Cecil's Saturday Puzzle - April 12, 2008

Berger - White to mate in three

Let's try the obvious idea, although I can easily see it mates in 4, not 3.
1. Qd3+ Kb4 2. Rb1+ Ka4 3. Qc2+ Ka3 4. Qb3 mate.

So let's cut off the king.
1. Rb1 e5 2. Qd3 mate
1. Rb1 Kd4 2. Qd3+ Ke5 3. e3 c4 4. f4 mate

So that doesn't work, perhaps we don't need the rook to the b file immediately.
Let's try
1. Rg1 Kxb5 2.Rb1+ Ka4 3. Qa8 mate
1. Rg1 Kd4 2. Qd3+ Ke5 3. Rg5+ Kf4 4. Qe3 mate

Hmm, let's look back at the line where we had to play 3.e3 to support 4.f4 mate.
What if we played 1.Rf1 to support a potential f4.

1.Rf1 e5 (takes away the square from its own king) 2.Rb1 Kd4 3. Qd3 mate
1.Rf1 Kxb5 2. Rb1+ Ka5 3. Qa8 mate
1.Rf1 Kd4 2.Qd3+ Ke5 3. f4 mate (and the point of Rf1 !)

Note: In the absolutely strange coincidences department - a few hours after posting the above - I was browsing through En Passant # 82 from December 1986 - and this problem is in there.



Also, here is the updated link to the Magnus Smith article.