Tabasco sauce: a mathematical diversion

Question

(Tabasco sauce: a mathematical diversion) My cousin Phil collects hot sauce. He has about a hundred different bottles on the shelf, and many of them, Tabasco for instance, have only three ingredients other than water: chilis, vinegar, and salt. What is the smallest number of bottles of hot sauce that Phil would need to keep on hand so that he could obtain any recipe that uses only these three ingredients by mixing the ones he had? (M. Artin Algebra)

My Thoughts: If we bring down a dimension, ie if we consider there are only 2 ingredients we use this lemma to show 2 sauces are enough to make any other

2-sauce 2-ingredient lemma: Given any 3 sauces which are made up of a combination of 2 ingredients, then there are 2 sauces which can make the third one.

Proof: Let $S_i=Sauce_i=(a,b)$ where $a$ is the amount of first ingredient and $b$ is the amount of second ingredient. Since $S_1,S_2,S_3$ are linearly dependent, there exists $x,y,z$ such that $xS_1+yS_2+zS_3=0$ and hence since the co-ordinates of $S_1,S_2,S_3$ are positive, we can show that $\exists m\ge0,n\ge0,i,j,k$ such that $S_i=mS_j+nS_k$ hence $Sauce_j$ and $Sauce_k$ can make $Sauce_i$

What we need is a n-sauce 3-ingredient lemma. I am not even sure what n should be and the lemma cannot be generalized in any obvious way.

Any help is highly appreciated!

Answer 1

I'm not even sure that the 2-ingredient version works completely. What if all the bottles he had contained at least 10% of one of the ingredients, and the recipe called for a mix with only 5% of that ingredient? Or similarly what if the recipe called for a higher fraction of one ingredient than any of the bottles has?

Assuming this is not the case,then for the 2-ingredient case you would choose the bottle with the highest fraction of the first ingredient (which is also automatically the lowest fraction of the second ingredient) and the bottle with the highest fraction of the second ingredient (and lowest fraction of the first). You can see this as two vectors in the first quadrant of the plane, and you can combine them in various proportions to get to any point that lies between them. The two chosen bottles are the ones with the lowest and the highest slope, so as to encompass as much of the coordinate space as possible.

For the 3-ingredient case I think things become messy. You could see each bottle as a vector in the positive octant of 3d space. With any 3 bottles you can mix them to reach any point in the space between those vectors, which is an infinite triangular cone. If you choose n bottles, the space their vectors generate is like their convex hull. They could be such that they generate an n-sided cone. If any of the vectors were removed, you would get an (n-1)-sided cone and hence lose some of the space, and hence lose the ability to produce some of the recipes.

Instead of visualising it in 3d, it might be easier to use a ternary plot. Each bottle is a point inside the triangle. Mixing two bottles generates everything on the line between the two corresponding points, and similarly mixing 3 bottles gives you averything inside the triangle that they form. If you have n bottles represented by n points, the convex hull of those points is the space of mixable recipes. If the n bottles form a convex n-gon, then none of those bottles is superfluous.

Answer 2

HINT

Converting this to a linear algebra problem, let a sauce be represented by how much it contains of the three ingredients as $(a,b,c)$

Now, to mix these sauces is to take a linear combination of these points, weighted by the proportions in which you mix them

So basically what you need is a basis for the 3 dimensional vector space, which is the smallest set of vectors which can be used to generate any vector in that space.

Answer 3

I think the answer of @JaapScherphuis is entirely correct. Here is a 3D "visualization". Suppose there are $n$ bottles and the $j$th bottle has the ingredients in this proportion:

$$v_j = (1, \cos \pi j /2n, \sin \pi j / 2n)$$

Then it is clear that none of the $v_j$ can be re-created or "emulated" by mixing the other $n-1$ bottles in any combination with positive coefficients. (Proof in brief: due to the $1$ in the first coordinate, any combination must be a convex combination with positive coefficients summing to $1$, which means the combination can only achieve points within the convex hull of the other $n-1$ points, and the $j$th point is outside that convex hull.)