Introduction
I learned how to solve puzzles from my ma: Do the edges first.
Accordingly, I start solutions with the perimeter. The perimeter is simple: each piece needs to have opposite parity from both its neighbors. Addressing the internal structure quickly gets tougher. This deceptively straightforward puzzle puzzle blooms into a deep and intriguing challenge.
My answer is quite long. Starting from pieces and perimeters, I prove 1-row, 2-row, and 3-row cases. I've been as concise as I know how, but there are many steps for many cases. I believe my proofs are complete. I'd be delighted for any corrections, compeletions, or streamlining suggestions.
See the Conclusion for a quick summary of my results.
Preliminary: The 1-Row Puzzle
The 1-Row Puzzle is easy. Still, it presents some useful concepts, so I'm including it as a preliminary case.
Notation. Each piece is represented by 2 characters: 1 or 0 indicates the parity of a face, and a bracket represents an end face. Since we're looking at 1-row puzzles, the top and bottom of each piece are assumed to be side faces.
List of All Pieces. With 3 options for 2 positions there are 3 x 3 combinations. There are 3 pairs that are equivalent under rotation. Therefore there are 6 unique pieces:
1. []
2. 00 -- this rotates, so is excluded
3. 11 -- this rotates, so is excluded
4. [0 (rotated: 0])
5. [1 (rotated: 1])
6. 01 (rotated: 10)
Solutions. Adjacent faces must have opposite parity. The solutions are small enough to be self-explanatory.
1-Piece solution
[]
2-Piece solution
[0 1]
3-Piece solution
[0 10 1]
4-Piece solution. None.
Notation
The original post poses the question: "What is the biggest puzzle in terms of number of pieces I can make?"
Therefore, our solution must be in terms of a set of pieces. Arrangements can help prove or disprove uniqueness, but are not an actual solution. A counterexample can be shown with two or more valid arrangements for a specific set of pieces. Arguments can show that an arrangement is possible or impossible.
Corner and side pieces have exactly two neighbors. Adjoining pieces must have opposite parity (described in the original post as tap and blank).
I represent parity as 1s and 0s to simplify calculations. Sometimes I use * or ? to represent parity we don't care about or don't know. I may refer to a piece by its label (e.g. "A" or "w") or simply by its parity ("00").
The perimeter and individual pieces are always read clockwise.
Puzzle Pieces
List of All Pieces.
- There are exactly 4 corner pieces
- There are exactly 8 side pieces -- I organize them in pairs which differ only in internal face parity
- There are exactly 3 internal pieces
Corners. With each connecting with two neighbors, there are exactly 4 different corner pieces.
Labels: A, B, C, D.
A. 00
B. 01
C. 10
D. 11
Sides. With two neighbor faces and one internal face, there are 2x2x2 distinct pieces. It's often helpful to pair them accordint to the lef and right parity, since the internal face does not affect constrcution of the perimeter.
Labels: w, x, y, z
1. w: 00 {w0, w1}
2. x: 01 {x0, x1}
3. y: 10 {y0, y1}
4. z: 11 {z0, z1}
Internal Labels: 𝛼, 𝛽, 𝛾
1. 𝛼: (0001)
2. 𝛽: (1110)
3. 𝛾: (0011)
Feasible Layouts
No larger layouts are possible because there are only 8 side pieces. The 4x4 layout above is not possible because there are only 3 internal pieces.
2x2 Puzzle
Claim. There is only one 2x2 solution:
Proof.
B (01) and C (10) are incompatible; they can never be neighbors. Thus,
there is only one 2x2 solution. See Diagram Below.
QED
2-Row Puzzles
Claim. The 2x2 solution is the only 2-row solution
Proof.
Since B and C can never be neighbors, there are only two possible "bookend" configurations for the ends of a 2-row puzzle.
The Diagram shows how the horizontal bookends and the vertical bookends present the same parity structure (00 and 11) to any candidate pieces. That means if a combination of pieces works with one set of bookends, you can remove the book ends, reassemble them into the other bookends. They will fit the original combination of pieces, giving two arrangements for the same set of pieces.
QED
3-Row Puzzles
3x3 Puzzle
Claim: There is no solution for the 3x3 layout.
Proof.
Up to rotation of the entire puzzle, there are 3! permutations of corners. This roster of 6 frameworks gives all possible layouts. Corner parity forces selection of side pieces (up to pairs w, x, y, z). See Diagram 1: Roster of Permutations.
Their structure lets us analyze them in two groups -- Group 1: Permutations 1 and 2 (two pairs of sides), and Group 2: Permutations 3, 4, 5, and 6 (no paired sides).
Group 1.
Having two pairs of sides means the internal piece must have two of each parity, which makes 𝛾 the only candidate. There are 4 ways parity can be selected.
Diagram 2 lays out candidate solutions for Permutation 1. The first two cases have like parity adjoining, which allows 𝛾 to fit as the internal piece. Transposing the x-pair and transposing the y-pair in effect rotates the 𝛾-piece. The second two cases show that any candidate internal piece is invalid because it would allow rotation.
The identical argument rules out Permutation 2, {w0, w1, z0, z1}.
Group 2.
Permutations 3, 4, 5, and 6 have one each of w, x, y, and z (no paired sides). That means any of the 2^4 parity combinations can be selected. However, the problem space is constrained because the only candidate internal pieces are 𝛼, 𝛽, and 𝛾.
Permutations 3 and 5 are related by horizontal reflection. Permutations 4 and 6 are related by horizontal reflection. These reflections are executed by permuting the corners and sides as indicated in Diagram 3, below. These reflections are represented as rotations of pieces 𝛼, 𝛽, and 𝛾.
Any set of pieces that fulfills this framework can be rearranged into at least one other arrangement, Therefore, none of these permutations represents any solution.
Group 1 and Group 2 represent all possible arrangements for the 3x3 puzzle. The arguments above show that any set of pieces in Group 1 or Group 2 that has one arrangement has at least on other arrangement. Therefore, there are no solutions for the 3x3 puzzle.
QED
3x4 Puzzle
There are 6 faces between the perimeter and the internal pieces. For internal pieces, we can have 𝛼 and 𝛽 (giving 3 faces of one parity and 3 of the other), or 𝛼 and 𝛾 (or 𝛽 and 𝛾) (giving 2 faces of one parity and 4 of the other). We can handle these groups separately because rotating or permuting any pieces cannot change the number of faces of each parity.
Group 1: 𝛼 + 𝛽
By rotating individually, 𝛼 + 𝛽 can give every parity combination.
In the perimeter there must be at least on pair from w, x, y, z. Such a pair can be transposed in the perimeter because they have the same left-hand and right-hand parity. This may change the pattern of the parity, but can't change count of 3 of each parity. Thus, every member of this Group 1 has at least a transpose partner and therefore is not unique.
Group 2: 𝛼+𝛾 or 𝛽+𝛾.
The 15 configurations of 𝛽+𝛾 are shown in the diagram. Two combinations are invalid, and marked in orange. One configuration remains unchanged with rotation of its pieces. The other 12 each have a partner where the rotation of a piece gives the same configuration as its partner.
That means, no matter what perimeter we have, the center pieces can be rotated without changing the perimeter. This gives two distinct arrangements for the same pieces. The same argument applies to 𝛼+𝛾. Therefore, no member of Group 2 has a unique arrangement.
Since all sets of pieces fall in either Group 1 or Group 2, there is no set of pieces with a unique arrangement. Therefore, no 3x4 can be made.
QED.
3x5 Puzzle
Claim: There is no 3x5 solution.
Proof: Counterexample
A solution requires a set of pieces that can be assembled in exactly one way. There are 4 corner pieces, 8 side pieces, and 3 internal pieces. A 3x5 puzzle must use all 15 pieces. Any two distinct ways to assemble a 3x5 prove there is no unique solution.
The diagram below shows two distinct 3x5 arrangments using the set of all 15 pieces. This shows that there is no 3x5 solution.
QED
Conclusion
The maximum solution is the 2x2 presented above.
The set of pieces is limited by excluding duplicates under rotation. In turn, this limits possible solutions to 8 different layouts. The 1-row layouts can be handled with exhaustive case-by-case analysis. The 2-row layouts are easily handled through a couple encompassing observations. The 3-row layouts can be addressed as 3x3, 3x4, and 3x5 cases: the 3x3 layout can be proved impossible by examining the cases allowed by the 6 perimeter permutations; the 3x4 layout can be proved impossible by examining the interior pieces; the 3x5 layout can be proved impossible by a counterexample. No larger layouts are possible, simply due to the limited set of pieces.