How to prove the sum of squares degrees $d^2_{1}+d^2_{2}+\cdots+d^2_{95}\le 300000$ in a graph

Question

Let $G$ be a simple graph with $95$ vertices,$2021$ edges,and vertex degrees $d_{1},d_{2},\cdots,d_{95}$,show that $$d^2_{1}+d^2_{2}+\cdots+d^2_{95}\le 300000$$

My attempt: Using Euler's theorem $$d_{1}+d_{2}+\cdots+d_{95}=2\cdot 2021=4042$$,and note $d_{i}\in N^{+}$.By the nature of the inequality, as if by an adjustment, but then How to prove $$d^2_{1}+d^2_{2}+\cdots+d^2_{95}\le 300000$$

Answer 1

A solution for a weaker inequality:

We know that $\sum d_i = 2\cdot 2021$, and $0\le d_i \le 94$. Now you can try to solve $\max \sum d_i^2$, where the conditions are as above. The numbers have to be as extreme as possible, since $a^2 + b^2 \le (a-\epsilon)^2 + (b+\epsilon)^2$, if $a\le b$. Therefore, we get as many numbers as possible to be $94$, and the rest $0$, and whatever is left. The fact is $4042 = 94 \cdot 43$, so the remainder is $0$.

We conclude $\sum d_i^2 = 43 \cdot 94^2 = 379948$

Now, the sequence of numbers $(0,0,\cdot, 0, 94, 94, \ldots 94)$ cannot be the degree sequence of a graph with $95$ vertices and $2021$ edges, so more analysis is needed.

$\bf{Added:}$

Consider a graph with $n$ vertices and $e$ edges that is a maximizer for $\sum d_i^2$. Let $k$, $l$ vertices that are not joined, and $i$, $j$ vertices that are joined. Let's alter the graph by moving the edge $(i,j)$ to edge $(k,l)$. The sequence of degrees changes only locally as

$$ d_k, d_k, d_i, d_j \mapsto d_k+1, d_l+1, d_i-1, d_j-1$$

and the objective changes from

$$ S + d_k^2 + d_l^2 + d_i^2 + d_j^2 \mapsto S+ (d_k+1)^2 + (d_l+1)^2 + (d_i-1)^2 + (d_j-1)^2$$

The change is $$\Sigma' - \Sigma = 2( (d_k + d_l) -(d_i+d_j) + 2)\le 0$$ that is $$d_k + d_l +2 \le d_i + d_j$$

An important conclusion:

if $i$, $j$ are joined, and $d_k + d_l \ge d_i+ d_j$, then $k$, $l$ are also joined.

How does the matrix of a graph with this property look like? Assume that the vertices are ordered such that $$d_1\ge d_2 \ge \ldots d_n$$

Then the matrix of the graph with the above property is such that the $1$ entries form a symmetric Young diagram (except the diagonal entries). Let's keep this in mind. Now, how to get $\sum d_i^2$ maximized? Build the graph matrix ( the Young diagram) by adding figure $L$ shapes of thickness $1$ and maximal size, starting from upper left corner and going down the diagonal ( basically because $a^2+b^2 < (a-1)+(b+1)^2$), skipping details). It's a fern growing in layers of thickness $1$, and maximal size, root being the top left corner.

Now, to our concrete case. We have $n=95$ and $e=2021$. Now, the optimal graph has the shape of $\Gamma$ with maximal thickness and edge $95$, and to it, from the inside, we add another $\Gamma$ shape of thickness $1$.

Now, the number of $1$ in an $\Gamma$ of edge $95$ and thickness $k$ is

$$2 \cdot k \cdot 95 - k^2 -k = k(189-k)$$

This should not exceed $2 e = 2\cdot 2021 = 4042$. For $k=24$ we get $24(198-24) =3960 < 4042$, while for $k=25$ we get $25(198-25)=4100>4042$. Hence $k=24$.

Now, there are $4042- 3960= 82$, $1$'s left, which will get us a $\Gamma$ shape of thickness $1$ and edge $42$ containing $2\cdot 41 = 82$ $1$'s.

Let's list the degrees of the graph (the left multiplier is the number of times it appears)

$$24 \cdot 94,\ 1\cdot (24+41=65),\ 41(24+1=25),\ 29\cdot 24$$

We get $$\sum_{i=1}^{95} d_i^2 = 24\cdot 94^2 + 65^2 + 41 \cdot 25^2 + 29\cdot 24^2= 258618$$

the maximum possible.

Answer 2

You can find the maximum via integer quadratic programming as follows. Let $n=95$ and $m=2021$. For $i\in\{1,\dots,n\}$, let integer decision variable $d_i\in[0,n-1]$ be the degree of node $i$. The problem is to maximize $\sum_i d_i^2$ subject to \begin{align} \sum_i d_i &= 2m \tag1\\ d_i &\ge d_{i+1} &&\text{for $i\in\{1,\dots,n-1\}$} \tag2\\ \sum_{i=1}^r d_i &\le r(r-1) + \sum_{i=r+1}^n \min(r,d_i) &&\text{for $r \in \{0,\dots,n-1\}$} \tag3 \end{align}

Constraint $(1)$ forces the graph to have $m$ edges. Constraint $(2)$ forces the degree sequence to be nonincreasing. Constraint $(3)$ is the Erdos-Gallai condition: https://mathworld.wolfram.com/GraphicSequence.html

The optimal objective value turns out to be $$24\cdot94^2+1\cdot65^2+41\cdot25^2+29\cdot24^2=258618.$$

Answer 3

List all degrees in decreasing order so that $94\ge d_1\ge d_2\ge\cdots\ge d_{95}$ For optimal graph, if there's edge (i,j) but no edge (j,k), (i,j,k are distinct vertexes) we have $d_i\gt d_k$, otherwise, replace edge (i,j) by edge (j,k) so could get another graph with larger degree square sum.

Now checking the adjacency matrix of the graph, we could find that for column i and column j of the matrix where $i\lt j$, except for elements in diagonal (i,i), if (j,x) is 1, (i,x) must be 1 too. A valid candidate adjacency matrix looks like below

0 1 1 1 1 1 0 0

1 0 1 1 0 0 0 0

1 1 0 1 0 0 0 0

1 1 1 0 0 0 0 0

1 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Since we have $A\begin{pmatrix}1\\1\\\cdots\\1\end{pmatrix}=\begin{pmatrix}d_1\\d_2\\\cdots\\d_n\end{pmatrix}$ (where n=95), so we have $\begin{pmatrix}1&1&\cdots&1\end{pmatrix} A^2\begin{pmatrix}1\\1\\\cdots\\1\end{pmatrix}=d_1^2+d_2^2+\cdots+d_n^2$ which is sum of all elements in $A^2$.

Now let's take a look at matrix $A^2$ and found element (i,j) of the matrix should be $min\{d_i,d_j\}$ or $min\{d_i,d_j\}-1$, and it is $min\{d_i,d_j\}-1$ iff the element (i,j) in A is 1, so we have $d_1^2+d_2^2+...+d_n^2 = 1d_1+3d_2+5d_3+...+(2n-1)d_n -2e$, where e is total number of edges

or

$\begin{cases}n-1\ge d_1\ge d_2\ge\cdots\ge d_n\ge 0\\ d_1+d_2+\cdots+d_n=2e\\ d_1^2+d_2^2+\cdots+d_n^2 = d_1+3d_2+5d_3+\cdots+(2n-1)d_n-2e \end{cases}$

More generally, some vertexes has same degrees we could merge them and generate new constrains $\begin{cases} 0=i_0\lt i_1\lt i_2\lt\cdots\lt i_{t-1}\lt i_t=n\\ n-1\ge d_1\gt d_2\gt\cdots\gt d_{t-1}\gt d_t\ge d_{t+1}=0\\ (i_1-i_0)d_1+(i_2-i_1)d_2+\cdots+(i_t-i_{t-1})d_t=2e\\ (i_1-i_0)d_1^2+(i_2-i_1)d_2^2+\cdots+(i_t-i_{t-1})d_t^2= \\ (i_1^2-i_1-i_0^2+i_0)d_1+(i_2^2-i_2-i_1^2+i_1)d_2+\cdots+(i_t^2-i_t-i_{t-1}^2+i_{t-1})d_t \end{cases}$ and find the maximal value of $(i_1-i_0)d_1^2+(i_2-i_1)d_2^2+\cdots+(i_t-i_{t-1})d_t^2$

Now let's allow $d_h$ to be real number since we only need get an upbound of it. Using Lagrange Multiplier, we could get $ \begin{cases} 2d_h-u\times(i_{h-1}+i_h)+u-v=0&(2\le h\le t-1)\\ d_{h+1}+d_h-2u\times i_h+u-v=0&(1\le h\le t-1) \end{cases} $ and the range of h could include 1 or t if $d_1$ or $d_t$ does not take boundary value (n-1 or 0)

Let $e_h=d_h+\frac {u-v}2, j_h=u\times i_h$, we have

$\begin{cases} 2e_h=j_{h-1}+j_h&(2\le h\le t-1)\\ e_{h+1}+e_h=2j_h&(1\le h\le t-1) \end{cases}$

and we could get $\begin{matrix} 2j_h=j_{h+1}+j_{h-1}&(2\le h\le t-2) \end{matrix}$

More generally, if at least three $d_h$ does not take boundary value (n-1 or 0), for example $n-1\gt d_2 \gt d_3\gt d_4 \gt 0$ , and let's only treat $d_2,d_3,d_4,i_2,i_3$ as variable and all others as constant. We could use uppercase character to represent constant and lowercase variable to make the equations more readable, we have

$\begin{cases} I_1\lt i_2 \lt i_3 \lt I_4 \\ D_1\gt d_2 \gt d_3 \gt d_4\gt D_5 \\ (i_2-I_1)d_2+(i_3-i_2)d_3 +(I_4-i_3)d_4 = A\\ (i_2-I_1)d_2^2+(i_3-i_2)d_3^2+(I_4-i_3)d_4^2 = (i_2^2-I_1^2)d_2+(i_3^2-i_2^2)d_3+(I_4^2-i_3^2)d_4+B \end{cases}$

We need find maximal value of $(i_2-I_1)d_2^2+(i_3-i_2)d_3^2+(I_4-i_3)d_4^2= (i_2^2-I_1^2)d_2+(i_3^2-i_2^2)d_3+(I_4^2-i_3^2)d_4+B$

By Lagrange Multiplier, we have

$\begin{cases} 2(i_2-I_1)d_2-u(i_2^2-I_1^2)-v(i_2-I_1)=0\\ 2(i_3-i_2)d_3-u(i_3^2-i_2^2)-v(i_3-i_2)=0\\ 2(I_4-i_3)d_4-u(I_4^2-i_3^2)-v(I_4-i_3)=0\\ d_2^2-d_3^2-u(2i_2d_2-2i_2d_3)-v(d_2-d_3)=0\\ d_3^2-d_4^2-u(2i_3d_3-2i_3d_4)-v(d_3-d_4)=0 \end{cases}$ That's equivalent to

$\begin{cases} 2d_2-u(i_2+I_1)-v=0\\ 2d_3-u(i_3+i_2)-v=0\\ 2d_4-u(I_4+i_3)-v=0\\ d_2+d_3-2ui_2-v=0\\ d_3+d_4-2ui_3-v=0 \end{cases}$

And finally we get $I_4-i_3=i_3-i_2=i_2-I_1=\frac{I_4-I_1}3=\Delta I\gt 0, d_2-d_3=d_3-d_4=\Delta d\gt 0$, where $\Delta I\gt 0$ is constant，and $\Delta d\gt 0$ is the a variable to be solved.

Let's use symbol $I_2=I_1+\Delta I, I_3=I_1+2\Delta I$, so where the local extreme value is reached, we have $i_2=I_2,i_3=I_3$ Next step we have $d_3=\frac{A}{3\Delta I}$ and

$(\frac{A}{3\Delta I}+\Delta d)^2+(\frac{A}{3\Delta I})^2+(\frac{A}{3\Delta I}-\Delta d)^2=(I_2+I_1)(\frac{A}{3\Delta I}+\Delta d)+(I_3+I_2)\frac{A}{3\Delta I}+(I_4+I_3)(\frac{A}{3\Delta I}-\Delta d)+\frac{B}{\Delta I}$

or $2\Delta d^2+4\Delta I \Delta d+H=0$, because $\Delta I\gt 0$, If the equation of $\Delta d$ have real roots, at least one of them is negative. And if it has positive root, the negative one has larger absolute value.

And the result of target function becomes $\Delta I (d_2^2+d_3^2+d_4^2)=(\frac{A}{3\Delta I}+\Delta d)^2+(\frac{A}{3\Delta I})^2+(\frac{A}{3\Delta I}-\Delta d)^2=\frac{A^2}{3\Delta I^2}+2\Delta d^2$ So it takes larger value for negative root of $\Delta d$.

If we add constrain $i_2=I_2,i_3=I_3$ to original problem and ignore inequality constrain $D_1\gt d_2 \gt d_3 \gt d_4\gt D_5$ to get a problem

under constrain

$\begin{cases} d_2+d_3 +d_4 = \frac{A}{\Delta I}\\ d_2^2+d_3^2+d_4^2 = (I_2+I_1)d_2+(I_3+I_2)d_3+(I_4+I_3)d_4+\frac{B}{\Delta I} \end{cases}$

to get maximal value of $(I_2+I_1)d_2+(I_3+I_2)d_3+(I_4+I_3)d_4$

The geometry meaning of it is to find range of plane of given direction to intersect with a circle in 3D space. So there must be exact one maxima and one minima which is corresponding to the two roots of above equation of $\Delta d$ and the positive root must be corresponding to the minima. So it means in the original problem, the positive root of $\Delta d$ could not be the maxima too (but it could not e minima too). So it means the maximal value could only be reached in boundary condition (Smaller t required).

Next we need to enumerate situations with at most two $d_h$ doesn't take boundary value (n-1 and 0). It takes a lot of effort but we could verify all of them does not exceed 300000. For example,

$n-1=D_1 \gt d_2 \gt d_3 \gt 0, 0=I_0\lt i_1 \lt i_2 \lt I_3=n$

We could result in $\begin{cases}D_1-d_2=d_2-d_3=\Delta d\\I_3-i_2=i_2-i_1=\Delta i\end{cases}$ and $\begin{cases}\Delta i\Delta d=\frac{n(n-1)-2e}3\\\Delta i^2-\frac35\Delta i-\frac{n(n-1)-2e}3=0\end{cases}$

The extreme value is $\frac53 (n(n-1)-2e)\Delta d-n(n-1)^2+4e(n-1)=246881.293\lt 300000$

And the global maxima is taken at $n-1=D_1 \gt d_2 \gt 0, 0=I_0 \lt i_1 \lt I_2=n$, we could get $d_2=\frac{2e-(n-1)i_1}{n-i_1}, i_1^2-(2n-1)i_1+2e=0$ and the global maxima value is $259781.511 \lt 300000$

Another approach is to use computer find exact solution:

Using the attribute of the adjacency matrix, we could easily get a formula to calculate the result efficiently by computer, assume f(n,E) is the max square sum of degrees of graph with n vertexes and E edges, By removing the first column of and first row of the adjacency matrix, we could find that there're still $E-d_1$ edge left in a graph with only $a=d_1$ vertexs available so that $f(n,E)=\max_{\begin{cases}a(a+1)\ge 2E\\a\le n-1\end{cases}} \{f(a,E-a)+4E-3a+a^2\}$. Using a simple C code below we could find f(95,2021)=258618.

#include <stdio.h>
#define MAXN 100
#define MAXE (MAXN*(MAXN-1)/2)

int ds[MAXN+1][MAXE+1];

int main()
{
    int n,e,a;
    ds[2][1]=2;
    for(n=3;n<=MAXN;n++){
        for(e=1;e<=n*(n-1)/2;e++){
            a=n-1;if(a>e)a=e;
            for(;a*(a+1)>=2*e;a--){
                int v=ds[a][e-a]+4*e-3*a+a*a;
                if(v>=ds[n][e])ds[n][e]=v;
            }
            printf("ds[%d,%d]=%d\n",n,e,ds[n][e]);
        }

    }
    return 0;
}

And for E small enough such as E<n, it is easy to show that the best result is achieved when all edges are from a same vertex. According to the formula $f(n,E)=\max_{\begin{cases}a(a+1)\ge 2E\\a\le n-1\end{cases}} \{f(a,E-a)+4E-3a+a^2\}$, it is likely that $f(n,E)=f(n-1,E-n+1)+4E-3(n-1)+(n-1)^2$ for some n,E in reasonable range. Computer calcuation shows it work for n=95, it holds for $94\le E \le 2206$ (but not for any (n,E))

It means we could construct a graph with 24 vertexes with degree 94, one with degree 24+41=65, and 41 vertex with degree 25 and others 29 vertex with degree 24 to reach the sum square of degree 258618.

And it could be found that for most combination of (n,E) there's only a unique solution to reach the optimal result while some have 2~6 different solutions. But only one combination have 6 solutions: (9,18)

ds[9,18]=192
        011111111 (quasi-star)
        101111111
        110111000
        111000000
        111000000
        111000000
        110000000
        110000000
        110000000

        011111111
        101111111
        110110000
        111010000
        111100000
        110000000
        110000000
        110000000
        110000000

        011111111
        101111000
        110111000
        111011000
        111101000
        111110000
        100000000
        100000000
        100000000

        011111110
        101111110
        110111110
        111000000
        111000000
        111000000
        111000000
        111000000
        000000000

        011111100
        101111100
        110111100
        111011100
        111100000
        111100000
        111100000
        000000000
        000000000

        011111100
        101111100 （quasi-complete)
        110111100
        111011000
        111101000
        111110000
        111000000
        000000000
        000000000