Mate in two (634)

sameez1

White in two    by I. Lacny 1st place, W. C. Composing Tourney, 1972-5

Arisktotle

A kaleidoscopic sort of problem with all of sorts of checkmates and temptations. Can't see a strong theme. Most action surrounds the choices of capturing on e7 and of course all the black pawns are very active. A very enjoyable problem to solve!

sameez1

Cyclic shift is what the author is calling it....."The extraordinary 634 shows a 5 fold cyclic shift of the defensive motives of five Black defenses  ( guard,capture of threat piece ,unpin,unblock,cutting White line of guard) as they defeat the different threats of try and key"......I seen the key on this one but still enjoyed finding the rest of it...If you were post the position with the key and stated mate in 1 move after Black moves what would you call the puzzle.

Arisktotle

That's the sort of modern "scheme" I am not strong at. It's like modern art. Once you know what it means, it starts to grow on you .... or not wink.png

If you create a problem where black moves first you should indicate it by adding half-a-move. Thus, #1.5. It is not common because they are easy to solve. However, when you state #1 and you can prove it is black's move then it is the same as #1.5.

 

Arisktotle

Edit: DEAD RECKONING

To illustrate the use of the half move numbers in the previous post, look at the following puzzle which consists of 2 parts. It's a cheeky puzzle, as you will have a hard time solving the 2nd part but I will explain it later. Note: "=2" means "forced draw in at most 2 moves" (might also be 1 move).

 

sameez1

this looks like a figure out how the W K got in there,...for the forced draw Pe4 white has one more move unless Black makes it none with E.P. capture or pe5  The 1.5 part I am trying to figure out

sameez1

I guess 1..... pf2  2.Pe4  f pawn captures e.p.

Arisktotle

Not quite! Black will not help by playing e.p. in the second part of the problem but instead promote: ... f1=Q 2. e5 Nc3#. But as said, the 2nd part is cheeky and actually quite tricky and advanced as well.

Let's first go back to part 1, the forced draw in 2 moves. This problem is a bit different from normal checkmate problems which always end with a white checkmate move. A stalemate problem can end with either a white or a black move while the assignment remains "=2" in both cases. That is because the optional black move is at the tail end of the solution. If black had been on move in this part then the assignment should have been "=2.5".  Note that in all cases white plays 2 moves, the 1st digit in the assignment. Something to remember!

The 2nd part is trickier. "=1.5" Indicates that black starts and white plays the next move. After that, an optional black move might follow at the tail end, but in all cases white plays just 1 move. However, there is a complication because you can prove that black cannot be on move in the diagram (white has no last move) which implies that white must start instead of black! This means that white must still play just 1 move, followed by an optional black move. You might say that "=1.5" has been changed to "=1" by the retrograde analysis (see my previous post). Q: Why does it not change to "=2"? A: Because the 1st digit always determines the number of white moves and it cannot change from "1" to "2". Conclusion: the 2nd part of the problem says you must draw in 1 move with white to start!

If you understand the above you have done your job and have a good understanding of full and half move assignments in common problems. What comes next is an advanced application of the rules and is taken serious by some and not by others. For me it is mostly fun though there are also actual chess rules involved.

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You can now see a contradiction between part 1 and part 2. How can there be different solutions for these 2 problems? If the "=2" problem has one 2-move solution then why wouldn't it also have the 1-move solution of the (adjusted) "=1" part? The reason it can't is because part 1 and part 2 are actually completely different problem types. Part 1 is a simple orthodox draw-problem but part 2 is a retrograde problem. And it is a retrograde problem because we had to do retrograde analysis to find out that black could not be on move and we changed that duty to white! In the first part that was no issue because there are many legal last moves for black for instance by capturing a white piece. Q; What does it matter that the 1st and 2nd part refer to different problem types? A: These problem types have slightly different rules. The difference is minimal but happens to be critical in this particular duo problem.

In the chess laws is an actual rule about dead positions and it applies to all chess games even the ones outside FIDE competitions. It says that a game terminates in a draw at the very moment it is sure that neither side can be checkmated even with the worst play. So, besides stalemates, that would count as a forced draw in problems with "=1", "=1.5" and "=2" assignments. Only it does not because the Composition Codex excludes the dead positions from "normal problems" with assignments such as given. However, the same Composition Codex includes the dead rule for retrograde problems. In casu, the dead rule applies to part 2 of my duo problem and not to part 1! How do I then solve the part 2 assignment? Not by starting with 1. Pe4 because the position is not yet dead after 1. ... Bd5! It is still possible that somebody gets checkmated and white must play 2. Pe5 to assure a draw. But that is one white move too many in part 2! On the other hand, after starting with 1. Pxf3! the game is certain to end in a draw whatever moves are played and therefore immediately ends as a dead position after 1. Pxf3! Note that part 1 cannot be solved in this way because there is no dead rule and it would take 3 moves to reach a stalemate position when black refuses to cooperate.

As said, you are forgiven if you don't understand the latter paragraphs but you may decide at some point in your life to look into the retrograde issues! Good luck!