Consider the following diagram of observers. If you're observer D, the only other observers you can see are B, C, F, J, and G, or five observers. The other observers are blocked. Observer B can see A, C, D, E, F, G, and I, or seven other observers. Observer A can see six other observers. If I gave the problem, "Plant 11 observers so that 4 of them can see exactly 7 other observers," then the following diagram would work (B E G I).

Cihan Altay sent me three Observer puzzles. As in the example above, observers can 'see' in any direction, not just along grid lines. The group of problems forms my puzzle of the week. A similar problem is to arrange 6 observers so that each one sees exactly 4 others. If you have n observers, and each one sees k or fewer others, how small can k be? At Cihan's site, he is currently running a competition problem: (It starts Asagidaki diyagramdaki toplam ...). Answer 1, Answer 2, Answer 3.

Koshi Arai

"If you have n observers and each one sees k or fewer others, how small can k be?"
Well, if they were all in a straight line then:
For any size n, k<=2

Here are the three Observer Puzzles.
Dick Saunders Jr.

Juha Saukkola
I didn't find puzzles from Cihan's site - maybe, because I have very poor knowledge of
Turkish. My solutions by hand (not really unique);

36 .. .. .. 40
02 06 .. 07 ..
03 .. 05 .. ..
01 .. .. .. ..
16 04 08 09 27

. . . . x .
x . . o . o
. o . . . .
. o o o o .
. . o o . x
. x . . . .

. . . . A . .
. . . x . . .
B . x . x . o
. . x . o . .
x . . . x . C
. . . . . . .
x . D . . . .

Aron Fay
If you have n observers, and each one sees k or fewer others, how small can k be?

This seems obvious. k=2. Put all observers in a line.

If each observer should see the same number of other observers, then I
think it will depend on the factors of n.

If n is divisible by 2, and (n>=6), then I can arrange them so each
observer can see exactly k=n-2 observers.

If n is divisible by 3, and (n>=9), then I can arrange them so that each
observer can see exactly k=n-3 observers. See examples for n=12 below.

n=12, k=10 n=12, k=9
.......X.. ..XX..
.X........ ......
....XX...X X.XX.X
...X..X... X.XX.X
X...XX.... ......
........X. ..XX..
..X.......

I don't know of any other generalizations that can be made.
I don't know how many solutions there are to A or B, but part C has
6 solutions. I really enjoyed doing these puzzles.

A:
X...X
4....
2...9
7.3.5
X168X


B:
....X.
X**.*.
..**..
....*.
.***.X
.X....


C:
....X..
...*...
X.*B*.A
.C*.a..
*.D.*.X
...D...
*.X.C..

For C, the obvious answer is to place to stones at 'a' and 'A'.

Another solution is to place stones at 'B' and the northern 'C'.

Four more solutions are to place one stone at a 'C' space, and
the other stone at a 'D' space.

-aron fay

p.s. Here is part C, but reformatted:
+-+-+-+-+-+-+-+
|.|.|.|.|X|.|.|
+-+-+-+-+-+-+-+
|.|.|.|*|.|.|.|
+-+-+-+-+-+-+-+
|X|.|*|B|*|.|A|
+-+-+-+-+-+-+-+
|.|C|*|.|a|.|.|
+-+-+-+-+-+-+-+
|*|.|D|.|*|.|X|
+-+-+-+-+-+-+-+
|.|.|.|D|.|.|.|
+-+-+-+-+-+-+-+
|*|.|X|.|C|.|.|
+-+-+-+-+-+-+-+

Michael J D Dufour

Problem A
36 -- -- -- 40
-- -- 06 -- --
-- -- 07 -- 09
-- 02 04 -- 03
16 01 08 05 27

Problem B
- - - - o -
o - * * - -
* - - * * -
- * - - * -
- * * - - o
- o - - - -

Problem C
- - - - A - -
- - - * - - -
B - * - * - -
- $ * - - - -
* - $ - * - C
* - D - - - -

Colin Bown
1)

8 sets add to 36, S_36:
3 sets add to 40, S_40:
18 sets adds to 16, S_16:
21 sets add to 27, S_27:

So I compare the topology of each of the 21!/9!/12! = 293930
point configurations against each of the 9072 allowed topologies
(actually only 3583 distinct ones) and see what works.

Here by topology I mean the 15-tuple which describes the way these sets of points intersect

( N(S_16), N(S_40), N(S_36), N(S_27),
N(S_16 V S_40), N(S_16 V S_36), N(S_16 V S_27),
N(S_40 V S_36), N(S_40 V S_27),
N(S_36 V S_27),
N(S_40 V S_36 V S_27),
N(S_16 V S_36 V S_27),
N(S_16 V S_40 V S_27),
N(S_16 V S_40 V S_36),
N(S_16 V S_40 V S_36 V S_27) )

I find that there are 1626 solutions. Here is a special one:

_36_ . . . _40_
8(9) 6 . 9(8) .
5 . 3 . 7
1 . 4 . 2
_16_ . . . _27_

This is the _least_ degenerate of the solutions, 8 and 9 can be
swapped, as indicated by the parenthesis, but no others can.
This pattern is unique in having only a two fold degeneracy.

24 fold degeneracy is not uncommon:

36 . 578 . 40
. . . . .
. 2 578 . 36
. 14 . . 14
16 36 9 578 27

Here 5 7 and 8 can all be swapped around (3!=6) as well as 3,6 and
1,4.

2)

7
. x . . . .
. . o o o x6
. . . o o .
. . o o o .
8x . . . o .
. . . . x .
5

3) General approach: draw all lines of vision, characturize each
remaining intersection by which observers counts would be
increased. Select a pair that increases each observers count.

4
. . . . a . .
. . . x . . .
5b . x . x . X
. . x . X . .
x . . . x . c6
. . . . . . .
x . d . . . .
7

Dick Saunders Jr. also sent solutions.