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Please see my previous question for more background.


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

enter image description here


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).


See also multiplication chart + calculator (under the graph)

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

1 Answer 1

3
+100
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There are 384 solutions. Here's one:

a=10, b=4, c=2, d=13, e=7, f=5, g=6, h=9, i=12, j=14, k=11, l=8, m=3, n=15

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.

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  • $\begingroup$ Brilliant! How many solutions do u think there are? 🤔 $\endgroup$
    – stargirl
    Dec 1, 2023 at 7:19
  • $\begingroup$ Also, if u don't mind sharing, how exactly did u go about tackling this? $\endgroup$
    – stargirl
    Dec 1, 2023 at 7:20

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