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MatPlus.Net Forum General 4*4 Is All Mate In Due Time
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(21) Posted by Siegfried Hornecker [Friday, May 24, 2019 09:01]; edited by Siegfried Hornecker [19-05-24]

The problem is easily cooked.

You must promote Pa7, otherwise I play 1.-Q:a7+. However, 1.a8B is no mate in 5 (1.-Qa7 mate). And 1.a8Q Qa7+ 2.Q:a7 mate (forced) also is mate earlier than five.
What does "play optimally" mean? Obviously it means that Black will play anything to make the mate earlier or longer than 5 moves. You need to re-phrase condition number 1.

Here is my suggestion: "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."
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(22) Posted by Hauke Reddmann [Friday, May 24, 2019 12:45]

Well, technically Siegfrieds last sentence is assumed by default in n#?!

@Rewan: Try to find a position where the only condition forced
against cooks is "White has to promote both pawns".

@else: Does anyone know a h# with different combinations of white
pawn promotions? About like this? R and B would be especially cool.
(= 3+4 )

(HR, scheme) h#1.5
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(23) Posted by Rewan Demontay [Friday, May 24, 2019 14:14]; edited by Rewan Demontay [19-05-24]

I got that in Siegfried just to make it clear, although Hauke is right in my opine in that Black naturally tries to stay alive as long as possible. And I’m not quite sure what you mean by your sentence directed to me.
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(24) Posted by Frank Richter [Saturday, May 25, 2019 14:52]; edited by Frank Richter [19-05-25]

@Hauke: Well, just yesterday I remembered the famous author Gideon Husserl ...:

Gideon Husserl
2011 Shahmat 02/1984
(= 4+3 )

*) 1. ... e8=T+ 2. Kd7 c8=D#
1) 1. Kd7 c8=T 2. Sa7 e8=D#

or P0553726 (only 5 pieces, but not so nice) or the promotion task P0508932:

Gideon Husserl
4776 feenschach 09/1986
[[Ka2 Ba6 Rh8 Bg7 Pc7 Pe7 Pf6 Pe4 - Kd7 Qc1 Ra7 Rd8 Sg8 Pd6 Pe5'>

1) 1. ... e8=L+ 2. Ke6 cxd8=S#
2) 1. ... e8=T 2. Dc6 cxd8=D#
3) 1. ... cxd8=L 2. Sh6 e8=D#
4) 1. ... c8=T 2. Tf8 exf8=S#
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(25) Posted by Hauke Reddmann [Saturday, May 25, 2019 15:08]

Rewan, whereas nothing in the codex forbids you to add
fairy or whatever conditions until your problem is correct,
chess is a form of art. And, as the famous quote by
Leonhard Frank says, "Art is leaving out". Just take
poor Pierre Drumare who wasted a life to show the Babson
theme (4 black promotions are answered by 4 white promotions
to the same piece). He never succeeded in a legal or orthodox
position, and then the completely unknown Jarosch came and
pwned the whole scene. Details:

Thus, a chess problem is an idea plus a form. We have your
idea "combined promotions (of White)", now to the form:
- your idea with lots of conditions: meh
- your idea with only one natural condition: okay
(there are ideas that resist to be shown in a decent form,
so this happens to all composers and is no shame),
- your idea in orthodox form but tons of material, short variants,
dual minor and whatnot - I'd still give it a prize.

If you like to hear some tips and tricks of the trade,
PM me via the "Notes" on the left or via my uni email
(I'm easily googled).

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(26) Posted by Rewan Demontay [Saturday, May 25, 2019 16:13]

I’ll be on it Hauke!

Also, Frank, I think that you forgot to include a diagram for yout second example. And the first one is very nice indeed!
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(27) Posted by Frank Richter [Saturday, May 25, 2019 16:27]

Well, you see the notation, but there seems to be a bug in my text, so no diagramm is shown, and I don't know how to correct this :((
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(28) Posted by Rewan Demontay [Saturday, May 25, 2019 16:45]

That sucks. : (

Maybe you could try to post the FEN alone? It won’t be hard for me to copy paste a FEN.
(Read Only)pid=17388
(29) Posted by Rajendiran Raju [Saturday, May 25, 2019 16:51]

Note: This text is not formatted due to unbalanced command strings!

Just you close with same kind of one bracket then diagram will be visible.

And also this one is H#1.5 4 Solutions
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(30) Posted by Rewan Demontay [Saturday, May 25, 2019 16:57]

Yep. Also, that’s an invalid FEN-it cotains a 9 in it.
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(31) Posted by Rajendiran Raju [Saturday, May 25, 2019 17:02]; edited by Rajendiran Raju [19-05-25]

9 is the Squares gap between two pieces ..

This format of FEN is accepted in this mat plus ...Pls just you enter what is posted FEN with one closed bracket...then you see the magic.

I keep this FEN deliberately without closing bracket because you people's need to learn.
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(32) Posted by Rewan Demontay [Saturday, May 25, 2019 17:06]; edited by Rewan Demontay [19-05-25]

(= 8+7 )

Here is a valid FEN for use elsewhere: 3r2nR/r1PkP1B1/B2p1P2/4p3/4P3/8/K7/2q5 w - - 0 1
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(33) Posted by Hauke Reddmann [Sunday, May 26, 2019 12:48]

@Rajendiran: I know there is a format magic help page
somewhere here on MPF but I can't find it anymore.
Would be nice to have it at the left sidebar,
someone better pester the admin :-)

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(34) Posted by Rajendiran Raju [Sunday, May 26, 2019 12:55]; edited by Rajendiran Raju [19-05-26]

@ Dear Hauke

It's in the name of Click here ? for help on diagram entry.

Below our typing area. Just you click that question mark it leads to how to entry diagram.
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(35) Posted by Hauke Reddmann [Sunday, May 26, 2019 19:56]

Oh *bother*. Dangled before my nose a thousand times. Must be my far-sighted eyes. :-/
Anyway, here is another but less impressive rendering of a h# promotion task
(got a 1st prize anyway :-):
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(36) Posted by Rewan Demontay [Sunday, May 26, 2019 21:28]; edited by Rewan Demontay [19-06-04]

Since we have talked about most queen and rook promotions in a row records, I recalled one that involves knight promotions from the dustbins of my head.

Here’s the Youtbube video:

This puzzle, credited to Andre Cheron Journal de Geneve 1964, is a mate in 8 with white to play, and it features 8 knight promotions by white in a row.

(= 11+13 )

Also, when I looked for it elsewhere online, I found this one that has 5 bishop promotions in a row:

This one is attributed to David Zimbeck Die Schwalbe 2003. It’s white to move and mate in 6.

(= 15+8 )

Now we have a complete list of promotions in a row records! Perhaps someone could try to do 6 bishop promotions in a row, or find if it has already been done?

Also also, in Sample Problems on, there’s this puzzle featuring 4 promotions in a row, and each type of promotion takes place.

Mate In 5-Stragtegems 2002

(= 11+9 )

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(37) Posted by Rewan Demontay [Monday, May 27, 2019 00:36]

Here’s another one from that same page. There’s 8 queen promotions, albeit not in a row.

Mate In 15-Strategems 2004

(= 12+11 )

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(38) Posted by Rewan Demontay [Friday, Jun 7, 2019 19:30]; edited by Rewan Demontay [19-06-08]

I will post all possible solutions to my fixed problem under a single fairy condition. I even managed to find more lines to my surprise!

And remember, this is a "Bedingungsaufgabe" or a mathematical construction, as Hauke Reddman has said before, not an actual chess problem with a key that you need to struggle to find. However, as naturally should be thought of, Black does attempt to live as long as possible with White playing to checkmate as fast as possible.

Bedingungsaufgabe-5#. White To Play
(= 7+9 )

Fairy Condition: White may only make moves that allow for 3 promotions of one pawn to work with at least 3 promotions of the other one, no matter what black plays.

All Possible Solutions (To My Knowledge):

1. a8=N+ or a8=Q or a8=R Qxa8+ 2. Kxa8 Ba6 3. Bd7

---3... Bb5 4. c8=Q+ Ka6 5. Qxb7#

---------4. c8=B+ Ka6 5. Bxb7#
---------4. c8=N+ ka6 5. Qb6#

---3... Kb5 4. c8=Q b6 5. Qxc6# or Bxc6#

---------4. c8=R b6 5. Bxc6#
---------4. c8=N b6. 5. Qxb6# or Na7# or Nd6#


1. a8=R Qa7+ 2. Rxa7 Ba6 3. Bd7 Kb5 4. c8=B b6 5. Bxa6#

---------4. c8=Q b6 5. Qxc6# or Qxa6# or Bxc6#
---------4. c8=R b6 5. Bxc6#
---------4. c8=N b6. Qxb7# or Nd6#

----3. Bxb7 Bxb7 4. c8=Q+ Kb5 5. Rxa5# or Qxa5# or Qxb7#
---------4. c8=R+ Kb5 5. Qxa5# or Qxb7#
---------4. c8=B+ Kb5 5. Qxa5# or Qxb7#
---------4. c8=N+ Kb5 5. Qxa5# or Qxb7# or Nd6#
---3… Kb5 4. Bxa6 Kb6 5. c8=N# or c8=R# or c8=Q# or c8=B#

----3. Be6 Through Bh3 Kb5 4. c8=Q Kb5 5. Qxa6#
---------4. c8=B Kb5 5. Bxa6#
---------4. c8=N Kb5 5. Qxb6# or Nd6#

After many minutes of branching out the possibilities using paper and pencil, I have been able to conclude that, counting different promotions as different moves, that there is a total of 63 mates possible mates in 5 in this simple position under my condition. However, many more exist without them. I will soon provide a count.

UPDATE: All Other 5# Not Under My Condition

1. a8=N+ or a8=Q or a8=R Qxa8+ 2. Kxa8 Ba6 3. Be6 Through Bh3 (These Do Not Count Because If 3... Kb5 By Black, Only A Horse Promotion Works)

---3... Bb5 4. c8=Q b6 5. Qxc6# or Bxc6#

---------4. c8=R b6 5. Bxc6#
---------4. c8=N b6. 5. Qxb6# or Na7# or Nd6#

---3... Kb5 4. c8=N b6 5. Qxb6#

Do you think that that could be a record for most mates in 5 in a position (to be determined after my count)? What are the records for 3 and 4? Of course, their mustn't be any earlier mates in these cases.

UPDATE: I have a count of 28, or at that's how my works is, of 5# not under my condition. This makes for a total of 91. World record or just a really high number? I'm counting different promotions as different moves here.

I'm honestly surprised on just how much variety exists in this construction. Heck, I'll admit that I sort of made the original version slightly on accident.

The Original, Heavily Cooked, Version (For The Sake Of Completeness)

(= 6+9 )

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