Download Search for silver cubes

Search for silver cubes
Scott Gildemeyer
Advisor: A. Khodkar
Department of Mathematics
University of West Georgia
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Introduction
Definition 1. An n × n is said to be silver if, for every i = 1, 2, . . . , n, each number
in {1, 2, . . . , 2n−1} appears either in the ith row or the ith column of A. For example,
in silver square of order 4 below, if i = 2, then we see numbers 3,1,2,6 in row i = 2
and numbers 5,1,7,4 in column i = 2. Hence, we see every number in 1,2,3,4,5,6,7
either in row 2 or column 2. Note that the diagonal entries need not to be the same.
Order 2
1 2
3 1
Order 4
1
3
6
2
5 4
1 2
7 3
4 1
Order 6
1 2 3 4
7 1 4 5
8 9 1 6
9 10 11 1
10 11 7 8
11 8 10 7
7
6
5
3
5 6
6 3
2 5
3 2
1 4
9 1
Figure 1: Silver squares of order 2, 4, and 6.
The next step is to study the silver squares of higher dimensions. In order to
define silver squares of dimension larger than 2 we fist need to define the concept
of independent cells in an n × n squares. In an n × n square two cells are called
independent if they are not in the same row or column. Hence, in and n × n square
the maximum size for a set of independent cells is n. Obviously, the diagonal cells
in a square form a maximum independent set. Now we need to define a maximum
independent set in n, n × n squares. First assume n = 2; so we have two 2 × 2 squares
(see figure 2). In order to build a maximum independent set in these two squares we
need a maximum independent set in the first square and a maximum independent set
in the second square. Moreover, these two sets must be disjoint. The shaded cells in
figure 2 form a maximum independent set in two 2 × 2 squares.
1
1
3
2
4
4
2
3
1
Figure 2: The unique three dimensional silver cube of order 2
Now pick a cell in this independent set, for example cell (2, 1) in the second
square. Cell (2, 1) is the cell in row 2 and column 1. We notice that every number in
{1, 2, 3, 4} appears either in row i = 2, column j = 1 or cell (2, 1) in the first square.
It is easy to check that every shaded cell has this property (the silver property).
Hence, we have a three dimensional silver cube of order 2.
Consider the squares shown in Figure 3. First note that the shaded cells form a
maximum independent set. Now if you pick a shaded cell, say cell (3, 2) in the first
square, the numbers 1, 6, 7 appear in row i = 3, numbers 1, 2, 5 appear in column
j = 2, and numbers 4 and 3 in cells (3, 2) of the second and third squares. Hence,
every number in {1, 2, 3, 4, 5, 6, 7} appears in row 3, column 2, or in cell (3, 2) of
the second or third squares. It is easy to check that every shaded cell has the silver
property. Hence, we have a three dimensional silver cube of order 3.
1 2
4 5
7 1
3
1
6
6 7
3 1
1 4
1
2
5
5
1
2
1 4
6 7
3 1
Figure 3: three dimensional silver cube of order 3.
Figure 4 displays a three dimensional silver cube of order 4. For a shaded cell
(i, j) in a square every number in {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} appears in row i, column
j, or in cells (i, j) of the other squares.
1
3
7
8
2 5
4 6
8 1
7 9
6
5
2
10
4
2
8
7
3 6
1 5
7 10
8 2
5
6
9
1
9 10 1
10 9 7
1 2 3
5 6 4
2
8
4
3
10 9
9 10
6 5
2 1
8 7
2 1
4 3
3 4
Figure 4: three dimensional silver cube of order 4
One of the successes I have reached in my research is with the 3 × 3 silver cube.
However the first example that I discovered appeared to be different than the example
2
given, but in reality ends up being the exact same because of the use of certain
permutations.
This is the original sample of a 3 × 3 silver cube:
6 7
3 1
1 4
1
2
5
1
4
7
2
5
1
3
1
6
5
1
2
1
6
3
4
7
1
Here is what I came up with:
1 4
2 1
3 7
5
6
1
6 2
1 3
5 1
1
7
4
7
4
1
1 3
5 1
6 2
Now in order to see the similarities, you need to rotate the cubes 90 degrees, and
then flip them over. Once you have completed this, you then replace the numbers
with these permutations: the numbers 1 and 5 stay the same, but change 2 to 4, 4 to
6, 6 to 3, 3 to 7, 7 to 2, and you should come up with the same cubes as the sample
answer.
After closely examining these results with the sample given, I was able to come to
the conclusion that in order to find a unique 3 × 3 silver cube, one would need to do
more than joggle around the silver cells themselves.
As my research concluded with the three-dimensional 4×4 silver cube, I discovered
the only way to select the correct numbers to be put in the silver cells. This can only
be accomplished by using either 4 of 2 numbers or 7 of 1 number, followed by the rest
of the 8 numbers in the cubes
1
1
1
For example:
3
1
7
4
2
2
2
8
5
2
9
6
10
Or:
1
1
1
3
1
4
1
1
1
7
8
5
2
9
6
3
10
It does not matter where the numbers go as long as you follow the rule about how
many of each number to use.
The only further research that this study suggests is a more complicated set of
orders using the 5 × 5, the 6 × 6, and the previously not solved 7 × 7. I believe that
if given an extensive amount of time and dedication, one can solve the ”impossible”
7 × 7 silver cube. This research is important because it can be used when dealing
with coding and other kept secret information and all you need is to create a key and
you should be able to create hidden messages and codes using the silver cubes.
References
[1] M. Blidia, M. Chellali, and L. Volkmann, Bounds on the 2-domination number of
graphs, Utilitas Math. 71 (2006), 209-216.
[2] F.J. MacWilliams and N.J. A. Sloane, The theory of error-correcting codes II,
North Holland Publishing Co., Amsterdam, 1977.
[3] M. Mahdian and E.S. Mahmoodian, The roots of an IMO97 problem, Bull. Inst.
Combin. Appl. 28 (2000), 48-54.
[4] P.J. Wan, Near-optimal conflict-free channel set assignments for an optical clusterbased hypercube network, J. Comb. Optim., 1 (1997), pp. 179-186.
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