I found that Lipinski's rule of five states that Log P (octanol-water partition coefficient, lipophilicity measure) usually should not exceed 5.

Many papers about drug discovery machine learning models tell about maximization of "penalized logP", but the paper they refer in the end does not contain any information on it. Let me show that.

Let's check 1, here we read about penalized log P:

The second task is to produce novel molecules with desired properties. Following (Kusner et al., 2017), our target chemical property y(·)is octanol-water partition coefficients (logP) penalized by the synthetic accessibility (SA)score and number of long cycles

y(m) = logP(m)−SA(m)−cycle(m) 

where cycle(m)counts the number of rings that have more than six atoms.

Then we go into Kusner et al., 2017:

For the second optimization problem, we follow (Gómez-Bombarelli et al., 2016b) and optimize the drug properties of molecules. Our goal is to maximize the water-octanol partition coefficient (logP), an important metric in drug design that characterizes the drug-likeness of a molecule.As in Gómez-Bombarelli et al. (2016b) we consider a penalized logP score that takes into account other molecular properties such as ring size and synthetic accessibility (Ertl & Schuffenhauer, 2009).

Here is Gómez-Bombarelli, 2016 and it seems to contain nothing on the topic (I can be mistaken, please point me then), except

The objective we chose to optimize was 5×QED−SAS, where QED is the Quantitative Estimation of Drug-likeness (QED),37 and SAS is the Synthetic Accessibility score.36Thisobjective represents a rough estimate of finding the most drug-like molecule that is also easy to synthesize.

Other examples of the papers also leading to the same papers are 2, 3, etc.

What does penalized logP mean for drug-likeliness and why it should be maximized?


2 Answers 2


A positive logP is good (hydrophobic), but high results in ADMET problems (cytochromes break overly hydrophobic compounds) and aggregation (gunk, tar, crud are words happily thrown around). However, in most cases the logP is not greater than +3 because good binding is result of hydrogen bonds etc —some bad algorithms will suggest a polyaromatic hydrocarbon that makes lots of pi-pi interactions but these are few.

Confusingly, a high synthetic accessibility score is bad (hard). 3 is the average and 10 is basically something impossible. This score is how hard it is to make (i.e. pricier), which is one of the major drivers.

The number of cycles is number of rings. Even if you get pi-pi, cation-pi and sulfur-pi bonds you also get aggregation due to stacking (a crud). So is somewhat bad.

So basically that score is

high=good minus high=bad minus high=bad

If difference in relative Gibbs energy of binding is thrown it, negative ∆∆G is good, so that would be subtracted. As its atom count dependent ligand efficiency is better, but that ranges from 0 to -1.2 kcal/mol. So would need scaling.


You were trapped by arXiv versioning convention that shows you the latest version of the preprint as default. It seems that this experiment was removed from the final publication.

If you have a look at the v1 (https://arxiv.org/abs/1610.02415v1) instead of the latest version (v3), you find the explanation:

As a simple example, we first attempt to maximize the water-octanol partition coefficient $logP$, as estimated by RDkit. [43] $logP$ is an important element in characterizing the druglikeness of a molecule, and is of interest in drug design. To ensure that the resulting molecules to be easy to synthesize in practice, we also incorporate the synthetic accessibility [44] (SA) score into our objective. Our initial experiments, optimizing only the $logP$ and SA scores, produced novel molecules, but ones having unrealistically large rings of carbon atoms. To avoid this problem, we added a penalty for having carbon rings of size larger than 6 to our objective. Thus our preferred objective is, for a given molecule m, given by: $J(m) = logP(m) − SA(m) − ring-penalty(m)$, (1)

where the scores $logP(m)$, $SA(m)$, and $ring-penalty(m)$ are normalized to have zero mean.


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