1

I had an idea for a type of stalemate: One side is stalemated with 8 pins in a legal position with as many pieces as possible, with all of the stalemating side’s units being “necessary.”

”Necessary” is defined as either pinning a piece, blocking a piece, or covering a king’s flight square. The king must be involved of course.

The best I came up with was 22 units. What is the optimal amount?

[FEN "b2r2q1/1R1BkR2/8/qN1K1Nq1/8/1P1N1P2/q2r2q1/2q1q1b1 w - - 0 1"]

As a bonus optional question, what is the maximum number of units in a legal position with 4 “natural” pins, which means no promoted pinning pieces?

  • 1
    I suppose you want to maximise the number of useful pieces, i.e. pieces that can't be removed without breaking the stalemate ? Hence it is not good enough to add a wPh4 or a BNa1 in your diagram ? – Evargalo Apr 5 '19 at 15:11
  • 1
    TBH, this task hurts a bit my sense of esthetics. It is usual to try to achieve a pattern with the least possible units, but to maximize the number of pieces is much less appealing to me. In this case, I wouldn't be surprised if you could include all 32 pieces (or 31, without the wQ), but with a lot of black pieces that could be replaced by one or two more effective ones, and useless blocked wPs. – Evargalo Apr 10 '19 at 20:06
1
 [FEN "b3r3/1R1p1p1q/1p1P1P1b/1Pk1BR2/1rN1K1Nq/p7/P1P1B1P1/1qn1q1nq w - - 0 1"]

Here is an (ugly) suggestion wth 29 units.

White is stalemated with 8 pins; removing any Black unit would break the stalemate. Black has promoted all of their four missing pawns (all into queens).

The three captures (white queen and 2 white pawns) ensure that this position is reachable. The three missing White pieces ensure that the position is legal and that Black could have promoted all four of their missing pawns.

The Black g-pawn captured the White h-pawn to produce doubled pawns that both promoted. The Black c-pawn captured the White queen to bypass the White pawn structure and promote. Finally, the White e-pawn was captured and then the Black e-pawn marched down and promoted.

This is optimal because 4 pawns must be promoted to create 8 pins, with requires a minimum of 3 units, thus leaving 29 left for the final position.

| improve this answer | |
  • @RewanDemontay I am afraid not... proof games are not my forte and it would take me way too much time. – Evargalo May 16 '19 at 13:02
  • Looking at it very fast, I cannot see either how I was intending to promote bPg7. To achieve 28 units, we just have to remove e.g. wPf2. – Evargalo May 16 '19 at 13:12
-1

Regarding the bonus question, for 4 “natural” pins, here is a position with 30 units. 31 is the theoretical maximum though, since the White queen would take to many units to stalemate.

[FEN "4r3/1b1pB3/1nRP1pkb/pPp4p/PrB1K1pP/1P2PpP1/2R2P2/Nqn4N w - - 0 1"]
| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.