Fault Lines in Common Pentomino Solutions Mike Beeler January 2026 Packing the 12 free pentominoes into a rectangle, or an 8x8 square with a center 2x2 hole, are puzzles whose solutions have been known for decades. Displaying these solutions can be interesting, such as in quilts and wall hangings, stained and fused glass, collage and decoupage, etc. In these media, it can be desirable to avoid (or to intentionally choose) solutions that contain a fault line. This analysis looks at solutions with fault lines. These days, it is hard to know whether this analysis is new. A brief internet search turned up nothing relevant. (See also the Appendix.) The Addenda briefly discuss some other properties of the solutions. This note uses ASCII art, best viewed in a fixed width font. The shapes to be packed considered here are: 6x10 rectangle -- 2339 distinct solutions, 16 have a fault line 5x12 rectangle -- 1010 distinct solutions, 264 have a fault line 4x15 rectangle -- 368 distinct solutions, 40 have a fault line 3x20 rectangle -- 2 distinct solutions, none have a fault line 8x8 square with 2x2 centered hole -- 65 distinct solutions, 12 have a fault line Here, we consider solutions distinct with respect to rotation and reflection. Re-packing, rotating or reflecting one or more sub-areas is considered as making a different, distinct solution. In all solutions, there is at most one fault line. In the rectangles with a fault line, that line is across the smaller dimension. The percentages with a fault line are: 6x10 -- 16 / 2339 = 0.68 % 5x12 -- 264 / 1010 = 26.1 % 4x15 -- 40 / 368 = 10.9 % 3x20 -- 0 / 2 = 0 % 8x8 with center 2x2 hole -- 12/65 = 18.5 % The following sections discuss each shape in more detail. ------------------------------------------------------------------------ The 6x10 has 16 of 2339 distinct solutions having a fault. In all 16, the fault splits the solution into two 6x5 rectangles, with I L T W Y Z in one and F N P U V X in the other. These pack as: I I I I I T N F F P P P N N N P P P Y Z Z T T T N N F F P P F F N N P P Y Y Z W W T V N F X U U V F F X U U Y L Z Z W W V N X X X U V F X X X U Y L L L L W V V V X U U V V V X U U The first set packs in only one way. The second set has two versions, by repacking the F,N unit. For now, consider just one of those two. The first 6x5 can be oriented in 4 ways, and the second in 4 ways. Either one of the 6x5 rectangles can be placed on top in the 6x10, for a factor of 2. This makes 4*4*2 = 32 solutions. Each of those solutions has 4 orientations, only one of which will be selected as the one to represent the four as a distinct solution. 32/4 = 8 distinct solutions. Now consider that the F,N unit can be repacked in each of these 8 distinct solutions, for a factor of 2, making the 16 distinct solutions that have a fault. ------------------------------------------------------------------------ The 4x15 has 40 solutions with a fault. All are split in one place, making a 4x10 and a 4x5. In all 40, the 4x5 has N T V Y, and these make the 4x5 in just one way. N T T T N N T Y V N T Y V N Y Y V V V Y There are 5 distinct ways to pack the 4x10: I P P P I P P P I P P P I P P P I U U U I Z P P I Z P P I Z P P I Z P P I U F U I Z Z Z I Z Z Z I Z Z Z I Z Z Z I F F F I W W Z I W W Z I W W Z I W W Z I W W F I F W W I F W W I X W W I X W W I X W W F F F W F F F W X X X W X X X W X X X W F X L L F X U U F X L L F X U U Z X L L X X X L X X X U F F F L F F F U Z Z Z L U X U L L X U U U F U L L F U U P P Z L U U U L L L L L U U U L L L L L P P P L Thus the factors are: 5 ways to pack the 4x10 4 orientations of the 4x10 1 way to pack the 4x5 4 orientations of the 4x5 2 ways to stack the 4x10 and 4x5 5*4*1*4*2 = 160 constructions = 40 distinct 4x15 solutions * 4 orientations. ------------------------------------------------------------------------ The 5x12 has 264 solutions with a fault. 224 are 5x1 + 5x11 8 are 5x3 + 5x9 16 are 5x5 + 5x7 16 are 5x6 + 5x6 ----- The 16 distinct solutions with two 5x6 rectangles are the same as the two 6x5s in the analysis of the 6x10 solutions, just turned 90 degrees. ----- The 5x3 + 5x9: The 5x3 is F P U, and the 5x9 is I L N T V W X Y Z. There is just one way to pack each rectangle. The factors are: 1 way to pack the 5x9 4 orientations of the 5x9 1 way to pack the 5x3 4 orientations of the 5x3 2 ways to stack the 5x9 and 5x3 1*4*1*4*2 = 32 constructions = 8 distinct solutions * 4 orientations. I Y Y Y Y P F F U U I Z Y L L P P F F U I Z Z Z L P P F U U I W W Z L I X W W L X X X W T V X T T T V N N N T V V V N N ----- The 5x5 + 5x7: The 5x5 is F L P U X, and the 5x7 is I N T V W Y Z. There is one way to make each rectangle. The factors are: 1 way to pack the 5x5 8 orientations of the 5x5 (it is square) 1 way to pack the 5x9 4 orientations of the 5x9 2 ways to stack the 5x5 and 5x9 1*8*1*4*2 = 64 constructions = 16 distinct solutions * 4 orientations. I Z V V V F F P P P I Z Z Z V L F F P P I W W Z V L F X U U I N W W Y L X X X U I N T W Y L L X U U N N T Y Y N T T T Y ----- The 5x1 + 5x11: 224 of the 1010 5x12 solutions split into a 5x1 and a 5x11. The 5x1 is of course I, and the 5x11 is the other 11 pieces. It is possible to analyze the 112 distinct ways to pack the 11 pieces into a 5x11, identifying some as related by reflecting or repacking subunits. However, there seems to be little benefit to that, and it does not further the analysis of solutions with faults. 1 way to pack the 5x1 1 orientation of the 5x1 112 ways to pack the 5x11 4 orientations of the 5x11 2 ways to stack the 5x1 and 5x11 1*1*112*4*2 = 896 constructions = 224 distinct solutions * 4 orientations. ------------------------------------------------------------------------ In the 8x8 with a centered 2x2 hole, 12 of its 65 distinct solutions have a fault. In all 12 cases, the fault splits the solution into two parts of equal area. Four ways to divide the pentominoes lead to solutions; the set of pieces containing F is any of these: F I L N U X (contains F,N unit) F I U V X Z F L N P T Y (contains F,N unit) F P T V Y Z The factors are (with, or without, the F,N unit): 2,2 -- divide pieces into 2 sets 2,1 -- repack F,N unit if it exists 2 -- reflect the part containing F 1 -- repack pieces in other part 2 -- reflect other part 4 -- rotate construction 90 degrees at a time those that have an F,N unit: 2*2*2*1*2*4 = 64 = 8 distinct solutions * 8 orientations those that lack an F,N unit: 2*1*2*1*2*4 = 32 = 4 distinct solutions * 8 orientations 8+4 = 12 distinct solutions with a fault line If we consider reflecting one part to be the same solution, then there are only 6 distinct solutions. If we also consider repacking the F,N unit to be the same solution, there are only 4 distinct solutions. These 4 differ in how the pieces are distributed between the two parts, so they are quite different. I I I I I F N L F I L N U X U U X F F F N L U X X X F N N L U U X N L L P P P V V V T P P W W Z Z V T T T Y W W Z V T Y Y Y Y W Z Z I I I I I F Z Z F I U V X Z U U X F F F Z V U X X X F Z Z V U U X V V V L L N P P P L N N W W P P T L N W W Y T T T L N W Y Y Y Y T I I I I I V V V F L N P T Y U U X W W Z Z V U X X X W W Z V U U X W Z Z L N F P P P L N F F F P P T L N N F Y T T T L L N Y Y Y Y T I I I I I N L L F P T V Y Z U U X W W N N L U X X X W W N L U U X W N L P P P F Z Z T P P F F F Z V T T T Y F Z Z V T Y Y Y Y V V V ------------------------------------------------------------------------ References Wikipedia, "Pentomino" Golomb, Solomon W., "Polyominoes", (1965), New York: Charles Scribner's Sons. R. L. Graham, Bell Laboratories, "Fault-free Tilings of Rectangles", https://mathweb.ucsd.edu/~ronspubs/81_01_fault_free_tilings.pdf a/k/a Graham, R. L., "Fault-free Tilings of Rectangles", in "The Mathematical Gardner: A Collection in Honor of Martin Gardner" (1981) (Ed. D. A. Klarner), Belmont, CA: Wadsworth, pp. 120-126. Weisstein, Eric W., "Fault-Free Rectangle." From "Mathworld" -- A Wolfram Resource. https://mathworld.wolfram.com/Fault-FreeRectangle.html A discussion and listing of solutions to many shapes is at https://isomerdesign.com/Pentomino/ I use the same solution numbering convention as at this site: cells are read left to right, top to bottom, as reading a page; the orientation that sorts first in alphabetic order represents that solution; solutions are then sorted in alphabetic order, and then numbered starting with 1. "Pentomino Relationships", an extensive, detailed grouping of solutions into families, in which the members of a family are related by operations such as rotating, reflecting, or re-packing a subunit, swapping subunits, etc.: http://abasmith.co.uk/pentanomes/pentanomes.html An excellent, comprehensive discussion of pentomino solutions is at ericharshbarger.org/pentominoes/ ------------------------------------------------------------------------ Appendix -- Asking AI systems for fault-free pentomino solutions October 14, 2025 Two AI systems performed very poorly when asked how many of the solutions are fault-free, or even whether there are solutions with faults. First, here are examples with the Google search web site, using its default AI system (Gemini 2.5 Pro, developed by Google). They are edited for brevity. Prompt: pentomino solutions with no fault line Response: The most common pentomino challenge is to tile a 6x10 rectangle with the 12 pentominoes. Among the 2,339 total solutions, several are fault-free. Below is a well-known example.  V V V V F F Y Y T T V I I P F P Y L T T V I U P F P L L X T W U U P F Z Z L X N W W U Q Z Z L N X N W W Q Q Q Q Q Q Q N [Comment: "Q" is not a usual pentomino name, and several are not even pentominoes. The N is not even connected.] Another try: Prompt: pentomino solutions with no fault line Response: There are 2,339 total solutions for the 6x10 rectangle. The vast majority of these solutions contain fault lines. The image below is an example of one of the few 6x10 "no fault line" solutions, first found by computer in 1968.  V V W P P P F F U U V V W P X F F U U U V W W X X F I I Y U W W X X L L I I Y Y T T N N N N I I Z Z T Y Y N N N Z Z L L [Comment: The names are pentomino names, but the pieces are mostly not. Only 0.68% of the 6x10 solutions have a fault, not "the vast majority".] More tries: Prompt: Of the 2339 distinct solutions to the 6x10 rectangle packed by the 12 pentominoes, how many contain a fault line?  Response: One solution has a fault line. This particular solution consists of two 5x6 rectangles placed side-by-side. [Comment: That's in the right direction; the 16 distinct solutions with a fault are split that way. The AI missed the F,N unit re-pack.] Prompt: Of the 2 distinct solutions for packing the 12 pentominoes in a 3x20 rectangle, how many are fault-free? Response: Only one is fault-free [1.1, 2.1]. [Comment: Wrong, both are fault-free. And what is "[1.1, 2.1]"?] After the above, clicking on "Dive deeper in AI Mode" gives: Response: There are no fault-free solutions for packing the 12 pentominoes in a 3x20 rectangle. Both of the two distinct solutions contain a fault. [Comment: Wrong again. The response on another try hints at why the AI system may have concluded this. The V pentomino forces a split across the 3-cell width of the rectangle. What that misses is that in both solutions, the V is at an end, thus no fault.] ----- The second AI system tried identified itself as "OpenAI ChatGPT (GPT-5 model, October 2025)". Prompt: Among the 2339 distinct solutions for packing a 6 by 10 rectangle with the 12 pentominoes, how many are free of faults? [Comment: It explored various approaches and sources, finally saying it was running a program using Knuth's DLX Dancing Links algorithm. It thought for 10m 38s.] Response: Short answer: I don’t have a reliably sourced number I can cite right now. I tried to look this up (the canonical list of the 2,339 distinct 6×10 solutions is available from the IsomerDesign site and others) and I attempted to compute the fault-free count by programmatically checking every one, but I couldn’t finish a trustworthy exact enumeration in this session (my automated search/solve attempts timed out). [Comment: It offered to parse the solutions at the IsomerDesign site and test each, to report the precise count; or to give me detailed instructions so I can do the same. I asked it to run the parse, test, and count. It thought for 1m 54s.] Response: Thanks — I tried to compute it here, but enumeration in this environment kept timing out. I can’t finish the full run reliably inside this chat session, so I don’t have a verified numeric answer I can produce from execution right now. [It gave me a Python script that it said should run in a few seconds on a typical laptop.] Prompt: Instead, please find the number of fault-free solutions to the 4 by 15 rectangle, again using all 12 pentominoes. [It thought for 36s.] Response: I tried to enumerate the 4×15 tilings here, but the solver kept timing out in this environment and I couldn’t finish the full run. I can’t produce a verified numeric answer from this session. [It gave a Python script that uses Dancing Links to generate all solutions, and tests them for being fault-free. I ran the script on my MacBook laptop. It ran for about 14 minutes, and reported that the total number of solutions is 11360, and the number that are fault-free is the same.] [I then asked about execution time, because my experience is that DLX is about 10 times slower when enumerating all solutions to a polyomino packing problem. The AI said the literature agrees.] Response: For complete enumeration, a compact recursive backtracking algorithm — especially one that keeps precomputed constraint maps or uses bitmasks — can outperform DLX by a factor of roughly 5–15x. For example: Hiroshi Kato (2003) and later Erich Friedman both reported their recursive pentomino enumerators (using simple grid arrays and pruning heuristics) outperforming their DLX equivalents by about that factor. Modern solvers (for exact-cover puzzles, Sudoku, polyominoes, etc.) often use DLX to get the first solution and bitmask recursion for counting all. In the literature and practical experience, Dancing Links (DLX) is indeed optimized for finding one (or a few) solutions quickly, but when you’re trying to enumerate all solutions, its advantage often narrows — or even reverses — compared to a well-optimized recursive backtracking search. [Next, I asked why the solution count of 11360 is so much greater than that in the literature, 368.] Response: The script we ran enumerates all exact-cover placements of the 12 pentominoes on a fixed 4×15 rectangle. A given board layout may allow the same set of 12 pentominoes to be arranged in multiple orders of piece placement. The exact-cover search treats these as distinct sequences of piece placements, even though the final tiling is identical. For example, placing pentomino “F” first versus placing “L” first can result in the exact same filled board. DLX sees these as separate solutions because the algorithm is solution-sequence sensitive, not board-sensitive. So the 11,360 reported by the script is actually “number of solution sequences”, not “number of distinct tilings.” [Then I had the AI system confirm that the total and fault-free counts being equal meant all 4x15 solutions are fault-free, and it did. It offered an "explanation" of why there cannot be any solution with a fault. It said that a fault across the 4-wide dimension is impossible because each pentomino contains 5 cells, so it must appear in at least two of the columns of 4, preventing a fault. It missed that the additional column(s) can be away from the proposed fault; they do not have to be across it. So, the explanation was insufficient.] [Then I said that I think there are solutions that split into a 4x5 rectangle and a 4x10 rectangle, and showed an example of the 4x5.] Response: This shows that faults in a 4x15 rectangle are possible in principle. Therefore, the conclusion that “all 4x15 tilings are fault-free” is not correct in general; it depends on how the 12 pentominoes are arranged. The Python script’s output reported all 11,360 solutions as fault-free, which suggests that the enumeration never produced a tiling like this one (with a 4x5 + 4x10 split). Likely reason: the script’s tiling generation never allowed this separation, either because of the exact-cover placement choices or the order of pieces. [Comment: In other words, its Python script has a bug.] ------------------------------------------------------------------------ Addenda Here are other ways to categorize solutions to packing the 12 pentominoes. They do not seem very useful in how or which solutions to show in an art or craft project, but they are occasionally mentioned in the literature. The same boxes used in the main discussion are analyzed here. The methods are: -- solutions with pieces not touching the border -- position of the X in solutions -- checkerboard parity of pieces ----- Solutions with "inside" pieces (not touching the border) Such pieces are sometimes called "landlocked". For each shape, a count of how many solutions have each number of landlocked pieces, followed by a brief discussion of which sets of pieces occur in such solutions. 8x8 minus center 2x2 (65 solutions): 0 landlocked pieces -- 0 solutions 1 landlocked piece -- 12 solutions 2 landlocked pieces -- 27 solutions 3 landlocked pieces -- 25 solutions 4 landlocked pieces -- 1 solution For 1 landlocked piece, that piece is either W (1 solution) or X (11). For 2 landlocked pieces, 13 sets of 2 pieces occur; the sets occur in from 1 to 6 (set TY) solutions. For 3 landlocked pieces, 13 sets occur; the sets occur in from 1 to 4 (set FWX) solutions. For 4 landlocked pieces, the one set is NUWX, in solution 58). 11 pieces occur as inside pieces; only I does not. Note: touching the 2x2 hole is not considered touching an edge! L L F Z Z V V V solution 58 L F F F Z U U V L F W W Z Z U V L W W U U T Y W X T T T Y X X X N N N T Y Y X N N P P P Y I I I I I P P 3x20 (2 solutions): 0 landlocked pieces -- 2 solutions 4x15 (368 solutions): 0 landlocked pieces -- 272 solutions 1 landlocked piece -- 96 solutions For 1 landlocked piece, that piece is L (6 solutions), N (46), P (38), or Y (6). 5x12 (1010 solutions): 0 landlocked pieces -- 40 solutions 1 landlocked piece -- 303 solutions 2 landlocked pieces -- 483 solutions 3 landlocked pieces -- 171 solutions 4 landlocked pieces -- 13 solutions For 1 landlocked piece, only I does not occur: F I L N P T U V W X Y Z (piece) 23 0 9 19 10 33 47 41 34 23 37 27 (solutions) For 2 landlocked pieces, 49 sets occur. The range is 6 sets occur in only 1 solution, to 1 set (TW) occurs in 41 solutions. For 3 landlocked pieces, 43 sets occur. The range is 10 sets occur in only 1 solution, to 1 set (TWX) occurs in 28 solutions. For 4 landlocked pieces, 4 sets occur: UWXZ occurs in 2 solutions, FTWX in 3, and both FUWZ and TWXZ in 4. Piece I occurs only in the set IY, which occurs in only one solution, solution 747. L L L L N N N F F P P P solution 747 L I I I I I N N F F P P V Z Y Y Y Y W T F X U U V Z Z Z Y W W T X X X U V V V Z W W T T T X U U 6x10 (2339 solutions): 0 landlocked pieces -- 2 solutions 1 landlocked piece -- 155 solutions 2 landlocked pieces -- 864 solutions 3 landlocked pieces -- 1111 solutions 4 landlocked pieces -- 207 solutions For 0 landlocked pieces, the 2 solutions are #1000 and #1105. For 1 landlocked piece, only I does not occur: F I L N P T U V W X Y Z (piece) 47 0 3 16 9 8 3 1 30 22 5 11 (solutions) The one solution in which V alone is landlocked, is solution 2169. For 2 landlocked pieces, 50 sets occur. The range is 3 sets (FI, LP, NU) occur in only 1 solution, to 1 set (FW) occurs in 79 solutions. For 3 landlocked pieces, 123 sets occur. The range is 14 sets occur in only 1 solution, to 1 set (FWZ) occurs in 61 solutions. For 4 landlocked pieces, 59 sets occur. The range is 17 sets occur in only 1 solution, to 1 set (UWXZ) occurs in 15 solutions. Piece 'I' occurs in 4 sets: FI (1 solution), FIW (2), IWX (2), and IYZ (6). The two solutions with no landlocked pieces are #1000 and #1105 I P P L U U X V V V solution 1000 I P P L U X X X F V I P N L U U X F F V I T N L L W W Z F F I T N N W W Y Z Z Z T T T N W Y Y Y Y Z I P W W L L F V V V solution 1105 I P P W W L F F F V I P P N W L Y F X V I Z Z N T L Y X X X I Z N N T Y Y U X U Z Z N T T T Y U U U The one solution with only V inside is #2169 P P I I I I I F Z Z solution 2169 P P V V V F F F Z Y P X L L V N F Z Z Y X X X L V N T W Y Y U X U L N N T W W Y U U U L N T T T W W The one solution with set FI inside is #2198 P P L L L L N N N T solution 2198 P P L F F N N T T T P X F F I I I I I T X X X F W W Z V V V U X U W W Y Z Z Z V U U U W Y Y Y Y Z V ----- Position of the X in solutions A solution can be classified by which cell holds the center of the X pentomino. In some programs, the position of a particular piece, such as the X, can be used to suppress duplicate solutions. And in puzzle analysis generally, determining the positions a piece may occupy can be a powerful tool in analyzing the puzzle. In this note, the solutions are ensured distinct by choosing the orientation that is first in alphabetic order. Note that the X may appear in various positions. In particular, positions that are symmetrically placed generally have different counts of X occupancy. In this section, a grid of numbers corresponds to the box that is packed with the 12 pentominoes. The number at each cell position is the number of solutions in which the center of the X occupies that cell. The right and bottom margins are subtotals of how many solutions have the center of the X in that row or column. It may be interesting to know the counts when symmetric cells are counted together as one count. For each shape to be packed, a second grid of numbers shows the result of folding each solution horizontally and vertically, to get the center of the X into the upper left quadrant of the rectangle. For the 8x8 square with the center 2x2 hole, each solution is further folded along a diagonal, so its second grid of numbers is the counts for each cell of the octant. 8x8 minus center 2x2 (65 solutions): 0 0 0 0 0 0 0 0 | 0 0 0 0 0 3 8 0 0 | 11 0 1 20 0 0 0 3 0 | 24 0 3 0 0 0 0 2 0 | 5 0 5 0 0 0 0 1 0 | 6 0 0 2 0 0 4 5 0 | 11 0 0 1 2 4 1 0 0 | 8 0 0 0 0 0 0 0 0 | 0 --- --- --- --- --- --- --- --- + ---- 0 9 23 2 7 13 11 0 | 65 0 0 0 0 | 0 0 19 20 | 39 26 0 | 26 0 | 0 --- --- --- --- + ---- 0 0 45 20 | 65 3x20 (2 solutions): 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- + ---- 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 2 0 0 0 0 0 0 0 0 0 0 | 0 0 0 2 0 0 0 0 0 0 0 | 2 --- --- --- --- --- --- --- --- --- --- + ---- 0 0 2 0 0 0 0 0 0 0 | 2 4x15 (368 solutions): 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 43 0 20 49 3 28 6 15 19 0 4 0 0 | 187 0 0 92 18 24 2 9 6 5 15 4 0 6 0 0 | 181 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- + ---- 0 0 135 18 44 51 12 34 11 30 23 0 10 0 0 | 368 0 0 0 0 0 0 0 0 | 0 0 0 145 18 67 81 23 34 | 368 --- --- --- --- --- --- --- --- + ---- 0 0 145 18 67 81 23 34 | 368 5x12 (1010 solutions): 0 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 20 19 33 101 39 27 53 36 0 0 | 328 0 28 54 28 36 8 17 25 45 3 54 0 | 298 0 0 64 55 40 54 44 35 52 40 0 0 | 384 0 0 0 0 0 0 0 0 0 0 0 0 | 0 --- --- --- --- --- --- --- --- --- --- --- --- + ---- 0 28 138 102 109 163 100 87 150 79 54 0 | 1010 0 0 0 0 0 0 | 0 0 0 160 179 135 238 | 712 0 82 57 73 61 25 | 298 --- --- --- --- --- --- + ---- 0 82 217 252 196 263 | 1010 example: the 3 solutions with X center at the third row, tenth column: (note: solutions are in the orientation first in alphabetic order) F Y Y Y Y W Z Z L L L L solution 80 F F F Y W W Z T L X U U P F V W W Z Z T X X X U P P V N N N T T T X U U P P V V V N N I I I I I F Y Y Y Y W Z Z U U U I solution 81 F F F Y W W Z T U X U I P F V W W Z Z T X X X I P P V N N N T T T X L I P P V V V N N L L L L I F Y Y Y Y W Z Z U U U L solution 82 F F F Y W W Z T U X U L P F V W W Z Z T X X X L P P V N N N T T T X L L P P V V V N N I I I I I 6x10 (2339 solutions): 0 0 0 0 0 0 0 0 0 0 | 0 0 0 35 67 55 92 67 149 0 0 | 465 0 69 106 90 85 78 83 34 177 0 | 722 0 64 78 41 106 68 62 44 132 0 | 595 0 0 78 102 75 69 106 127 0 0 | 557 0 0 0 0 0 0 0 0 0 0 | 0 --- --- --- --- --- --- --- --- --- --- + ---- 0 133 297 300 321 307 318 354 309 0 | 2339 0 0 0 0 0 | 0 0 0 389 342 291 | 1022 0 442 262 276 337 | 1317 --- --- --- --- --- + ---- 0 442 651 618 628 | 2339 ----- Checkerboard parity of pieces A common technique in puzzle analysis is to color each solution as a checkerboard. 11 of the pentominoes split 2-3, that is, 2 cells will have one color, and 3 will have the other color. The X splits 1-4. And in all the shapes discussed in this note, the overall solution will split 30-30. This forces 7 pieces to have their majority color (the color of 3 of their cells) to be the same color as the center of the X, while 4 pieces have their majority color the same as the arms of the X. Therefore, solutions can be categorized by which 4 pieces have their majority color the same as the arms of the X. For example: I P P P U U U V V V solution 1019 I T P P U X U Z Z V I T T T X X X Z F V I T W N N X Z Z F F I W W Y N N N F F L W W Y Y Y Y L L L L has the checkerboard coloring I p P p U u U v V v i T p P u X u Z z V I t T t X x X z F v i T w N n X z Z f F I w W y N n N f F l w W y Y y Y l L l L where upper and lower case is the checkerboarding. The arms of piece X are upper case. The 4 pieces with 3 upper case cells are F, I, N and T. So this solutions's set is FINT. 8x8 minus center 2x2 (65 solutions) ----------------------------------- 37 different sets of 4 occur. All 11 pieces (X is excluded) occur. Solutions per set range from 1 (23 sets) to 6 (1 set, NTYZ). First and last 4: 1 FITU 1 FITZ 1 FLNP 1 FLNV ... 4 LTVY 4 PTWY 4 PTWZ 6 NTYZ No set occurs in exactly 5 solutions. U occurs in the 4-piece set for 13 (the least) of the 65 solutions; Z occurs in the set for 35 (the most) solutions. 3x20 (2 solns) -------------- The set FNYZ occurs in both solutions. 4x15 (368 solns) ---------------- 100 different sets of 4 occur. All 11 pieces occur. Solutions per set range from 1 (28 sets) to 32 (1 set, FNPV). First and last 4: 1 FILZ 1 FITU 1 FIUV 1 FLVY ... 12 FVWY 20 FPTY 26 FNYZ 32 FNPV No set occurs in exactly these numbers of solutions: 11, 13-19, 21-25, 27-31. U occurs in the 4-piece set for 46 (the least) of the 368 solutions; F occurs in the set for 249 (the most) solutions. 5x12 (1010 solutions) --------------------- 190 different sets of 4 occur. All 11 pieces occur. Solutions per set range from 1 (36 sets) to 26 (1 set, PTWZ). First and last 4: 1 FILZ 1 FITY 1 FLPT 1 FLPV ... 23 FPYZ 24 FNPZ 24 NTUW 26 PTWZ No set occurs in exactly 18 solutions, or 22 or 25. U occurs in the 4-piece set for 245 (the least) of the 1010 solutions; Z occurs in the set for 496 (the most) solutions. 6x10 (2339 solutions) --------------------- 287 different sets of 4 occur. All 11 pieces occur. Solutions per set range from 1 (32 sets) to 64 (1 set, PTYZ). First and last 4: 1 FINT 1 FIPW 1 FITU 1 FITY ... 41 FLNV 48 TWYZ 49 FINP 64 PTYZ There are sets that occur from 1 to 22 times; 23 is the smallest count for which no set occurs. U occurs in the 4-piece set for 421 (the least) of the 2339 solutions; N occurs in the set for 1012 (the most) solutions. Observations ------------ PTWZ, the most common set for the 5x12 box, occurs only 12 times for the 6x10 box. It may seem strange that U occurs the least in the sets of 4, for each shape. (In the 3x20, I L P T U V and W occur zero times.) The U is convenient to surround an arm of the X, and in that position it is not in the set of four. This may explain its rarity. The maximum possible number of 4-piece sets selected from the 11 pieces excluding X is C(11,4) = 330. For the 6x10 box, 330-287=43 do not occur, such as FILN. ----- A final word: the 3x20 solutions fold! This is an odd tidbit that does not fit with the rest of this note, but it seems good to include it here anyway. Many years ago, my sister made a felt applique art piece of one of the 3x20 solutions. After studying it for a while, my brother exclaimed, "Oh! It folds!" Consider the 12 flat pentominoes as made from cubes instead of squares. In a 3x20 solution, put a hinge pin through the arms of the X and its neighbors, and similarly with the T. This shows where the pins go: v v v v U U X I I I I I N N N F T W Y Y Y Y Z V U X X X P P L N N F F F T W W Y Z Z Z V U U X P P P L L L L F T T T W W Z V V V ^ ^ ^ ^ The assembly can then be folded into a long rectangular donut, with the U end abutting the V end. The shape is a 3x3x9 box with a 3x1x7 hole through the middle. A nice fact is that the part that flips to make the other solution is symmetric around the T, so the other solution also folds in the same way: v v v v U U X I I I I I Z W W T T T F L L L L V U X X X P P Z Z Z Y W W T F F F N N L V U U X P P P Z Y Y Y Y W T F N N N V V V ^ ^ ^ ^ ------------------------------------------------------------------------ End of Note.