praicanNo.1146 
Paul Rãican
(Romania)

JF-LOGO-1

Original Problems
JF Retro and PG problems 2015-2016

   →Previous ; →Next→List 2016(II)

Please send your original fairy problems to: julia@juliasfairies.com


The versions below, No.1146.1;2, are published after the end of tournament, 28.10./12.11.2017, but before the award is made, accepted by the judge of PG/Retros 2015-16.


Definition: (click to show/hide)


No.1146 Paul Rãican
Romania

original – 24.10.2016

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 16.5                                     (8+11)
Masand


No.1146.1 Paul Rãican
Romania

version of No.1146 – 28.10.2017

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 17                                      (10+9)
Masand


No.1146.2 Paul Rãican
Romania

version of No.1146 – 12.11.2017

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 17                                      (13+6)
Masand


Subscribe
Notify of
guest
26 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
Luce Sebastien
Luce Sebastien
October 25, 2016 12:09

Julia, the n° of this diag is 1146 !

Nikola Predrag
Nikola Predrag
October 25, 2016 14:10

Dmitri’s Py2Web is priceless for such problems.
It makes me aware that I don’t actually see why not 11.Rxg1/17.Kd3.

Joost
Joost
October 25, 2016 16:04
Reply to  Nikola Predrag

Because after 11. Rxg1 Sc6, 12. Qd8 results in a recolouring of Sd1.

Nikola Predrag
Nikola Predrag
October 25, 2016 17:15

Thanks Joost!
Nice line-closing and of course unexpected for those (it’s me) who unduly but stubbornly look for the answer only on e,f,g files.
Quite a misconception 🙁

Juraj Lörinc
June 18, 2017 17:16

Quite an achievement, do I see it right as Schnoebelen AUW by both sides? Great choice of fairy condition right for the idea…

Joost de Heer
Joost de Heer
June 19, 2017 23:43
Reply to  Juraj Lörinc

No white queen promotion.

Personally, I think the chance of this being correct is less than 1%, but it takes too much time to find the cook.

François Labelle
François Labelle
July 22, 2017 21:36
Reply to  Joost de Heer

Cooked in 15.5 moves:

1. d4 e5 2. Bg5 Qxg5 3. dxe5 Qxe5 4. a4 Qc3 [c7,g7=w][b2,c2=b]+ 5. Sxc3 Ke7 6. Ra3 cxd1=Q [d7=w][e2,a4=b]+ 7. Sxd1 Sc6 8. Kd2 exf1=R 9. Rh3 Sd8 10. gxh8=B Rxf2 [f7=w][g2=b]+ 11. Kd3 Bh6 12. Bf6 [b2=w]+ Rxf6 13. dxc8=B Rf1 14. f8=R Rf3 [h3,f8=b]+ 15. Sxf3 g1=B 16. Rxg1

This solution was found by an experimental fairy PG solver I’m working on. The program uses heuristics to search the most likely moves first, and this problem seemed like a good test given Joost’s comment. In 25 min it found cooks in both 16.5 and 16.0 moves. Letting it run for 13h, it found the cook shown above in 15.5 moves.

My program looks really promising so I’ll keep working on it and I plan to make it available. Timescale: a few months.

Juraj Lörinc
July 23, 2017 15:21

Astounding.
Hats off!

Paul Rãican
Paul Rãican
July 25, 2017 15:47

Even if the problem has been broken, I am glad that we finally have such a versed solver in Canada. Congratulations, Francois! A little question: If the last white move is dropped, is the problem still cooked?

Paul Rãican
Paul Rãican
July 25, 2017 15:59
Reply to  Paul Rãican

Yes, of course: 15….g1=S!

François Labelle
François Labelle
October 31, 2017 19:15

Version 1146.1 is cooked by Jacobi in “PG demolition mode”:

1.Sf3 e6 2.a4 Qf6 3.d3 Qc3 [c7,g7=w][b2,c2,d3=b]+ 4.Sxc3 dxe2 5.Be3 f6 6.Bxe2 Ke7 7.gxf8=R cxd1=Q [d7=w][a1,e2,a4=b]+ 8.Sxd1 Rc1 9.Kxe2 Rc2 [b2=w][c7=b]+ 10.Kd3 Rxf2 11.Rf1 Rxg2 12.Bg1 Sc6 13.Rxf6 Sh6 14.d8=Q [c7,c8,h8=w]+ Sxd8 15.Bxe6 Rg3 16.Bc8 Rh3 17.Rg8 Sxg8

Finding this cook was harder for my program than for 1146. It took 15h to find this cook on my computer. Trying in 16.5 moves, it hasn’t found any cook after 43 hours.

Paul Rãican
Paul Rãican
November 1, 2017 17:41

A very hidden alternative game, virtually impossible to discover with the human brain. Chapeau!

François Labelle
François Labelle
November 17, 2017 22:03

1146.2 isn’t C+, contrary to what the caption says. At publication time the problem had merely survived 4 days of “demolition mode” on my computer. In an unfortunate game of “Chinese whispers” between me, Paul and Julia, this information turned into “checked for 4 days” and then into the label “C+”.

Today the problem has survived 14 days of demolition mode, but all this means is that Jacobi v0.1 finds no obvious cook so far. The problem is still C-.

Juraj Lörinc
November 17, 2017 22:48

Actually I prefer three-state logic:
C+ – fully tested and considered sound by at least one program
C- – unsound
C? – anything else = no defect found yet, but not yet fully tested by any program yet

In this logic, 1146.2 is still C?.

François Labelle
François Labelle
November 18, 2017 07:23
Reply to  Juraj Lörinc

I followed the JF notation which seems to be

C+: sound
cooked: unsound
C-: unknown

The letter C comes from the word “computer”, so maybe you could write “C-” for a cook found by a computer, but I don’t see why you’d write “computer minus” for a cook found by a human. You’d need four-state logic.

Vlaicu Crisan
Vlaicu Crisan
November 18, 2017 10:37

When the problem is fully tested and considered sound by two or more programs/computers, then should we use C++ or C#? 🙂

François Labelle
François Labelle
November 18, 2017 16:59
Reply to  Julia

About the choice of notation for problems with unknown status (C- vs C?), with Google I found two definitions in favor of C-:

C- : problem was not tested by computer program or was tested only partially (Award of Marián Križovenský 55 JT)
C-. Computer-tested only partially (WCCI 2013-2015. Fairies)

and one in favor of C?:

C- means unsound problem. C? means not computer tested problem. (c2_text_en.doc on http://www.jurajlorinc.com)

Maybe someone subscribed to every chess magazine can tell us which convention each magazine is using?

dupont
dupont
November 19, 2017 16:30

In order to simplify the situation, there is another option followed e.g. by WinChloe:

C- = cooked
Nothing = not computer tested.

My favorite is clearly in favor of C?, so that each problem has some “C status”:

C+ = correct
C- = cooked
C? = not computer tested

Obviously C+ and C- are definitive (relatively to a given program and version), while C? may be changed to C+ or C- some day.

The big question is whether C- should denote a cooked or a not computer tested problem. François said this is 2-1 in favor of not computer tested, but if we add WinChloe’s opinion, this is 2-2!

Nikola Predrag
Nikola Predrag
November 19, 2017 17:09

Wouldn’t it be absurd using C- to indicate an ORIGINAL which a computer has PROVED to be cooked?

Nikola Predrag
Nikola Predrag
November 20, 2017 02:25
Reply to  Julia

Yes Julia, that’s what I say.
C- or C? could mean “not tested” or perhaps that a computer didn’t find the solution which the humans are able to see.
A trivial example, Popeye can’t solve Loyd’s #2: Ke6,Qa6/Ke8,Ra8,Pa7c7, so it’s not C+ 🙂

dupont
dupont
November 20, 2017 15:15
Reply to  Nikola Predrag

Of course, as this is a retro problem… An amount of retro problems are 100/100 correct because the reasoning is enough easy to avoid any human mistake, but are nevertheless C? because computers can’t handle such retro problems, even trivial for human brains.

François Labelle
François Labelle
November 19, 2017 17:26

More votes (notation of problems with unknown status, retros section):

feenschach: nothing
PDB: nothing
Phenix: nothing
Problem Paradise: nothing
Julia Fairies: C- (prior to 1146.2)
Strategems: C-
Problemist: C?

François Labelle
François Labelle
December 10, 2017 02:05

Version 1146.2 is cooked by Jacobi after about 30 days in “PG demolition mode”:

1.a4 d5 2.Ra3 Bf5 3.e4 dxe4 4.Qh5 Sh6 5.Qxf7 [f5,e7,g7,f8=w]+ Kxf7 6.Ke2 Rxf8 7.g8=Q [h7,f8=w][g2=b]+ Kxe7 8.Rh3 Sxg8 9.f3 Qd3 [e4=w][c2,d2,f3=b]+ 10.Kxd3 d1=S 11.Sxf3 gxf1=S 12.Be3 Sg3 13.Rf1 c1=R 14.Bg1 Rc3 [c7=w]+ 15.Sxc3 Sxe4 16.Sxd1 Sef6 17.Rxf6 Sd7

Paul Rãican
Paul Rãican
December 10, 2017 10:51

This is a spectacular alternative solution. Thank you, François, for your patience!