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MatPlus.Net Forum General Construct the longest ser-x with a Black homebase

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(377) is still cooked:
1.Bc2 2.a4 3.a5 4.a6 5.Qb5 6.Qxb7 7.Qxc7 8.Qa5 9.c7 10.cxd8=B 11.Bxe7 12.Bxf8 13.e7 14.Rdg6 15.Rxg7 16.Rxg8 17.Qe5 18.Qxh8 19.Kb2 20.Kc3 21.Kd4 22.Ke5 23.Kf6 24.Kg7 25.f6 26.Bxh7 27.Rg6 28.g4 29.Bf4 30.Bh6 31.g5!=

Wow!
What about the ser-r= ? This cook doesn't work there because of Sb6.

Never mind, miscounted the moves.

Is even shorter with
1.Sd5 2.Sxe7 3.Sxg8 4.Bc2 5.Rxd7 6.Rxc7 7.Rxb7 8.Rxa7 9.Rxf7 10.c7 11.cxd8=B 12.Be7 13.Bxf8 14.e7 15.Qd4 16.Qxg7 17.Qxh8 18.Kb2 19.Kc3 20.Kd4 21.Ke5 22.Kf6 23.Kg7 24.f6 25.Bxh7 26.Rg6 27.g4 28.Bf4 29.Bh6 30.g5 Rxa2=

Well done, Arno!
(376) can serve as a backup record for ser-r=, waiting for possibly more(?).

(= 10+16 )
ser-r=30

After Arno's cook demonstrated that bPf7 has not to be captured in the previous scheme, here is a "corrected" scheme where bPf7 has to be captured(?!)

(= 9+16 )
ser-!=30

1.Bç2 2.a4 3.a5 4.a6 5.a×b7 6.b×ç8=Q 7.Q×ç7 8.Qg3 9.ç7 10.ç×d8=B 11.B×é7 12.B×f8 13.é7 14.Rf6 15.R×f7 16.f6 17.B×h7 18.Kç2 19.Kd3 20.Ké4 21.Kf5 22.Kg6 23.K×g7 24.K×h8 25.Q×g8 26.Rg3 27.Rgg7 28.g4 29.g5 30.g6 !=

The last scheme can be used for a ser-s= :

(= 10+16 )
ser-s=34

1.Bç2 2.Rb6 3.Sb5 4.a4 5.a5 6.a6 7.a×b7 8.b×ç8=Q 9.Q×ç7 10.Qg3 11.ç7 12.ç×b8=B! 13.Bd6 14.B×é7 15.B×f8 16.é7 17.Rf6 18.R×f7 19.f6 20.B×h7 21.Kç2 22.Kd3 23.Ké4 24.Kf5 25.Kg6 26.K×g7 27.K×h8 28.Q×g8 29.Rg3 30.Rgg7 31.g4 32.g5 33.g6 34.Sç7+ Q×ç7=

1.Rb3-b6 could not be included in the ser-!= because of cook 1.Rg3 2.R×g7 3.R×f7...

Not so immediate to get a ser-r= from this scheme...

The Ser-s=34 is cooked in 33 moves with
1.f6 2.fxg7 3.gxf8=B 4.Bg7 5.Bxh8 6.g4 7.g5 8.g6 9.g7 10.Sc4 11.Sb6 12.Sxd7 13.Sb6 14.Sxc8 15.Bc4 16.Rxb7 17.Rxb8 18.Rxa8 19.Kb2 20.Kb3 21.Ka4 22.Ka5 23.Ka6 24.Kb7 25.Kb8 26.Ra3 27.R3xa7 28.Ba6 29.Bb7 30.a4 31.a5 32.a6 33.exf7+ Kxf7=

Oups. Again well (and quickly) done, Arno! Indeed, there are considerably more final positions to calculate for ser-s= or ser-r= than for ser-!=...
So the 2 first moves in (387) have to be deleted...

(= 10+16 )
ser-s=32

(376) is cooked by
1.Bxc7 2.Bd6 3.Bxe7 4.Bf6 5.e4 6.e5 7.e6 8.e7 9.Se5 10.Sxd7 11.Sxf8 12.Sxh7 13.Sf8 14.Sg6 15.Sf4 16.g4 17.Bd4 18.Bxa7 19.Bxb8 20.a5 21.a6 22.a7 23.b4 24.b5 25.b6 26.c4 27.c5 28.c6 29.c7 Qxd2=

This still keeps the 30-mover with wBf3 unharmed but I feel…

(391) Posted by Arno Tungler [Wednesday, May 18, 2022 13:47]

Yes, the (385) did not survive…
1.Se5 2.Sxd7 3.Se5 4.Sxf7 5.Be4 6.Bxh7 7.Bd3 8.Kg6 9.Sh6 10.Sf5 11.g4 12.g5 13.e4 14.e5 15.e6 16.Bxc7 17.Bb6 18.Bxa7 19.Bxb8 20.a5 21.a6 22.a7 23.b4 24.b5 25.b6 26.c4 27.c5 28.c6 29.c7 Qxd3=

For (389) just one move is too much for the beneath. Maybe you have an idea how to shorten?
1.Sb5 2.Sxa7 3.a4 4.a5 5.a6 6.axb7 7.bxc8=Q 8.Qxc7 9.Qg3 10.Qxg7 11.Qxh8 12.Sc8 13.Sxe7 14.Sxg8 15.c7 16.cxd8=B 17.Be7 18.Bxf8 19.e7 20.Rg3 21.Kb2 22.Kc3 23.Kd4 24.Ke5 25.Kf6 26.Bh6 27.Kg7 28.f6 29.Bxh7 30.Rg6 31.g4 32.g5

And a second with 33 moves…
1.f6 2.fxg7 3.gxf8=S 4.Sg6 5.Sxh8 6.Sxf7 7.Sxd8 8.g4 9.g5 10.g6 11.g7 12.Bd3 13.Ba6 14.Bxb7 15.Bxc8 16.Rxb8 17.Sc2 18.Sb4 19.Sa6 20.Rxa8 21.Sb8 22.Ra3 23.R3xa7 24.Kb2 25.Kb3 26.Kb4 27.Ka5 28.Ka6 29.Kb7 30.a4 31.a5 32.a6 33.exd7+ Kxd8=

Hi, Arno. Thanks again. (376) and (385) are corrected by adding a white Pawn on h3.

With wPh3 you add other possibilities…
1.Se5 2.Sxd7 3.Se5 4.Sxf7 5.Kh5 6.Sg5 7.Bd5 8.e4 9.e5 10.e6 11.g4 12.h4 13.d4 14.Bxc7 15.Bxb8 16.Bxa7 17.Bb8 18.a5 19.a6 20.a7 21.b4 22.b5 23.b6 24.c4 25.c5 26.c6 27.c7 Qxd5=

Deleted

Ah! I should be ashamed as last cook by Arno is very similar to the one I found in (370B) by Miodrag.
A psychological bias I often experienced : it is easier to find cooks in problems by other composers than in one's own problems.
When adding wPh3 I noticed that the position included almost 8 wPawns. Now with Arno's last cook, this "dream" becomes true :

(= 12+16 )
ser-r=31

The anti-cook Ph3 also allows new possibility of adding one move as it prevents Qh2xh7
Solution ending 29.g3 30.Qa2 31.Qd5 Q×d5=

I am still wondering whether something like the beneath could lead to a correct ser-h= in this realm. Similar like the ser-= it would make full use of the 16 black pieces...

(= 14+16 )

ser-h=19

1.axb6 2.Rxa4 3.Rc4 4.Rc6 5.d5 6.d4 7.d3 8.dxc2 9.Qd2 10.Sd7 11.Rc3 12.Rxe3 13.Rxg3 14.Qxh6 15.Qg6 16.h6 17.Qh7 18.Rg6 19.f6 Bxc7=

This is most likely cooked but if we could shorten the solution somehow, it could maybe become a correct problem?!

EDIT: Added a wRg3 to avoid the cook 1.axb6 2.Sc6 3.d5 4.d4 5.d3 6.dxc2 7.Qd4 8.Qf4 9.Bd7 10.Rd8 11.Bc8 12.Rd7 13.Qxh6 14.Qg6 15.h6 16.Qh7 17.Se5 18.Sg6 19.f6 Bxc7=

Here is also a longer version - as indicated by the theme of this thread. However, I feel that with this stipulation we should try to have it realized with the minimum number of moves…

(= 16+16 )

Ser-h=22

1.axb6 2.Rxa6 3.Rxa5 4.Rxa4 5.Rd4 6.Rxd6 7.Rc6 8.d5 9.d4 10.d3 11.dxc2 12.Qd2 13.Sd7 14.Rc3 15.Rxe3 16.Rg3 17.Qxh6 18.Qg6 19.h6 20.Qh7 21.Rg6 22.f6 Bxc7=

You can get countable infinite moves with a few fairy conditions, obviously.

(= 8+16 )

Series double stalemate in 24
Alphabet Chess
No capture

(= 1+16 )

Series selfstalemate in 27
Single time use wormholes a6->b3, b6->c3, c6->d3, d6->e3, e6->f3, f6->g3, g6->h3, h6->a3
No capture

(= 1+16 )

Series selfstalemate in 3+24*n
n-times use wormholes a6->b3, b6->c3, c6->d3, d6->e3, e6->f3, f6->g3, g6->h3, h6->a3
No capture

Or even easier:
(= 1+16 )

Series selfstalemate in 3+3*n
n-times use wormhole a6->a3
No capture

Sorry, this is the wrong thread for this stuff.