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I'm reading the book Introduction to Data Mining by Tan, Steinbeck, and Kumar.
In the chapter on Decision Trees, when talking about the "Methods for Expressing Attribute Test Conditions" the book says :

"Ordinal attributes can also produce binary or multiway splits. Ordinal
attribute values can be grouped as long as the grouping does not
violate the order property of the attribute values. Figure 4.10
illustrates various ways of splitting training records based on the
Shirt Size attribute. The groupings shown in Figures 4.10(a) and (b)
preserve the order among the attribute values, whereas the grouping
shown in Figure a.10(c) violates this property because it combines the
attribute values Small and Large into the same partition while Medium
and Extra Large are combined into another partition."

Why ordinal attribute values can be grouped as long as the grouping does
not violate the order property of the attribute values?

I guess the reason is clear. We usually split things into specified parts which are not contradictory. A special thing can be small and medium, as one group, and large, as the other group. But it cannot be small and large at the same time. The point is that you have a sequence in your data. If there was no such thing you could have different combinations of attribute values. Suppose you have a set of attribute values for a fruit. It can be apple, pineapple and watermelon. Due to the fact that there is no ordinal, you can have all possible combination for binary splits; in the previous case, you can not because your binary split somehow violates the logical sequence.

I'd say the distinct handling of the ordered and unordered factor in decision trees is more convention and
implementation detail than a necessity.

But it is also an important optimization feature. See the documentation of the rpart here

We have said that for a categorical predictor with $m$ levels, all $2^(m-1)$ different possible splits are tested..

and

Luckily, for any ordered outcome there is a computational shortcut that allows the program to find the best split using only $m-1$ comparisons.

As you see, the ordered factor may be processed much effectively.

My advice therefore - as a part of the feature ingeneering decide whether to use a factor ordered or unordered:

Use ordered factor only if it is highly correlated with the output variable, otherwise fall back to an unordered factor

Bellow is a simple example, how can a scattered output variable with an ordered factor as a feature fool the decision treee to be very deep and ineffective.

> df
 X Y
1 1 0
2 2 1
3 3 0
4 4 1
> str(df)
'data.frame': 4 obs. of 2 variables:
 $ X: Ord.factor w/ 4 levels "1"<"2"<"3"<"4": 1 2 3 4
 $ Y: num 0 1 0 1

Notice that the output variable $Y$ is highly uncorrelated with the ordered factor.

> fit <- rpart(Y ~ X, method="class", data = df, control=rpart.control(minsplit = 1)) 
> print(fit)
n= 4 

node), split, n, loss, yval, (yprob)
 * denotes terminal node

 1) root 4 2 0 (0.50000000 0.50000000) 
 2) X=1 1 0 0 (1.00000000 0.00000000) *
 3) X=2,3,4 3 1 1 (0.33333333 0.66666667) 
 6) X=3,4 2 1 0 (0.50000000 0.50000000) 
 12) X=1,2,3 1 0 0 (1.00000000 0.00000000) *
 13) X=4 1 0 1 (0.00000000 1.00000000) *
 7) X=1,2 1 0 1 (0.00000000 1.00000000) *

Which leads to a deep (and unscalable) decision tree.

Making the factor unordered results in the optimal decision tree.

> df
 X Y
1 1 0
2 2 1
3 3 0
4 4 1
> str(df)
'data.frame': 4 obs. of 2 variables:
 $ X: Factor w/ 4 levels "1","2","3","4": 1 2 3 4
 $ Y: num 0 1 0 1
> 

> fit <- rpart(Y ~ X, method="class", data = df, control=rpart.control(minsplit = 1)) 
> print(fit)
n= 4 

node), split, n, loss, yval, (yprob)
 * denotes terminal node

1) root 4 2 0 (0.50000000 0.50000000) 
 2) X=1,3 2 0 0 (1.00000000 0.00000000) *
 3) X=2,4 2 0 1 (0.00000000 1.00000000) *