Source: 
Beat the world champion and win a prize!
See details below.
Want to know a simple strategy that would beat the world champion? Get in line. But in an even game, the existence of such a strategy is unlikely, to say the least. But what if you, as Black, had some free-placement handicap stones? How many would you need?
Now go a step further. What is the MINIMUM number of handicap stones that would always GUARANTEE a win for a rank beginner, i.e., someone who just learned the rules? In the position above with 181(!) handicap stones, Black always wins, because White doesn't even have a legal move. But surely, a rank beginner with some simple strategy could guarantee a win with a LOT fewer than 181 handicap stones.
I'll simplify the problem scope even more by REQUIRING that Black, after placing the initial free-placement handicap stones, mirrors every White play. Mirror means that Black rotates the position of White's last play by 180 degrees about the board's center point, and then plays on that point. See the diagram below for a game where Black only uses mirror plays. White is always free to make any legal play.
SO THE PRIZE QUESTION IS: What is the minimum number of free-placement handicap stones and their corresponding locations, for which Black's mirror strategy always wins? Obviously, you must also make a reasonably convincing argument as to why the mirror strategy always works against perfect play, given your initial Black handicap stones.
The first correct answer will win 2 back issues of Go World magazine! Submit your response to the Problem of the Week Editor . Submissions will remain open for at least 2 weeks, but may remain open longer depending on the level and quality of responses. The Problem Editor's judgement is final.
Please read ALL the following important notes BEFORE emailing a solution!!
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If Black cannot, or does not, mirror a White play, Black
immediately loses. Assume though that Black can pass after
White passes, in order to end the game.
-
The following diagram shows why Black cannot simply place a
handicap stone on the center point, and then mirror White's
moves to win.
White 1 through 13 sets up a trap to capture Black's center
stones. After Black 14, White captures the 4 center stones.
But after Black mirrors White's capture play (below White 1),
White plays on the center point. Black obviously cannot
mirror the center point play, so Black loses!
Black needs more handicap stones to beat White!!
-
Assume Japanese rules. Note that the Japanese rules disallow
suicide plays, which White could use to prevent Black
from mirroring.
-
Assume territory scoring.
- Assume no komi. Also assume no jigo. Specifically, Black (not White) wins in the event of a tie, e.g., triple ko, eternal ko.
We have a winner!!
Only 3 people provided proposals that understood how to create
workable solutions. Dan Tashjian, along with a friend,
provided the minimal solution, while Jim Gonnella submitted a
proposal that, although not minimal, captured the essence of
the problem.
Dan's 157 handicap stone solution:
Many people sent in a proposal similar to the following:
While the two "wings", extending almost to the edge, prevent the
standard anti-mirror White strategy, White can simply play the
following line:
After Black captures 2 White stones with 12, White has not made
a corresponding capture! Next, White can play back into where
White 1 was. Black cannot mirror this move, and hence loses!
The main point of this problem shows how weak a mirror-only strategy is for Black, i.e., to be effective, Black has to almost smother the entire board with handicap stones.
Email the Problem Of The Week editor at potw@usgo.org
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