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the strange island of two truths

Last time we had a brief encounter with the island of two truths, invented by Karin Cvetko-Vah. See her posts:

On this island, false statements have truth-value $0$ (as usual), but non-false statements are not necessarily true, but can be given either truth-value $Q$ (statements which the Queen on the island prefers) or $K$ (preferred by the King).

Think of the island as Trump’s paradise where nobody is ever able to say: “Look, alternative truths are not truths. They’re falsehoods.”

Even the presence of just one ‘alternative truth’ has dramatic consequences on the rationality of your reasoning. If we know the truth-values of specific sentences, we can determine the truth-value of more complex sentences in which we use logical connectives such as $\vee$ (or), $\wedge$ (and), $\neg$ (not), and $\implies$ (then) via these truth tables:

\downarrow~\bf{\wedge}~\rightarrow & 0 & Q & K \\
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\downarrow~\vee~\rightarrow & 0 & Q & K \\
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \]
\downarrow~\implies~\rightarrow & 0 & Q & K \\
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\downarrow & \neg~\downarrow \\
0 & Q \\
Q & 0 \\
K & 0

Note that the truth-values $Q$ and $K$ are not completely on equal footing as we have to make a choice which one of them will stand for $\neg 0$.

Common tautologies are no longer valid on this island. The best we can have are $Q$-tautologies (giving value $Q$ whatever the values of the components) or $K$-tautologies.

Here’s one $Q$-tautology (check!) : $(\neg p) \vee (\neg \neg p)$. Verify that $p \vee (\neg p)$ is neither a $Q$- nor a $K$-tautology.

Can you find any $K$-tautology at all?

Already this makes it incredibly difficult to adapt Smullyan-like Knights and Knaves puzzles to this skewed island. Last time I gave one easy example.

Puzzle : On an island of two truths all inhabitants are either Knaves (saying only false statements), Q-Knights (saying only $Q$-valued statements) or K-Knights (who only say $K$-valued statements).

The King came across three inhabitants, whom we will call $A$, $B$ and $C$. He asked $A$: “Are you one of my Knights?” $A$ answered, but so indistinctly that the King could not understand what he said.

He then asked $B$: “What did he say?” $B$ replies: “He said that he is a Knave.” At this point, $C$ piped up and said: “That’s not true!”

Was $C$ a Knave, a Q-Knight or a K-Knight?

Solution : Q- and K-Knights can never claim to be a Knave. Neither can Knaves because they can only say false statements. So, no inhabitant on the island can ever claim to be a Knave. So, $B$ lies and is a Knave, so his stament has truth-value $0$. $C$ claims the negation of what $B$ says so the truth-value of his statement is $\neg 0 = Q$. $C$ must be a Q-Knight.

As if this were not difficult enough, Karin likes to complicate things by letting the Queen and King assign their own truth-values to all sentences, which may coincide with their actual truth-value or not.

Clearly, these two truth-assignments follow the logic of the island of two truths for composed sentences, and we impose one additional rule: if the Queen assigns value $0$ to a statement, then so does the King, and vice versa.

I guess she wanted to set the stage for variations to the island of two truths of epistemic modal logical puzzles as in Smullyan’s book Forever Undecided (for a quick summary, have a look at Smullyan’s paper Logicians who reason about themselves).

A possible interpretation of the Queen’s truth-assignment is that she assigns value $Q$ to all statements she believes to be true, value $0$ to all statements she believes to be false, and value $K$ to all statements she has no fixed opinion on (she neither believes them to be true nor false). The King assigns value $K$ to all statements he believes to be true, $0$ to those he believes to be false, and $Q$ to those he has no fixed opinion on.

For example, if the Queen has no fixed opinion on $p$ (so she assigns value $K$ to it), then the King can either believe $p$ (if he also assigns value $K$ to it) or can have no fixed opinion on $p$ (if he assigns value $Q$ to it), but he can never believe $p$ to be false.

Puzzle : We say that Queen and King ‘agree’ on a statement $p$ if they both assign the same value to it. So, they agree on all statements one of them (and hence both) believe to be false. But there’s more:

  • Show that Queen and King agree on the negation of all statements one of them believes to be false.
  • Show that the King never believes the negation of whatever statement.
  • Show that the Queen believes all negations of statements the King believes to be false.

Solution : If one of them believes $p$ to be false (s)he will assign value $0$ to $p$ (and so does the other), but then they both have to assign value $Q$ to $\neg p$, so they agree on this.

The value of $\neg p$ can never be $K$, so the King does not believe $\neg p$.

If the King believes $p$ to be false he assigns value $0$ to it, and so does the Queen, but then the value of $\neg p$ is $Q$ and so the Queen believes $\neg p$.

We see that the Queen and King agree on a lot of statements, they agree on all statements one of them believes to be false, and they agree on the negation of such statements!

Can you find any statement at all on which they do not agree?

Well, that may be a little bit premature. We didn’t say which sentences about the island are allowed, and what the connection (if any) is between the Queen and King’s value-assignments and the actual truth values.

For example, the Queen and King may agree on a classical ($0$ or $1$) truth-assignments to the atomic sentences for the island, and replace all $1$’s with $Q$. This will give a consistent assignment of truth-values, compatible with the island’s strange logic. (We cannot do the same trick replacing $1$’s by $K$ because $\neg 0 = Q$).

Clearly, such a system may have no relation at all with the intended meaning of these sentences on the island (the actual truth-values).

That’s why Karin Cvetko-Vah introduced the notions of ‘loyalty’ and ‘sanity’ for inhabitants of the island. That’s for next time, and perhaps then you’ll be able to answer the question whether Queen and King agree on all statements.

(all images in this post are from Smullyan’s book Alice in Puzzle-Land)

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the L-game

In 1982, the BBC ran a series of 10 weekly programmes entitled de Bono’s Thinking Course. In the book accompanying the series Edward de Bono recalls the origin of his ‘L-Game’:

Many years ago I was sitting next to the famous mathematician, Professor Littlewood, at dinner in Trinity College. We were talking about getting computers to play chess. We agreed that chess was difficult because of the large number of pieces and different moves. It seemed an interesting challenge to design a game that was as simple as possible and yet could be played with a degree of skill.

As a result of that challenge I designed the ‘L-Game’, in which each player has only one piece (the L-shape piece). In turn he moves this to any new vacant position (lifting up, turning over, moving across the board to a vacant position, etc.). After moving his L-piece he can – if he wishes – move either one of the small neutral pieces to any new position. The object of the game is to block your opponent’s L-shape so that no move is open to it.

It is a pleasant exercise in symmetry to calculate the number of possible L-game positions.

The $4 \times 4$ grid has $8$ symmetries, making up the dihedral group $D_8$: $4$ rotations and $4$ reflections.

An L-piece breaks all these symmetries, that is, it changes in form under each of these eight operations. That is, using the symmetries of the $4 \times 4$-grid we can put one of the L-pieces (say the Red one) on the grid as a genuine L, and there are exactly 6 possibilities to do so.

For each of these six positions one can then determine the number of possible placings of the Blue L-piece. This is best done separately for each of the 8 different shapes of that L-piece.

Here are the numbers when the red L is placed in the left bottom corner:

In total there are thus 24 possibilities to place the Blue L-piece in that case. We can repeat the same procedure for the remaining Red L-positions. Here are the number of possibilities for Blue in each case:

That is, there are 82 possibilities to place the two L-pieces if the Red one stands as a genuine L on the board.

But then, the L-game has exactly $18368 = 8 \times 82 \times 28$ different positions, where the factor

  • $8$ gives the number of symmetries of the square $4 \times 4$ grid.
  • Using these symmetries we can put the Red L-piece on the grid as a genuine $L$ and we just saw that this leaves $82$ possibilities for the Blue L-piece.
  • This leaves $8$ empty squares and so $28 = \binom{8}{2}$ different choices to place the remaining two neutral pieces.

The $2296 = 82 \times 28$ positions in which the red L-piece is placed as a genuine L can then be analysed by computer and the outcome is summarised in Winning Ways 2 pages 384-386 (with extras on pages 408-409).

Of the $2296$ positions only $29$ are $\mathcal{P}$-positions, meaning that the next player (Red) will loose. Here are these winning positions for Blue

Here, neutral piece(s) should be put on the yellow square(s). A (potential) remaining neutral piece should be placed on one of the coloured squares. The different colours indicate the remoteness of the $\mathcal{P}$-position:

  • Pink means remoteness $0$, that is, Red has no move whatsoever, so mate in $0$.
  • Orange means remoteness $2$: Red still has a move, but will be mated after Blue’s next move.
  • Purple stands for remoteness $4$, that is, Blue mates Red in $4$ moves, Red starting.
  • Violet means remoteness $6$, so Blue has a mate in $6$ with Red starting
  • Olive stands for remoteness $8$: Blue mates within eight moves.

Memorising these gives you a method to spot winning opportunities. After Red’s move image a board symmetry such that Red’s piece is a genuine L, check whether you can place your Blue piece and one of the yellow pieces to obtain one of the 29 $\mathcal{P}$-positions, and apply the reverse symmetry to place your piece.

If you don’t know this, you can run into trouble very quickly. From the starting position, Red has five options to place his L-piece before moving one of the two yellow counters.

All possible positions of the first option loose immediately.

For example in positions $a,b,c,d,f$ and $l$, Blue wins by playing

Here’s my first attempt at an opening repertoire for the L-game. Question mark means immediate loss, question mark with a number means mate after that number of moves, x means your opponent plays a sensible strategy.

Surely I missed cases, and made errors in others. Please leave corrections in the comments and I’ll try to update the positions.

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The forests of the unconscious

We start from a large data-set $V=\{ k,l,m,n,\dots \}$ (texts, events, DNA-samples, …) with a suitable distance-function ($d(m,n) \geq 0~d(k,l)+d(l,m) \geq d(k.m)$) which measures the (dis)similarity between individual samples.

We’re after a set of unknown events $\{ p,q,r,s,\dots \}$ to explain the distances between the observed data. An example: let’s assume we’ve sequenced the DNA of a set of species, and computed a Hamming-like distance to measures the differences between these sequences.

(From Geometry of the space of phylogenetic trees by Billera, Holmes and Vogtmann)

Biology explains these differences from the fact that certain species may have had more recent common ancestors than others. Ideally, the measured distances between DNA-samples are a tree metric. That is, if we can determine the full ancestor-tree of these species, there should be numbers between ancestor-nodes (measuring their difference in DNA) such that the distance between two existing species is the sum of distances over the edges of the unique path in this phylogenetic tree connecting the two species.

Last time we’ve see that a necessary and sufficient condition for a tree-metric is that for every quadruple $k,l,m,n \in V$ we have that the maximum of the sum-distances

$$\{ d(k,l)+d(m,n),~d(k,m)+d(l,n),~d(k,n)+d(l,m) \}$$

is attained at least twice.

In practice, it rarely happens that the measured distances between DNA-samples are a perfect fit to this condition, but still we would like to compute the most probable phylogenetic tree. In the above example, there will be two such likely trees:

(From Geometry of the space of phylogenetic trees by Billera, Holmes and Vogtmann)

How can we find them? And, if the distances in our data-set do not have such a direct biological explanation, is it still possible to find such trees of events (or perhaps, a forest of event-trees) explaining our distance function?

Well, tracking back these ancestor nodes looks a lot like trying to construct colimits.

By now, every child knows that if their toy category $T$ does not allow them to construct all colimits, they can always beg for an upgrade to the presheaf topos $\widehat{T}$ of all contravariant functors from $T$ to $Sets$.

But then, the child can cobble together too many crazy constructions, and the parents have to call in the Grothendieck police who will impose one of their topologies to keep things under control.

Can we fall back on this standard topos philosophy in order to find these forests of the unconscious?

(Image credit)

We have a data-set $V$ with a distance function $d$, and it is fashionable to call this setting a $[0,\infty]$-‘enriched’ category. This is a misnomer, there’s not much ‘category’ in a $[0,\infty]$-enriched category. The only way to define an underlying category from it is to turn $V$ into a poset via $n \geq m$ iff $d(n,m)=0$.

Still, we can define the set $\widehat{V}$ of $[0,\infty]$-enriched presheaves, consisting of all maps
$$p : V \rightarrow [0,\infty] \quad \text{satisfying} \quad \forall m,n \in V : d(m,n)+p(n) \geq p(m)$$
which is again a $[0,\infty]$-enriched category with distance function
$$\hat{d}(p,q) = \underset{m \in V}{max} (q(m) \overset{.}{-} p(m)) \quad \text{with} \quad a \overset{.}{-} b = max(a-b,0)$$
so $\widehat{V}$ is a poset via $p \geq q$ iff $\forall m \in V : p(m) \geq q(m)$.

The good news is that $\widehat{V}$ contains all limits and colimits (because $[0,\infty]$ has sup’s and inf’s) and that $V$ embeds isometrically in $\widehat{V}$ via the Yoneda-map
$$m \mapsto y_m \quad \text{with} \quad y_m(n)=d(n,m)$$
The mental picture of a $[0,\infty]$-enriched presheaf $p$ is that of an additional ‘point’ with $p(m)$ the distance from $y_m$ to $p$.

But there’s hardly a subobject classifier to speak of, and so no Grothendieck topologies nor internal logic. So, how can we select from the abundance of enriched presheaves, the nodes of our event-forest?

We can look for special properties of the ancestor-nodes in a phylogenetic tree.

For any ancestor node $p$ and any $m \in V$ there is a unique branch from $p$ having $m$ as a leaf (picture above,left). Take another branch in $p$ and a leaf vertex $n$ of it, then the combination of these two paths gives the unique path from $m$ to $n$ in the phylogenetic tree, and therefore
$$\hat{d}(y_m,y_n) = d(m,n) = p(m)+p(n) = \hat{d}(p,y_m) + \hat{d}(p,y_n)$$
In other words, for every $m \in V$ there is another $n \in V$ such that $p$ lies on the geodesic from $m$ to $n$ (identifying elements of $V$ with their Yoneda images in $\widehat{V}$).

Compare this to Stephen Wolfram’s belief that if we looked properly at “what ChatGPT is doing inside, we’d immediately see that ChatGPT is doing something “mathematical-physics-simple” like following geodesics”.

Even if the distance on $V$ is symmetric, the extended distance function on $\widehat{V}$ is usually far from symmetric. But here, as we’re dealing with a tree-distance, we have for all ancestor-nodes $p$ and $q$ that $\hat{d}(p,q)=\hat{d}(q,p)$ as this is just the sum of the weights of the edges on the unique path from $p$ and $q$ (picture above, on the right).

Right, now let’s look at a non-tree distance function on $V$, and let’s look at those elements in $\widehat{V}$ having similar properties as the ancestor-nodes:

$$T_V = \{ p \in \widehat{V}~:~\forall n \in V~:~p(n) = \underset{m \in V}{max} (d(m,n) \overset{.}{-} p(m)) \}$$

Then again, for every $p \in T_V$ and every $n \in V$ there is an $m \in V$ such that $p$ lies on a geodesic from $n$ to $m$.

The simplest non-tree example is $V = \{ a,b,c,d \}$ with say

$$d(a,c)+d(b,d) > max(d(a,b)+d(c,d),d(a,d)+d(b,c))$$

In this case, $T_V$ was calculated by Andreas Dress in Trees, Tight Extensions of Metric Spaces, and the Cohomological Dimension of Certain Groups: A Note on Combinatorial Properties of Metric Spaces. Note that Dress writes $mn$ for $d(m,n)$.

If this were a tree-metric, $T_V$ would be the tree, but now we have a $2$-dimensional cell $T_0$ consisting of those presheaves lying on a geodesic between $a$ and $c$, and on the one between $b$ and $d$. Let’s denote this by $T_0 = \{ a—c,b—d \}$.

$T_V$ has eight $1$-dimensional cells, and with the same notation we have

Let’s say that $V= \{ a,b,c,d \}$ are four DNA-samples of species but failed to satisfy the tree-metric condition by an error in the measurements, how can we determine likely phylogenetic trees for them? Well, given the shape of the cell-complex $T_V$ there are four spanning trees (with root in $f_a,f_b,f_c$ or $f_d$) having the elements of $V$ as their only leaf-nodes. Which of these is most likely the ancestor-tree will depend on the precise distances.

For an arbitrary data-set $V$, the structure of $T_V$ has been studied extensively, under a variety of names such as ‘Isbell’s injective hull’, ‘tight span’ or ‘tropical convex hull’, in slightly different settings. So, in order to use results one sometimes have to intersect with some (un)bounded polyhedron.

It is known that $T_V$ is always a cell-complex with dimension of the largest cell bounded by half the number of elements of $V$. In this generality it will no longer be the case that there is a rooted spanning tree of teh complex having the elements of $V$ as its only leaves, but we can opt for the best forest of rooted trees in the $1$-skeleton having all of $V$ as their leaf-nodes. Theses are the ‘forests of the unconscious’ explaining the distance function on the data-set $V$.

Apart from the Dress-paper mentioned above, I’ve found these papers informative:

So far, we started from a data-set $V$ with a symmetric distance function, but for applications in LLMs one might want to drop that condition. In that case, Willerton proved that there is a suitable replacement of $T_V$, which is now called the ‘directed tight span’ and which coincides with the Isbell completion.

Recently, Simon Willerton gave a talk at the African Mathematical Seminar called ‘Looking at metric spaces as enriched categories’:

Willerton also posts a series(?) on this at the n-category cafe, starting with Metric spaces as enriched categories I.


Previously in this series:

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