From Lee Sallows, a geomagic square with a staircase theme:

(Thanks, Lee!)

From Lee Sallows, a geomagic square with a staircase theme:

(Thanks, Lee!)

From Lee Sallows:

(Thanks, Lee!)

By Lee Sallows: If the letters BJFGSDNRMLATPHOCIYVEU are assigned to the integers -10 to 10, then:

J+A+N+U+A+R+Y = -9+0-4+10+0-3+7 = 1 F+E+B+R+U+A+R+Y = -8+9-10-3+10+0-3+7 = 2 M+A+R+C+H = -2+0-3+5+3 = 3 A+P+R+I+L = 0+2-3+6-1 = 4 M+A+Y = -2+0+7 = 5 J+U+N+E = -9+10-4+9 = 6 J+U+L+Y = -9+10-1+7 = 7 A+U+G+U+S+T = 0+10-7+10-6+1 = 8 S+E+P+T+E+M+B+E+R = -6+9+2+1+9-2-10+9-3 = 9 O+C+T+O+B+ER = 4+5+1+4-10+9-3 = 10 N+O+V+E+M+B+E+R = -4+4+8+9-2-10+9-3 = 11 D+E+C+E+M+B+E+R = -5+9+5+9-2-10+9-3 = 12

Similarly, if -7 to 7 are assigned SROEMUNFIDYHTAW, then SUNDAY to SATURDAY take on ordinal values. See Alignment.

(David Morice, “Kickshaws,” *Word Ways* 24:2 [May 1991], 105-116.)

From Lee Sallows:

Can you complete the ‘self-descriptive crossword puzzle’ at left below? As in the solution to a similar puzzle seen at right, each of its 13 entries, 6 horizontal, 7 vertical, consists of an English number name folowed by a space followed by a distinct letter. The number preceding each letter describes the total number of occurrences of the letter in the completed puzzle. Hence, in the example, E occurs thirteen times, G only once, and so on, as readers can check. Note that the self-description is complete; every distinct letter is counted.

Though far from easy, the self-descriptive property of the crossword enables its solution to be inferred from its empty grid using reasoning based on orthography only.

Another remarkable contribution by Lee Sallows:

(Thanks, Lee!)

Another amazing contribution by Lee Sallows:

“The picture above shows a 4×4 geomagic square, which is to say a magic square using geometrical shapes that can be fitted together so as to form an identical target shape, in this case a 4×6 rectangle, rather than numbers adding to a constant sum. In addition, the square is also panmagic, meaning that besides the usual 4 rows, 4 columns, and both main diagonals, the shapes occupying each of the so-called ‘broken’ diagonals, afkn, dejo, cfip, bglm, chin, belo, are also able to tile the rectangle. Lastly, the 4 shapes contained in the corner cells of the four embedded 3×3 sub-squares, acik, bdjl, fhnp, egmo, are also ‘magic’, bringing the total number of target-tiling shape sets to 20, a small improvement over the 16 achieved by a panmagic-only square. With that said, it is worth noting that 4×4 geomagic squares have been found achieving target-tiling scores as high as 48.”

Click the image to expand it. Thanks, Lee!

From Lee Sallows:

Earlier this year, Futility Closet featured a puzzle based upon the well-known 7-segment display. Less well known is the 15-cell display shown in Figure 1, in which each decimal digit appears as a pattern of highlighted cells within a 3×5 rectangle. Call these the

smallrectangles. Observe also that the digit 1 is represented as the vertical column of 5 cellsin the centreof its small rectangle rather than as either of the two alternative columns immediately to left and right, a detail that is important in view of what follows.Figure 1.

Figure 2 shows a pair of readouts using 15-cell displays each arranged in the form of a (large) 3×5 rectangle that mirrors the smaller rectangles just mentioned. The two readouts describe each other. The top left cell in the right-hand readout contains the number 18. A check will show that the number of highlighted top left cells appearing in the left-hand readout is indeed 18. Take for instance the top left-hand cell in the left-hand readout. It contains the number 17, which employs two digits, 1 and 7. None of the 5 cells forming the digit 1 is in top left position within its small rectangle. But the leftmost cell in digit 7 is indeed in top left position. Proceeding next to the left-hand readout’s top centre cell we find two cells in top left position: one in digit 2 and one in digit 4. The score of top left cells so far is thus 1+2 = 3. Continuing in normal reading order, a list of the left-hand readout numbers followed by their top left cell scores in brackets is as follows: 17 (1), 24 (2), 17(1), 13(1), 9(1), 15(1), 17(1), 25(2), 17(1), 8(1), 9(1), 14(1), 15(1), 24(2), 17(1). The sum of the scores is 18, as predicted.

In the same way, a number occupying position

xin either of the readouts will be found to identify the total number of cells occurring in positionxwithin the digits of the other readout. That is, the two readouts are co-descriptive, they describe each other.Figure 2.

Recalling now the solution to the earlier mentioned 7-segment display puzzle, some readers may recall that it involved an iterative process that terminated in a loop of length 4. Likewise, the pair of readouts in Figure 2 are the result of a similar process, but now terminating in a loop of length 2. In that case we were counting segments, here we are counting cells. An obvious question thus prompted is: What kind of a readout would result from a loop of length 1? The answer is simple: a description of the readout resulting from a loop of length 1 would be a copy of the same readout. That is, it will be a self-descriptive readout, the description of which is identical to itself. Such a readout does indeed exist. Can the reader find it?

From Lee Sallows:

The drawing at left above shows an unusual type of 3×3 geomagic square, being one in which the set of four pieces occupying each of the square’s nine 2×2 subsquares can be assembled so as to tile a 4×8 rectangle. The full set of subsquares become more apparent when it is understood that the square is to be viewed as if drawn on a torus, in which case its left-hand and right-hand edges will coincide, as will its upper and lower edges. In an earlier attempt at producing such a square several of the the pieces used were disjoint, or broken into separated fragments. Here, however, the pieces used are nine intact octominoes.

(Thanks, Lee!)