This reminds me of a rather surprising method for direction-finding without a compass. It goes like this: point a stick in the direction of the sun, at any time during the day, and wait a short moment, then the direction of the new (very short, infinitesimal) shadow from the base of the stick points to the east. Here’s a short “proof”: Let the sun’s original location be called Q, let the base of the stick be called A and let the tip of stick be called B. So the vectors AB and AQ are collinear. At any given moment, the earth’s surface instantaneously rotates into the eastern direction at A, so in the earth’s local reference frame (the one we’re interested in) the sun’s apparent location moves slightly to the west to a new point Q’. Now, the shadow cast by the stick has a tip at a new point C (on the ground). Since A, B, C, Q, Q’ all must be coplanar, and the vector AC has no elevation component (same as QQ’), then the vector AC must be parallel to QQ’, and BQQ’ and BAC being similar triangles, AC must point to the east.

The trick is, you can’t wait too long, or else all the linear approximations and Euclidean geometry break down and you won’t get the right direction.

Three surgeons (A, B, C) participate in an operation involving three patients (a, b, c). For simplicity, each surgeon operates with just one hand, and operates just once on each patient. There are a bunch of sterile surgical gloves that stop contamination from one side to the other. What’s the minimum number of gloves needed to ensure that no body part ever comes into contact with another, even indirectly? Note, surgeons do not want to contaminate each other either.

Clearly, the upper bound is 9, for the 9 pairings. The lower bound is 3, for the 3 patients, or alternatively the 3 surgeons. Is the answer something in between?

Here’s a hint, they are allowed to wear multiple gloves or flip them inside out or do any such thing as needed.

It turns out 3 is not sufficient, but 4 is. There are many ways to achieve 4. Here is one:

First label the 4 gloves 1, 2, 3, 4. The notation “X-m-n-y” indicates surgeon X wears glove n outside glove m and operates on patient y.

A-1-3-c
A-1-2-b
A-1-a
B-4-3-c
B-4-2-b
B-4-1-a
C-3-c
C-3-2-b
C-3-1-a

But no matter how you try, 3 is not possible. You are always short by 1 or 2 sterile surfaces, depending on how you count. And this counting argument lies behind the following proof that 3 is not sufficient. The proof is rather short:

First, two facts.

1. Clearly, once a sterile surface is used by a surgeon or patient, it cannot come into contact with somebody else, so each surface is somehow “assigned” to a specific person.
2. After all the operations involving that person is finished, only then can that person’s assigned surface be contaminated

So, suppose 3 were sufficient. Then, there are 6 persons in total, and 6 sterile surfaces initially. That means there can be no sterile-to-used surface contact until all 6 surfaces are assigned, or there would be a shortage.

There are three types of operations, one type involves contaminating sterile surfaces by direct use, let’s call this an “allocation” operation; a second type involves already contaminated surfaces cross-contaminating each other, let’s call this a “contamination” operation; a third type reuses already contaminated surfaces but in a different yet contact-compatible combination, without introducing additional contamination, let’s call this a “reuse” operation.

There can be at most 5 “allocation” operations to allocate the 6 sterile surfaces to the 6 persons. (The first operation necessarily contaiminates 2 sterile surfaces.)

There can be at most 1 “reuse” operation. This is due to a kind of parity constraint. 4 surfaces have to already be allocated, with 2 sterile surfaces left for this operation.

Now there are 9 operations in total, so there are at least 3 more to go. These remaining operations are “contamination” operations. That means any operation involving surgeons or patients associated with any of the contaminating “inside” glove surfaces must not occur again.

Now, without loss of generality, suppose the glove surface assignments are

A-1-a
B-2-b
C-3-c

(It has to be surgeon on one side, patient on the other, because any other way is necessarily more wasteful.) Now note that any 2 operations involves at least 3 distinct people, and simultaneously “disable” the 3 other sides of the gloves (and the associated people) from further operations. Further, for any 3 people, there can be at most 2 operations between them. That means we can only perform 2 “contamination” operations. As an example, the two operations A-1-2-b, A-1-3-c forbids a, B, and C from being in any further operations. That leaves A, b, c, but between them, there are only the two pairs that have already occurred (A-b and A-c).

Since there is 1 of the 9 operations that cannot be performed in safe conditions, that’s a contradiction. Q.E.D.

Now only if this can be generalized to larger numbers of patients and surgeons. One key number series in making bounds is the least number of distinct people that must be involved for \(N\) operations. We know that a full bipartite graph between \(n\) nodes and \(k\) nodes has \(n \times k\) edges, so \(N\) operations will involve at least about \(2\sqrt{N}\) people. The problem is this approximate lower bound grows too slowly and becomes loose for larger numbers of pairings so that doesn’t guarantee as simple a proof for those cases. So maybe another day for the generalization.