What does the $i < j$ mean in $ E = -(\sum_{i

Question

This is the energy function that is defined for Boltzmann Machines:

$ E = -(\sum_{i<j} w_{ij}s_i s_j + \sum_i \theta_i s_i)$

What does the $i < j$ part mean for the running variables under the sum? How does a sum $\sum_{i<j} x_{ij}$ look like?

Answer 1

You can think of it as a double sum, say the upper index is $n$. Then: $\sum_{i<j} "=" \sum_{i=1}^n\sum_{j=i+1}^n$

So you really just sum over all pairs $(i,j)$ with the property that $i<j$. For instance,

$\sum_{i<j} x_{ij} = x_{12}+x_{13}+\cdots+x_{1n}+x_{23}+\cdots$.

Hope this helps.

Answer 2

Let the set

$$S=\{(i,j)\mid 1\le i,j\le n\}$$ and we can write $S$ as a partition : union of disjoint sets

$$S=\{(i,i)\mid 1\le i\le n\}\cup\{(i,j)\mid 1\le i<j\le n\}\cup \{(i,j)\mid 1\le j<i\le n\}$$ so for example we can write using this partition

$$\sum_{1\le i,j\le n} x_{ij}=\sum_{i=1}^n x_{ii}+\sum_{1\le i<j\le n} x_{ij}+\sum_{1\le j<i\le n} x_{ij}=\sum_{i=1}^n x_{ii}+2\sum_{1\le i<j\le n} x_{ij}$$

Answer 3

$\sum_{i<j} x_{i,j} = x_{1,2}+x_{1,3}+\cdots+x_{1,n}+x_{2,3}+\cdots + x_{n-1,n}$

The above implies that

$\sum_{i<j} "=" \sum_{i=1}^{n-1}\sum_{j=i+1}^n$

Note the only difference with Mankind's answer is the first sum for $i$ is only allowed to go up to $n-1$, to comply with $i<j$ and $j=n$

Interpretations

-Matrix

Another way to visualise this, is to have a two dimensional matrix, where $i$ denotes the row and $j$ the columns. Then the $\sum_{i<j} x_{i,j}$ will sum all the elements from the upper triangle of your matrix without the diagonal (for each $i^{th}$ row you need to pick all the columns greater than $i$).

-Pairs from n-uple, choosing to the right

Imagine a n-uple with $n$ elements (you label each element from $1$ to $n$), and you want to pair the elements up, under some constraint. Let your pairs be $(i,j)$, where i is the left element of your pair and j the right element. Then to choose your pairs, you set the constraint that once you chose the left element of your pair in the ith position, then you are forced to pick the right element of your pair to the right of the ith position. And for that to be possible you need $1<j$ and $i<n$.