# The Earth Mover's Distance (EMD)

We begin some intuition, examples, and informal definitions.

## Intuition and Examples

The Earth Mover's Distance (EMD) is a distance measure between discrete, finite distributions

 x = { (x1,w1), (x2,w2), ..., (xm,wm) } and y = { (y1,u1), (y2,u2), ..., (yn,un) }.

The x distribution has an amount of mass or weight wi at position xi in RK, i=1,...,m, while the y distribution has weight uj at position yj, j=1,...,n. An example pair of distributions in R2 is shown below.

Here x has m=2 masses and y has n=3 masses. The circle centers are the points (mass locations) of the distributions. The area of a circle is proportional to the weight at its center point. The total weight of x is wS=sumi=1..m wi, and the total weight of y is uS=sumj=1..n uj. In the example above, the distributions have equal total weight wS=uS=1.

### Equal-Weight Distributions

Although the EMD does not require equal-weight distributions, let us begin our discussion with this assumption. The EMD between two equal-weight distributions is proportional to the amount of work needed to morph one distribution into the other. We morph x into y by transporting mass from the x mass locations to the y mass locations until x has been rearranged to look exactly like y. An example morph is shown below.

The amount of mass transported from xi to yj is denoted fij, and is called a flow between xi and yj. The work done to transport an amount of mass fij from xi to yj is the product of the fij and the distance dij=d(xi,yj) between xi and yj. The total amount of work to morph x into y by the flow F=(fij) is the sum of the individual works:

WORK(F,x,y) = sumi=1..m, j=1..n fij d(xi,yj).

In the above flow example, the total amount of work done is 0.23*155.7 + 0.51*252.3 + 0.26*316.3 = 246.7.

The flow in the previous example is not an optimal flow between the given distributions. A minimum work flow is shown below.

The total amount of work done by this flow is 0.23*155.7 + 0.26*277.0 + 0.25*252.3 + 0.26*198.2 = 222.4. The EMD between equal-weight distributions is the minimum work to morph one into the other, divided by the total weight of the distributions. The normalization by the total weight makes the EMD equal to the average distance travelled by mass during an optimal (i.e. work minimizing) flow. The EMD does not change if all the weights in both distributions are scaled by the same factor. In the previous example, the total weight is 1, so the EMD is equal to the minimum amount of work: EMD(x,y)=222.4.

### Unequal-Weight Distributions

When the distributions do not have equal total weights, it is not possible to rearrange the mass in one so that it exactly matches the other. The EMD between unequal-weight distributions is proportional to the minimum amount of work needed to cover the mass in the lighter distribution by mass from the heavier distribution. An example of a flow between unequal-weight distributions is given below.

Here wS=1 and uS=0.74, so x is heavier than y. The flow covers or matches all the mass in y with mass from x. The total amount of work to cover y by this flow is 0.51*252.3 + .23*292.9 = 196.0. In this example, 0.23 of the mass at x1 and 0.03 of the mass at x2 is not used in matching all the y mass.

In general, some of the mass (wS-uS if x is heavier than y) in the heavier distribution is not needed to match all the mass in the lighter distribution. For this reason, the case of matching unequal-weight distributions is called the partial matching case. The case when the distributions are equal-weight is called the complete matching case because all the mass in one distribution is matched to all the mass in the other distribution. Another way to visualize the partial matching case is to imagine pile of dirt and holes in the ground. The dirt piles are located at the points in the heavier distribution, and the the holes are located at the points of the lighter distribution. The volume of a dirt pile or the volume of dirt missing from a hole is equal to the weight of its point. Matching distributions means filling all the holes with dirt.

The flow between the unequal-weight distributions in the previous example is not optimal. The minimum work flow is shown below.

The total amount of work for this flow to cover y is 0.23*155.7 + 0.25*252.3 + 0.26*198.2 = 150.4. The EMD is the minimum amount of work to cover the mass in the lighter distribution by mass from the heavier distribution, divided by the weight of the lighter distribution (which is the total amount of mass moved). As in the complete matching case, the normalization of the minimum work makes the EMD equal to the average distance mass travels during an optimal flow. Again, scaling the weights in both distributions by a constant factor does not change the EMD. In the above example, the weight of the lighter distribution is uS=0.74, so EMD(x,y)= 150.4/0.74 = 203.3.