Question

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final thoughts Let us start with a vector (or just specify the distinct entries) and then try to build a graph. With only 2 distinct entries the partition does need to be almost equitable. I think that it is easy to create high irregularity if some entries are 0. Here is an example cobbled up ad hoc (goals three distinct entries, none being zero. every vertex has neighbors with the other two labels, every label showing two types of vertices): Suppose that the entries are $1$,$2$,$-3$ (how many times each to be determined later). We can ignore edges between vertices with the same label. Suppose $u,v$ are two vertices labeled $1$ , that $u$ has $a$ and $b$ neighbors labeled $2$ and $-3$. Then the eigenvalue is $1(a+b)-2a+3b=-a+4b$ I'll pick $a=1$ and $b=5$ giving eigenvalue $19.$ Then $-c+4d=19$ where $c$ and $d$ are the number of edges from $v$ to vertices labelled $2$ and $-3$ I'll say that some have $c=5$ and $d=6$. Similar calculations and choices lead to this possible situation

$[p;1 \rightarrow 2^1,-3^5]$ meaning $p$ vertices labeled 1 each having 1 neighbor labeled 2 and 5 labeled -3

$[q;1 \rightarrow 2^5,-3^6]$

$[r;2 \rightarrow 1^3,-3^7]$

$[s,2 \rightarrow 1^8,-3^6]$-

$[t,-3 \rightarrow 1^{13},2^1]$

$[u,-3 \rightarrow 1^8,2^5]$ meaning $u$ vertices labelled -3 with 8 and 5 neighbors labeled 1 and 2 respectively

Now we have 6 equations in the (positive integer) variables which will need to be satisfied. Counting in two ways edges with ends labelled 1 and -3 gives $5p+6q=13t+8u$If my calculations are correct then one possible solution is $[p,q,r,s,t,u]=[21m,4m,3m,4m,5m,8m]$ I do not think that there would be any problem (other than tedium) in assigning edges to make that work (say with m=10 or 20). At any rate, the idea is clear.