20 April 2012

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.

  1Andrew Buchanan  
Internet, 2011
[2k5/Pp1p4/P7/p7/p7/p1P5/PP1P4/1KB5]
  Add one unit(8+6)  

  2Andrew Buchanan  
Internet, 2011
[8/PPP2k2/P7/P5K1/p4P2/pp6/pp6/8]
  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.

  3Andrew Buchanan  
Internet, 2011
[4k3/Pp6/P7/P7/p7/p7/pP1P4/b3K3]
  Add one unit(6+6)  

  4Thierry Le Gleuher  
Internet, 2011
[8/2p4p/2p5/1Pp5/bpp5/qkPP4/p1pPP3/K7]
  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.

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