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MatPlus.Net Forum General 4*4 Is All Mate In Due Time

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4*4 Is All Mate In Due Time

Are there any known problems in which two or more pawns promote and all combinations of promotions lead to checkmate in a certain number of moves?

I managed to make one in which 3 promotions of the first pawn works with 3 other promotions of the second pawn, and all 9 combos are a mate in 5. Try to find all of the solutions without computer assistance! Each side plays optimally of course.

(= 6+9 )

White to play and mate in 5 moves in 9 different ways.

Incorrect, as there are several 3rd moves: 3. Kb8, 3. Qd7, 3. Qb8, 3. Qd8 (the variation with the three promotions).

[ deleted -- misunderstanding ]

Sure there are several third moves, but those are besides the point.

Also, I did use Stockfish and the Aprons computer to validate my position, so there’s that. Sure computers aren’t exactly entirely accurate but I’ll give it a pass this time.

Also, also, White would not make any secondary moves as they are to checkmate here in 5 moves specifically here.

BLANK

> Sure there are several third moves, but those are besides the point.

I can't see any point here. This is simply an incorrect #5 with several cooks.

The I’m no expert in chess problems yet. Maybe there are more mate in 5s then I thought. This is a themed problem. White shall go along with the theme. You still can’t at least try? It’s really not that hard.

And the theme here is specific, so it’s not too hard to figure out what White needs to do. White shall not make those side moves and as such there are no cooks for at least white.

Perhaps there are other mates in 5. Heck, Shinkman’s mate in 8 was found to have an alternative. Find those if you want.

Also, I did find another mate in 5 that works, and is the only one that works, depending on if black makes another move. But since that move by black would not be the strongest, it wouldn’t count in a sense. Feel free to find it though.

> ... White shall not make those side moves and as such there are no cooks for at least white.

Sorry, Rewan, but chess is an objective play and not a request show.
Please refer to the Codex of Chess Compositions (www.wfcc.ch/1999-2012/codex/" target=_blank>https://www.wfcc.ch/1999-2012/codex/), Article 10 – Dual:
"A dual is said to occur if, after the first move, there is more than one method of satisfying the stipulation."

In given position White may play, f.ex.:
1.a8Q Q:a8 2.K:a8 Ba6 3.Kb8,Qd7,Qb8,Qd8 with mate in 5th move.
These moves are clear duals.
And you CAN NOT simply prohibit them.

Furthermore a chess problem should have some kind of surprise, a trick, an effect, a clear idea etc. Here White simply moves the a-pawn (what else due to 1.- Q:a7#)?) and then almost all moves lead to mate.

Well the Qb8 and Kb8 are the only line one that don’t involve two pawn promotions-the Qd7 and Qd8 lines do. So Qb8 and Kb8 are the only cooks. Also, there more options then a8=Q.

So, yes, it is cooked, I must agree, but I still like my problem.

And Qd8 was the line I chose for my 9 solutions. Qd7 isn’t a dual-it’s just something that I had not thought of.

Why not try to find all the mate in 5s within the given stipulations though for fun?

You should give Rewan the "newbie" bonus, this is not a #5
but rather a "Bedingungsaufgabe" or a mathematical construction.
I seriously doubt that 4*4 (and *only* that) is constructible,
but still, it would interest me what is possible, say, in
the h# genre.
@Rewan: You surely be interested in all kinds of promotion
by task master Sir Jeremy Morse. Or the rook promotion orgy
by Karlheinz Bachmann in the SCHWALBE.

That’s an interesting name. And what “newbie bonus?”
Plus, whaf is the #h theme exactly, and is there a website with chess problem terms somewhere?

Is this the rook problem that you speak off?: www.schachbund.de/problem/problemschachaufgabe-352.html" target=_blank>https://www.schachbund.de/problem/problemschachaufgabe-352.html

And where may I find the 6 queens one?

a) Problem Chess, as any puny hobby-horse :-), has loads of
conventions. (Main: The codex quoted by Frank.) The newbie bonus
is that we don't immediately flame you to hell if you are unaware
of it as a newcomer to problem chess (in the sense of: not just
math chess constructions, which is a sub-area).
But be aware that even in this forum sometimes the temperature gets hot ;-)

b) I'm not enough familiar with h#, but it seems easier to me
to show a combination of promotions. I also don't know specific
theme names.

c) About all online chess problem databases are searchable.
E.g., try http://www.yacpdb.org and enter "promotion, task" in
the theme field to get 128 promotion tasks. (There are more,
tasks as well as databases :-)

d) Yes, that's *a* R task, although AFAIK he increased the record to 5 or 6.
Would have to research that one longer.

c) 6Q: http://chesscomposers.blogspot.com/2012/12/december-10th.html
scroll down somewhat.

Hauke

6 rook promotions in #8: https://pdb.dieschwalbe.de/search.jsp?PROBID=P1008344

Thanks for all of that info Hauke!

PDB sometimes plays silly bugger. I post it directly:

(= 11+7 )

K. Bachmann, Schwalbe 9952 (170 April '98) #8

Thanks!

Also, here is the all of the mates in 5 that I know of for my position. I’m counting different promotions as different lines, so the total racks up to 39 here.

The = signs indicate the type of promotion made by the pawn promotion on white’s first move. The annotations Qb8, Qd8, Qd7, and Kb8 that are by themselves indicate white’s third key move as a manner of organization.

The lines with Qd8 are the lines that I originally spoke of when I started this thread.

Qd8

=R

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=N+ Ka6 5. Qb6#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=Q+ Ka6 5. Qxb7#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=B+ Ka6 5. Bxb7#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Na7#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Nd6#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Qxb6#

=Q

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=N+ Ka6 5. Qb6#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=Q+ Ka6 5. Qxb7#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=B+ Ka6 5. Bxb7#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Na7#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Nd6#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Qxb6#

=N

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=N+ Ka6 5. Qb6#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=Q+ Ka6 5. Qxb7#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Bb5 4. c8=B+ Ka6 5. Bxb7#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Na7#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Nd6#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd8 Kb5 4. c8=N b6 5. Qxb6#

Qd7

=Q

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Na7#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Nd6#

=R

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Na7#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Nd6#

=N

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Nd6#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qd7 Kb5 4. c8=N b6 5. Na7#

Kb8

=R

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Kb8 Kb5 4. Ka7 b6 5. Qxa6#

=Q

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Kb8 Kb5 4. Ka7 b6 5. Qxa6#

=N

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Kb8 Kb5 4. Ka7 b6 5. Qxa6#

Qb8

=Q

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. Ka7 b6 5. Qxb6#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Na7#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Nd6#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Qxb6#

=R

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. Ka7 b6 5. Qxb6#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Na7#

1. a8=R Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Nd6#

1. a8=Q Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Qxb6#

=N

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. Ka7 b6 5. Qxb6#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Na7#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Nd6#

1. a8=N+ Qxa8+ 2. Kxa8 Ba6 3. Qb8 Kb5 4. c8=N b6 5. Qxb6#

Correct link as intended by Joost de Heer:
https://pdb.dieschwalbe.de/search.jsp?PROBID=%27P1008344%27

EDIT: Strange, now the link above in Joost de Heer's posting also works for me, but it did not five minutes ago. Must have been a hiccup serverside.

QUOTE
1 - P1008344
Karlheinz Bachmann
9952 Die Schwalbe 170 04/1998 (= 11+7 )
(K. Bachmann). 1. e8=T! Txf5 2. f8=T! (f8D? Tf3!) Tc5 3. c8=T! (c8D? Tc6+!) Ta5! 4. a8=T Td5! 5. d8=T Tb5 6. b8=T T beliebig 7. TxT/KxT Txg3 8.Sxg3 (1.- T:e2 2.T:e2 T:g3 3.D:f2 Tg2 4.b8D(4.b8T, ,a8D,L) S:e2 5.Df3 Sg1 6.a8D S:f3 7.D:f3 Kg1 8.Db1). Das wurde nur von wenigen gelöst: "Konsekutive sechsfache T-Umwandlung im orthodoxen Mattproblem! Super!" (MWa). "Interessante Idee, aber ein Wust von verwirrenden Varianten, ist nicht mein Fall!" (KHS) [Entsprechende schwarze Turmzüge können 14 weitere dualfreie T-UW-Folgen auslösen] KEw: "(..) ein erstaunlicher Task und trotzdem nicht nach meinem Geschmack. Wäre die Verhinderung der Nebenvariante nicht einen sBe3 wert gewesen?" [NL 1.e8D s.u.] MOe: "(..) Die Konstruktion ist einfach sagenhaft, zweitklassige Aufgaben solcher Art brauchen ein riesiges Kohlebergwerk um den sK. 5/IV". - Nicht 1.e8D!? Te4! 2.d8D Tc4 3.c8D Tb4 4.b8D Tg4! 5.Dcc7(D:g4??; 5.f6 Th4+!) Te4! 6.Df4 T:e2! 7.De:e2 T:g3 #9 bzw. (1.e8D Te4) 2.f8D Ta4! 3.Dh5 [4.D:h3] Ta1! 4.D:a1 f1D 5.D:f1 T:e2 6.Df:e2 S:e2 7.Df3+(D:e2 Kg1 8.?) Kg1 8.Dc5+ Sd4! #9. - Im Diagramm zwei Auftritte aus dem Vorprogramm; Kraemer/Grasemann: 1.b8T Td2 2.d8T Te2 3.e8T Tc2 4.c8T Ta2 5.a8T T beliebig 6.T:T S beliebig 7.Sh3; itshape Duale 3.e8D und 5.T8b..e1; Bachmann: 1.e8T Td6 2.d8T Tc6 3.c8T Tb6 4.b8T Ta6 5.a8T Ta beliebig 6.T:T/K:T T beliebig 7.S:g3. endwindow
play all play one stop play next play all
SBD: Nach 4. a8=T muss Schwarz nicht 4. .. Td5 spielen, es geht auch 4. .. Ta5:

5. b8=D Td5 6. d8=D Td1 4. Ddxd1 Txg3 5. Sxg3#.

und auch 4. .. Tb5:

1. b8=T Ta5 2. a8=T Td5 3. d8=D Td1 4. Ddxd1 Txg3 5. Sxg3# (2012-1-10)
comment
Keywords: konsekutive Umwandlungen 6 (TTTTTT)
Genre: n#
Input: Gerd Wilts, 2003-6-29
Last update: Olaf Jenkner, 2012-3-4 more...

Thx Siedfried! The link doesn’t seem to work for me though still. Eh well though!

(19) Posted by Rewan Demontay (Real Name: J. Malcom) [Friday, May 24, 2019 03:14]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-24]

Also, I believe thaf I may have found a fix for my problem if I fix the given stipulations and the position a little, both of which I have done. I shall use the computer for verification.

Here is my revised problem. Consider a “Bedingungsaufgabe,” as Hauke Reddmann argued, instead an actual mate in 5 problem.

As far as I can tell, from reading the Codex, if White is stated to have to do something in the stipulations, then they have no other choice but to follow it. I believe that the fact maximummers exist back me up here (if Black can be stipulated to move in such a fashion, what’s preventing me from making a few of my own restrictions sp long as they are clear?)Now there will be no more cooks. I even managed to add in more 3*3 solutions.

The Position

(= 7+9 )

Stipulations:

1. Black only can play variations in which Black tries to be checkmated as late as possible or not at all and White tries to checkmate Black in the smallest number of moves Black also plays their best moves as well.

2. White must promote both pawns,

3. The only moves that are allowed for White are ones that allow the three promotions of the first pawn to all work with three promotions of the second pawn.

4. All possible solutions must be mates in 5. No more, no less.

Now that the stipulations are done with, try to find as many of the possible solutions as you can without using a computer. I have calculated that there are 27 possible solutions for this, as I count different promotions as different solutions. I argue that since I know all of the possible solutions there are thus no duals.