A
Quad-Omino (not to be confused with a
tetromino) is a square-shaped tile with numbers in each corner. Like the game of dominoes, the object is to place them next to each other so their numbers match.
The standard set I own with numbers 0 to 5 has 125 tiles. Does the set really contain every possible tile?
Find a formula for the total number of different tetrominoes in a set numbered from 0 to n.
Quad-Ominoes with only one number repeated four times: n+1
Quad-Ominoes with two different numbers:
Split one and three: (n+1)*n
Split two and two with like numbers adjacent: (n+1)*n/2
Split two and two with like numbers diagonally opposite: (n+1)*n/2
Quad-Ominoes with three different numbers:
With the two like numbers adjacent:
(n+1)*n*(n-1) if order (orientation) counts
(n+1)*n*(n-1)/2 if order (orientation) does not count
With the two like numbers non-adjacent:
(n+1)*n*(n-1)/2
Quad-Ominoes with four different numbers:
C(n+1,4)*3! if orientation counts
C(n+1,4)*3 if orientation does not count
(in this case orientation is not the same as order.)
If orientation counts, that's
(n+1)*(2*n+1) + (n+1)*n*(n-1)*3/2 + C(n+1,4)*3!
If orientation doesn't count, it's
(n+1)*(2*n+1) + (n+1)*n*(n-1) + C(n+1,4)*3
For n=5, the former is 336 and the latter is 231.
The respective OEIS numbers are
A006528 (n^4 + n^2 + 2*n)/4
Number of ways to color vertices of a square using <= n colors, allowing only rotations.
A002817 n*(n+1)*(n^2+n+2)/8
Number of inequivalent ways to color vertices of a square using <= n colors, allowing rotations and reflections.
where in each case the n is one more than the n used in the current problem, as the current n is the range 0 to n, while the OEIS n is the count ot this range, which also counts the zero.
Here are the 231 disregarding orientation:
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
0 1 1 1
0 2 2 2
0 3 3 3
0 4 4 4
0 5 5 5
1 0 0 0
1 2 2 2
1 3 3 3
1 4 4 4
1 5 5 5
2 0 0 0
2 1 1 1
2 3 3 3
2 4 4 4
2 5 5 5
3 0 0 0
3 1 1 1
3 2 2 2
3 4 4 4
3 5 5 5
4 0 0 0
4 1 1 1
4 2 2 2
4 3 3 3
4 5 5 5
5 0 0 0
5 1 1 1
5 2 2 2
5 3 3 3
5 4 4 4
0 0 1 1
0 1 0 1
0 0 2 2
0 2 0 2
0 0 3 3
0 3 0 3
0 0 4 4
0 4 0 4
0 0 5 5
0 5 0 5
1 1 2 2
1 2 1 2
1 1 3 3
1 3 1 3
1 1 4 4
1 4 1 4
1 1 5 5
1 5 1 5
2 2 3 3
2 3 2 3
2 2 4 4
2 4 2 4
2 2 5 5
2 5 2 5
3 3 4 4
3 4 3 4
3 3 5 5
3 5 3 5
4 4 5 5
4 5 4 5
0 0 1 2
0 1 0 2
0 0 1 3
0 1 0 3
0 0 1 4
0 1 0 4
0 0 1 5
0 1 0 5
0 0 2 3
0 2 0 3
0 0 2 4
0 2 0 4
0 0 2 5
0 2 0 5
0 0 3 4
0 3 0 4
0 0 3 5
0 3 0 5
0 0 4 5
0 4 0 5
1 1 0 2
1 0 1 2
1 1 0 3
1 0 1 3
1 1 0 4
1 0 1 4
1 1 0 5
1 0 1 5
1 1 2 3
1 2 1 3
1 1 2 4
1 2 1 4
1 1 2 5
1 2 1 5
1 1 3 4
1 3 1 4
1 1 3 5
1 3 1 5
1 1 4 5
1 4 1 5
2 2 0 1
2 0 2 1
2 2 0 3
2 0 2 3
2 2 0 4
2 0 2 4
2 2 0 5
2 0 2 5
2 2 1 3
2 1 2 3
2 2 1 4
2 1 2 4
2 2 1 5
2 1 2 5
2 2 3 4
2 3 2 4
2 2 3 5
2 3 2 5
2 2 4 5
2 4 2 5
3 3 0 1
3 0 3 1
3 3 0 2
3 0 3 2
3 3 0 4
3 0 3 4
3 3 0 5
3 0 3 5
3 3 1 2
3 1 3 2
3 3 1 4
3 1 3 4
3 3 1 5
3 1 3 5
3 3 2 4
3 2 3 4
3 3 2 5
3 2 3 5
3 3 4 5
3 4 3 5
4 4 0 1
4 0 4 1
4 4 0 2
4 0 4 2
4 4 0 3
4 0 4 3
4 4 0 5
4 0 4 5
4 4 1 2
4 1 4 2
4 4 1 3
4 1 4 3
4 4 1 5
4 1 4 5
4 4 2 3
4 2 4 3
4 4 2 5
4 2 4 5
4 4 3 5
4 3 4 5
5 5 0 1
5 0 5 1
5 5 0 2
5 0 5 2
5 5 0 3
5 0 5 3
5 5 0 4
5 0 5 4
5 5 1 2
5 1 5 2
5 5 1 3
5 1 5 3
5 5 1 4
5 1 5 4
5 5 2 3
5 2 5 3
5 5 2 4
5 2 5 4
5 5 3 4
5 3 5 4
0 1 2 3
0 1 3 2
0 2 1 3
0 1 2 4
0 1 4 2
0 2 1 4
0 1 2 5
0 1 5 2
0 2 1 5
0 1 3 4
0 1 4 3
0 3 1 4
0 1 3 5
0 1 5 3
0 3 1 5
0 1 4 5
0 1 5 4
0 4 1 5
0 2 3 4
0 2 4 3
0 3 2 4
0 2 3 5
0 2 5 3
0 3 2 5
0 2 4 5
0 2 5 4
0 4 2 5
0 3 4 5
0 3 5 4
0 4 3 5
1 2 3 4
1 2 4 3
1 3 2 4
1 2 3 5
1 2 5 3
1 3 2 5
1 2 4 5
1 2 5 4
1 4 2 5
1 3 4 5
1 3 5 4
1 4 3 5
2 3 4 5
2 3 5 4
2 4 3 5
The set of 125 is not complete.
There's a website containing a claim that a complete set would have 126 tiles so that only one is missing from a set of 125. The basis of this is:
Tiles with one number only: 6
Tiles with two different numbers:
1-3 split: 6*5 = 30
2-2 split: 6*5/2 = 15 (ignores whether like numbers are adjacent or diagonally opposite)
Tiles with three different numbers:
6*5*4/2 = 60 (again ignores whether like numbers are adjacent or diagonally opposite)
Tiles with four different numbers:
C(6,4) = 6*5*4*3/(4!) = 15
This discounts not only orientation, but also order (adjacency), as 1234 counts the same as 1324 and 1243, where the lowest digit, 1, is bounded by 2 and 4, 3 and 4, or 2 and 3 respectively.
Of course the actual set lacks all the variety this methodology lacks, but in addition leaves out one tile that even this methodology includes.
The total is 6+30+15+60+15 = 126, so 125 is still not a "full" set.
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Posted by Charlie
on 2021-10-20 11:41:12 |
solution
