How many components of memory are there?

Question 1

Your formulas show that you can calculate R from S using the equation: R=exp(-kt/S). Should this not mean that there is only one component of memory? You also speak of DSR model, which one might read as three components of memory (Difficulty D, Stability S, and Retrievability R).

Answer 1

Two components of long-term memory are needed to describe the status of memory at any time (t). Those components are stability (S) and retrievability (R). You might correlate those component with a large number of molecular or behavioral expressions of the two variables. For example, instead of stability, you might use the level of certain synaptic proteins. Instead of retrievability, you might measure synaptic conductivity, etc. Retrievability might also be expressed as time. If you know S and t, you know R. You cannot compute R from S without t. You need both Stability and Retrievability to describe the status of a unitary memory at any time (t).

In DSR model, Difficulty is a property of macroscopic memory. It is associated with items, not with synapses. You might illustrate the information that describes the memory status of each items as a set of pairs S:R, where each pair corresponds with a single synapse. SuperMemo algorithms show stellar performance if your items are easy/simple. You can imagine those items as involving very few synapses (e.g. connecting the concept of a horse with the word horse in learning vocabulary). Once you start piling up S:R pairs, difficulty increases, chaotic changes in memory status follow.

Preliminary simulations shows that for high difficulties, it might be impossible to grow intervals beyond certain value (e.g. 3-4 years). Leeches (difficult items) waste your time by clogging the learning process. Moreover, at some point, they stagnate and repetitions do not result in increasing intervals. This is a treacherous process that you need to be aware early in your SuperMemo adventures.

There are some indications that there might be a third component of memory that shows up in association with the spacing effect. It can be shown that in cramming, frequent review does not change S, while keeping R close to the maximum. At the same time, it might accelerate forgetting resulting in non-exponential forgetting curves. In other words, unchanging S and R cannot effectively describe the differences between the status of memory in prolonged cramming. This might look like a process that requires a third memory variable. Possibly though, the process can also be explained with formulas for composite memories.

Question 2

Once you say there are two components of memory. In other places you mention three components of memory. How many components are there in the end?

Answer 2

It depends which level of complexity we consider:

• synaptic level: memory variables stability (S) and retrievability (R) suffice to describe the status of long-term memory (e.g. as observed in SuperMemo)
• network level: item difficulty (D) is a variable that results from using composite memories in which many synapses with various S:R are involved

Unitary long-term memories are then described by two variables (S and R), while items in SuperMemo need three variables to compute optimum intervals (D, S and R). See: Algorithm SM-17