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MatPlus.Net Forum General The six-fold dentist
 
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(1) Posted by Geoff Foster [Saturday, Oct 17, 2009 13:27]

The six-fold dentist


In the March 2006 issue of "The Problemist", Stephen Rothwell presented some notorious unsolved tasks. One of these was the six-fold Dentist in a selfmate two-mover. The rest of this post is taken from his article.

Rudolf Prytz, Chemnitzer Tageblatt, 1925
(= 4+8 )

S#2
1.Bd2! (>2.Qg2+)
1...Rd5 2.Qd3+
1...Rb7 2.Qb1+
1...Rc6 2.Qc4+
1...Sd5 2.Qxf4+
1...f3 2.Qe2+

The above problem is one of my all-time favourite selfmate twomovers, a highly elegant and economical fivefold setting of the Dentist theme, which can be defined as follows: in a selfmate Black parries a threat by unpinning a white piece (usually the wQ) by setting up a masked battery on the pin-line, which White can exploit by means of a check on the previously pinned piece, forcing the battery to open, giving mate. One is inclined to think that, if the five-fold setting can be done with just 12 men, it should be easy to construct a six-fold setting. But, indeed, this is so fiendishly difficult that nobody has been able to achieve it. A close try is the following:

Hrvoje Bartolovic, 3HM, Probleemblad, 1961
(= 7+15 )

S#2
1.Qxd5! (>2.Sd1+)
1...Bc5 2.Qd4+
1...Bg5 2.Qxd2+
1...Sc5 2.Qb3+
1...Se5 2.Qc4+
1...Sg5 2.Qf3+

In the above problem you could easily add a sixth variation by moving the bSh7 to f7 and the bBf1 to g8. Now, you would get the additional variation 1...Sfe5 2.Qd3+ with a nice pair of dual avoidances by black line opening after 1...Sde5/Sfe5 2.Qc4+/Qd3+. Unfortunately, the threat 2.Sd1+ has disappeared by shifting bBf1 to g8 and it is apparently impossible to construct an alternative threat in this scheme. You will find an excellent article on the subject by Hartmut Laue and Albrecht Colditz in "feenschach", August 1982, pp. 478-488, in which the many vain attempts at a six-fold Dentist are described in detail. Therefore, be warned; if you want to have a go, many have failed before!
 
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(2) Posted by Geoff Foster [Saturday, Oct 17, 2009 13:28]

The close attempt at a six-fold Dentist described in the above post can be achieved by setting the problem as a helpselfmate with multiple solutions, as follows:

(= 7+9 )

HS#1.5 (6 solutions)
1...Bc5 2.Qd4+ Bxd4
1...Bg5 2.Qxd2+ Bxd2
1...Sc5 2.Qb3+ Sxb3
1...Sde5 2.Qc4+ Sxc4
1...Sfe5 2.Qd3+ Sxd3
1...Sg5 2.Qf3+ Sxf3

Is this acceptable?
 
 
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(3) Posted by Joaquim Crusats [Saturday, Oct 17, 2009 17:50]; edited by Joaquim Crusats [09-10-18]

With an additional unit maybe the “nice pair of dual avoidances” can still be achieved in a five-fold dentist:

Hrvoje Bartolovic, 3HM, Probleemblad, 1961
(version)

(= 7+16 )


S#2
1.Qxd5! (>2.Sd1+)
1...c5 2.Qd4+
1...Sc5 2.Qb3+
1...Sde5 2.Qc4+
1...Sfe5 2.Qd3+
1...Sg5 2.Qf3+

Again, if only a second bBh6 could be used…
 
   
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(4) Posted by Hauke Reddmann [Sunday, Oct 18, 2009 18:20]

Just curious, what were the other notorities?
(As an OTB player, I don't do s#, or I might compose
your %&$ tooth extractor in ten minutes :-)

Hauke
 
   
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(5) Posted by Geoff Foster [Sunday, Oct 18, 2009 22:48]

The "millennium" problems were:

1. Six-fold Stocchi-block on a single square in the orthodox two-mover
2. The full BS-wheel of interferences yielding 8 distinct white mates
3. Double castling square vacation with both white castlings in a more-mover
4. The six-fold Dentist in a selfmate two-mover
5. Consecutive Allumwandlung in the orthodox selfmate-miniature
6. The Oudot-Theme in the orthodox helpmate
7. The 100 dollar theme

In relation to the first task, I was amazed that a five-fold Stocchi block had been done (Lev Loshinsky, 1 Prize, Sverdlovsk Chess Club, 1940)!

If you want sample problems from the article then I'm happy to provide them, but please start a new thread, because I don't want to distract attention away from the Dentist theme.
 
   
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(6) Posted by Geoff Foster [Tuesday, Oct 20, 2009 00:12]; edited by Geoff Foster [09-10-24]

Here is a more economical setting of the above helpselfmate.

(= 5+8 )

HS#1.5 (6 solutions)
1...Bc5 2.Qd4+ Bxd4
1...Bg5 2.Qxd2+ Bxd2
1...Sc5 2.Qb3+ Sxb3
1...Sde5 2.Qc4+ Sxc4
1...Sfe5 2.Qd3+ Sxd3
1...Sg5 2.Qf3+ Sxf3

[Position edited on 24 October to further improve economy]
 
   
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(7) Posted by Steven Dowd [Tuesday, Oct 20, 2009 17:30]

Is that the maximum number that can be built into a hs#?
 
   
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(8) Posted by Geoff Foster [Tuesday, Oct 20, 2009 23:03]

It is certainly the maximum for this matrix. The only other square that the queen could move to off the pin line is c6, and a forced battery mate could never be arranged for that square.

I am reminded of the following problem, which shows an apparently unachievable #2 task (6 interference unpins of the white queen) as a H#1.

Michael McDowell, Special HM, The Problemist, 2002
(= 6+9 )
H#1 (6 solutions)

1.Sg3 Qf2
1.Rg3 Qf1
1.g3 Qf3
1.Se5 Qe3
1.Sd6 Qd2
1.Rc7 Qxc4
 
   
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(9) Posted by Steven Dowd [Wednesday, Oct 21, 2009 01:08]

I was thinking that because of course, whatever could be done in helpplay would logically be more than what could be done in direct or self play (although that is not necessarily true, just an assumption and certainly helpplay would do no less than equal the max) so if six were the maximum possible in a hs#, it would be logical to conclude that it would be a near impossibility in regular self play....
 
   
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(10) Posted by Marcos Roland [Thursday, Oct 22, 2009 05:36]; edited by Marcos Roland [09-10-22]

In the late eighties, Felix Sonnenfeld and I composed the following problem:

F. Sonnenfeld & M. Roland
Boletim da UBP (don't remember the exact number)
(= 7+12 )

s#2
1.Sf3 th. 2.Df3+
1...Te6 2.De4+
1...Tf7 2.Df5+
1...c4 2.Dd3+
1...e6 2.Df5+
1...b3+ 2.Dxb3+
1...Se6 2.Dxc5+
1...Sxd5(!) 2.Se3+
1...Lh7(!)2.Dxc5+

There are eight variations (seven if 1...b3+ doesn't count), but only five of them are typical dentist variations, so this problem surely doesn't solve the "six-fold" puzzle. But there are two other variations which, from my point of view, enriches the problem:
1...Sxd5 and 1...Lh7. I don't remember seeing in other problems the capture of the checking white piece or the abandonment of the pinning line by the black pinner.

But the most "problematic" variation is this one: 1...Sxe6 2.Dxc5+, reaching the following position:

(= 7+11 )


Now both 2.Sc4 and 2.Sc5 mate. Petko Petkov says that this is a dual. Of course, I respect Petko Petkov very much, but I don't fully understand this. In that time, I thought that any alternative move for Black, even in the mating move, should count as an additional variation, not as a dual.

Any way, I remember Felix and me being almost "crucified" in the annual meeting of the UBP in 1988 (or 1989), because we sustained that this was a variation, not a dual...
 
   
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(11) Posted by Marcos Roland [Thursday, Oct 22, 2009 05:46]

In our problem, the threat, of course, is 2.Db3+, not 2.Df3+.
 
   
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(12) Posted by Geoff Foster [Thursday, Oct 22, 2009 06:17]

In the Sonnenfeld/Roland problem, 1...Rf7 and 1...e6 are both answered by 2.Qf5+. Surely this doesn't count as 2 Dentist variations?
 
   
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(13) Posted by Marcos Roland [Thursday, Oct 22, 2009 14:05]

You see, 1...e6 and 1...Tf7 involve anti-dual effects: the first opens a diagonal for the bLf8, so 2.Dxc5+ could be answered by 2...Lxc5; the second opens a diagonal for the bDh8, so 2.Dxc5+ could be answered by 2...Dc3. Besides that, the mating moves in both cases are different precisely because Pe6 and Tf7 are different pieces. So, I think that, in the whole, they are different variations. But I understand your point in this sense: it would be better if the W2 moves were different.
 
   
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(14) Posted by Frank Richter [Thursday, Oct 22, 2009 20:30]; edited by Frank Richter [09-10-23]

I would say, only different W2 moves are in the sense of the theme.

It is possible to extend the hs# to 9 unpins, but again only 6 different Queen moves. Qd1+ is only theoretically a possibility, or not!?

(= 7+10 )

hs#1,5, 9 sol.
(Edited: Position without bQ, see below)

1...Tc2-b2 2.Dd4*f2 + Tb2*f2 #
1...Tc2-c3 2.Dd4-e3 + Tc3*e3 #
1...Sa4-b2 2.Dd4-d3 + Sb2*d3 #
1...Sa4-c3 2.Dd4-e4 + Sc3*e4 #
1...Tg6-f6 2.Dd4-f4 + Tf6*f4 #
1...Tg6-g7 2.Dd4-g4 + Tg7*g4 #
1...Sd7-e5 2.Dd4-d3 + Se5*d3 #
1...Sd7-f6 2.Dd4-e4 + Sf6*e4 #
1...e7-e5 2.Dd4-f4 + e5*f4 #

On the other side - we have three s#-specific "changed mates" ...
 
   
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(15) Posted by Geoff Foster [Friday, Oct 23, 2009 00:34]; edited by Geoff Foster [09-10-23]

This is a very clever construction Frank! I especially like how the bishop on b1, which is needed to guard a2, also stops 2.Dd4-d3+ after 1...Tc2-c3.

There is one bit of chess blindness -- the black queen is not needed! Perhaps in an earlier version the black queen was there to stop 2.Dd4-g4+ after 1...Sd7-e5 or 1...Sd7-f6, but the black rook on g6 does that job.

Of course, if you only wanted the six variations with different W2 moves then the black pawn on e7 and the black knight on a4 could be removed.
 
   
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(16) Posted by Frank Richter [Friday, Oct 23, 2009 08:29]

Dear Geoff,

thank you very much for the hint - yes, initially I placed the bQ on c8 to avoid 2.Qg4+.
I corrected the diagram above.
And of course we have a 2nd sixfold dentist without the additional black units (7+8 pieces).
 
   
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(17) Posted by Geoff Foster [Saturday, Oct 24, 2009 00:32]

The following helpselfmate only has 5 different Dentist solutions, but the first one is a cross-check!

(= 5+7 )

hs#1.5 (6 solutions)

1...Lh3-f5 2.Dd5-f3 + Lf5-d3 #
1...c7-c5 2.Dd5-d4 + c5*d4 #
1...Sd7-c5 2.Dd5-b3 + Sc5*b3 #
1...Sd7-e5 2.Dd5-c4 + Se5*c4 #
1...Le7-c5 2.Dd5-d4 + Lc5*d4 #
1...Le7-g5 2.Dd5-d2 + Lg5*d2 #

Has this "cross-check Dentist" been done before in a s#2?
 
 
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MatPlus.Net Forum General The six-fold dentist