Proof of non-existence of a (1-1) hexagram


Pieces are numbered as shown in the first figure, and cells of the board are lettered as shown in the following figures. When a piece is placed, its 'corners' correspond with 'nodes' of the board. We shall need a shorthand as follows. A cell letter followed by a piece number means that the specified piece is allocated to the specified cell. For example, A5 means that piece 5 is allocated to cell A, Q18 means that piece 18 is allocated to cell Q, and so on. Piece orientation is uniquely determined by the context, and we do not need a notation for orientation. The external node common to a pair of cells is denoted by the two cell letters, and its color is indicated by placing them in brackets, round for black and square for white. Thus, [RS] means that the external node common to cells R and S must be white. A number in round brackets means that a piece has at least that number of adjacent black corners, and square brackets are used for white corners. A cell letter followed by a number in brackets means that the cell requires a piece of the specified type. For example, L(4) means that cell L requires a piece with at least 4 adjacent black corners. Note that a piece with property (n) or [n] also has property (m) or [m] provided m is less than n. We use -> to mean 'implies', + to mean 'and', and v to mean 'or'. Finally we use the expression 'Insuf(n)' or 'Insuf[n]', to mean that a condition cannot be satisfied because there are Insufficient pieces with property (n), or [n], remaining to be allocated to cells.

Note that there is one piece with property (6) or [6], two pieces with property (5) or [5], three pieces with property (4) or [4], and six pieces with property (3) or [3].

The pieces 1 to 7 are the set of restricted hexagram pieces which, incidentally, can be arranged to give a (1-1) hexagram. There is no piece 8 because piece 7 is self-dual. Amongst the remaining pieces the edge-congruent pairs are (9, 11), (10, 12), (13, 15), (14, 16), (17, 19), and (18, 20). Any piece numbered 9 or higher may be replaced by its edge congruent partner without affecting the argument.

Suppose there exists a (1-1) hexagram. Then the exterior edge will comprise at most one line of consecutive black nodes, and at most one line of consecutive white nodes. If this were not true the board would be divided by one or more bands of one color which would not result in a (1-1) hexagram. One of these lines will be 15 or more nodes in length, and we assume it is black without loss of generality. We consider each of the cases (a) to (e) illustrated on the right. The color of a node is indicated by a dot. Note we have assumed that one end of the line of exterior black nodes occurs where the color changes, but not necessarily the other end, as illustrated. The five cases are sufficient for the proof. All others are equivalent to the illustrated cases after a rotation or mirror reflection.


Before considering the five cases in full detail we prove that piece 2 must be placed on a peripheral cell. After that the argument will be centred on which cell is occupied by piece 2.

(a) C(3)+G(3)+L(4)+P(3)+S(4)+R(3) -> 2 is peripheral
{The six (3) pieces include piece 2}
(b) C(4)+L(4)+S(4) -> 2 is peripheral
{The three (4) pieces include piece 2}
(c) As case (b)
(d) B(3)+C(4)+G(3)+L(4)+P(3)+S(3) -> 2 is peripheral {As case (a)}
(e) B(3)+C(4)+G(3)+L(4)+P(3) -> B(4)vG(4)vP(4) -> 2 is peripheral


Case (a)
G2 -> C(4)+L(5)+S(4) Insuf(4)
L2 -> G(4)+P(4)+S(4) Insuf(4)
P2 -> L(5)+S(5) Insuf(5)
S2 -> L(4)+P(4)+R(4) Insuf(4)
R2 -> C(3)+G(3)+L(4)+P(3)+S(5) -> L6+S4+(C7vG7vP7) -> Q5
Q5 -> A[4]+H[4]+M[4] Insuf[4]

Case (b)
C2 -> G(4)+L(4)+S(4) Insuf(4)
G2 -> C(5)+L(5) Insuf(5)
L2 -> C(4)+G(4)+P(4)+S(4) Insuf(4)
P2 -> C(4)+L(5)+S(5) Insuf(5)
S4 -> C(4)+L(4)+P(4) Insuf(4)

Case (c)
All conditions of case (b) apply to case (c)

Case (d)
B2 -> C4+G(3)+L6+P(3)+S(3) (Feasible and gives [RS])
B2 also -> A[3]+A(2) -> A5
A5 -> D[4]+H[4]+Q[4] Insuf[4]

C2 -> B(4)+G(4)+L(4) Insuf(4)
G2 -> C(5)+L(5) Insuf(5)
L2 -> C(4)+G(4)+P(4) Insuf(4)
P2 -> C(4)+L(5)+S(4) Insuf(4)
S2 -> C(4)+L(4)+P(4) Insuf(4)

Case (e)
B(2) -> C4+L6+A(3)+A[2] -> A7
but C4+L6 -> G7vP7vS7 Insuf 7
C2 -> B(4)+G(4)+L(4) Insuf(4)
G2 -> C(5)+L(5) Insuf(5)
L2 -> C(4)+G(4)+P(4) Insuf(4)
P2 -> C(4)+L4+S(3) (Feasible)
L4+C(4) -> C6
S(3) -> S7
C6 -> B10vB12 (Edge congruent so we select B10 and G12)

D1 -> A5+H[5]+Q[4] Insuf[4]
H1 -> D[4]+M[4]+Q[4] Insuf[4]
M1 -> H[5]+Q[5] Insuf[5]
Q1 -> H[4]+M[4]+R[4] Insuf[4]
R1 -> Q[5] -> Q3
H[4] -> H5 -> {D9+M11}v{D11+M9} -> A17
But 9 and 11 are edge-congruent so we select D9 and M11
We now have the edge cell assignments shown in the last figure. If a (1-1) hexagram exists these assignments are unique apart from trivial interchanges of edge-congruent pairs.

We now consider assignments to the internal cells. Only two assignments are possible to cell E. These are E18 and E20 which are edge-congruent pieces. But whichever is assigned we find that we need a similar edge-congruent piece in both cell F and cell I. This is sufficient to prove that a (1-1) hexagram cannot be realised.





























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