The puzzle

During the fourth of five puzzles in the final installment of Destiny 2's series of DLCs, in the raid called Salvation's Edge, fireteams of 6 enter yet another of The Witness' monolithic puzzles and find themselves in an antechamber with 6 statues of themselves.

Upon interacting with a statue displaying a prompt to "free Guardian," three of the players will be teleported into the shadow realm colloquially referred to as "inside."

Inside

The three lucky mensa candidates will find themselves each in their own instances of the starting chamber, this time with only three statues instead of six. Each of the statues –representing a player that has been thusly banished– holds one of three, distinct two-dimensional shapes: a circle, triangle, or a square.

Each "inside" player must relay the order of the shapes they see to the "outside" players. "Inside" players have access to only two of the shapes held by their statues, easily attainable by killing the hive knights roaming around their private chamber.

The overarching goal for each "inside" player is to attain a key comprised of one of each shape. That is, the player whose guardian statue is holding a square needs to trade with the other two "inside" guardians as needed to get a circle and a triangle as well, and so on for the other two inside players as well for their own statue's held-shape.

Outside

The remaining three "outside" guardians need to organize themselves into a protective vanguard with highest IQ player being the designated-dissector. Similar to the "inside" shadow realm, the "outside" will have three statues representing the "inside" players. However, the outside statues will be holding three-dimensional volumes instead of two-dimensional shapes.

The designated-dissector needs to collect the 2D shapes that are dropped by the hive knights roaming around the "outside" and use them to dissect the volumes held by the statues by dunking a 2D shape into it. This will do two things:

1. Dunking a 2D shape will stage that shape for removal from the 3D shape
2. It will also stage that removed shape for a swap

After a successive dunk, the two staged dissections will be swapped, and the resultant volume will now appear in the respective guardian statues' clutches.

Shapes

The shapes have dumb names, but I can make them dumber. I've labeled the elements of the set of 2D shapes according to the formula $2^{i}$ where $i$ is $0, 1,$ or $2$.

• 2D shapes: $\mathcal R^2 = \{1,2,4\}$
• 3D shapes: $\mathcal R^3 = \{2,3,4,6,8, 5\}$

(the peculiarly numbered elements of the set of volumes $\mathcal R^3$ will be clear when we see how we combine shapes to form volumes).

Dunking and Dissecting

"Inside" players can trade shapes with each other by dunking shapes into the guardian statue of their intended recipient within their instance of the shadow chamber.

As mentioned before, shapes are collected by killing knights. Guardians are able to hold more than one shape at a time... sort of. Picking up two 2D shapes will fuse them into a three-dimensional volume.

I'll introduce a function $\phi: \mathcal R^2 \times \mathcal R^2 \rightarrow \mathcal R^3$ which defines how two-dimensional shapes combine into volumes, given by:

where the bracket notation, rather than meaning "the index of $\mathcal R^3$" can instead be interpreted as "the element of $\mathcal R^3$." We can enumerate all possible combinations visually as follows:

Thus, the numerical values of the three-dimensional volumes are purely derived from their two-dimensional constituents.

Dissection, given by the function $\delta: \mathcal R^3 \times \mathcal R^2 \rightarrow \mathcal R^2$ is defined by the function:

If the designated-dissector attempts to dunk a shape into the guardian statue holding a volume which does contain that shape, they will receive a warning that their offering has been denied (denoted "undefined" above), therefore the range of possible dissections are:

Note each volume is the sum of two shapes, both of which must be powers of two.

These two operations can be thought of together as multiplication and division, though the underlying operations are more grotesque.

Swapping

Now, my notation thus far has been perverse if not outright misleading. Reasoning about how the shapes are combined via "multiplicative" dunking and "divisive" dissection is straightforward enough without the above explication. The rub of the puzzle as it's presented is that these functions $\phi$ and $\delta$ are not pure. The act of dissection carries a stateful "swap" which must be implicitly tracked by guardians during the trial.

For example, dissecting a sphere "stages" the residual circle for a swap with whatever the residual shape of the next dissected volume is. The domain of the swap-space is the union of the set of two-dimensional shapes with the initial "empty" state: $\mathcal R^2 \cup \varnothing$.

Note also that after a swap is executed and the resultant volumes are fused together (according to $\phi$), the staging area is cleared.

You cannot swap with the volume previously dissected. I'm not sure how to concisely express this stateful constraint with mathematical notation so I'm putting it here in bold. If you find yourself in this position (holding a shape which has no legal dissections available), the only way to soft-reset your "outside" puzzle state is by picking up another 2D shape. This will fuse it into a 3D shape according to $\phi$. Dunking a 3D shape will despawn all your shapes and return them to the knights.

All of this is happening on a 3 minute timer btw, and the knights will try to kill you, and there's ads as well.

An example configuration can be traced as follows.

Note that the solution will never contain one of the perfect volumes (sphere, tetrahedron, or cube) because $2^n - 1$ will never be a perfect square for any non-negative $n$. Each initial "inside" shape is uniquely solved by whichever of the three remaining volumes that is comprised of the other two 2D shapes:

• circle is solved by prism,
• square is solved by cone,
• triangle is solved by cylinder.

We can verify whether a given puzzle state is a solution if it satisfies the two following constraints:

1. the logarithmic difference between the volume and the "inside" shape is a not a power of two
2. the volume is not "perfect"

Mathematically, we can express these constraints as follows:

Visualized alternatively with the ~logarithmetic computations made explicit, we can see that each 2D shape has a unique and corresponding volumetric solution indicated by the rows where both of the rules are satisfied:

You can play around with the simulation here: https://escapevi.github.io/verity-simulator/

Optimal Strategy

Inside

The inside strategy is straightforward: kill knights and trade shapes with other "inside" players until each guardian has a key.

Outside

For the "outside" mentat, proceed from left to right by feeding each statue precisely the shape corresponding to that guardian's "inside" shpe to effectively remove it from the "outside" volume.

Remember that the goal is to execute a series of dissections, swaps, and fusions such that the resultant volume is the combination of the other two 2D shapes in $\mathcal R^2$ than the one corresponding to that guardian's statue.

The initial pass of feeding each statue its own corresponding 2D shape will necessarily result in one swap between the left and middle statues, clearing the staging area, and priming the third shape in the far right staging area.

After defeating the knights to receive a fresh batch of 2D shapes, consider which other unsolved volume contains the 2D shape that the far right statue needs. Unsolved volumes are easily identifiable by comparing them to their 2D, "inside" representations.

It is possible, though unlikely for the initial configuration to be solved. The worst case is when each held-volume is the perfected version of the 2D shape.

Looking for Group

We can consider an alternative representation of the problem through the lens of group theory. If we define our collection of objects as all of the shapes $\mathcal R^2 \cup \mathcal R^3$, with a single group operation $\circ$ which captures the behavior of both $\phi$ and $\delta$, we might be able to prove that the strategy outlined above is optimal with some cool properties of groups.

Recall from any number of previous posts about group theory, that a group is defined by a collection of objects $G$ together with some operation $\star$ which satisfy the following axioms:

1. For any $x, y \in G$, $x \star y \in G$. That is, the result of the operation between two elements in our group is another element which is also in our collection; a group is closed under its operation.

2. $\exists \epsilon \in G \; s.t. \; \forall x \in G, \epsilon \star x = x = x \star \epsilon$. Every group must have a "no-op" element which is called the identity of the group $G$. The identity of group formed by the integers with addition $G(\Z, +)$ is $0$.

3. The group operation $\star$ is associative. That is, $\forall x,y,z \in G, (x \star y) \star z = x \star (y \star z)$ which allows us to write $x \star y \star z$ unambiguously

4. Every element in $G$ has an inverse: $\forall x \in G \exists x^{-1} \in G \; s.t. \; x \star x^{-1} = \epsilon = x^{-1} \star x$

For example, $(\Z, +)$ form a group, but $(\Z, \times)$ does not, because $\not\exists x \in Z \; s.t. \; 2x = 1$.

In order to satisfy the first axiom, we must introduce another object into our collection to represent those combinations which result in an undefined "denied offering", such as when we try to dissect a square from a sphere. We can call this element $\varepsilon$, commonly reserved for the identity element of an abstract group like this, since we'll also use it for the second axiom.

with our group now being defined as

where the objects of our group are the three 2D shapes, the six 3D volumes, an identity element; and the god-operator $\circ$. We can verify that $G$ satisfies the first axiom:

We can tabularize all possible relations in a Cayley table:

so the first axiom is satisfied, since every combination of elements results in another one of the elements of our group.

The second axiom is satisfied with the identitive element $\varepsilon$.

The fourth axiom is satisfied as well per the Cayley table: each element $x \in G$ has an inverse $x^{-1}$ such that when they are combined they result in the identity $\varepsilon$.

However, I'm afraid out journey ends when we consider the third axiom of associativity. Our operation as defined does not compose:

Therefore, $G$ as presented is not a group, sad!