Václav Kotěšovec (Czech Republic)

Original Fairy problems
January – June’2019

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No.1372 Václav Kotěšovec
Czech Republic

original – 18.02.2019

Solutions: (click to show/hide)

white Kc1 Sa3 black Ke5 EAb5 EAd3 EAe1

hs=5            4 solutions           (2+4)
Eagle b5, d3, e1

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February 18, 2019 21:43

four amazing echoes with knight pinnned,

Geoff Foster
Geoff Foster
February 18, 2019 22:46

But the bK has a choice of 6 squares each time, which is especially bad because it is just a waiting move (White is already in stalemate).

February 19, 2019 07:03

Yes. Big weakness

Juraj Lörinc
February 19, 2019 21:22
Reply to  Seetharaman

I dare to disagree with the automatic downgrading of helpself problems having multiple final black moves. More specifically, in this case – the point is in the critical positions, i.e. in the positions after the last move of White. In spite of many moves available, there is no way to disturb stalemate net – that is a kind of paradox.

I know some helpself experts disagree and indeed, in hs# where final black move is included in the main idea, it should be unique. But every composition should be considered separately. Automatic rule in this case is not justified in my opinion.

Vaclav Kotesovec
February 19, 2019 11:26

In the selfmate (selfstalemate) problems are moves of Black a variants, not duals. See my another hs= with last move K~,, p.42-43, 3. Prize and a comment by judge.

shankar ram
shankar ram
February 19, 2019 16:04

Last move duals could be eliminated by setting problem as a h==4.5 without BK, but BK is utilised in the previous moves of all the solutions!

shankar ram
shankar ram
February 19, 2019 16:07

Such duals also possible in a hs#, but I’ve not seen an example yet!

Sergey Shumeiko
Sergey Shumeiko
February 19, 2019 20:56

I propose to write down the condition “4 solutions” as “4.1..6”, for example:
Sergey Shumeiko
original – 19.02.2019
hs=5 ..1.5 1+13
1.Ke2 Rf5+ 2.Kхe3 g4+ 3.Ke2 Rg5 4.Kf2 Kh3 5.Kg1
5…e3, g3, Kg3, Rf5, Rf6 =

Nikola Predrag
Nikola Predrag
February 19, 2019 22:20

-“But every composition should be considered separately.”-
Yes Juraj, in principle. Sergey’s example nicely demonstrates the point – 5 black pieces can move and can’t avoid to attack the existing wK’s flight.

Multiple utterly banal and boring TEMPO MOVES indeed should be “considered separately”!
Of course, as a TERRIBLE FLAW that HUMILIATES the delicacy of tempo-play.

Juraj Lörinc
February 19, 2019 22:52
Reply to  Nikola Predrag

OK, the second paragraph is your opinion and perhaps it is even majority opinion, based on the above comments. I can accept that.

Nikola Predrag
Nikola Predrag
February 20, 2019 00:27

Juraj, just to make it clear – my ‘opinion’ was not about Vaclav’s problem but about ‘what should be considered separately’ and what should not.

What is a paradox and what is not?
White Kd3; Black Pb4,Pa2,Pb2; H==2
Put bK e.g. on d5 and there’s “paradoxical” hs=2.5 with bK’s “star+cross”.
Or, should we add more black pieces to “increase the paradox” of many final moves which “cannot disturb the stalemate”?

I simply don’t understand what you’re talking about. I hope at least YOU know what is ‘your opinion’.

Stephen Emmerson
Stephen Emmerson
February 21, 2019 02:08

Interesting debate about H# vs S# conventions when applied to HS#. There does seem to be a difference between a final move such as a random battery opening as sometimes happens in the last move of a S#, and a purposeless move in an already-achieved stalemate net as here.
In any case…I would also consider as flawed the stalemate net itself: it is not a model (b5 guarded by two eagles in the first line). It’s an inevitable feature of the pin; without the need to pin the EAd5 could stand on e5.

Marjan Kovacevic
Marjan Kovacevic
March 5, 2019 19:46

Going to the roots of hs#
YACPDB >>351059
Seidemann, Felix
Deutsche Märchenschachzeitung 1931
1st-2nd Prize

A nice example of multiple mates:
1.Qb6 Sxh2 2.Bxf3+ Rg4,Sg4,Sxf3,g4,Rxf3#

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