- FIDE Album 2010-2012 - a successful participation of the bulgarian composers! The upcoming album is for 2013-2015
- Olympic Gold medal for IM Diyan Kostadinov!
- Bulgarian successes (Part 3)
- Great success for Bulgaria at the World Championship!
- Bulgarian successes (Part 2)
- Bulgarian successes (Part 1)
- Petko Petkov - world champion again!

The Wonderful Fairy options of WinChloe by IGM Petko Petkov |

(19.01.2019) Here you can see an interesting article by Petko Petkov about the new fairy options in WinChloe which looks very promising for a future work. It was published in Phenix 288-289 in September-October 2018
THE WONDERFUL FAIRY OPTIONS OF WINCHLOEBy IGM Petko PetkovDedicated to Christian PoissonI. FOREWORDI write this article because of a very interesting new phenomenon in the world of fairy composition : the special options
Conditionsfromply=n and Conditionsuntilply=n, invented by the world-known programmer Christian Poisson and practically demonstrated in version 3.36 of his wonderful program WinChloe.More than a year has passed since the discovery of Christian Poisson, but to my surprise there are still few published problems of this type. But I think that WinChloe’s options open up very rich and interesting possibilities for the composition of fairy problems of the hypermodern form ! Therefore, I decided to write this article with the hope that it will be of interest for composers and solvers all over the world ! In fact, the beginning of this very thrilling story happened with the traditional 12th Belgrade Problem Chess Festival 2017 (26th-28th May 2017). The Serbian organisers, who have always been authors of very interesting events and competitions, this time offered in the Belgrade Internet Tourney 2017 the following very fresh, non-standard theme :
Group C – hs# maximummer 2,5 & 2 solutions ; Thematic condition : Help-selfmate Maximummer with help-play before the last two half-moves, presenting s#1 Maximummer (Black has to play the geometrically longest move). No other fairy conditions or pieces are allowed. The theme was illustrated by the following example (note that the number of solutions must be exactly two, and the number of moves is also strictly defined - 2,5 moves !) :
Solutions :
I. 1…e1B 2.Rf8 Bh4 and now Black are required to play using the condition Maximummer : 3.Rf6+ the finale is the geometrically longest move by Black : 3…Bxf6# II. 1…e1R+ 2.Rc1 Rh1 and then follows : 3.Rb1 Rxb1# We will demonstrate other examples from the thematic tournament, which turned out to be quite interesting.
After the tournament was announced, I asked WinChloe’s author, Christian Poisson, whether it was possible to create a computer program to solve problems of type hs#2,5 Maximummer ? I was extremely surprised when Christian wrote to me a day later that he had already created a program that had much more potential and after a few days of testing he would include it in the new version of WinChloe, at that time 3.36 !
Above all, I want to express my great gratitude and respect to the world-known author of WinChloe Christian Poisson ! My suggestion is that this kind of fairy problem, with WinChloe’s special options, be named TYPE POISSON !Of course, Christian`s new program can also solve the hs#2,5 Maximummer problems in just a few seconds ! After the new version of 3.36 (in May 2017) was released, I immediately announced this in the forum on the world-known Serbian site
MatPlus, for the benefit of all participants in the 12th Belgrade T.T. (section C - Fairies). Marjan Kovachevic and Borislav Gadjanski have made a great contribution by organising this tournament - two great names in the world of chess composition that do not need a detailed presentation ! The first journal to publish Poisson-type problems was the United States
StrateGems - in its issue n°79 (July-September 2017). Later, new originals of this type appeared also in this magazine. The Bulgarian site KoBulChess also published in 2017 the detailed definition and explanations of these options and an original hs# problem by P.A. Petkov.II. FEATURES OF THE NEW OPTIONS IN WinChloeIn brief, the essence of the program was explained by its inventor - Christian Poisson, in «Readme» of the new (in May 2017), version 3.36 :
«Two new options are programmed, allowing to indicate from which ply, or until which ply, fairy conditions are taken into account. These options,
to write in the «twin» field, are : Conditionsfromply=n (or in French : Conditionsapartirde=n) and Conditionsuntilply=n (or in French : Conditionsjusqua=n), «n» being the number of plies.For example :
Conditionsfromply=5 means that the conditions are taken into account only from the 5th ply (the 3rd white move for a direct, a self or a help-self problem) and the 3rd black move for a help - problem. Conditionsuntilply=6 means that the conditions are taken into account only until the 6th ply (the 3rd black move for a direct, a self or a help-self problem and the 3rd white move for a help - problem).It is possible to combine both options (write them on to different lines). IMPORTANT : THESE OPTIONS WORK WITH «SIMPLE» CONDITIONS (Maximummer, Minimummer, Circe, Sentinelles…), BUT CAN (MAYBE) GIVE UNEXPECTED RESULTS WITH CONDITIONS ASKING MORE COMPLICATED CALCULATIONS. THEY DON’T WORK WITH CONDITIONS WHOSE EFFECTS ARE NOT CALCULATED AT EACH MOVE.
For more information you can read about the version 3,36 (the Readme file, named “readme.txt” is in the “C:\WinChloe” folder).
Of course, some additional explanations are needed here, which I wrote with the help and control of Christian, who on this occasion answered me very kindly and comprehensively on my long letter sent to him on May 6th, 2017.
What does the word «ply» mean ? The answer is elementary - «ply» is a half-move. Consequently, each problem has a certain number of «X» «plies» (half-moves) = sum from all single half-moves by White and Black. Examples :
a) Orthodox problems and direct non-orthodox problemsA #2 has X=3 plies, they are : 1st half-move (key move) by White + 1st half-move by Black (defence) + 2nd move by White (mating move). A #3 has X=5 plies, a #4 has X=7 plies, a #5 has X=9 plies and so on. Here it is not difficult to define a simple formula : if in an orthodox problem the condition is #n, the sum «X» of plies (half-moves) = 2n-1. For example, a #11 has 2x11-1 = 21 plies. In the general case, this formula also applies to unorthodox direct-problems of the form #n, =n, s#n, s=n r#n, r=n etc.
b) Helpmates (help stalemates, help double-stalemates, help-checks, help castlings, etc) and also Help-selfmates (hs#, hs=, hs==, etc) :A helpmate in «n» moves where «n» is a whole number (an integer) has X=nx2 half-moves (plies). For example, a h#2 has X=2x2=4 plies, they are : 1st move by Black + 1st move by White + 2nd move by Black + 2nd (mating) move by White ; a h#3 has 3x2=6 plies, a h#4 has 8 plies etc… The wonderful fairy options of WinChloe 3
Very Important ! If «n» is a fractional number, the number of plies is also X=2n+1 half-moves ! For example, h#1,5 has X=2x1,5+1=4 half-moves (plies) H#2,5 has X=2,5x2+1=6 (plies), H#3,5 has 8 plies etc. The same situation occurs with hs#n problems in which «n» is also a fractional number, etc !At first sight these words appear to contain an error. The logical question can be : if we have for example a h#1,5 this problem has de facto only 3 plies : the first move by White + the second half-move by Black + the last-third half-mating move by White.
BUT! By using WinChloe’s options here, we must conditionally assume that such problems also have 4 plies but the first (missing) half-move of Black is assumed as a «null ply». In other words, we must know that according to WinChloe the play begins here with the second ply (de facto - the first half-move of White) followed by the 3rd ply (the second half-move by Black) and the end follows the 4th ply (the third mating half-move by White). These rules have a very important practical value and they should be applied accurately ! Errors in this respect lead to erroneous results by the program. Let’s see a simple example, scheme n°2 :n°2. Here it is clear that the condition Sentinels must be applied only on the last half-move - the mating move of White. But since we conditionally accept that this problem also has 4 plies, with the first ply being the absent first half-move of Black («null ply»), the play starts here on the 2nd and ends on the 4th ply ! For this reason, we must write under the diagram ConditionsFromPly=4.
Solution :
1…Rb5 2.Ba7 Rb8(+Pb5)#But if we write erroneously ConditionsFromPly=3, WinChloe claims that the problem does not have a solution ! The reason is clear - the program perceives here as a third half-move (ply) the black move
2.Ba7(+Pb6) and the Sentinels condition is mistakenly applied (before we need it). As a result, a black Pb6 closes the line for the White rook and 2…Rb8(+Pb5) is an impossible move !n°3. This is an analogous example with a hs#2,5. Here the condition Anti-Andernach is applied only on the first (key-move) by Black but WinChloe perceives this move as a second ply (the first missing halfmove by White is a «null ply») and under the diagram we must write ConditionsUntilPly=2.
Solution :
1…g1=Q(wQ) 2.Qg5 c6 3.Qb5+ cxb5#Of course, here too it is erroneous to write ConditionsUntilPly=1 - WinChloe does not give a solution !
c) Series-movers (help-play) - Ser.H#n, Ser.S#n, Ser.R#n , etc.Here, in a series-problem in «n» moves the number X of all plies is n+1. In other words, we have n-1 half-moves (plies) and only the last nth move is a compound move : it includes the last (nth) half-move (ply) of the party carrying out the series and the n+1th last half-move of the other party, which obliges the mate, the stalemate, etc…
d) Series- Autostalemate in «n» moves (ser-!=n)The number X of all plies is also n.
The basic principle of applying WinChloe’s options is very interesting : the play in each problem with X plies in the general case contains two parts :
A) Orthodox part, where the number of plies = AHere the play is carried out only according to orthodox rules.
Very important : in each problem the presence of the orthodox part A is obligatory ! In other words : a problem with conditionsfromply (or conditionsuntilply) cannot exist without the orthodox part A !B) Fairy part - here the play containing B pliesHe is carried out according to the rules of one or more (two, three, four, etc…) fairy conditions, for example : B= Circe, or B= Circe + Maximummer, etc…
Consequently, in accordance with the structure of these problems, the common sum X of the half-moves (plies) can be :
X = A + B or X = B + A
In other words, the play can begin with the orthodox part A, followed by the fairy part B (Option = ConditionsFrom Ply) or vice versa (Option = ConditionsUntil Ply).
A more complex variant is possible using both options as we shall see later.
1.Use of option CounditionsFromPlyAs we already noted above, the first part here has A half-moves (plies). Then follows the second fairy part of B half-moves (plies).
For example : a h#3 has 6 plies : 1st ply = first half-move by Black, 2nd ply = first half-move by White, 3rd ply = second half-move by Black, 4th ply = second half-move by White, 5th ply = third half-move by Black and 6th ply = third white mating move. The composer himself chooses a fairy condition (or combination of fairy-conditions) for the fairy part of his problem. After that, he chooses from what half-move (ply) the use of this fairy-condition commences. If we have 4 orthodox plies (A = 4) plus two fairy-plies (B = 2) then this means that at first we make 4 orthodox half-moves and only with the 5th ply does the execution of the fairy condition commence. In this case we must write : ConditionsFromPly = 5.
Solution of n°4 :
1.Rg5! Bc4 2.Rgb5! Ba2 and then follows : 3.Ka4(+Pa3) Bb3(+Pa2)# Returning to the scheme n°1, we can note that here we can write the stipulation under the diagram in a new way :
hs#2,5 & ConditionsFromPly=5 Maximummer. After solving this problem by computer, you need to write below the diagram : 2 solutions :
I. 1…e1=B 2.Rf8 Bh4 and here, from the 5th ply, we must apply the condition Maximummer : 3.Rf6+ followed by the last (mating) move (6th ply) : 3…Bxf6# ; analogically : II. 1…e1=R+ 2.Rc1 Rh1 and here follows : 3.Rb1 Rxb1# Under the diagram of each problem from Belgrade Internet Tourney (BIT) 2017 Fairy T.T., one can write exactly the same text as seen in scheme n°5.
We stress that the option ConditionsFromPly = 5 should be written in the «Twin» field of the program WinChloe, but the condition Maximummer must be marked in the field «Conditions» using the ordinary practice of such a choice ! It is clear that under the diagram of Scheme n°6 we can also write down the option ConditionsFromPly = 6, since de facto here the 5th ply is also an orthodox half-move by White (the condition Maximummer applies only to Black !).
Demonstrating the enormous possibilities of the WinChloe’s options, first of all I want to show several of the most interesting problems from the T.T. Belgrade Internet Ty 2017.
Solutions of n°6 :
I. 1…Rg1 2.Qxg1 Bf8 3.Qg8 Qe7# II. 1…Rf1 2.Qxf1 Rg4 3.Qh3 Qf4# Comment by Emanuel Navon in Variantim (n°72/2017) : «A complex combination of sacrifice and pinning (of attacking pieces), two characteristic elements that prevent longer moves by Black. The whole content is harmonized in ODT form, with Bristol clearance, black halfpin, and anticipatory self-pins.»
Solutions of n°7 :
I. 1…Bh4 2.Se3 Qd8! and then follows the s#1 Maximummer : 3.Rg8! Be7# Analogically :
II. 1…Rh3 2.Sf6 Qb3! 3.Rb1 Rc3# A very interesting play where both sides prepare the final mates 3…Be7(I) and 3…Rc3(II), depriving Black of the unnecessary longest moves after the reciprocal blocks and line-closings by the black Bishop and the black Rook. Active role of the black Queen who blocks as a pinned piece !
Solutions of n°8 :
I.1…Qxg7! 2.Rb4 Qg8!! 3.Rf4! Qxd5# II. 1…Qxb5! 2.Rg4 Qa6!! 3.Re4! Qxf6# Extreme activity of the black Queen which performs all six(!) black half-moves ! Here are particularly interesting the annihilation captures of Rg7 and Rb5 that lead to pins of the white Knights, the blocks realized by the white Rooks on e4 and f4, the reciprocal change of functions in both solutions. The finales are beautiful model pin-mates, realised in Meredith form.
Solutions of n°9 :
I.1…Rxc2 2.cxf4 Bc6! 3.Rd3 Qxf4# II. 1…Bxf3 2.cxb2 Rc6! 3.Bd3 Qxb2# A very non-standard and beautiful demonstration of the Black Grimshaw theme after critical moves of R-B ! This beautiful idea is complemented by the sacrifices by white Knight, white blocks on d3 and the surprising mates by the black Queen.
Solutions of n°10 :
I. 1…h1=Q+! 2.Ke3 e1=R+! 3.Kxd2 Qd5# II. 1…e1=B! 2.Ke2 h1=S! 3.Kf1 Sg3# A real sensation - White Rex Solus + black AUW ! This is probably one of the few possible positions of this kind ! On the other hand, it is obvious that there are no further strategic elements. But nevertheless the author deserves praise ! It is necessary to note that so far only a small number of problems of WinChloe’s options have been published. The world-known American magazine StrateGems was the first that very quickly (two months after the publication of the results from Belgrade Internet Ty 2017) began popularising this novelty from issue n°79/2017. Here follows a small series of problems of WinChloe’s options published in StrateGems. We start with two beautiful miniatures by the Australian maestro Geoff Foster. Solutions of n°11 :
I.1.Sc6 Bd1 2.Kd5 Kf4 3.d4 Bh5 and follow : 4.Rf7+ Bxf7# ! II.1.Ra4 Kg3 2.Ke3 Sd6 3.Re4 Bc4 and follow : 4.Sf5+ Sxf5# In a miniature form is realized a very nice complex : ideal mates, reciprocal change of functions between two duets of pieces : Ra7/Se7 and Bb3/Se4, self-blocks.Solutions of n°12 :
I. 1.Sf2 Sb4 2.Kc3 Se2+ 3.Kd2 Sg3 4.Ke1 Kc1 5.Sd3+ Sxd3# II. 1.Sf4 Ka1 2.Kc2 Sf3 3.Kd1 Kb2 4.Se2 Kb1 5.Sc3+ Sxc3# Another interesting Tanagra problem (only 5 pieces on the board !) in aristocratic form and with chameleon-echo ideal-mates ! The use of the fairy condition Alphabetic Chess seems very fruitful but so far this opportunity has been demonstrated only in two problems. The second opus is the following hs#7.Solutions of n°13 :
1.e4 Sa1 2.Kd4 Kd2! 3.Be3+ Ke2! 4.Sc1+ Ke1! 5.Sd3+ Kd1 6.Sb2+ Kc2 7.Sc4 Sb3# Round-trip Kc2-d2-e2-e1-d1-c2 and switchback Sb3-a1-b3. A surprising and beautiful finish with model mate ! In the following problem, the condition Minimummer is applied for the first time. The play here also seems very non-standard. Solutions of n°14
: I 1.Bf5! d5 2.Bxh7! d4 3.Bg8 dxe3 4.Bb3! e2# II. 1.Sf5! h5 2.Sxd6! hxg4 3.Se4 g3 4.Sc3! g2# Here the reciprocal change of functions is realised in the play of two duets of figures : Bg4-Se3 and Ph7-Pd6. White captures one of the enemy Pawns to prepare the final mate by the other. This annihilation + Zilahi is combined with taking away a field from the black King. Model-mates in Meredith form. Here follow two very interesting problems by the world-known American composer Kostas Prentos - the editor of the «Retro and PG» section in the American magazine StrateGems.Solutions of n°15 :
a) 1…Bxc4! 2.Sd3! Bf7 3.Kb5 Bg6 4.Kc4 Bxd3# b) 1…Rxd4! 2.Sd7! Rh4 3.Kd5 Rh7 4.Kd4 Rxd7# Double realisation of the «Rundlauf» theme, reciprocal change of functions between T-F and of course, a very interesting demonstration of the Umnov theme after the key moves : In a) 1…Bxc4! 2.Sd3! and in b) 1…Rxd4! 2.Sd7! Excellent !Solutions of n°16 :
a) 1.Bb2 Bxb3 2.c3 Bf7 (2…Be6?) 3.Kc2 Be8 4.Kb3 Ba4# b) 1.Be1 Bxa3 2.Kc3 Be7 (2…Bd6?) 3.Kb2 Bg5 4.Ka3 Bc1# A similar theme to n°12 but here the double «Rundlauf» is performed only by the black Bishop, the finales are model mates and the form is Meredith. This light problem, n°17, demonstrates a nonstandard opportunity in this arena - a twin with the replacement of the fairy-condition : a
) 1…Bb1! 2.Nf1 Ba2 3.Ng3 Bd4# b) 1…Be3! 2.Ng5 Bc1 3.Na2 b2# Model-mates in both twins.The following Tanagra problem is the first one in which the fairy part B contains two fairy conditions.
n°18. We assume conditionally that the absent 1st half-move of Whitе is a «zero-ply». Therefore here the orthodox part A has 2 half-moves (plies) - the 1st black half-move as a 2nd ply and the 2nd white move as a 3rd ply. But from the 4th ply the fairy part B starts, which is a combination of two fairy conditions : Disparate and Anti-Andernach !
I. 1…Rb6 2.c4 end of the Orthodox part A ! Now follows the fairypart B : 2…Qd7(wQ)!! This is not a self check by Black because the white Queen is paralysed for White ! 3.c5(bP)+!! This is a check to the black King from the unparalysed white Queen plus black block on c5! 3…Kxd7# ! This is mate to the white King because the black King paralyses the white King ! Similarly : II. 1…Kc5 2.Kh5 Qd6(wQ)!! 3.c4(bP)+!! Kxd6#! Chameleon-echo model mates after rich strategic play. I have no doubt that every reader, looking at problems n°1-14, will be surprised at the fact that in almost all cases (with little exceptions) the composers have used only the stipulation hs#. The explanation is simple : this is probably so because the first fairy T.T. of Belgrade Internet Ty 2017 had the same stipulation and then probably «by some inertia» many authors began, also only with hs#, using conditions Minimummer or Maximummer.Of course, we can use WinChloe’s options with all other stipulations but so far there have not been any such published problems.
2. Use of the option ConditionsUntilPly=nHere we have : a fairy half-moves (plies) + B orthodox half-moves (plies). As far as I know, the first problem of this type by me was published on September 17th 2017 in the famous Bulgarian problem site KoBulChess. The problem is dedicated to Christian Poisson ! We will analyse this problem in detail, but first of all let us consider some easy examples of this type.
n°19. Here option ConditionsUntilPly = 1 means that condition AntiAndernach is applied only on the first ply (white key-move). After this, two orthodox half-moves of Black and White follow.
Solution:
1.b8=S(bS)! blocusThe end of fairy condition is followed by two defences by a black Springer :
1…Sc6 2.Sab7#1…Sd7 2.Scb7#A small thematic try : 1.b8=B(bB)? blocus 1…Bc7 2.Be7# but 1…Ba7!
n°20. Here after the A fairy part (two half-moves) follows the B orthodox part (only one half-move which is a mating move). Because every #2 has only three half-moves (plies), it means that the 1st ply (white key-move) and 2nd ply (black defence) are under the condition Anti-Andernach, followed by the last 3rd orthodox ply (white mating move).
It is obvious that here we have to start (as the 1st ply) with the white Pawn b7, which can immediately promote into a black figure ! The black defence (2nd ply) is a move with this black piece that instantly turns into a white figure. So ends the first fairy phase of the problem, as the condition Anti-Andernach works only to this point (until ply = 2). Next we must give mate with the new white figure on the board. Of course : 1.b8=Q(bQ)? blocus is not a good idea, for example : one refutation 1…Qxe5+!
But another attempt : 1.b8=R(bR)? blocus deserves attention ! Now Black is in zugzwang and can only play with his Rook. But it’s interesting that this figure has a lot of opportunities for defence : 1…Rb8(wR)~ (along the 8th rank) ends the Anti-Andernach phase and is followed by the orthodox move : 2.Rd8# But there are here many additional black defences that show «Black correction» :
1…Rb7(wR)! 2.Rd7#
1…Rb6(wR)! 2.Rd6#
1…Rb5(wR)! 2.c6#
1…Rb4(wR)! 2.Rd4#
1…Rb2(wR)! 2.Rd2#
1…Rb1(wR)! 2.Rd1#
But 1…Rb3(wR)!! and no mate.
Correct is
1.b8=S(bS)! blocus and now follows :1…Sd7(wS) 2.Sb6#1…Sc6(wS) 2.Se7#1…Sa6(wS) 2.Sc7#Of course, using this matrix you can compose a lot of direct problems with similar contents with multiple promotions of white Pawns(s) !
n°21 : As we have already said, each problem with stipulation sh# (=, ==, etc…), ss# (=, ==, etc…) etc… in n moves, has n + 1 halfmoves (plies), since all the moves (except the last, combined move of both sides) are single moves. In other words, here we have in total seven half-moves, five of them under the condition of Sentinels. After this follow two orthodox halfmoves. Here’s how it works :
1.Rg2(+Pg7) 2.g8=Q 3.Qa2 4.Qb2(+Pa2) 5.Qxc2 (+Pb2) This is the end of the fairy-part with Sentinels condition. Now follows the orthodox finale : 6.Qc1+ Rxc1#The 22 is the best problem of all those I created using WinChloe’s options ! Here ConditionsUntilPly=4, means that condition AntiAndernach is valid only for the first four half-moves (plies) i.e. until the second move of White and Black. Then the remaining two halfmoves (the 3rd move of White and Black) are performed as orthodox half-moves !
Some explanations about the solutions : here alphabetic symbols (X, Y, Z for the key-moves and A, B, C for the second white moves) are needed to make easier to understand the cyclical motives in anti-dual attempts at the formation of black batteries.
Thematic complex :
1 - Super AUW, realised in three phases after cyclic six-fold promotions of the white Pawns ?7 and f7, that form black batteries ! This idea is combined with a double cyclic anti-dual after promotions of the rear battery pieces ! This idea and mechanism are shown for the first time ! Here’s the cyclic formula :
a) 1.X! - 2.A! B? C?
b) 1.Y! - 2.B! C? A?
c) 1.Z! - 2.C! A? B?
2 - The black moves : a) 1…Qxf3! ; b) 1…Qxc2! ; c) 1…Qxg2! aim to keep the black colour for the black Queen ; to open a line for the white Locust.
3 - During the play the black Queen makes a record number of moves - a total of 9 in the three phases ! I think that this form of Super AUW - with six different promotions to black figures from only two white Pawns (in addition with thematic cyclical anti-dual motives connected with the promotions of the rear figures of the black batteries !), is shown here for the first time in a fairy chess composition !
3. Combined play with both optionsThis is perhaps the most complex and interesting combination of options. Unfortunately, probably there are no published examples of this type so far. So we will briefly explain the definition with a light scheme.
For example, if the total number of half-moves is X, we can note :X = A orthodox half-moves + B fairy half-moves + C orthodox half-moves (combining ConditionsFromPly and ConditionsUntilPly).If we have a problem with total X = 10 half-moves, a possible variant can be the following : X = 3 orthodox half-moves + 4 fairy half-moves + 3 orthodox half-moves.
If we use the condition «Circe» below the diagram in the twin field of WinChloe we must write:
ConditionsFromply=4
ConditionsUntilPly=7
And in the conditions field we must of course mark «Circe».
In other words here our condition Circe «works» only between the 4th ply and 7th ply. All other halfmoves (from 1st to 3rd and from 8th to 10th) are orthodox plies !
Of course, it is not easy to compose sufficiently interesting problems with such a unique «formula» because it is rather difficult to motivate the necessity (expediency) from such a difficult structure ofhalf-moves (plies). I think that drawing up such tasks is to a large extent a test for the composer’s fantasy, or more precisely to say, a test of his ability for a non-standard mission !
Now follows a light example. Here we have : two orthodox halfmoves (plies) + three fairy (Sentinels) half-moves (plies) + two last orthodox half-moves (plies).
n°23 :
1.Rh2 2.Ra2End of first orthodox part which was intended to prepare for the next «blockade cannonade» of the condition Sentinels :
3.Rc2(+Pa2) 4.Rf2(+Pc2) 5.Rh2(+Pf2)End of fairy part. Now follows the second orthodox part :
6.Rh8+ Kxh8=Pay attention to the very important feature of the final part : after 6.Rh8+ the Sentinels condition no longer applies and a white Pawn is not put on the square h2 !
III. CONCLUSIONIn my opinion, almost certainly problems of WinChloe’s options have a great future. Of course, in this respect very thorough and comprehensive work is needed. Many tests are needed for the various features offered by the options. I am convinced that play with change of conditions within a single solution or single variant is a very avant-garde approach in Fairy Chess, although almost similar ideas have been defined and slightly experimented with in the past.
It would be very interesting to know if Christian is going to work in the future in this arena, since his options are certainly a very rich opportunity for a new discovery on the same basis !? I have a positive expectation in this regard !
Speaking of these non-standard forms, I would like to briefly mention that in the year 1999, in the German magazine, feenschach, (Heft 134, Band XXVI, November-December 1999) was published my article : «Problems and endgames with change of conditions (BW)». BW in German means BedingungsWechsel (change of conditions). In this long article (nine pages) dedicated to the 50th anniversary of the magazine, I offered different versions of BW, but my approach was quite different - a change of the fairy condition on every move by both sides (change after every two half-moves). Since I’m not a programmer, I cannot say at all whether it was possible to program my «BW» ideas. On this question, I cannot give an answer to this day…
Thus, the idea of BW is still not very popular and I have composed few originals of this type (some of them can be found in the databases of WinChloe).
But I must repeat : Christian Poisson’s approach to solving the problem about change of fairy conditions is fundamentally new and already works well ! So the name of such problems WinChloe’s options seems to me quite natural !
Petko Petkov, September 2018,with the help of Geoff Foster for translation in english |

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*Seetharaman Kalyan*

- Problem 436: Alexey Oganesyan - Selfmate (6 comments) More economic version saving a one White Pawn: Re5->e6, Pg6->h5, Pg7->...
- New on KoBulChess (2 comments) If you want to hookup with a girl on the internet, you'll find that th...
- Women chess composers (Updated) (18 comments) РђР·Р°СЂС‚ - РїСЂРёСЏС‚РЅР°СЏ СЌРјРѕС†РёСЏ, РґРµР»Р°СЋС‰Р°СЏ Р¶РёР·РЅС...
- Sad news: IM Krassimir Gandev passed away today (2 comments) Sad news. I always remember his miniature h#s with neutral Ps and Circ...
- Problems 978-980: Petko Petkov - Fairy (Anda, Anda Inverse) (2 comments) I really like the specific fairy effects which are possible. 978 and 9...
- Problem 976: Vladimir Pankov - Helpmate (4 comments) [quote name="Kostas Prentos"]This is practically a series-h# idea (wit...
- Zoran Gavrilovski 50 JT - Award (1 comments) I've great admiration for the composer Z. Gavrilovski. He composes and...
- Seetharaman - 64, Superguards Tourney 2013 Award (3 comments) Last year I sent book prizes to all the prize winners. Excuse the dela...