Place six points around a circle of circumference 27 such that every integral distance from 1 to 26 is represented as the circumferential distance between two of these points.
There are 28 distinct solutions, that is, not rotations or reflections of one another:
We assume that one point is located at 0, and since at least one distance of 1 is required, another at 1. All the solutions that satisfy this convention are:
0 1 2 5 13 22 0 1 5 16 19 26
0 1 2 6 9 16 0 1 5 17 19 25
0 1 2 6 10 13 0 1 5 18 20 21
0 1 2 6 17 20 0 1 6 8 11 15
0 1 2 7 16 24 0 1 6 9 11 13
0 1 2 9 12 23 0 1 6 14 18 25
0 1 2 13 20 23 0 1 6 15 23 26
0 1 2 16 19 23 0 1 6 16 20 25
0 1 3 5 11 18 0 1 7 8 10 23
0 1 3 5 13 21 0 1 7 8 12 25
0 1 3 5 14 20 0 1 7 9 23 24
0 1 3 7 11 16 0 1 7 15 23 25
0 1 3 7 12 20 0 1 8 10 22 24
0 1 3 8 12 22 0 1 8 11 13 17
0 1 3 8 14 18 0 1 8 11 22 26
0 1 3 8 17 23 0 1 8 12 14 17
0 1 3 9 11 23 0 1 8 14 23 25
0 1 3 10 14 22 0 1 8 16 21 25
0 1 3 13 19 23 0 1 9 17 22 24
0 1 3 16 20 21 0 1 10 13 21 23
0 1 4 5 11 13 0 1 10 14 20 25
0 1 4 5 19 21 0 1 10 17 23 25
0 1 4 6 11 19 0 1 11 14 16 20
0 1 4 6 18 20 0 1 11 15 17 20
0 1 4 8 10 15 0 1 11 16 18 24
0 1 4 9 11 15 0 1 12 17 21 25
0 1 4 10 12 17 0 1 12 19 22 26
0 1 4 12 21 26 0 1 13 17 19 24
0 1 5 7 15 18 0 1 13 17 20 22
0 1 5 8 15 26 0 1 13 18 20 24
0 1 5 9 12 26 0 1 15 17 19 22
0 1 5 9 15 25 0 1 15 17 23 24
0 1 5 11 20 25 0 1 15 18 22 26
from
DEFINT A-Z
DIM tk(27)
CLS
p(1) = 0: p(2) = 1
FOR p3 = 2 TO 23
p(3) = p3
tk(p3) = 1
FOR p4 = p3 + 1 TO 24
p(4) = p4
IF tk(p4) = 0 THEN
tk(p4) = 1
FOR p5 = p4 + 1 TO 25
p(5) = p5
IF tk(p5) = 0 THEN
tk(p5) = 1
FOR p6 = p5 + 1 TO 26
p(6) = p6
IF tk(p6) = 0 THEN
tk(p6) = 1
REDIM dist(26)
FOR a = 1 TO 5
FOR b = a + 1 TO 6
d = (p(b) - p(a)) MOD 27: IF d < 0 THEN d = d + 27
dist(d) = 1
d = (p(a) - p(b)) MOD 27: IF d < 0 THEN d = d + 27
dist(d) = 1
NEXT
NEXT
good = 1
FOR d = 1 TO 26
IF dist(d) = 0 THEN good = 0: EXIT FOR
NEXT
IF good THEN
rw = ct MOD 33 + 1: cl = ct \ 33
LOCATE rw, cl * 40 + 1
FOR i = 1 TO 6
PRINT p(i);
NEXT
ct = ct + 1
END IF
tk(p6) = 0
END IF
NEXT p6
END IF
tk(p5) = 0
NEXT
END IF
tk(p4) = 0
NEXT
tk(p3) = 0
NEXT
However, some of these are rotations and reflections of others. To get a better handle on that, the following is a list of the spacing between successive points for each of the presented solutions:
1 1 3 8 9 5 1 4 11 3 7 1
1 1 4 3 7 11 1 4 12 2 6 2
1 1 4 4 3 14 1 4 13 2 1 6
1 1 4 11 3 7 1 5 2 3 4 12
1 1 5 9 8 3 1 5 3 2 2 14
1 1 7 3 11 4 1 5 8 4 7 2
1 1 11 7 3 4 1 5 9 8 3 1
1 1 14 3 4 4 1 5 10 4 5 2
1 2 2 6 7 9 1 6 1 2 13 4
1 2 2 8 8 6 1 6 1 4 13 2
1 2 2 9 6 7 1 6 2 14 1 3
1 2 4 4 5 11 1 6 8 8 2 2
1 2 4 5 8 7 1 7 2 12 2 3
1 2 5 4 10 5 1 7 3 2 4 10
1 2 5 6 4 9 1 7 3 11 4 1
1 2 5 9 6 4 1 7 4 2 3 10
1 2 6 2 12 4 1 7 6 9 2 2
1 2 7 4 8 5 1 7 8 5 4 2
1 2 10 6 4 4 1 8 8 5 2 3
1 2 13 4 1 6 1 9 3 8 2 4
1 3 1 6 2 14 1 9 4 6 5 2
1 3 1 14 2 6 1 9 7 6 2 2
1 3 2 5 8 8 1 10 3 2 4 7
1 3 2 12 2 7 1 10 4 2 3 7
1 3 4 2 5 12 1 10 5 2 6 3
1 3 5 2 4 12 1 11 5 4 4 2
1 3 6 2 5 10 1 11 7 3 4 1
1 3 8 9 5 1 1 12 4 2 5 3
1 4 2 8 3 9 1 12 4 3 2 5
1 4 3 7 11 1 1 12 5 2 4 3
1 4 4 3 14 1 1 14 2 2 3 5
1 4 4 6 10 2 1 14 2 6 1 3
1 4 6 9 5 2 1 14 3 4 4 1
So for example 1 1 3 8 9 5 is the same as 1 3 8 9 5 1, and both 1 1 5 9 8 3 and 1 5 9 8 3 1 are the reversal of the first two.
If the rotations and reflections are eliminated, this becomes
1 1 3 8 9 5
1 1 4 3 7 11
1 1 4 4 3 14
1 1 4 11 3 7
1 2 2 6 7 9
1 2 2 8 8 6
1 2 2 9 6 7
1 2 4 4 5 11
1 2 4 5 8 7
1 2 5 4 10 5
1 2 5 6 4 9
1 2 5 9 6 4
1 2 6 2 12 4
1 2 7 4 8 5
1 2 10 6 4 4
1 2 13 4 1 6
1 3 1 6 2 14
1 3 2 5 8 8
1 3 2 12 2 7
1 3 4 2 5 12
1 3 5 2 4 12
1 3 6 2 5 10
1 4 2 8 3 9
1 4 13 2 1 6
1 5 2 3 4 12
1 5 3 2 2 14
1 7 3 2 4 10
1 7 4 2 3 10
for 28 unique solutions.
In terms of positions, this is:
0 1 2 5 13 22
0 1 2 6 9 16
0 1 2 6 10 13
0 1 2 6 17 20
0 1 3 5 11 18
0 1 3 5 13 21
0 1 3 5 14 20
0 1 3 7 11 16
0 1 3 7 12 20
0 1 3 8 12 22
0 1 3 8 14 18
0 1 3 8 17 23
0 1 3 9 11 23
0 1 3 10 14 22
0 1 3 13 19 23
0 1 3 16 20 21
0 1 4 5 11 13
0 1 4 6 11 19
0 1 4 6 18 20
0 1 4 8 10 15
0 1 4 9 11 15
0 1 4 10 12 17
0 1 5 7 15 18
0 1 5 18 20 21
0 1 6 8 11 15
0 1 6 9 11 13
0 1 8 11 13 17
0 1 8 12 14 17
The solution presented in Leming's comment 1 was the reversal of the spacing
1 2 5 9 6 4
or positioning
0 1 3 8 17 23
(Leming's spacing was 1 4 6 9 5 2)
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Posted by Charlie
on 2006-05-21 20:05:35 |
computer solution
