Another puzzle from Dragoslav Andric's 1981 book “Matni Udar”.

**White to play and win**

Alekhine v Kussman, simul New York 1924

**Solution**

The first move, and Black's best reply are obvious. 1 Qb5+! Nd7

It is then not a hard task to see that there is no forced mate, but after 2 Rfe1 White's advantage is overwhelming, and White needn't calculate further. Black has a few replies. If say 2…Be7 then 3 Ned6+ and if 3…Kf8 then 4 Re7 1-0. Or if say 2…Bb4 then 3 Nf6++ and 4 Nd7+ 1-0

A break for a day. A few weeks ago, whilst on holiday in Turkey, I posted this position on my Instgram, Facebook and Twitter (@allanbeard) feeds.

Below, I show the position in more convenient format.

**Black to play and mate in 2**

NN v Kokshal, Prague 1928 (source a 14/5/2003 article by Mark Dvortesky on Chesscafe.com)

I would have posted the position sooner, except that I typically do my daily posts a week or two in advance, and to then insert a new posting is time consuming (a new daily posting: perhaps I should have just posted it same day).

The solution is pretty: 1…Rd4+! and after either 2 Kd4 Qc4 is a smothered mate, or 2 Qd4, 2…Qf5 is a different smothered mate.

The initial position is also instructive. It is part of a series of positions Dvortesky gives on a theme of 'double pin' or he says 'Maltese cross'.

**Black to play and win**

1…Qd2+! 2 Kf3[] Rd3+ 3 Re3 (3 Kg4 Qf2 4 Qe5 Rg3+! 5 Qg3[] Qf5 mate) 3…Qe1!! The Maltese Cross.

4 Qe5 Qf1+ 5 Ke4 (5 Kg4 Re3 6 Qe3 Qf5 mate) reaches the initial diagram position.

If instead 4 g4 then there is another Maltese Cross: 4…Qf1+ 5 Kg3 Qe2!!

(5..Rd2 wins more prosaically). If (after 5…Qe2) 6 Kf4 then 6….Qf2+ 0-1.

Yesterday, I posted this problem, from Dragoslav AndriÄ‡'s book Matni Udar, the puzzle book that I am presently working through.

**White to play and mate Black **

Hartlaub v Walle, Bremen 1923

I gave the solution, as in the book, 1 Re7 Qe7 2 Rf6 gf 3 Nf5 Qe4 4 Qg5+!! 1-0. The book also gives the alternative line 3…Qe6 4 Qf4 Kh8 5 Qh6 1-0).

Alas, something is wrong. Stockfish, the iPad app I use for posting the diagrams, immediately flashed up 1 Nf5! with a +5 assessment; giving only a +2 assessment to 1 Re7, and swinging to a -4 assessment after 2 Rf6?.

Once I saw the assessment, I looked to understand the position after 3…Qe6. It immediately became clear that 4…Rd8! gives Black's king sufficient luft: there is no mate or N fork after 4 Qg4+ Kf8 5 Qg7+ Ke8[].

Instead, better, after 1 Re7 Qe7 is 2 Bf6! gf[] 3 Nf5 so that after 3…Qe4 or 3…Qe6 White plays 4 Qh6 and mates unless Black plays 4…Qf5 5 Rf5, after which White has an easy win in the endgame.

But best of all, simply play 1 Nf5 and after say 1…Nc6 (to defend the Be7 and develop) 2 Qg5 is overwhelming. Black loses a piece after 2…g6 (or, in fact, it is mate in 6, per my engine: 3 Nh6+ Kg7[] 4 Re7! etc).

So the problem isn't cooked, but the solution was, but the spoiler was well hidden.

Another puzzle from Dragoslav Andric's 1981 book “Matni Udar”.

** White to play and mate Black (with a twist)**

Hartlaub v Walle, Bremen 1923

**Solution **

'Fairly' standard today: just a bit of calculation required to put the moves into order, and one pretty move at the end. For what I mean by 'fairly', see tomorrow's post.

1 Re7! Qe7[] 2 Rf6!

2… gf[] 3 Nf5

3…Qe4

4 Qg5+! 1-0 since 4…gh 5 Nh6 mate.

Very pretty.

**Puzzle**

What's the twist?

Another puzzle from Dragoslav Andric's 1981 book “Matni Udar”.

**Black to play and mate White**.

Serbasin v Sozin, Novogorod 1923

**Solution**

The first problem in the book that I totally failed with: 1…Qb6!! is the stunning first move. Of course, I had similar motifs in my mind (the b line, the jump-check Rb1, the jump-check Bh6-c1) but didn't have the imagination to see this move.

The move is a double attack. On the LPDO Be3 and LPDO Pb2: so 2 Bb6 Ne2 mate.