# Nimber mnemonic combinatorial puzzle

Please see my previous question for more background.

The following represents an unfolded version of PG(3,2) with 1 as the center point:

Given that each number must be an end point of a line which passes thru the origin at 1, how can the numbers 2-15 be added such that each line thru the origin has as it's endpoints numbers whose nim product is 1 (eg. the pairs 2/3, 4/15, 5/12, 6/9, 7/11, 8/10 & 13/14)

Hint The problem reduces to solving the 3 triangles with 1 in the middle (the pieces that fold into the center). It is easy to randomly fill in the lines, but the goal is to get each of the 3 triangles to satisfy being valid nimonics (eg. straight lines have as their midpoint their product).

• Can you please label the unknowns in the figure so that it is clear which ones need to take the same value? Nov 30, 2023 at 22:30
• @RobPratt Updated! Nov 30, 2023 at 22:59
• Thanks. For the dotted triangles, which entry is the midpoint? Nov 30, 2023 at 23:04
• @RobPratt Corners are strange. j*h=b (eg. 2 points on the dotted have their shared corner as the sum) Nov 30, 2023 at 23:06
• *product not sum Nov 30, 2023 at 23:16

I used integer linear programming as follows. Let $$P=\{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o\}$$ be the set of positions, where position $$o$$ must take value $$1$$. Let $$V=\{1,\dots,15\}$$ be the set of values. For $$i,j\in V$$, let $$m(i,j)$$ be the nim product of $$i$$ and $$j$$. Let $$L$$ be the set of lines $$(p_1,p_2,p_3)$$ for which the value assigned to position $$p_2$$ must be the nim product of the values assigned to positions $$p_1$$ and $$p_3$$. For $$p\in P$$ and $$v\in V$$, let binary decision variable $$x_{p,v}$$ indicate whether position $$p$$ is assigned value $$v$$. The constraints are: \begin{align} \sum_{v \in V} x_{p,v} &= 1 && \text{for p \in P} \tag1\label1 \\ \sum_{p \in P} x_{p,v} &= 1 && \text{for v \in V} \tag2\label2 \\ x_{o,1} &= 1 \tag3\label3 \\ x_{p_1, v_1} + x_{p_3, v_3} - 1 &\le x_{p_2, m(v_1,v_3)} &&\text{for (p_1,p_2,p_3) \in L, v_1 \in V, and v_3 \in V} \tag4\label4 \end{align} Constraint \eqref{1} assigns one value to each position. Constraint \eqref{2} assigns each value to one position. Constraint \eqref{3} assigns position $$o$$ the value $$1$$. Constraint \eqref{4} enforces the logical implication $$(x_{p_1, v_1} \land x_{p_3, v_3}) \implies x_{p_2, m(v_1,v_3)}$$ that expesses the requirement that the midpoint is the nim product of the two endpoints.