The 6-Square Pieces

For listings and diagrams of all the six-square pieces see: Enumeration by Binary Method or Key to FCR Notation.

General Results

The 6-square pieces were first enumerated by H. D. Benjamin in Problemist Fairy Chess Supplement, December 1934. In fact the very first dissection problem (¶1597) in PFCS/FCR is Benjamin's question whether the 35 pieces of 6 squares can be arranged to cover a rectangular area. The text reads as follows:

"A BIG CUTTING-OUT DESIGN — AND A PRIZE OFFER. Here is the special poser, a real Christmas puzzle:— 1597. H. D. Benjamin, Calcutta."
"Cut a board 15 by 14 into 35 pieces each of 6 squares attached to one another by their edges, all cuts along the lines, the 35 pieces to be all fundamentally different in shape. Rotation or reflection of a piece is NOT to be considered a difference."
"For a solution of this, which he is not sure can be done, Mr. Benjamin has placed a prize of 10 shillings at my disposal. Solvers can attack it in either of two ways, of course—by actually cutting or marking off the pieces, or by discovering the 35 possible 6-square shapes and then assembling them in a 15 by 14 rectangle. In the latter method, which seems more useful, any piece can be turned over, which amounts to reflection, at will, so long as no two pieces of the same shape occur in the 35 initially; or stated in another way, it does not matter in which aspect any piece is used for its one appearance in the design."

The solution of this problem appears in the next issue, as follows:

"Dr. F. Kadner has sent two elegant proofs that a 15 x 14 rectangle can not be dissected into the 35 different 6-square shapes. Hs first proof is geometrical and commences by showing that the 35 pieces will asemble to a 15 x 14 rectangle except for one single square overlapping an adjacent one. Thus to achieve the complete assembly, translations, rotations, and reflections of THIS array must end in a resultant displacement of 1 square 1 step. But each of the 35 pieces in any of these motions alters the array by an EVEN number of squares x steps. Hence the resultant of any number of such motions can never give the required ODD unit change."
"The second proof is even more simple—if the 35 pieces are chequered, it is found 24 of them must have 3 white and 3 black squares, but the remaining 11 (the piece abcde1, b2 is an example) MUST have 2 squares one colour and 4 the other. Now the 15 x 14 rectangle has equal numbers of each colour, so the 11 pieces just noted must be coloured half white and half black—and this is clearly impossible. Thus, again, the problem cannot be solved."
"These beautiful bits of reasoning amount to a complete solution, of course, and Dr. Kadner easily secures the prize offered."

The argument shows it is impossible to use the 35 hexominoes to cover any rectangle of 210 squares, 2×105, 3×70, 5×42, 6×35 , 7×30, 10×21 or 14×15. More generally it follows that any chequered area covered by the 35 hexominoes must show an excess of squares of one colour over squares of the other colour. This colour difference cannot be zero but must take one of the values 2, 6, 10, 14, 18 or 22.

Another general result about rectangles appeared much later:

F. Hansson (FCR 1950 ¶8555). "Two rectangles are joined along one edge to form a figure of 210 squares. Typical examples of joining are abc1 joined to bcd2, or cde2 or a234, and so on, but the rectangles need not be equal. What are the relative dimensions and joining squares for formation of the 210-square shape by the 6-pieces?"
Solution: "For double rectangles A,B to be possible with the 6-sq. pieces, A and B must both be odd and have the same corner colour. A and B cannot both be primes. Following are impracticable: 11, 199; 13, 197; 17, 193; 19, 191; 29, 181; 31, 179; 37, 173; 41, 169, 43, 167; 47, 143; 53, 157; 59, 151; 61, 149; 67, 143; 71, 139; 73, 137; 79, 131, 83, 127; 89, 121; 97, 113; 101, 109; 103, 107. Very doubtful is 9, 201."

The following diagrams show some of the shapes that have been devised to make use of all 35 of the hexominoes. They are classified broadly according to the symmetry of their outline shape. This of course is only one way they might be classified, but one I have found helpful. Within each section those with fewest holes come first.

Biaxial Symmetry

F. Hansson (FCR 1950, ¶8558) "From a SQUARE cut out a centrally situated rectangle and and form the shape left with the 6-pieces." The solution is this 15×15 square with central 3×5 hole; it has no crossroads, CD = 2.

the 6-square pieces in square 15 by 15 with rectangular hole

W. Stead (FCR 1954 ¶30). This 15×17 rectangle with a Greek cross-shaped 45-square hole appears in the illustrated article, CD = 2.

the 6-square pieces in rectangle 15 by 17 with cross-shaped hole

H. D. Benjamin (FCR 1945 ¶6375-6). Two rectangles 13×24 with holes in the shape of two versions of the Cross of Lorraine; both have CD = 2.

the 6-square pieces in rectangle 13 by 24 with cross-shaped hole: two versions

It could be argued that these should be classified under those with an axis of symmetry, since the holes form the design rather than the outer rectangle, but sticking to my (admittedly rather arbitrary) rule I only take account of the outer outline.

G. P. Jelliss (Chessics #28 1986 p.145). This 'rhombus' (or 'battleship' or 'aircraft carrier') pattern with serrated edges in a regular stepped form was composed specifically as an example of CD = 18. It has no crossroads (the diagram is rotated a quarter turn from that in Chessics). The containing rectangle is 13×37.

the 6-square pieces in rhombus with rhomb-like hole

G. P. Jelliss (Chessics #28 1986 p.145). This 'diamond' or square with serrated edges of 11-square length within a 21×21 area and with a 1×11 hole was composed to show the CD = 22 case.

the 6-square pieces in diamond with hole 1 by 11

W. Stead (FCR 1952 ¶9331). A 15x15 square with three cross-shaped holes. Dedicated to H. E. Dudeney and his work on Greek Crosses.

the 6-square pieces in square 15 by 15 with 3 cross-shaped holes

G.P.Jelliss (Chessics #28 1986 p.144). A cross with 6×7 arms on a 7×7 centre, within a 19×19 area, and having 7 holes regularly spaced down the centre axis, CD = 6.

the 6-square pieces in cross with 7 holes

W. Stead (FCR 1954 ¶35). Rectangle 5×45 with 15 holes. CD = 2.

the 6-square pieces in rectangle 5 by 45 with 15 holes

Rotational Symmetry

F. Hansson (FCR 1950 ¶8556). "Place two 3×35 rectangles in the longest possible shape formable by the 6-pieces." [This is the 105, 105 case of the double rectangle problem.]

the 6-square pieces in two rectangles 3 by 35 joined

A well-known arrangement with CD = 14 is this parallelogram with sloping edges serrated. It appears in W. Stead's notebooks with his initials alongside, and it was first published in his article in FCR December 1954 ¶32. He notes there that the central stepped piece is pre-set to echo the outline. The containing rectangle is a doubled square, 14×28, and there is a vertical fault line between the two squares, except where the stepped piece crosses it.

the 6-square pieces in parallelogram

Diagonal Axis of Symmetry

This serrated triangle with 20-unit sides was given by H. D. Benjamin (FCR Feb/Apr 1937 ¶2622) it has no crossroads, and is the first arrangement of the hexominoes in a form without holes, diagonal axis, CD = 10.

the 6-square pieces in triangle

H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4985) 15×15 square with 15 cells removed, diagonal axis CD = 2.

the 6-square pieces in modified square

H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4986) 15×15 square with 15 cells removed, with diagonal axis CD = 6.

the 6-square pieces in modified square

H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4987) 15×15 square with 15 cells removed, CD = 14.

the 6-square pieces in modified square

H. D. Benjamin (FCR Dec/Feb 1941/2 ¶4988) 16×16 square with 46 cells removed, CD = 18.

the 6-square pieces in modified square

H. D. Benjamin (FCR Dec/Apr 1941/2 ¶4989) 17×17 square with 79 cells removed, CD = 22.

the 6-square pieces in modified square

Lateral Axis of Symmetry

Double Rectangles

H. D. Benjamin (FCR 1950 ¶8912). "Arrange the 35 6-pieces in a symmetrical figure, a rectangle plus a square. There are only 3 figures possible.": Solution: "The Sombrero Hat." This is 3×43 + 9×9 and has colour difference = 2.

Sombrero: the 6-square pieces in rectangle 3 by 43 with square 9 by 9 in middle of long side

"Other cases are 11×19 with 1×1 and 19 ×11 with 1×1." Solutions for these cases were not given. The following are my own solutions (G. P. Jelliss May 2006). The 'suitcase' has one crossroad,

the 6-square pieces in rectangle 11 by 19 with single square in middle of long side

the 'clipboard' has three.

the 6-square pieces in rectangle 19 by 11 with single square in middle of short side

Trees

H. D. Benjamin (FCR Apr/Aug 1952 ¶9235). 'The Spreading Chestnut Tree'.

the 6-square pieces depicting a spreading chestnut tree

W. Stead (FCR Dec/Apr 1953/4 ¶9783). Christmas Tree.

the 6-square pieces depicting a christmas tree