### March 27, 2017

# On the Nomenclature of Complexity Reductions

I think it’s fair to say that there is a general confusion about reduction in complexity theory, and I think a lot of it comes from the terminology used. To underscore the point, I recently had a discussion about the time complexity of prime factorization with a friend, during which neither of us were able to consistently use the terminology correctly, leading to confusion in a discussion of an already complicated subject.

Our goal in using reductions is as follows: For two given problems, *A* and *B*, we want to prove that *B* ≥ *A*, that is; problem *B* is at least as hard as a problem *A*. To prove *B* ≥ *A*, we show that *A* can be represented as an instance of *B*. It must be the case that *B* ≥ *A* then^{1}, because if *A* can be represented as *B*, any easy solution to *B* could be used as an easy solution to *A*. We say that we *reduce* *A* to *B*.

Intuitively reducing *A* to *B* means we make *A* as small as *B*. While It makes sense that if we can show *A* to be as small as *B*, *B* is at least as big as *A*, the directionality of reduction is inverse to our intuition, in that to prove that *B* is hard, we show that *A* is easy. When learning about reductions in complexity theory, many students accidentally reduce in the wrong direction. That is, in an attempt to prove *A* ≥ *B*, they reduce *A* to *B*, getting them into trouble. My claim is that the unintuitive directionality of the term *reduce* is a part of the problem. In fact, one of my algorithms teachers, an associate professor of algorithm engineering, told me he specifically remembered the direction to reduce in, by telling himself to reduce in the *opposite* direction of what he would think made sense. That, to me, seems like a problem.

We can help this counter-intuitiveness by always describing the problem as “showing that *A* is at least as small as *B*” (simply inverting the left-/right-hand sides of the inequality), in which case the directionality of reduction from *A* to *B* might sound right, but perhaps it is better to do away with the term *reduction* altogether.

Saying that *A* can be reduced to *B* is the same as saying that *B* is sufficient for *A*, or that *B* is large enough to cover *A*. In these formulations, the analogy to a simple less-than relation between scalars is more direct, in that an amount *B* is sufficient when needing a smaller amount *A*.

With this formulation, we could say that showing *B* to be *sufficient* for *A* proves that *B* ≥ *A*. This might be clearer than the equivalent formulation; showing that *A* is reducible to *B* proves that *B* ≥ *A*.

### Footnotes

For such a proof to hold in general, the cost of transforming problem

*A*into problem*B*must be taken into account. Normally, we use*B*≥_{p}*A*to denote that the cost of the transformation may be polynomial in the problem size. The transformation is then called a polynomial-time reduction. I omit such terminology and notation, as it is immaterial to the overall nomenclature of reductions, which is what we are concerned with here.↩