In our previous post about Circe we said that it is an interesting Condition of the Fairy chess.
The world of the orthodox (directmate) problems has been described, many years ago, as a continent having very few unknown regions. Around this continent a magical sea is stretched, having many islands, some of which are uncartographed or unexplored. Many of these mythical islands are growing, just like the volcanic islands of the real world, and sometimes new islands emerge.
This is the case with “chess problem composition” (or “artistic chess” as we called it in Greece), where the fantasy of the composers endlessly creates new worlds, which the new composers are exploring, trying to map their limits.
In Homer's Odysseia, the island Aiaia (eea) of the nymphe witch Circe, doughter of Helios and Persa, was only one, but in Fairy Chess a whole archipelago of islands has been discovered with various kinds of the Circe Condition.
Let us see problem-202, by John Rice, which is composed in Zagoruiko frame, (that is, at least two defenses of the Black, always the same, are answered with different way in at least three Phases of the solution).
 | (Problem 202)
John Rice,
Third Price, Phoenix 1994
Mate in 2 moves. Circe.
#2 Circe (9+7) |
| [2Q1S2B/1q1BS3/3p2P1/3r1p2/3sk3/4P3/2P2Kb1/8] |
There are a few tries in this problem : {1.Bxd4(+bSb8)? Rxd4(+wBc1)!}, {1.Bxf5(+bPf7)+? Rxf5(+wBf1)+!}, {1.Sxd6+? Rxd6(+wSg1)!}, {1.Sf6+? Ke5!}.
Let us make a special note for the next two tries on the square c6 :
{1.Sc6? (is guarding e5 and is threatening 2.Sf6#).
If 1...Qxc6(+Sb1) 2.Sd2#. (Defence
a, Mate
A).
If 1...Sxc6(+Sb1) 2.Sc3#. (Defence
b, Mate
B).
But there is the defence 1...Sxc2!}.
{1.Bc6? (is pinning Rd5 and is threatening 2.Sxd6(+d7)#).
If 1...Qxc6(+Bf1) 2.Bd3#. (Defence
a, Mate
C).
If 1...Sxc6(+Bf1) 2.Bxg2#. (Defence
b, Mate
D).
If 1...Sxc2 2.Qxf5(+f7)#. (Defence
c, Mate
G).
If 1...Qc7 2.Bxd5(+Ra8)#. (Defence
d, Mate
H).
But there is the defence 1...Qxe7(+wSg1)!}.
Key: 1.Qc6! (is pinning Rd5 and is threatening 2.Sxd6#).
If 1...Qxc6(+wQd1) 2.Qd3#. (Defence
a, Mate
E).
If 1...Sxc6(+wQd1) 2.Qxd5(+bRa8)#. (Defence
b, Mate
F).
If 1...Sxc2 2.Bxf5(+bPf7)#. (Defence
c, Mate
I).
If 1...Qc7 2.Qxd5(+bRa8)#. (Defence
d, Mate
J).
Other variations : 1...Sb5 2.Bxf5(+bPf7)#, 1...Qb7-b4 / Qb8 2.Qxd5(+bRa8)#, 1...Qxd7(+wBf1) 2.Qxd5(+bRa8)# / Bd3#.
We see that in three (3) phases of the problem there are answers A-B C-D E-F for the two (2) defences a-b, that is we have a
Zagoruiko 3x2.
There are, also, changed mates G-H I-J for the defences c-d, as an additional asset of the problem.
We will see next a help-stalemate problem with condition Circe.
 | (Πρόβλημα 206)
Harry G. Polk,
Mat 1982,
Black plays and helps white to stalemate in 6 moves. Circe.
h=6 Circe (15+16) |
| [S1s1BQ2/2RpP1rp/PB1pPpPp/1S1sPpPK/4PrP1/5p2/p1b3k1/2q3b1] |
From the initial position only a white Rook is missing.
The Pawns can have this peculiar setting, because the captured pieces were not lost but regenerated, or reborn by Circe.
During the solution all black pieces are trapped and Black is stalemated.
Key : 1.Bxe4(+wPe2) exf3(+bPf7)
2.Qxc7(+wRal) Sbxc7(+bQd8)
3.Bxb6(+wBcl) Bxf4(+bRh8)
4.Rxf8(+wQdl) Qxd5(+bSg8)
5.Kxf3(+wPf2) Rxa2(+bPa7)
6.Rxe8(+wBfl) Sxb6(+bBf8) =
The following twenty captures are illegal, as the black King is exposed to check : a7xb6(+wSg1)+, Sxb6(+wSg1)+, Qxc7(+wSg1)+, Bxd5(+wQd1)+, Scxe7(+wPe2)+, Qxe7(+wPe2)+, Rxe7(+wPe2)+, Bxe7(+wPe2)+, Sgxe7(+wPe2)+, d7xe6(+wPe2)+, f7xe6(+wPe2)+, d6xe5(+wPe2)+, f6xe5(+wPe2)+, Kxf4(+wBc1)+, f7xg6(+wPg2)+, Rxg6(+wPg2)+, h7xg6(+wPg2)+, f6xg5(+wPg2)+, h6xg5(+wPg2)+, f5xg4(+wPg2)+.
Ending this post, we will see a Circe with theme Zappas.
 | (Problem 207)
Byron Zappas,
Die Schwalbe 1987,
Mate in 2 moves. Circe.
#2 Circe (12+13) |
| [1B2R2Q/P2p1P1p/1p5r/1P1kSB2/2psS1p1/1Pp1ps1p/4b3/3R3K] |
In theme Zappas, one flight is guarded by three white pieces and in two moves the flight is unguarded, and this is repeated cyclically. There must be three thematic tries.
Here we observe that the square e5 is guarded by Bb8, Re8 and Qh8.
Thematic tries :
{1.Sxc4(+bPc7)? Already Bb8 is blocked,
and the threat [2.Sxe3(+bPe7)#] will block Re8,
so there is a defense 1...Rf6! to block Qh8.
It is no good to play 1...Bxc4(+wSb1)? 2.Sxc3(+bPc7)#}.
{1.Sxg4(+bPg7)? Already Qh8 is blocked ,
and the threat [2.Sxe3(+bPe7)#] will block Re8,
so there is a defense 1...Rd6! to block Bb8}.
{1.Qf8? The Qh8 has stopped guarding e5 in order to hold c5
and now is threatening [2.Sxc3(+bPc7)#], but this blocks Bb8
so there is a defense 1...Re6! to block Re8}.
There are many more tries, making the problem difficult to solve : {1.f8=B? Rd6!}, {1.a8=Q+? / a8=B+? / a8=S? Rc6!}, {1.Sd3? / Sg6? / Sc6? cxb3(+wPb2)!}, {1.Sxf3(+bSg8)? Bxf3(+wSb1)+!}, {1.Sxc3(+bPc7)+? Kc5!}, {1.bxc4(+bPc7)+? Bxc4(+wPc2)!}.
Key : 1.f8=Q! (which is guarding c5 and is threatening [2.Sxc3(+bPc7)#] )
If 1...Rd6 2.Qxd6#
If 1...d6 2.a8=Q/B#
(
This post in Greek language).
