On the traces of Giegold (2)

Let's have another look into my own history book. Here some more of my compositions. Enjoy!

1 Gerson Berlinger   32er 7/1988   #4 (10+4)

2 Gerson Berlinger   32er 12/1988   #5 (10+4)

No. 1 has the try 1. Be4? h6? 2. Bb1 hxg5 3. Rc2 Ke4 4. Rc4# that is naturally refuted by 1. - dxe4! The solution goes 1. Be2! h6 2. Ba6 hxg5 3. Bb7 Ke4 4. Rc4#. Too bad that there is no way to convert the try into another variation. No. 2 is a nice puzzle, I think. 1. Ka4! Kd4 2. Rc1 Ke3 3. Rd1 Kd4 4. Nc4 Kxc5 5. d4#

3 Gerson Berlinger   Schach-Echo 12/1988   #5 (8+4)

4 Gerson Berlinger   ?   #3 (8+9)

The third chess problem shows two white pieces "going home": 1. Bh3! h5 2. Bf1 h6 3. Ng1 Kd4 4. Nf3+ Kc5 5. d4#. The last diagram is a bit of a mystery. When I saw it some time ago here, I immediately knew that it was composed by me. For example, I remembered how I tried to do better with those white pawns on h5 and h6. The really strange thing is that I have no other memory or any notes about when and where it was published for the first time. I only know it was reprinted in a newspaper in 2004 (Stadtzeitung Rieselfeld 12/2004 Nr.34). The comment for that diagram states it was published in a German TV magazine somewhere in the 1990s. But I am sure that also was a reprint. Can you help me with that issue? The solution is 1. Ra4! (threat 2. Bd4+ Kxd4 3. Qc3#); 1. - c5 2. Bb8+ Kxf6 3. Ra6# or 2. - Kd4 3. Qc3#; 1. - f3 2. Qxe3+ Kxf6 3. Bd4,Qd4,Qe7#.

I am sorry

Once again, I didn't manage to write more than this ... it's too hot and I am too busy, excuse me.

Enzo Minerva
The Problemist Supplement 09/2009
#1   How many solutions?	(8+9)

Okay, did you see that the position is illegal? Good. You have to remove a black pawn to make it legal. But you have to be careful. Taking away bPd5 in order to play 1. d5# or removing bPf5 with the intention 1. f5# are no options — the position is still illegal! But you can dispose of any of the remaining pawns:
I) -bPd7: 1. Qc8#
II) -bPd6, 1. Nc5#
III) -bPe7, 1. Qe8#
IV) -bPe5, 1. Re1#
V) -bPf7, 1. Qg8#
VI) -bPf6, 2. Ng5#
So there are six solutions.

Not much

Sorry, today I have just some moremovers that I find interesting. Maybe, you like them as well.

1 Kay Soltsien   Kieler Nachrichten, 1956   #4 (8+5)

2 Wilfried Neef   Badische Neueste Nachrichten, 1999    #5 (15+2)

3	V. Savchenko
Shakhmatnaya Moskva, 1970
1st Prize
#8	(13+12)

Solutions

1	1. Rd5! c6 2. Ra5 c5 3. Bxc5 Kxg5 4. Be3# 1. - c5 2. Bc1 c4 3. Rd2 Kxg5 4. Rd5# Twice the Indian theme with changing funcions of rook and bishop.
2	1. Ng7! Bxg7 2. Rf6 Bxf6 3. Be5 Bxe5 4. Rc7 5. Bb7# 2. - Bh6 3. Bf4 Bxf4 4. Rc7 5. Bb7# White must prevent the black bishop from going to b6, so that the wPa7 would not be protected anymore. A funny composition.
3	1. Rd1? Nd2! 2. Kxd2 gxh5 3. Nxh5 h1=Q 4. Rxh1 Ng5 5. Kd3 Bf1+ 6. Rxf1 Nf3 7. Bd4+ Nxd4 8. Re1+ Ne2 9. Rxe2# and White is just one move too late. 1. 0-0-0! axb2+! 2. Kc2! b1=Q+ 3. Kc3!! Qb2+! 4. Kd3!! Qe2+! 5. Kxe2 f3+ 6. Ke1 Nd2 7. Nd7+ Ke4 8. Bc2# Fantastic journey of the white king back to his original square.

Colliding with the masters

This one is about three ideas which not only I had.

The first is the flight task which I already mentioned in a post last year. My aim was to compose a threemover miniature where the black has eight flight squares after the key move. The great Shinkman even managed to use a white queen to accomplish this. The keymove of his 1a gives five more flight squares. Depending on Black's play queen and rook change their functions. Unfortunately, there are several duals on move three. You may judge yourself whether and how much this adversely affects the chess problem. 1. Rc2! (threats 2. Rxg2+ 3. Q#) g1=Q 2. Qxg1+ 3. R# is the solution. My 1b is probably easier to solve, but there are no duals: 1. Rc2! ~ 2. Rg2(+) Kf~/Kh~ 3. Rf1/Rhxh2#. I didn't know Shinkman's composition at that time, that's all I can say.

1a William A. Shinkman   Checkmate 12/1901   #3 (4+3)

1b Gerson Berlinger   Der Tagesspiegel, 1994   #3 (5+2)

The Hinterstellung has always been one of my favourite themes. I once had found the very same position that you see in No. 2. Too bad. 1. Rh2! Kxa4 2. Bc6+ Kxa3/Ka5 3. Bc5/Rh5#.

2	William A. Shinkman
Checkmate 07/1903
#3	(7+2)

The third part of this post has quite a long history. It starts back in the early 1990s when I bought my first PC. Until then, already a bunch of my chess problems had been published. And there was still a lot in the queue or in the making. All this stuff only existed on paper and I started to collect the data in text files and databases. Of course, from time to time, you buy a new personal computer and transfer some of the files that have accumulated to the new machine.

Apparently, you don't always know what stuff you keep throughout the years, right? So, rather by chance, in Summer 2010, I detected some old ChessBase databases with which I had worked in 1995 for the last time! Quite a nice surprise, as some of them contained ideas and composing attempts where I got stuck. I hope I can further develop those rudementary orthodox moremovers and endgame studies. I've already started doing so with the aim to show a march of the white king from one corner to another, e.g. from a1 to h1 or a8 but not to h8. For some months, I experimented with diferent setups and also discussed the subject with the young Dutch problemist Jokim van den Bos who published his best result in Probleemblad in April 2011. I still was not satisfied yet with my own approaches but had less time on my hand to search further. Some months later, though, I made an important discovery. Look at the following moremover which could be an endgame study as well.

3	Jonathan Levitt
? 1996
#18	(7+10)

You can also find this composition online on his web site. First the black king is driven across the bottom rank. 1. Bd5+! Kg1 2. Bc5+ Kf1 3. Nd2+ Ke1 4. Nf3+ Kd1! 5. Bb3+ Kc1 6. Ba3+ Kb1 7. Nd2+ Ka1. In the second phase the white king must also make a corner to corner king march. 8. Bc2 Rxh7+ 9. Kg8 Rh8+ 10. Kf7 Rh7+ 11. Ke8 Rh8+ 12. Kd7 Rh7+ 13. Kxd8 Rh8+ 14. Kc7 Rh7+ 15. Kb8! Rh8+ 16. Kxa7 Ra8+ 17. Kxa8 Now, there is nothing left that can delay the mate 18. Nb3#.

After having learnt this, it doesn't seem to make much sense to further look for a corner-to-corner king march, does it?! Anyway, that really discouraged me for a while. Nevertheless, I won't give up until I have found my version of that theme, I promise.

Go All The Way

Wow, this is already my 100th post. Time goes by. 100 ... reminds me of the neutral pieces which were introduced in 1912. They are those chessmen that belong to that side which chooses to use them. Do you remember? It was exactly one year ago today that I mentioned them in a blog post.

Now the question of the day: Is it such a bad idea to Go All The Way? That depends, right? How about chess problems with everything neutral? As for me, I am absolutely fascinated by them. Others, even hard-core fairy problemists, prefer not to touch such compositions at all. But are they really so alien? Let's start with a simple diagram to see what is allowed and what is not.

Legal moves? (0+0+3)   a) White to move b) Black to move

Though there are only neutral pieces on the board, we still have to think in black and white. That may be the first psychological barrier. It's important to know which side has the move.

a)

The nK must not go to e6 or g6 due to an illegal self-check (the nPf7 has to be considered to be black)! For the same reason nBh7 is not allowed. Neutral pieces can capture each other. Neutral pawns promote to neutral pieces. nKe4,e5,f4,f6,g4,g5; nBxf7; nPf8=nB,nN; nPxg8=nQ,nR,nB,nN are legal.

b)

This time, the nK can go anywhere he likes. The moves nKe6,g6+ are no self-checks! Of course, they are checks to the neutral king thereafter. He is then considered to be white and the nPf7 to be black. But that's not illegal, for it is White's turn and so the check is parried by moving the king away (e.g. nKf6, but not nKxf7) or by making a legal pawn move (nKe6: nPxg8=nN,nR; nKg6: nPf8=nQ,nR,nB or nPxg8=nB,nN). Legal moves are nK~(+); nBxf7; nPf6.

I hope you are prepared for some real chess problems with neutral pieces only. All of them result in checkmates given by a pawn. I recommend that you take a closer look at the mate positions. Observe how and why it is impossible for those pawns to move away and thus spoil the checkmate in each case.

1 Henning Müller      The Problemist, 1988   h#3* (0+0+5)

2 Henning Müller   Hans-Peter Reich   Sinfonie Scacchistiche 12/1988   h#3   2.1... (0+0+6)

3 Jerome Auclair   harmonie 04/1993         h#3   Duplex (0+0+6)

4 Manfred Rittirsch   The Problemist 05/1988 Dedicated to Walter Wittstock 1st Prize   h#4   0.1... (0+0+7)   a) Diagram b) nNe3 → f8

Solutions

1	*1. - Ka5+ 2. b3 Nb4 3. c3 cxb4# 1. Qb8 Ka5+ 2. Ka6 Nc3 3. Nb5 cxb5# A perfect echo that results from the fact that it is not possible to make a tempo move. Also notable are the two ways to answer the check. In the setplay the pawn moves, in the solution the king does.
2	I) 1. nPd1=nR nPf7 2. nRd7 nPf8=nB 3. nBd6 nPxd7# II) 1. nPd1=nQ nPf7 2. nQd6 nPf8=nN 3. nNd7 nPxd7# Sure, the ever-present Allumwandlung mustn't miss! A very nice composition.
3	b) 1. nPd5 nQa7 2. nPd4 nBd5 3. nPd3 nPxc4# w) 1. nBd5 nPc2 2. nRc6 nPc1=nQ 3. nQa3 nPxc6# Once again the pawns are starring! In both solutions the nPd7 gives mate. The other pawn is also very active.
4	a) 1. - nNeg4 2. nPf5 nKg6+ 3. nKh7 nBd4 4. nPg5 nPxg6 e.p.# b) 1. - nQb8 2.nKf6+ nKe7 3.nPg5 nBh4 4.nPf5 nPxf6 e.p.# Beautiful ideal mates with en passant capture.

Logical

I am sure you've heard about the Logical or New German school — and hopefully it was in connection with chess problems. Searching the Internet in order to get more information about this subject might rather lead you to articles that refer to music. And adding the search word "Problem" will not really be helpful as this gives you links to sites dealing with certain aspects of social science. In fact it's quite hard to get the desired information at all, the more if you have no concrete idea what you're searching for. Anybody interested in Web Content Mining?

The chess problems I have chosen for this post are logical. That term refers to the way in which the solution is structured. In the initial position, White has a so-called main plan, a series of moves with which he wants to checkmate. At first, the execution of this plan fails to a refutation that Black has up his sleeve. Therefore, White first executes a foreplan, whereby Black's defence to the main plan is negated in some way. Roughly speaking, that's what logical problems are about.

Let's start with a lightweight example.

Jörgen Möller Skakbladet, 1920   #3 (4+5)

The main plan is to play the queen to the b-file and give checkmate on b8. But the attempt 1. Qb1? Bg3! is premature. Another idea that fails is 1. Qg2+? d5!. 1. Qg7! (2. Qxd7 3. Qc6/Nb6#) Be7 2. Qb2 Bd6 3. Qg2# The bishop was lured away to another diagonal and thus was forced to replace the good defence Bg3 with the bad defence Bd6. This caused the interference with the pawn d7. This theme is called Interference Roman (translation found here).

Stefan Schneider Austria-Switzerland Match, 1977 1st Place   #5 (5+15)

1. Re7? with the idea 2. Qxe2+ Bxe2 3. Rxe2+ shows that queen and rook are the wrong way around. Hence, the main plan must be Qe8, followed by Re7 and Rxe2+. But how to do it? 1. Qe8? is much too slow, for the black queen can leave c1 not being bound to defend against Qxd2# anymore. We must find a way to play Qe8 with gain of time — very hard to imagine. Here we go:

1. Ka8!!

Wow! True, an experienced solver might consider this move even without exactly knowing how to proceed thereafter — it's the feel. Otherwise, deep thinking is required.

White now threatens 2. Rh7! 3. Rxh2 4. Ng2#. The point of going to a8 is to avoid 2. Rxh7 being check. In doing so the king has to avoid b8 because of the potential pin by the black bishop coming to h2. Why 1. Kc8? is wrong will be explained later.

1. - Rh8+

Black is so tied down that this is the only way to challenge White's idea. White continues with another incredible move:

2. Qe8!!

2. Rxe8+ 3. Ka7!
Not 3. Kb7? Rb8+ 4. Kxb8 Bh2! This also makes clear that in case of 1. Kc8? the white king would not have been able to escape the h2-b8 diagonal in time.
Black can now no longer guard g2, so that after
3. - Ra8+ 4. Kxa8
there is no way to prevent 5. Ng2# or (when Nf2 moves away) 5. Nd3#.

Alternatively, the rook can only return.

2. - Rh2

By now, we have achieved our goal, so that the main plan can be executed.

3. Re7 ~ 4. Rxe2+ Bxe2 5. Qe2#

Y. Vladimirov Macleod Memorial Tourney, 1994 1st Prize (Version)   #17 (10+7)

This one is probably easier to solve despite its length. White's main plan is Rf8 followed by Rc8#, but Black is stalemated. The foreplan is to build a bishop-rook battery on g1,f2 with the black king on c5 and play Rf8+. But this requires the protection of the pawn d6, so that playing the pawn e2 to e5 is another foreplan. Important things to observe: White always has to a) protect d6 and b) give a check when the king is on c5. See and enjoy how all this is accomplished. 1. Bc1 Kc5 2. Be3+ Kc6 3. Bf4 Kc5 4. Rf5+ Kc6 5. Be5 Kc5 6. Bh2+ Kc6 7. Rf6 Kc5 8. Bg1+ Kc6 9. e3 Kc5 10. e4+ Kc6 11. Bh2 Kc5 12. Rf5+ Kc6 13. e5 Kc5 14. Bg1+ Kc6 l5. Rf2! Kc5 16. Rf8+ Kc6 17. Rc8# We see a variety of self-interferences employed by White to relieve the stalemate. A very nice puzzle.

A.Lobusov & A.Spirin E.Zepler Memorial Tourney, 1985 1st Prize   #6 (9+11)

Black to move allowed White to mate on d4 or d5. But there is no waiting move. The goal of the foreplan is to lose a move. 1. Ne8! (threat 2. Nf6 3. Ng4#) Ng8 2. N6c7! Nfe7 3. Ng7! White now threatens 4. Nxd5+ and 5. Nf5# or vice versa. 3. - Nf6 4. Nce6! and 5. Nxd4# or 5. N(x)f5#. 3. - Nxh6 4. Nge6 Nhf5 5. h6! There is the waiting move! The diagram position, without the pawn h5, has been repeated. It is Black to move and mate follows on the next move. Did you notice? During the solution, the two white knights have swapped places. Moreover, the black knights also performed such a Platzwechsel!

Not enough

This is a continuation post of Who's missing. Again, I invite you to have a look at some diagrams that ask you to add one unit. All of them are legal positions of the so-called type A, which means that no king is in check and that it is not stated who has the move. Naturally, such chess problems are always a constructional challenge with the main goal to find the most economic position.

1 Andrew Buchanan   Internet, 2011   Add one unit (8+6)

2 Andrew Buchanan   Internet, 2011   Add one unit (7+6)

In No. 1, the white pawns have captured nine black pieces and the black bishop from c8 never left this square. Therefore, the piece to add has to be white. Looking at the pawns on the g- and h-file, we see that the black pawns must have promoted, so that the white pawns could capture them on their way to a6 and a7. The black pawns on the a-file have captured six white pieces, so one white pawn from g2 or h2 must have promoted to be captured, the other had to be captured by a black pawn on these files. The only white piece that could not have been captured is the white rook from a1. It can only be added on a1: +wRa1.

A closer look at the second problem reveals that there have been seven obvious captures by white pawns and at least eight captures by the black pawns. Considering the pawn formations, we see that there must have been two more captures by one side. Counting the missing pieces tells us that this must have been white captures. Two white and one black pawn(s) on the g- and h-file have promoted, the fourth (black) was captured by a white pawn. As all captures are now resolved, we see that only the black f-pawn is left. It has not captured, thus never left its file. The only possibility to add it is: +bPf5.

3 Andrew Buchanan   Internet, 2011   Add one unit (6+6)

4 Thierry Le Gleuher   Internet, 2011   Add one unit (6+11)

As the logic of No. 3 is quite similar to that of the first two, I leave it up to you to prove that the solution is +wBc1.

The last one is a little bit different. The black pawns captured all the missing white pieces. This means that the white h-pawn had to promote and, as a consequence, it had to capture once to bypass the black pawn h7. Moreover, there are the three captures which you immediately can spot by looking at the diagram. Hence, there is one black unit left that can be added. To resolve the position, it is necessary to put this unit on b1. Of course, it's not a black bishop, as all eight black pawns and another bishop on white square are already there. Thus, we have to add a black knight on b1. Then, we can play back for example: 1. - d7xQ,R,B,Nc6 2. ~ Qc1-a3 3. ~ Ka3-b3 etc.