My interpretation of the question is that you want a checkmate position where none of the pieces can be removed and it remain a checkmate position. In that case I present the following:
In this setup the black king is in check by the white knights which cannot be captured and the king has no valid moves due to being blocked by black pieces or the white ...
4: (can probably be improved by a ton)
Chris in the comments claims to describe an 11-piece solution but I don't fully understand his comment so I hope he posts a solution with his answer. I'll update this answer in the future if I find a better answer (which I may soon)...
An alternate phrasing of the movement rules, which took a lot of toil to get a decent description of the Bishop: (Movements ending at the starting square are implicitly forbidden)
A bonk is a unit move (knight leap for the Knight, diagonal step for the bishop), except if this move would take them to a square off the board, the piece moves to the closest on-...
We can also have rotational (180° only, though) and mirror symmetry at the same time:
Please note that only four squares really use Swiss rules (labeled s in the diagram). All others are covered by plain chess moves.
Explanatory 'showcase' answer provided by OP: All three of the existing answers here made great contributions to solving the puzzle (@Graylocke, @samm82, and @berkeleybross - to whom I have awarded the checkmark for being first to find the final answer), but none of them quite described the intended solution path in full (without making any assumptions), ...
One possible answer is
It doesn't look very symmetric at first glance, but if you look more closely,
I think it's likely that there isn't any answer which does not follow this pattern. I'm pretty sure this property holds for all answers, with the following reasoning:
Let's classify the cells into four categories.
A B C D
C D A B
B A D C
D C B A
And let's ...
must be enough, and also required.
white has two unavoidable captures coming up: on white's next move, the
Black can prevent only one of those moves, so
the first objective is completed.
Continuing from there, white will play
next. It threatens black's F-pawn, and also the queen's rook or some other piece in front of it. All those pieces ...
Another proof of Albert's lemma (and one that I believe is much more elegant than the others presented):
I will prove a stronger lemma instead. Namely:
(A proof of the reduced statement follows. This proof is similar to Gareth's answer to the same question, but does not rely on an arbitrary choice of "leftmost".)
Here's (what I think is) a simpler proof of Albert's lemma than the one in loopy_wall's answer. We'll find either a king-path of 0-squares connecting N and S sides, or a king-path of 1-squares connecting W and E sides. The basic idea is to walk along the boundary between 0-squares and 1-squares until we reach an edge of the board. So here's an example board; ...
Every pair of zeros and every pair of ones are connected via some King chain
is confusing. If you mean that every $1$ can be reached from every other $1$ by the traversal rules (similarly for $0$), then I think 8 is a minimum. If we can have disconnected pairs,
gets you to 7.
As for a proof, there might be fertile ground in looking at what can stop a path (...
(Note: I have treated the question as a classical covering problem, while OP apparently intended that the occupied squares need to be attacked as well. I'm leaving the answer up anyway, since this interpretation yields an interesting puzzle too.)
Here's the biggest one I got:
It took surprisingly long to fiddle the placements so that everything fit, ...
This answer was a group effort between @Graylocke, @berkeleybross, and myself, with a little confirmation from @Enigma, so I'm more than happy to make this a community wiki answer (unless there are any objections or @Stiv wants the +2 rep for accepting XD).
The answers to the questions:
What you should have done is
@PaulPanzer made a summary of existing answers here (including some they gave themselves).
I want to provide a clearer visualization for each of the existing solutions:
Source code for the above program
You can solve the problem via integer linear programming as follows. Define a graph with one node per cell and an edge for each pair of cells that are a knight's move away from each other. For node $i\in N$, let $N_i \subset N$ be the neighbors of $i$, and let binary decision variable $x_i$ indicate whether node $i$ is selected. The problem is to minimize ...
There are 6 missing black pieces, and since the black pawns haven't moved, the bishop at c8 must have been captured by white knight. That gives 5 pawn captures, and so by counting, the third white knight must have promoted at g7.
That means the white knight at h8 must have been there before black pawn got into g6.
Now, consider white knight at h1. It must ...