Twinning by exchange of two statical units 1

a) 1.Bd6? th. 2.Rxc4#, 1...Ra5!

1.Qxh4! th. 2.Qxe4#
1...cxd3, exf5, Sxd3
2.Bd6#, Rxd5#, c3#

b) 1.Qxh4? th. 2.Qxe4#, 1...Rh6!

1.Bd6! th. 2.Rxc4#
1...exd3, Rc5, Sxd3, b5
2.Qxh4#, Be5#, c3#, Sc6#

Both positions contain the try refuted by pinning the threat piece by black rook that is unpinned in try. The question is, which role plays the wp that is exchanged with wK. Do you see it? In b) position it avoids a cook 1.Ba3. In a) position it can be omitted.

a) 1.Rxf6! zz
1...S~, Sf7!, Sxe6!, Sxf3+
2.Qh7#, Rxh7#, Rxe6#, Rxf3#

b) 1.Qc5! th. 2.Qxg5#
1...S~, Sxh3!, Sxe6!, Sxf3+
2.Bh5#, Bxh3#, Bxe6#, Bxf3#

Perfect use of two batteries. Also, the use two exchanged pawns is very interesting. At the first place, any pawn on f6 avoids 1.Qxg5#. In a) wpg6 avoids 2.Rh6#, in b) bpg6 avoids 1.Rxf6. Although these two pawns aren’t in the middle of the problem idea, the twinning is well used.

a) 1.Bf2! th. 2.Qe2#
1...Qe6, Qxe5, Rxf2, Se3+, Se4
2.Sc5#, Sxe5#, Sxf2#, Qxe3#, Qxe4#

b) 1.Qc5! th. 2.Qc2#
1...Qc6, Qxc5, Rxf2, Se3+, Sc4
2.Se5#, Sxc5#, Sxf2#, Qxe3#, Qxc4#

Very symmetrical example of twinning that is inherent to the matrix. Black pawn in both positions prevents defence by bQ that would keep her eye at the square, where after defence on the 6th row follows the knight mate. White pawn then prevents defence by bR on 8th row. I think the idea of the problem could be expressed without twinning.

a) 1.Rff2! th. 2.Kd4#
1...Qxf2 2.Qa3#
1...axb4 2.Kxb4#

b) 1.Kd4! th. 2.Qa3#
1...Qxf2 2.Rfxf2#
1...axb4 2.Rxa7#

Djurasevic cycle with thematical defence 1...Qxf2 is powered by other change after defence 1...axb4. The role of exchanged pieces is not the same, but they are well used. In a) Sa7 prevents dual threat 2.Kb5# and cook 1.Kd4, in b) Sh6 attacks bishop battery line, forcing doublecheck variation mate. In a) ph6 prevents defences by Rh2, while in b) pa7 removes cook 1.Kb5. It means moving knight is a part cyclical mechanism while pawn is well used for making the twomover sound.

a) 1.Sf1+!
1...Kxf3, Kxf5, Qe3
2.Sxh2#, Sg3#, Sd2#

a) 1.Sc4+!
1...Kxf3, Kxf5, Qe3
2.Se5#, Qe5#, Sd2#

Checking keys lead to 3 pin mates in both positions. Bishop guards f4 from both sides of b8-h2 diagonal. Moreover it guards e5 in a) and g3 in b), effectively forcing the only possible key. And why there is ph2? It is acting after exchange in b) position - it guards d6.

a) 1.Qf7! th. 2.Qf4#
1...Bxc7 2.Sc5#
1...Bf5 2.Qxf5#

b) 1.Qb5! th. 2.Qd5#
1...Bxc7 2.Qxc4#
1...Bc5 2.Sxc5#

White must choose the key leading to the threat that guards the square initially blocked by bQ. Then follows the doublepin mate. The lines of wB and wR pinning the bQ must be closed in any case. And why not by bS? That’s why there is exchange of places of bQ and bS. Well done

a) 1.Qxc6! th. 2.Qe4#
1...Ra4 2.Qd5#
(1...Re5 2.Sh4#)

b) 1.Qb2! th. 2.Sh4#
1...Ra4 2.Qe5#

Here the situation is very similar. In b) position White must in key guard f6 to prevent strong defence 1...Qxf7. But in a) there is a difference - 1.Qb2 would be refuted by 1...c6+!

a) 1.Rg5? th. 2.Rxf5#, 1...Qe4!

1.Re7! th. 2.Bxe5
1...Rxc7 2.e3# (Qf3?)
1...Rxb4 2.Qf3# (e3?)

a) 1.Re7? th. 2.Bxe5#, 1...Qxc7!

1.Rg5! th. 2.Rxf5#
1...Rxc7 2.Qf3# (e3?)
1...Rxb4 2.e3# (Qf3?)

The mechanism of this problem is based on exchange of places of two pieces (bQ, bS) and as such it is probably the best example for 10th TT CCM of all given on this page. The square f3 is guarded by bS after its unpins from both d4 and e5, the same is true for e3 and bQ. The mates can follow only after opening a8-f3 line and hence we have dual avoidance in every variation.