Predictive classification when only ranks are observable

Question

I am taking part in a classification challenge (classes are 0 and 1) where the inputs are encrypted (because these are expensive financial data). As the encryption is order-preserving I can only use the fact that e.g. $$ x_1 > x_2 $$ but not $$ d = x_1-x_2 $$

Besides trees, which machine learning algorithms give sound models under these circumstances?

EDIT: I assume that neural nets, SVM or logistic regression are not appropriate in this setting as they use linear transformations $b \cdot x$ which I can not apply as I don't have the "numerical structure" for this.

EDIT 2: I am given data of the following form: $$ (0.2,0.1,0.5,0); (0.1,0.2,0.3,1); (0.02,0.7,0.33,1) $$ and thousands of rows of them (and in my application more columns). In this example the first 3 entries are inputs and the 4th one is the target. All clumns consist of 1001 unique values in the range [0,1]. So I really think that only comparisons are possible.

I am sorry if my question was not formulated precisely enough ... I hope now the problem is clearer!

Answer 1

In order to better understand the problem I think it is worth explaining the main idea of how order-preserving cryptography works.

Imagine that we have column vector $\mathbf{x}$ than we want to encrypt. The encryption function is $f(\cdot)$ and it is monotonically increasing, and potentially with varying slope.

The cipher text of $\mathbf{x}$ is $\mathbf{y}=f(\mathbf{x})$, and has the following properties

• if $x_1 \geq x_2$ then $y_1 \geq y_2$
• if $x_2 = x_2$ then $y_2 = y_2$ (this is usually prevented)
• $d(x_1,x_2) \neq d(y_1,y_2)$
• if $d(x_1,x_2) \geq d(x_3,x_4)$ does not give information about $d(y_1,y_2)$, and $d(y_3,y_4)$

The second property can be prevented by adding noise to $\mathbf{x}$. For example if $x_i$ is an integer between $[0,9]$. Then, $\hat{\mathbf{x}}=10\times\mathbf{x}+\mathbf{n}$, where $\mathbf{n}$ is vector a of random integer between $[0,9]$. This transformation preserves the ordering in $\hat{\mathbf{y}}=f(\hat{\mathbf{x}})$, but removes the identification of attributes with the same value.

Given that I think you should stick to tree based classifiers such as random forest.

Answer 2

I don't think you should use order to predict the binary outcome since if a model trained on your data is applied to totally unseen data which was collected in a slightly different way (not sure how that data was collected) then it's going to perform very badly.

Moreover, such "golden features" based on data ordering are sometimes a result of data leaks and you do not want to train your model on such data for the reason given above.