
The inequality (G-G' > r_1) at the previous chapter can be used to determine whether the biggest gote play is better than the reverse sente play. Marking G' ' = g_3 + g_5 + .. = G - g_1 we get g_1 + G' ' - G' > r_1, which can be rearranged to g_1 > r_1 + G' - G' '. Because g_1 is the value of the biggest gote play, it makes sense to call the other side the value of the reverse sente play. If we examine that value, we'll notice that G' - G' ' is actually the value of sente after g_1 is played.
In other words, the value of a reverse sente play is the value of the sente play it prevents, plus the value of playing first after the biggest gote is taken.
Here are some examples. It is assumed that any sente plays will actually get answered, so that they can be removed from the calculations. The "opp. sente" column lists the value of the opponent's biggest sente move, the next column lists all mutually gote plays.
Example | opp. sente | Gote plays | after next gote | sente worth then | Value of reverse sente | Correct move |
Example 1 | 3 | 4, pass | pass | 0 | 3 + 0 = 3 | 3 < 4, Biggest gote |
Example 2 | 1 | 1, 1, pass | 1, pass | 1 | 1 + 1 = 2 | 2 > 1, Reverse sente |
Example 3 | 3 | 4, 2, pass | 2, pass | 2 | 3 + 2 = 5 | 5 > 4, Reverse sente |
Example 4 | 2 | 6, 5, pass | 5, pass | 5 | 2 + 5 = 7 | 7 > 6, Reverse sente |
Example 5 | 1 | 2, 2, 2, pass | 2, 2, pass | 0 | 1 + 0 = 1 | 1 < 2, Biggest gote |
Example 6 | 2 | 4, 3, 2, pass | 3, 2, pass | 3 - 2 = 1 | 2 + 1 = 3 | 3 < 4, Biggest gote |
Use miai counting
In miai counting, there is no such thing as doubling the values for sente or reverse sente plays, so this may be less confusing for many go players. Read more on its proper page.
Bass: In miai counting there is halving the value of gote play instead. Miai counting alone will not give an answer to the question presented in the topic of this page, since it cannot handle tedomari at all.
Some reader discussion
Reverse sente plays are not more valuable than sente plays of the same size, or gote plays, either. In fact, the general rule is to play sente first, gote second (unless the gote are miai), and reverse sente third.
-- BillSpight (whose further comments were pasted into the last big point section
NickGeorge Ok, If I understand reverse sente (which would be a bit of a shocker to me) it means the gote play prevents your opponent from making a sente play there. Thus, reverse sente > gote. But it seems I have to also say that sente > reverse sente > gote, because I can go from a sente play to a reverse sente play. Therefore, how can one say reverse sente value = sente value? Is it something like, it's gote, so half the value, but it prevents a sente play for an opponent, so double it again? I feel like you end up with something around 3/4.
Oh! I think I get it. You assess the value of all the moves and then go from highest sente down, and then any time you have sente go highest reverse sente down, and then highest gote down. But 7 points in gote > 3 points in reverse sente, so your reverse sente is not universally greater than gote.
Should I erase all this now that I understand? (if I do in fact understand)
Tirian: I don't think it's quite right to say that sente > reverse sente. Or perhaps it's just that "sente" might not mean what you think it would when there are reverse sente plays on the board.
I'm probably not good enough to get this right the first time, so hopefully some master player can see what I'm trying to say and correct the diagram if need be. In fact, the whole concept may be half-baked.
You might think to play the sente move at a before the reverse sente move at c, but when you play a you might find that White plays d instead and trumps your sente with a larger threat. If Black a, White d, Black b, then White plays e and you have lost more than you have gained. I think Black's only response is to finish the bottom edge allowing White to play b afterward.
Although I haven't seen it discussed outside CGT circles, I think that the value of a sente endgame move needs two values, the value of the sequence you expect to be played out, and the value of the follow-up move if the opponent plays elsewhere. Some moves are "more sente" than others even if they gain the same profit in the game.
Black likes reverse sente, because the sente it prevents white from playing is bigger than the sente black could have played. Now I wonder, what if black had 2 sente moves, which combine to be greater than white's sente move.
I spent half an hour trying to make a diagram to demonstrate this, but failed. It seems, at any rate, that it would be possible to compile hard and fast rules for what order one should play these moves. Is this done anywhere?
Bill: - gains 3 points, while gains 1 point. My reverse sente is bigger than yours. :)
Mef: Miai Counting gives a value for moves that takes into account sente values already, also most of the CGT pages deal with proper order of endgame moves. If all else fails you can play a difference game to give yourself a surefire answer.