From Lee Sallows:
(Thanks, Lee!)
From Lee Sallows:
(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 small rectangles. Observe also that the digit 1 is represented as the vertical column of 5 cells in the centre of 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 x in either of the readouts will be found to identify the total number of cells occurring in position x within 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?