Chinese Checkers — the laws

Every playable hole is identified by an axial hex coordinate $(q,r) \in \mathbb{Z}^2$, with the third cube coordinate implicit as $s = -q-r$.

Six directions connect adjacent holes:

$$D = {(1,0),\ (1,-1),\ (0,-1),\ (-1,0),\ (-1,1),\ (0,1)}$$

Two holes $u, v$ are adjacent exactly when $v - u \in D$. The listing above is in rotational order, so consecutive directions are $60^\circ$ apart. Only the set matters to the rules; the order is fixed so that direction indices are stable.

In the implementation, directions are an enumeration rather than a collection of vectors, which makes "there are exactly six directions" a structural fact: no seventh direction is representable.

The board's centre is a hexagon of radius four:

$$H_4 = {(q,r) \in \mathbb{Z}^2 : |q| \le 4,\ |r| \le 4,\ |q+r| \le 4}$$

It contains $1 + 6(1+2+3+4) = 61$ holes.

All three constraints are required. Dropping the bounds on $q$ and $r$ leaves $|q+r| \le 4$, which describes an unbounded diagonal strip rather than a hexagon.

Six triangular camps of ten holes each surround the hexagon. Camp $C_0$ is seated flush against the hexagon's $q = 4$ edge:

$$C_0 = {(q,r) \in \mathbb{Z}^2 : 5 \le q \le 8,\ -4 \le r \le -(q-4)}$$

Its columns hold $4, 3, 2, 1$ holes, decreasing outward to a single apex at $(8,-4)$. The triangle therefore points away from the centre, and its four-hole base lies against the hexagon.

The orientation is the whole content of this chapter, and it is easy to get wrong. The inward-pointing variant

$$C_0^{\text{bad}} = {(q,r) : 5 \le q \le 8,\ -q+5 \le r \le 0}$$

also has ten holes per camp and also yields a 121-hole board, so every cardinality check passes. But it meets the hexagon at the single hole $(5,0)$ instead of along a four-hole edge, leaving the six camps dangling from the hexagon's corners. It is not a six-pointed star.

Cardinality alone cannot detect this. The distinguishing property is contact with the hexagon: a correct camp has four holes adjacent to $H_4$, contributing eight camp-to-hexagon adjacent pairs, whereas the degenerate one has a single contact hole and a single adjacent pair.

The remaining camps are rotations of $C_0$. A $60^\circ$ rotation in axial coordinates is

$$R(q,r) = (-r,\ q+r)$$

and $C_i = R^i(C_0)$ for $i = 0, \ldots, 5$.

$R$ has order six, and $R^3 = -\mathrm{id}$. The second fact is the important one: it means the camp three positions away is the point reflection of the original,

$$C_{(i+3) \bmod 6} = -C_i = {-x : x \in C_i}$$

which is why a player's target is camp $i+3$ — it is geometrically opposite, directly across the centre.

Whether $R$ reads as clockwise or counter-clockwise depends on the rendering convention and is not fixed here. Nothing in the rules depends on the choice, only on $R$ having order six and the camps being indexed consistently.

Note that $R$ sends $(1,0) \mapsto (0,1)$, stepping backwards through the direction order of chapter 1. That is harmless but worth knowing when comparing indices.

The board is the disjoint union of the hexagon and the six camps:

$$V = H_4 \mathbin{\dot\cup} C_0 \mathbin{\dot\cup} \cdots \mathbin{\dot\cup} C_5,\qquad |V| = 61 + 6 \cdot 10 = 121$$

A correct construction satisfies more than its cardinality. Each camp must meet the hexagon in exactly eight adjacent pairs; the board graph must be connected; and the whole board must be centrally symmetric, $V = -V$.

A useful end-to-end check is that all six players have the same number of legal moves in the initial position — fourteen on the standard board. This follows from six-fold symmetry, so it fails loudly if a camp is misplaced.

Rendering the construction is the fastest way to catch a mistake:

               5
              5 5
             5 5 5
            5 5 5 5
 4 4 4 4 . . . . . 0 0 0 0
  4 4 4 . . . . . . 0 0 0
   4 4 . . . . . . . 0 0
    4 . . . . . . . . 0
     . . . . . . . . .
    3 . . . . . . . . 1
   3 3 . . . . . . . 1 1
  3 3 3 . . . . . . 1 1 1
 3 3 3 3 . . . . . 1 1 1 1
          2 2 2 2
           2 2 2
            2 2
             2

Each row $r$ is drawn at horizontal offset $2q + r$. Opposite camps sit diametrically across the centre, as chapter 4 requires.

Six players each own ten indistinguishable pieces. A position is an occupancy function

$$s : V \rightarrow P \cup {\varnothing},\qquad P = {0,\ldots,5}$$

with every hole holding at most one piece. Player $i$ starts with all ten pieces in camp $C_i$ and every other hole empty, so a valid position always has 60 occupied and 61 empty holes.

Player $i$'s target is the opposite camp, $O_i = C_{(i+3) \bmod 6}$.

A turn moves exactly one piece belonging to the active player, and is either one adjacent step or a sequence of jumps — never a mixture.

A piece at $x$ may step to $y$ exactly when

$$s(x) = i \ \land\ y - x \in D \ \land\ s(y) = \varnothing$$

that is, the destination is adjacent and empty. The piece vacates $x$ and occupies $y$; nothing else changes.

A piece at $x$ may jump in direction $d$ to $x + 2d$ exactly when the intervening hole is occupied and the landing hole is an empty board hole:

$$x+d \in V \ \land\ s(x+d) \neq \varnothing \ \land\ x+2d \in V \ \land\ s(x+2d) = \varnothing$$

The jumped piece is never captured or removed, and may belong to any player — legality depends only on the hole being occupied, not on who occupies it. The only occupancy changes are that $x$ becomes empty and $x+2d$ becomes the mover's.

Since $d \neq 0$, a jump displaces by $2d \neq 0$ and so can never land on its own origin. The jumped hole is exactly the midpoint of the hop.

A turn may chain arbitrarily many jumps, all performed by the same piece, changing direction freely between hops. The player may stop after any jump; continuing is optional.

Legality is evaluated against the position produced by the preceding jumps. That statement is true but routinely over-read, so it is worth being precise about what does and does not depend on the evolving position.

Because a turn moves only one piece and jumps never capture, the occupancy of every other hole is fixed for the whole turn. Writing $\Omega$ for the occupied holes excluding the moving piece, a jump from the piece's current hole $x$ is legal exactly when

$$x+d \in \Omega \ \land\ x+2d \in V \setminus \Omega$$

Only $x$, $d$, and the fixed set $\Omega$ appear. Within a single turn, therefore, occupancy is a function of the moving piece's position, and the available jumps depend on that position alone. The moving piece can never block itself, since it is excluded from $\Omega$.

The consequence is practical: the set of reachable destinations is the forward closure of a directed graph fixed once per turn, so a breadth-first search over positions, with a single visited set, computes it exactly and always terminates.

It is tempting to conclude instead that the search must be keyed on the pair (position, board state), on the grounds that the board changes after each hop. That is a mistake. Within a turn the state is determined by the position, so such a key can never distinguish two visits that the position alone would not — while making the search appear to need unbounded state. A search keyed that way does not terminate.

Jump sequences genuinely may revisit holes, including the starting hole: with a blocker adjacent, a piece can hop out and straight back. The space of jump paths is therefore infinite, and any procedure enumerating paths needs an explicit guard — forbidding repeats within the current path is the natural choice. The space of destinations is finite regardless, which is all move generation requires. A turn is conventionally required to end somewhere other than where it began, since ending at the origin is indistinguishable from not moving.

A move is identified by its kind, origin, and destination — not by the route taken. Distinct jump routes to the same hole produce the same resulting position, so they are the same move; counting them separately inflates move counts and any search built on them.

A route may still be recorded for animation or notation, but it must be excluded from equality and hashing.

The complete legal move set for player $i$ is the adjacent steps from each of their pieces, together with one jump move per reachable destination.

Applying a move vacates the origin and occupies the destination. For a jump sequence, no intermediate hole is modified and nothing is captured, so

$$s'(x) = \varnothing,\qquad s'(y) = i,\qquad s'(z) = s(z)\ \ \forall z \notin {x,y}$$

Replaying a route hole-by-hole and applying this net effect therefore agree, which is what justifies identifying moves by destination rather than by route.

Turns proceed cyclically through the six players.

The rules do not guarantee the active player has a legal move. The situation is reachable: if a player's ten pieces fill a camp, opponents can occupy that camp's frontier holes and the holes beyond them, leaving no step and no jump available. Such a player still holds all ten pieces, so this is neither a win nor a loss.

This specification resolves it by passing: a player with no legal move forfeits the turn and play continues. If all six players pass in succession the position cannot change, and the game is a draw.

One consequence is that the active player is not simply the turn number modulo six, since passing advances the player without consuming a turn in that sense. Implementations should track the active player as explicit state rather than deriving it.

Rule sets differ here. Passing is the least intrusive convention; others forbid the blocking configuration outright, or oblige the blocking player to move aside.

Player $i$ wins by occupying every hole of the opposite camp:

$$\forall x \in C_{(i+3) \bmod 6}:\ s(x) = i$$

Since the target camp has ten holes and the player has exactly ten pieces, this is equivalent to all of their pieces having arrived. The game ends at the first position satisfying this for some player.

Every completed move preserves the following. Each player owns exactly ten pieces; exactly 60 holes are occupied and 61 empty; every hole holds at most one piece. No move creates, destroys, or transfers ownership of a piece — jumping is not capture.

Two points are deliberately left open, and an implementation must choose explicitly rather than letting the choice emerge from its geometry code.

Camp restrictions. This specification uses the unrestricted convention: camp membership imposes no additional movement constraint, so a piece may enter, leave, or move within any camp subject only to the ordinary rules. Some rule sets restrict occupation of an opponent's camp. Such a restriction belongs in a separate legality predicate, not embedded in the movement rules.

Blocked players. See chapter 12. This specification passes; other conventions exist.