Dominoes are small rectangular tiles with dots called spots or pips embossed at both halves of the tiles. A standard "double-six" domino set has 28 tiles: one for each of the unordered pair of integers from (0,0) to (6,6). In general, a "double-n" domino set would consist of domino tiles for each ordered pair of integers from (0,0) to (n, n). Determine all values of n for which one constructs a ring made up of all the tiles in a double-n domino set.

For the given problem, the algorithm is as follows:

The value of dominoes pip is between 0 to n

So, the total count of pips is n + 1

Compute the total count of tiles (t) = {(n + 1) * (n + 2)}/2

Each tile consists of two pips, therefore total number of occurrence of pip = t*2

Therefore, the occurrence of each of the pip = Total count of occurrence of pips/total number of pips = (t * 2) / (n + 1)

If the occurrence of individual pip is even in number, then ring is arranged otherwise not.

Example for odd number of n.

Let the value of n = 3

The value of pip is 0, 1, 2, 3

Therefore, the total number of pips = n + 1 = 4

The total number of tiles (t) = {(n + 1) * (n + 2)}/2

                                      = {5 * 4}/2

                                      = 10

Therefore, the tiles would be (0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)

Total occurrence of pips = 10 * 2

                                         = 20

Therefore, the count of each pip = (t * 2)/ (n + 1)

                                                     = 20/4

                                                     = 5

Since, the number of individual pip are odd in number; therefore, ring can’t be formed.

Explanation:

Since, for changing the adjacent tile, the pip value must be same.

Therefore, the occurrence of pip must be in pairs.

Example for even number of n.

Let the value of n = 2

The value of pip is 0, 1, 2

Therefore, the total number of pips = n + 1 = 3

Total number of tiles (t) = {(n + 1) * (n + 2)}/2

                                      = {4 * 3}/2

                                      = 6

Therefore, the tiles would be (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

Total occurrence of pips = 6 * 2

                                       = 12

Therefore, the count of each pip = (t * 2)/ (n + 1)

                                                     = 12/3

                                                     = 4

Since, the total number of individual pip are even in number; therefore, ring can be formed.

Explanation: