Let \(x^{x^{x\cdots}} = 2\). Then \(x^{x^{x\cdots}} = x^{x^{x^{x\cdots}}} = 2\). Substituting, we get \(x^2=2\) and \(x=\sqrt{2}\). Now let \(x^{x^{x\cdots}} = 4\). Similarly, \(x^{x^{x^{x\cdots}}} = 4\), \(x^4=4\) and \(x=\sqrt{2}\). But then \(2=x^{x^{x\cdots}}=4\) and \(2=4\).

Obviously 2 cannot equal 4.

The problem seems to occur with solving for \(x\) by inverting the function \(f(x)=x^{x^{x\cdots}}\) via the recursion \(x^{f(x)}=f(x)\), \(x=f(x)^{1/f(x)}\). While we can set \(f(x)\) to any value \(s\) to solve \(x=s^{1/s}\), we mustn’t neglect also the condition that \(s=f(x)\) should have a solution.

Here we plot \(x=s^{1/s}\) for \(s\) from 0 to 5. (Let us only consider \(x\ge 0\), where everything is real.) Both \(s=2\) and \(s=4\) give \(x=\sqrt{2}\), but only the former satisfies \(s=f(x)\), while no \(s\) to the right of the peak has a solution. The reason is that \(f(x)\) for \(x>1\) has to be increasing by the monotonicity of exponentiation in each of its two arguments (\(a^b\) increasing in both \(a,b\) if \(a,b>1\)), and so the proper inverse (where it is defined) can only be taken from the left branch of the plot. The peak occurs at \(s=e\), \(x=K=e^{1/e}\). For every \(x>K\), \(f(x)\) diverges to \(\infty\). For \(1< x\le K\), \(\overbrace{x^{x^x\cdots}}^n\) is increasing in the tetration depth \(n\) and has an upper-bound (\(e\)), so \(f(x)\) is a well defined limit which is exactly \(s\).

The other half is more complicated. When \(x=0\), \(0^s=0\) for all \(s>0\) and \(0^s=1\) for \(s=0\), so there is no solution to \(s=f(x)\). This is a consequence of where \(a,b<1\), \(a^b\) is increasing in \(a\) while decreasing in \(b\). From \(x<1\), we get \(x^x>x\), \(x^{x^x} < x^x\), and so on in this alternating fashion. We also get from \(x^x < 1\) that \(x^{x^x} > x\), so in fact we have two non-crossing monotonic subsequences — the odd-depth tetrations increasing and the even-depth ones decreasing — that must therefore both converge. \(f(x)\) is well defined iff they converge to the same value \(s\). Suppose one subsequence has limit \(s_1\) and the other \(s_2\). Either of the cases \(s=s_1\) or \(s=s_2\) satisfies \(x^{x^s}=s\), so where there is a unique solution for a given \(x\), the limits must equal and \(f(x)\) must be well defined. From the plot (no proof) this occurs from \((1,1)\) down to the intersection at \(s=1/e\), \(x=L=(1/e)^e\). So for \(L \le x < 1\), \(f(x)\) is well defined. For all \(x < L\) though, multiple solutions for \(s\) actually correspond to the curved branch of this \((s_1,s_2)\) plot, which shows solutions to \(x^{s_1}=s_2\) and \(x^{s_2}=s_1\), i.e., \(s_1^{1/s_2}=s_2^{1/s_1}\), so \(f(x)\) is not defined there. (*)

In summary, \(f(x)\) is well defined between \(1/e\) and \(e\) for \((1/e)^e\le x \le e^{1/e}\), goes to \(\infty\) over the range, and flips between two values under the range. Since \(x^{x^{x\cdots}}=4\) can never happen, it is a vacuous assumption that can logically lead to any conclusion.

For any non-negative random variable \(X\) having finite first and second moments, \(\mathbb P(X>0) \ge (\mathbb EX)^2/\mathbb EX^2\).

The proof isn’t difficult. Here are three different ones.

Proof 1. We already know from Jensen’s inequality that \(\mathbb E f(X) \ge f(\mathbb E X)\) if \(f\) is convex. This gives \((\mathbb EX)^2/\mathbb EX^2 \le 1\) for any \(X\). The trick to make it be \(\le \mathbb P(X>0)\) is to note that the density at \(X=0\) contributes nothing to any moment. In particular, if \(F_X(t)\) is the distribution function of \(X\), then define a random variable \(Y\) that is \(X\) without the probability at zero, that is, according to the distribution function \(F_Y(t)=(F_X(t)-F_X(0))/(1-F_X(0))\). Then Jensen’s gives \((\mathbb EY)^2/\mathbb EY^2 \le 1\). However, \(\mathbb EY = \mathbb EX / (1-F_X(0)) = \mathbb EX / \mathbb P(X>0)\), and \(\mathbb EY^2 = \mathbb EX^2 / (1-F_X(0)) = \mathbb EX^2 / \mathbb P(X>0)\), so \((\mathbb EX)^2/\mathbb EX^2 = [(\mathbb EY)^2 \mathbb P(X>0)^2] / [\mathbb EY^2 \mathbb P(X>0)] \le \mathbb P(X>0)\). \(\blacksquare\)

The statement would also work for non-positive \(X\), of course; and an analogous statement can be made for arbitrary \(X\) comparing \(\mathbb P(X\ne 0)\) with some combination of moments for the positive and negative parts of \(X\).

Proof 2. Apparently this problem can also be proved by an application of the Cauchy-Schwarz inequality. Assume the probability space \((\Omega, \mathcal F, \mathbb P)\). The space of finite second-moment real-valued random variables \(L_2(\Omega)=\{X(\omega):\Omega \to R\}\) with the inner product \(\langle X,Y\rangle_{L_2(\Omega)}=\mathbb E XY\) and induced norm \(\Vert X\Vert_{L_2(\Omega)}=\sqrt{\mathbb EX^2}\) is a Hilbert space (modulo \(L_2\) equivalence). Given this, let us apply Cauchy-Schwarz on the two random variables \(X\) and \(\mathbf 1_{X>0}\):

This is a special case of something called the Paley-Zygmund inequality. I didn’t know such a thing existed.

Proof 3. This one only proves the discrete case. It is well known that for positive discrete random variables \(X\), \(\mathbb EX = \sum_{k=0}^\infty \mathbb P(X>k) = \mathbb P(X>0)+\mathbb P(X>1)+\cdots\). Basically \(\mathbb P(X=1)\) is counted once, \(\mathbb P(X=2)\) is counted twice, and so on. The analogous thing can be derived for \(\mathbb EX^2\), except now we need to count in squares. Happily we also know that squares accumulate by odd integers, i.e. \(n^2=1+3+5+\cdots+(2n-1)\), so \(\mathbb EX^2 = \sum_{k=0}^\infty \mathbb (2k+1) \mathbb P(X>k) = \mathbb P(X>0)+3\mathbb P(X>1)+5\mathbb P(X>2)+\cdots\).

Let’s simplify the notation a bit. Put \(q_k=\mathbb P(X>k)\), then \(q_0\ge q_1\ge q_2 \ge \cdots\). We just need to prove that \(q_0\ge (q_0+q_1+q_2+\cdots)^2 / (q_0+3q_1+5q_2+\cdots)\), which is to say, \((q_0+q_1+q_2+\cdots)^2 \le q_0(q_0+3q_1+5q_2+\cdots)\). The two sides both have limits, so this just requires some accounting. On the left hand side, \((q_0+q_1+q_2+\cdots)^2\) expands to \(q_0^2+(q_1^2+2q_0q_1)+(q_2^2+2q_0q_2+2q_1q_2)+\cdots = Q_0+Q_1+Q_2+\cdots\), where \(Q_k \triangleq q_k^2 + 2 \sum_{i=0}^{k-1} q_iq_k \le (2k+1)q_0q_k \triangleq R_k\). But \(R_0+R_1+R_2+\cdots\) is exactly the right hand side. So the left hand sum is dominated by the right hand sum. \(\blacksquare\)

With some real analysis, this proof could be made to work for random variables that are not discrete, but it might also turn into a special case of Proof 1. In any case, it’s interesting in its own right.

There is an island upon which 1000 people with various eye colors live. 900 have brown eyes, and 100 have blue. Each resident can (and does) see the eye colors of all other residents, but is forbidden by custom to try to discover one’s own (no talking about it, etc.). If (and only if) one does discover one’s own eye color somehow, then one commits suicide the following day for all to witness.

One day a visitor unaware of island customs comes to the island and announces (truthfully) to everyone: I see at least one blue-eyed person among you. What happens next?

One might be tempted to say that nothing happens, since every islander already sees either 99 or 100 blue-eyed people, so the visitor seemingly brought no new information.

But this is not true. In fact, every islander eventually commits suicide. We can show this by mathematical induction on the number of blue-eyed persons K that a particular islander can see.

First note that the only relevant information in this problem are the visitor’s statement and the visible actions of other islanders. Furthermore, committing suicide on some day k is equivalent to discovering one’s eye color on exactly day k-1.

Proposition 1 (base case): Seeing 0 blue-eyed persons is equivalent to discovering one’s own eye color to be blue on day 0 (i.e. the moment) of the visitor’s statement, and committing suicide on day 1.

Proof: The only situation in which I can discover information on day 0 is if I see 0 blue-eyed persons, making the visitor’s truthful statement a contradiction if I were not blue-eyed. (The information obtained is exactly my own eye color.) ∎

Proposition 2: Seeing 1 blue-eyed person is equivalent to discovering one’s own eye color on day 1, and committing suicide on day 2.

Proof: The only situation in which I can discover information on day 1 (but not before) is if the visitor’s statement is uninformative, but the action of those discovering information on day 0 is informative. The latter information, by Proposition 1, is whether there is exactly 1 blue-eyed person (if he commits suicide on day 1) or not (if he does not). Therefore, I must be deciding whether there are 1 or 2 blue-eyed persons, i.e. I see 1 blue-eyed person. (The information obtained is again my own eye color.) ∎

From these we can figure out what the induction step needs to be:

Proposition 3 (induction step): If it is true for all k from 0 to K that seeing k blue-eyed persons is equivalent to discovering one’s own eye color on day k and committing suicide on day k+1, then it is also true that seeing K+1 blue-eyed persons is equivalent to discovering one’s own eye color on day K+1 and committing suicide on day K+2.

Proof: Suppose I see K+1 blue-eyed persons, then I have two hypotheses, namely that there are K+1 or K+2 blue-eyed persons. Distinguishing between these two cases is equivalent to discovering my own eye color. By assumption, I do not discover anything on days 0 through K, nor do I commit suicide on days 1 through K+1, for the reason that it would be equivalent to seeing anywhere from 0 to K blue-eyed persons — a contradiction.
On day K+1, suppose I still did not discover my eye color. Here there would be two cases: if there were exactly K+1 blue-eyed persons, then each of them would see K blue-eyed persons, but by assumption, they would commit suicide on day K+1, so I would discover my own eye color to be brown — a contradiction; so there must be K+2 blue-eyed persons, then again, I would discover my own eye color (blue) — a contradiction. Therefore, it must be the case that I discover my own eye color on day K+1 and commit suicide the following day.

In the converse, suppose I discover my eye color on day K+1 but not before. Then the visitor’s statement and all my observations from days 1 to K are not informative, but the one on day K+1 is informative. By assumption, on each day k for k from 1 to K+1, those who see k-1 blue-eyed persons commit suicide. In particular, consider the action of blue-eyed persons on day k: it communicates whether there are k blue-eyed persons in total (if they commit suicide) or not (if they do not).* Now, anybody who sees B blue-eyed persons has two hypotheses: that there are B or B+1 blue-eyed persons. So the only way that day K+1 can be the first informative day to resolve these two cases, is if B=K+1 — for a smaller B, the question would be resolved on an earlier day; and for a larger B, the information up to day K+1 is insufficient. ∎

* It is easy to show that the action of brown-eyed persons on each day provides no additional information, conditioned on the action of blue-eyed persons on the previous day.

Taken together, we conclude that: on day 100 after the visitor’s statement, all blue-eyed islanders commit suicide; and on day 101, all brown-eyed islanders commit suicide.

This isn’t the end of the story, however, because while Proposition 1 is obvious enough, it remains perplexing that the visitor’s statement of something publicly known by all of the islanders since time immemorial results in anything at all. Why couldn’t the islanders have “filled in” for the visitor’s role somehow? What form of information did the visitor provide?

The answer is that the visitor brought a synchronization signal. There was no time origin in the islanders’ “history,” which goes back to negative infinity. Since the entire inference exercise depended on everybody counting days from a common “day 0,” they couldn’t align themselves without a common signal.

(Some further explanations of this puzzle here
and here, where the synchronization can also be interpreted as between all “levels” of inferential belief by each islander — this is the mechanism by which a statement about the world prompts each islander to start “counting” in the first place; after all, the visitor never explicitly asks the islanders to synchronize to anything, nor do the islanders have an agreed upon protocol to coordinate.)

There is a similar problem of this flavor:

Suppose there are some beasts who like to eat meat, but if they eat any meat, they themselves turn into a piece of meat (still “alive,” but liable to be eaten by other beasts). Presuming the beasts would all like to survive foremost (even as meat), and eat meat if they can get away with it, what would happen if there were, say, 100 beasts and 1 piece of meat?

The answer is that when there is an odd number of beasts, one of the beasts will eat the meat; when there is an even number, all of them will refrain.

The usual adage is that the rules of “Go” are simple, but the strategies are difficult. Sometimes they cite the fact that computers can’t play “Go” very well (compared to, say, chess). I’m not in a position to defend or refute this sentiment yet, but I think this stuff is easier than it first appeared to me years ago, if you think about it the right way. “Go” seems a very mathematical game, almost a counting and topological reasoning game. And all the difficulties arise out of the fact that the counting and reasoning is done in 2D (rather than 1D, which would be simple).

So there is this list of terminology like “liberties,” “eyes,” “false eyes,” “alive and dead regions,” etc., but sometimes they greatly confuse the matter, so I found it easier to recast these basic concepts in more general terms.

There are really only two axiomatic rules to play Go.

Def. 1: A group of stones is some stones of the same color.
Def. 2: A liberty of a group is any of the cardinal neighbor points of the stones in the group that is unfilled.
Rule 1: A group can be captured if all the liberties of the group are filled by the opposing color.
Rule 2: A point that when filled would result in an immediate capture of a group to which the point also belongs cannot be filled, unless it results in a capture of a group of the opposing color and if in the immediate previous move it was not itself captured.

Of course, nobody plays “Go” using only axioms. Instead we use the axioms and properties of the board* to derive new results:

I. The concept of an “alive group” by means of forming “two eyes.” This group of black stones is said to be alive idiomatically because it has “two eyes.” Well, turns out the real proposition is this:
Def. 3: A “connected group” is a group for which there exists no subgroup with all the liberties unfilled by a stone of the same color. (We’re neglecting some issues related to stones on edges and corners.)
Prop. 1: A connected group must be captured together.
Proof: To capture any subgroup of the connected group, its liberties must be taken, but some of its liberties is occupied by a same-color stone not belonging to the subgroup, therefore the subgroup cannot be taken. Since all subgroups of a connected group are like this, no subgroup of a connected group can be captured alone.
Def. 4: An “eye” is a minimal subgroup of a connected group that, if alone on the board, has exactly one liberty for which filling it would trigger Rule 2. To “fill” an eye is to fill all its liberties.
Prop. 2: A connected group with two or more “eyes” cannot be captured.
Proof: Each player can only make one move per turn, and therefore two or more “eyes” cannot be filled simultaneously. Then each eye in the group must be filled separately, in some order. But that would involve invoking Rule 2 at some point for the first of these eyes, when none of the other eyes have been filled. Since the group is connected, it must be captured together, yet we just said it does not have all of its liberties filled at the time that Rule 2 is invoked, and therefore filling the first eye is an unallowed move. Therefore, a connected group with two or more “eyes” cannot be captured.

II. Next, the “false eye” is simple, like the top one here. It’s just not an eye by Def. 4, since the circled black stone and the one above it are not in a connected group with the rest of the black stones.

Corollary 1: A group that is a disjoint union of connected subgroups cannot be captured if each subgroup has two or more liberties that would trigger Rule 2.
(Formations around these liberties are also called “eyes” usually, but I don’t consider them eyes.)

III. Usually the next step is learning how some groups that don’t have two or more eyes are nevertheless “alive,” “dead,” or “unsettled” based on how further moves could necessarily, necessarily not, or possibly — dependent on move order — allow the capture of the group. This is usually part of some rote memorization drill in Go, but it doesn’t have to be this way.

Def. 5: A region is a group along with some additional unfilled space.
Def. 6: A critical point in the unfilled space of a region is a point that if and only if filled, would make the extended group (original group of the region, along with the newly filled point) a connected group with two or more eyes.
Prop. 3: A group in a region is “alive,” if it has two or more critical points.
Proof: Assume it’s the opposing color to move. The best move would be to deny one of the critical points. Yet there are other critical points, so therefore a response at one such point would create a connected group with two or more eyes, which cannot be captured by Prop. 2.

Prop. 4: A critical point, if it exists, has at least two liberties.
Proof: A point with zero or one liberty does not create new eyes upon filling the point, so it cannot be a critical point.

Thus, immediately, the enclosing group of L-shape and Z-shape spaces-of-four are alive, because there are two critical points in the unfilled space. The enclosing group of straight-line spaces-of-five-or-more are also alive, since there are even more critical points.

Def. 7: A critical tree in the unfilled space of a region is a set of points that can be arranged into a hierarchical tree, such that making moves along any complete path in the tree results in a connected group with two or more eyes, and making any other moves does not.
Prop. 5: A region with a critical tree of branching factor at least two at each level (a good tree), is “alive.” If all branches of the root have some child node with a single branch (a bad tree), then the region is “dead.” If only one subtree of the root is good and the remaining subtrees are bad, then the region is “unsettled.”
Proof: In the first case, since the branching factor is at least two, no matter what the opposing color does at each move, there is always another subtree remaining to complete a path. In the second case, then even if the group’s color makes the first move, at some level, the tree does not branch, and the opposing color would play the corresponding non-branched child, making it impossible for a complete path to be made by the group’s color. In the third case, if it’s the group’s color to move, then the “good” subtree would be chosen; if it’s the opposing color to move, also the “good” subtree would be chosen, but in this case, the opposing color occupies the root of the good subtree and denies a complete path to the group’s color.

Corollary 2: Prop. 3 is a special case of Prop. 5 with a one-level tree.

So we see, for instance, that the enclosing group of the spaces-of-four in the square shape is “dead,” since the branches of the critical tree consists of the two internal diagonals of the square shape, the two subtrees of which (both single-branching) are both “bad.”

Corollary 3: A group in a region may be “unsettled,” if the region has exactly one critical point.
Proof: This is a special case of Prop. 5 with a single “good” subtree of depth 1, and possibly some remaining subtrees from the root that may be good or bad. If the group’s color is first to move, then the critical point can be filled and the extended group cannot be captured by Prop. 2. However, if it is the opposing color to move, then the best move would be to deny the critical point with an opposing color stone. In order to capture that stone, the group’s color would have to exhaust all the liberties of the original critical point, thereby making it not a critical point in the resulting configuration by Prop. 4. Whether the new configuration is alive is undetermined as it depends on whether at least some of the other subtrees from the root are good.

In the case of the enclosing group of straight-line and L-shape spaces-of-three, as well as the pyramidal spaces-of-four, for example, there is only one branch from the root and it is a “good” subtree of depth 1. Therefore, by Prop. 5, these groups are “unsettled.”

* Incidently, “Go” potentially has an infinite number of variations, because it can be played under changes of the board topology and handicap stones. The axiomatic rules don’t need to change in that they are topologically agnostic. There are even 3D “Go” boards. Additionally, if multiple moves can be made per turn, the entire game also changes — for instance, you’d need to create three or more eyes to be alive if two moves can be made per turn, etc., all the more reasons to prove general results. (See also this.)

Our synthetic genomic approach stands in sharp contrast to a variety of other approaches to genome engineering that modify natural genomes by introducing multiple insertions, substitutions, or deletions (18–22). This work provides a proof of principle for producing cells based upon genome sequences designed in the computer. DNA sequencing of a cellular genome allows storage of the genetic instructions for life as a digital file.

This seems significant, equivalent to booting up the first stored-program computer.

Scientists who were not involved in the study are cautioning that the new species is not a truly synthetic life form because its genome was put into an existing cell.

That’s sour grapes, because the original cell cytoplasm decays to zero exponentially fast in the number of replications, a point well made in the paper. It’s only needed for booting. What’s more useful to know is how much of the 1.08Mbp genome consists of existing genes. The paper says it’s a close copy of M. mycoides:

The synthetic genome described in this paper has only limited modifications from the naturally occurring M. mycoides genome. However, the approach we have developed should be applicable to the synthesis and transplantation of more.

The next step will be a basic cell with a minimal genome, a barebones cell OS, if you will. Then, on to synthetic functions. Pretty soon we’ll have cell API’s, fancy-pants programming frameworks, and bugs and viruses. I mean real ones.

\(\min_x \max_y f(x,y) \ge \max_y \min_x f(x,y)\).

The proof is almost trivial. Let \((x^*, y^*)\) optimize the left hand side and let \((x_*, y_*)\) optimize the right hand side. Now note two facts: for any given \(x\), \(f(x, \arg\max_y f(x,y)) \ge f(x,y) \forall y\); for any given \(y\), \(f(x, y) \ge f(\arg\min_x f(x,y), y) \forall x\).

So in particular, \(f(x^*, y^*) \ge f(x^*, y_*)\) and \(f(x^*, y_*) \ge f(x_*, y_*)\). So putting everything together, \(f(x^*, y^*) \ge f(x_*,y_*)\).

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.