# An illusionist and their assistant are about to perform the following magic trick

Let $$k$$ be a positive integer. A spectator is given $$n=k!+k−1$$ balls numbered $$1,2,\dotsc,n$$. Unseen by the illusionist, the spectator arranges the balls into a sequence as they see fit. The assistant studies the sequence, chooses some block of $$k$$ consecutive balls, and covers them under their scarf. Then the illusionist looks at the newly obscured sequence and guesses the precise order of the $$k$$ balls they do not see.

Devise a strategy for the illusionist and the assistant to follow so that the trick always works.

(The strategy needs to be constructed explicitly. For instance, it should be possible to implement the strategy, as described by the solver, in the form of a computer program that takes $$k$$ and the obscured sequence as input and then runs in time polynomial in $$n$$. A mere proof that an appropriate strategy exists does not qualify as a complete solution.)

Source: Komal, October 2019, problem A $$760$$.
Proposed by Nikolai Beluhov, Bulgaria, and Palmer Mebane, USA

I can prove that such a strategy must exist:

We have a set $$A$$ of all permutations (what assistant sees) and a set $$B$$ of all possible positions of a scarf (mark it $$0$$) and remaining numbers (what the illusionist sees).

We connect each $$a$$ in $$A$$ with $$b$$ in $$B$$ if a sequence $$b$$ without $$0$$ matches with some consecutive subsequence in $$a$$. Then each $$a$$ has degree $$n-k+1$$ and each $$b$$ has degree $$k!$$. Now take an arbitrary subset $$X$$ in $$A$$ and let $$E$$ be a set of all edges from $$X$$, and $$E'$$ set of all edges from $$N(X)$$ (the set of all neighbours of vertices in $$X$$). Then we have $$E\subseteq E'$$ and so $$|E|\leq |E'|$$. Now $$|E|= (n-k+1)|X|$$ and $$|E'| = k!|N(X)|$$, so we have $$(n-k+1)|X| \leq k!|N(X)|\implies |X|\leq |N(X)|.$$ By Hall marriage theorem there exists a perfect matching between $$A$$ and $$B$$...

...but I can not find one explicitly. Any idea? At least a strategy for $$k=3$$.

• In the case $k=3$ there are $6$ possible positions for the scarf. Those correspond to six possible mappings $f_i, i=1,2\ldots,6$, from $S_8$ to $S_3$ (how are the 3 balls under the scarf shuffled). If the "pile" of a permutation $\sigma\in S_8$ is the product of $f_i(\sigma)$s in $S_3$, that would be a somewhat generalizable idea. But I have no idea whether it works. Just tossing an idea in the air. – Jyrki Lahtonen Jul 24 at 19:27
• – Aqua Jul 24 at 20:55
• Regarding the .pdf to which you have just linked, there was a company with a booth at the last Joint Math Meetings which gave several examples using a similar scheme (e.g. they showed four cards in some order, and a "hidden" card). The goal was to figure out how to infer the value of the hidden card from the four cards shown. I got a really nice insulated water bottle for coming up with a solution (though, I must admit, a colleague provided me with a great deal of help). :) – Xander Henderson Jul 24 at 21:24
• I have found an explicit solution for $k=3$, however it is not very enlightening which is why I post this as a comment and not an answer. You can view it here. The solution consists of (permutation, cloth index) pairs, with the cloth index and ball numbering both starting at $0$. This does not rule out the existence of a more structured matching, as the matching is not unique. Needless to say this strategy is near-superhuman as the required memory feat is on the order of the world record of recalling digits of pi. – orlp Jul 28 at 2:01
• @rtybase when $k=3$ it could be $8,1,7,2,6,3,5,4$. I don't see a monotone block of three. – Jyrki Lahtonen Aug 9 at 21:02

This is a strategy that works in every case I checked, but its validity for all k needs to be proved. By his choice of placing the scarf, the assistant conveys information to the illusionist. There are $$k!$$ ways of placing the scarf, to the assistant.

Consider a random permutation ($$a_1,a_2,...,a_n$$) of {$$1,2,...,n$$}, that the audience chooses. The assistant looks at this arrangement and assigns a parity to it as follows:

1. He takes the $$k$$-touple ($$a_1,a_2,...,a_{k}$$). He constructs a $$k$$-digit number (concatenation) from it as $$a_1a_2...a_{k}$$ He then calculates the rank of this number among all of its $$k!$$ permutations (lowest number gets rank 0 and highest gets rank $$(k!-1)$$). He assigns it a variable $$\boxed{x_1=(rank)}$$

2. Assign $$x_2$$ to {$$a_2,a_3,...,a_{k+1}$$}. Similarly assign $$x_3,x_4,...,$$ and $$x_{k!}$$. Take that $$a_i=a_{i+n}$$

Compute the parity: $$p=mod\left(\sum_{i=1}^{k!} x_i,k!\right)$$

There are $$k!$$ values of $$p$$ possible: {$$0,1,2,...,k!-1$$}. Depending on the value of $$p$$ the assistant chooses where to place the scarf. (When $$p=0$$ he places over first $$k!$$ balls. When $$p=1$$ he places over second $$k!$$ balls,.. etc)

When the illusionist sees the obscured arrangement, he knows the value of $$p$$ by the position of the scarf. Although I haven't proved it yet, only one arrangement of the obscured balls gives this value of $$p$$. This strategy works for all cases of $$k=2$$ and many cases of $$k=3$$. But it is a behemoth to prove for general $$(k,n)$$.

• I assume you mean $k$-tuples rather than $k!$-tuples? Also, I would write the parity equation as $p=\prod x_i$ to distinguish it from the concatenation $a_1 a_2 ... a_{k}$ (assuming that's what you meant). Also, can we get away with only computing $x_1, ..., x_{k!}$? I'm not convinced of your bold statement, but we can perhaps weaken it to "exactly one of the $k!$ choices yields a parity equal to $p$", which I am more inclined to believe. In any case, nice idea. I'm currently checking to see if it works for small $k$. – Andrew Szymczak yesterday
• I found a counterexample for k=3. Both 1,2,3,[4,6,5],8,7 and 1,2,3,[5,4,6],8,7 have parity $p=3$ (using all $x_1, ..., x_n$). – Andrew Szymczak yesterday
• @AndrewSzymczak I have changed the definition of parity:$$p=x_1+....+x_{k!}$$It gives no counter example for $k=3,p=3$. Maybe the strategy is complete now? – user635640 yesterday
• Then it doesn't work for $k=2$ (both 2,[1,3] and 2,[3,1] have parity 1). There are many other counterexamples for $k=3$ as well (for both definitions of parity). I think your solution is on the right track, but it's currently unclear to me how to fix it. – Andrew Szymczak yesterday

Whoops, I just found a bug in my code and it seems my algorithm is unfixably wrong. I'm leaving the answer up and will fix the algorithm section when I find a working one.

## Notation and Remarks

Let $$S_n$$ denote the set of permutations of length $$n$$ and let $$C_{n,k}$$ be the set of covered permutations. For example, the permutation $$12345678 \in S_8$$ and $$123\cdot\cdot\cdot78 \in C_{8,3}$$. For brevity I often drop the subscripts.

The act of covering gives us a relation $$\sim$$ between the two sets, i.e. we say that $$\pi \in S$$ and $$c \in C$$ are compatible if $$c$$ is a covering of $$\pi$$. We can visualize this compatibility relation as a bipartite graph. For $$k=2$$ and $$n=3$$ we have:

$$\qquad \qquad \qquad \qquad \qquad \qquad$$

We need to find an injective function $$f : S_n \rightarrow C_{n,k}$$ such that $$\pi$$ and $$f(\pi)$$ are always compatible. The assistant performs the function $$f$$ and the illusionist performs the inverse $$f^{-1}$$. As per the problem requirements, both $$f$$ and $$f^{-1}$$ need to be computable in poly(n) time for a given input.

For $$n = k! + k - 1$$ we note that $$|S_n| = |C_{n,k}| = n!\,$$. We also note that each permutation is compatible with exactly $$k!$$ coverings and each covering is compatible with exactly $$k!$$ permutations. As OP mentioned, the Hall marriage theorem thus implies that a solution exists. We can find such a solution using a maximum matching algorithm on the bipartite graph. This is how @orlp found a solution for $$k=3$$ in the comments. However, the maximum matching algorithm computes $$f$$ for every permutation in poly(n!) time, where we instead need to compute $$f$$ for a single permutation in poly(n) time.

## Algorithm (does not work)

The following is inspired by @yukelid's answer. Given a permutation $$\pi \in S_n$$, we cover the $$k$$-segment that has the smallest rank (lexicographical index) among its permutations ($$\cong S_k$$). For example, for permutation $$132$$, the segment $$13$$ has rank 1 among $$\{13, 31\}$$ and $$32$$ has rank 2 among $$\{23, 32\}$$, so the assistant would cover segment $$13$$, i.e. $$f(132) = \cdot \cdot 2$$. If multiple segments have minimum rank, then we take the first such segment.

The inverse is straightforward. Given a covering $$c$$, we consider the compatible permutations. Among these, we choose the $$\pi$$ such that $$f(\pi) = c$$. It remains to be shown that when this $$\pi$$ exists, it is unique.

We must prove $$f$$ is injective. (Note that this suffices for bijectivity as well, since $$|S| = |C|$$.) Assume not, i.e. $$f(\pi_1) = f(\pi_2) = c$$ where $$\pi_1 \neq \pi_2$$. The rest of the proof is left as an exercise to the reader...

Finally, it is clear that the algorithm for $$f$$ and $$f^{-1}$$ have time complexity $$\mathcal{O}(n^2)$$ and $$\mathcal{O}(n^3)$$, respectively.