I am a chemist working on a project where I synthesize Mixture Libraries. Using different starting materials in solution mixtures, all possible combinations can theoretically form as products.

To identify the formed products, I look at the molecular weights. To identify each product separately, I do not want double molecular weights, or at least keep the number of doubles to a minimum. The only method to prevent this so far is to refrain from using starting materials with the same molecular weight.

When I calculated the products of my 10.000 mixture (100 different products A times 100 different products B = 10.000 products AB), I used Excel to analyze and saw that I have on average 2 products that have the same molecular weight and the highest frequency of a molecular weight was 23 times. Is there a way to use mathematics to determine which starting materials I have to use to create a library of the same size, but with less double molecular weights?

So me trying to make it a maths problem: If I use these two sets of numbers, and look at all the possible sums between these sets, I get a library of size X with a minimum amount of sums with the same value.

I hope that you as mathematicians understand this problem, and to maybe make it easier: think of the "molecular weights" as numbers, and "products" are chemical products, so in this problem they are sums!

  • $\begingroup$ So to be clear: all of the reactions in question are of the form $A + B \rightarrow AB$ with no side products, no reactions of the form $2A + B \rightarrow A_2 B$, etc? $\endgroup$ – Peter Taylor Jul 18 '17 at 8:33
  • $\begingroup$ This is correct $\endgroup$ – Lubberink Jul 24 '17 at 12:20
  • $\begingroup$ a bit confusing; seems to me to understand that, given two sets of numbers $A$ (for example$\{1,3,7\}$) and $B$ (for example$\{2,4,5\}$), in making all the possible sums $A_{k}+B_{j}$ you want to minimize the number of occurrences that $A_{k}+B_{j}=A_{m}+B_{n}$ (e.g. $1+4=3+2$ ), is that correct ? $\endgroup$ – G Cab Jul 24 '17 at 14:07
  • $\begingroup$ You are exactly right $\endgroup$ – Lubberink Jul 24 '17 at 14:41

Since this is a strange setup anyway you can do the following:

Assume that the molecular weights of the A substances are numbers of the form $$w=\sum_{k\geq0} a_k\,2^{2k},\qquad a_k\in\{0,1\}\ ,$$ and the molecular weights of the B substances are numbers of the form $$w=2\sum_{k\geq0} b_k\,2^{2k},\qquad b_k\in\{0,1\}\ .$$ The first few $A$-numbers are $$0, 1,4,5,16,17,20,21,64,65,68,69,80,81,84,85,\ldots\ ,$$ and the $B$-numbers are the doubles of these.

Since any number has a unique binary representation you can say the following: Any number $N\geq1$ can be written as sum of an $A$-number and a $B$-number in a unique way, e.g., $93=85+8$.

  • $\begingroup$ As I am no mathematician, I have a hard time understanding this answer. Does it say that I need to use only even MW's for starting material A, and only odd MW's for starting material B? $\endgroup$ – Lubberink Jul 24 '17 at 13:01
  • $\begingroup$ good point (+1): I took the liberty to "illustrate" it $\endgroup$ – G Cab Jul 24 '17 at 15:39

Since you confirmed the meaning of your question, the Christian answer is fully correct. To facilitate you in understanding the "core" of it have a look to the following binary coding tables: wish they are self explicative.



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