A peculiar sequence of events

There is a graph related to the geometry of a chess board. The vertices are the squares of the board. We are ignoring for the moment the knights, and are joining two vertices if there is a line of attack (horizontal, vertical, or diagonal) between them, i.e. if a queen can (reciprocally) reach one square from an other. In this graph, between two connected vertices there is a notion of distance, the cartesian distance between the centers of the corresponding squares.

Note that only the last check may be a knight check.

Let $j, J$ be the two kings. One of them is the King on $d1$, but we do not know which one. The colors are a while separated by using lower and capital letters. We do not know so far which of them got checked first. At any rate, the story must have been as follows:

• A piece $S$ moves, discovers the line of attack of the piece T, which gives a check to $j$.
• A piece $u$ interposes between $T$ and $j$, and gives a check to $J$.
• A piece $V$ interposes between $u$ and $J$, and gives a check to $j$.
• A piece $w$ interposes between $V$ and $j$, and gives a check to $J$.
• A piece $X$ interposes between $w$ and $J$, and gives a check to $j$.

It is given that S, the first piece moving, is either V or X, since each piece on the board gave a chess in the sequence of the five half-moves. All the time there were exactly $2+5=7$ pieces on the board.

In a picture using the lines of attack mentioned above, we have topologically a subgraph:

[T] (1)
 |
 |
[u] (2) 
 | \ 
 |  \ 
 |   \
 |    \
 |     \
 |      \
 |       \
 |        \
 |         \
 |          [V] (3)
 |         /  \
 |        /    \
 |       /      \
 |     /         \
 | (4)/           \
 | [w]----[X]-----[J]
 | /    .::(5)
[j]:::::

There is a final dotted line connecting $Xj$, which is either a connection in the graph, or not, and in the latter case it is a "night check".

• Let $\Bbb H$ be the half-plane delimited by the line $jJ$ of the two kings in the plane of the board, which contains the first piece $T$ that gives the first check. It is clear by convexity, that all pieces live in their final position in $\Bbb H$. And even more in the "solid" triangle $\Delta_T$ with $T,j,J$ as vertices, which is the convex hull of these points.

• Let $\Delta_u$ be the solid triangle (on the plane of the board) with vertices $u,j,J$. Then the "next" pieces $V,w,X$ live in their final position in this triangle, including boundaries.

• We may further continue with smaller solid triangle regions to restrict the next points.

• Which is the angle between the vectors $\overrightarrow{Tj}$ $\overrightarrow{uJ}$, measured by rotating $\overrightarrow{uj}$ (normed to length one) in the direction of $\overrightarrow{uJ}$ (normed to length one), seen as positive direction? It is a multiple of $45^\circ$ since two lines of attack in chess are so. It is easy to see that $135^\circ$ is excluded, because else in the triangle $\Delta ujJ$ the angle in $u$ is $135^\circ$, and no point $V$ can be furthermore chosen on $uJ$ to make an allowed angle $$ \widehat{uVj}< \widehat{uJj}< 180^\circ- \widehat{Juj}=180^\circ-135^\circ=45^\circ\ . $$

• So only angles of $45^\circ$ and $90^\circ$ are allowed in the process.

• Consider now the following configuration only:

    [u] (2) 
     | \ 
     |  \ 
     |   \
     |    \
     |     \
     |      \
     |       \
     |        \
     |         \
     |          [V] (3)
     |         /  \
     |        /    \
     |       /      \
     |      /        \
     |     /          \
     |    /           [J]
     |   /
     |  /
     | /
    [j]
  

  and in the triangle $\Delta ujJ$ the angle in $u$ is bigger than the angle in $V$ in $\Delta VjJ$.

• With this argument all vectors $\overrightarrow{Tj}$ $\overrightarrow{uJ}$, $\overrightarrow{Vj}$, $\overrightarrow{wJ}$, moved parallely to have the same origin, say $T$, stay in the half-plane $\Bbb H$, including its boundary, and have (strictly) increasing consecutive angles (with alternating orientationa), which are multiples of $45^\circ$.

• So it is not possible to have one more half-move, and in our situation

  • the angle from $\overrightarrow{uj}$ to $\overrightarrow{uJ}$ is the minimal possible angle $45^\circ$,
  • the angle from $\overrightarrow{Vj}$ to $\overrightarrow{VJ}$ is the minimal possible angle $90^\circ$,
  • the angle from $\overrightarrow{wj}$ to $\overrightarrow{wJ}$ is the minimal possible angle $135^\circ$.
  • the piece that interposes now and gives a check must be a knight, so a knight $S=X$ was the piece moved first, discovering the line of attack from $T$ to $j$.

• If $j$ is the black king, then the first line of attack was not a diagonal, since a night on a light square (initially on this diagonal) cannot move twice and attack also a light square like d1. (Initially it was on a light square, which attacks only black squares. The sames happens in two moves.) If it was the vertical, then $T$ is a rook or a queen on the $d$-file, without restriction of generality it is $Rd8$. Then $u$ interposes and we must have a diagonal check, so it is a light square, as $d1$. In particular $u$ is also a diagonally working piece, without restriction it is a queen on one of the squares $d7$, $d5$, $d3$. Without restriction, we can reflect the position if needed w.r.t. the $d$-line so that the white king is on the king side. The last piece $w$ gives a check and a knight interposes with check. This is only possible with a knight on the king side on $f2$. (The other square $e3$
  is easily excluded.) The only possibility for $w$ is in the final position $Re2$ or $Qe2$. So the white king is on the second line, and there is only one light square beyond $f2$, which is $g2$. So $u$ is a $Qd5+$ (or $Bd5$), $V$ is the $Qf3+$. And the only final position that may occur is:

$\hspace{2cm}$

But how can a $N$ move starting from the $d$-file twice, reach $f2$ in two moves, and not interfere with other lines of attack? It must start on a black square. The path $d4-(?)-d2$ is not possible, the remained black squares are $d2,d6$, and $f2$ can be reached only through $e4$. Which interferes. No such solution. (We would have one with a $-1$-line, the first knight move would put it on $e(-1)$, from where $f2$ can be reached.

• If $j$ is the black king, and the first line of attack is a horizontal, then we may assume that everything happens on the king side, and we deduce successively: A rook $T$ gives check from the first line, a bishop $u$ interposes, so the white king sits diagonally on its line of attack. A queen $V$ (or a bishop, but it is harder to come there as bishop, so in such situations we take the queen,) interposes and checks the $Kd1$, so it sits on a light square, so the white king is also on a light square. So the bishop $u$ is on a light square. Only $f1$ is available, but this leads to no solution. We would have a solution on a board with nine columns, when the final position is...

$\hspace{2cm}$

Which can be reached by $1.Ne1-c2+$ (from $Ri1$) $1\dots\ Qh2-h1+\ 2.Qd4-d3+ \ Qd2-e2+\ 3.Nc2-e3+$ - but we need that $i$-file.

• So $j$ is the white king, and the first piece giving the check is a black piece, and there is a knight move that discovers the line of attack. The same parity reason forbids a diagonal as first line of attack. (The black knight and the white king would be on a square of same color, after two (or after and even number of) moves the knight attacks the opposite color.) We can realize a position with a first check on a horizontal line as follows:

$\hspace{2cm}$

$$1\dots\ Nd4-b3+\ 2.Bh3-g4+\ Qe1-e2+\ 3.Qc3-d3+\ Nb3-d2+\ .$$ What about a vertical line? No, we would need a line below the black king, in fact even two lines for the first knight move, in order to realize

$\hspace{2cm}$

To conclude, only a solution with a horizontal first check is possible, the third diagram shows a possible realization.