Tetromino tilings of a 4x5 rectangle with minimal diversity

Question

This is a relatively easy manual tiling puzzle. In fact the tiling is all done for you, you just have to specify how many of each of the 26 given tilings to use. The puzzle is: Using N complete sets of tetrominoes, tile N 4x5 rectangles in such a way as to minimise diversity of pieces across the N rectangles. Score your set of rectangles by calculating the average number of distinct tetrominoes in each rectangle. Find the smallest N that results in the smallest possible average score. Here is a list of all the ways to tile a 4x5 with tetrominoes (only one tiling shown for each distinct set of tetrominoes):

A parity consideration prevents a complete tiling for all odd N, so N can only be even.

Here is an example for N=4, ie tiling four 4x5 rectangles with four sets of tetrominoes. The diversity scores are 2,2,3,2 summing to 9 which is an average score of 2.25 (NB this is not minimal for N=4)

A computer program will make short work of this, but I suspect that many people will find the answer by hand quicker than by writing a program.

Answer 1

I suspect the answer is

Score = 29/16 = 1.8125 at N = 16

with a possible tiling of

• 5x I (bottom row rightmost col) x3
• 4x T + 1x S (2nd row 2nd col) x4
• 2x L + 3x S (1st row 1st col) x4
• 2x L + 3x O (1st row 4th col) x4
• 1x I + 4x O (bottom row 3rd col) x1

First of all, I found that achieving the score of exactly 2 wasn't very hard with N = 4, so I focused on maximizing the ratio of the 5x I board (which is the only way to pull down the score strictly under 2). Initially I considered spending every single I on the board of that type, but it seems impossible because most of the tilings are very heavy on L. Indeed, the only boards available (costs of 1 or 2, excluding those containing I except for the 5xI one) were

L I T S O
0 5 0 0 0
0 0 4 1 0
2 0 0 3 0
2 0 0 0 3 
4 0 0 0 1
3 0 2 0 0
4 0 0 1 0

To minimize the use of L and make the number of tiles of each type equal, I did the following calculations:

L I T S O
0 5 0 0 0 x4
0 0 4 1 0 x5
2 0 0 3 0 xA
2 0 0 0 3 xB

At this point, A+B is fixed to 10 (because I is used 20 times), but then there would be 20x L while there are only 70x I+T+S+O in total. Therefore there are too many L's for this to be an answer.

Now, assuming we want to add one of the boards containing an I, let's do the calculations in a different way.

L I T S O
0 5 0 0 0 xA
0 0 4 1 0 xB
2 0 0 3 0 xC
2 0 0 0 3 xD

T=S; 4B = 1B + 3C; B = C
T=L; 4B = 2C + 2D; B = C = D

It is possible to match L, T, and S, but at this point O is too low by 25%. To fix this, I add a 1xI 4xO board:

L I T S O
0 5 0 0 0 xA
0 0 4 1 0 xB
2 0 0 3 0 xC
2 0 0 0 3 xD
0 1 0 0 4 xE

T=S; 4B = 1B + 3C; B = C
T=L; 4B = 2C + 2D; B = C = D
T=O; 4B = 3D + 4E; B = C = D = 4E

At this point, setting B = C = D = 4 and E = 1 gives 1xI and 16x everything else, so I could fit three I-only boards to complete the collection.