Problem of the Month (April 2018)

Given an n × m rectangle, how can we pack the 9 polyominoes below into it to maximize the total of the numbers packed?

We break ties in the total by preferring larger polyominoes.

ANSWERS

George Sicherman, Maurizio Morandi, and Joe DeVincentis sent improvements.

The best known solutions are shown below.

Rectangles
n \ m123456789101112
5
1
2
3
4
5
6
1
2
3
6
7
8
7
1
2
3
6
8 (GS)
9
11 (GS)
8
1
2
4
8
9
12
14 (GS)
16
9
1
2
4
8
12
13 (GS)
17
18 (JD)
21
10
2
4
6
11
13
18
20
22
24
29
11
2
4
6
11
15
19
21
23
27 (GS)
31
34 (GS)
12
2
4
6
12
16
20
22
25 (MM)
29 (GS)
36
38
40 (GS)
13
2 (GS)
4 (GS)
7 (GS)
13 (GS)
17 (GS)
21 (GS)
23 (GS)
26 (MM)
31 (GS)
38
40
43 (GS)
14
2 (GS)
4 (GS)
7 (GS)
14 (GS)
18 (GS)
24 (GS)
26 (GS)
30 (GS)
34 (MM)
40 (GS)
44 (MM)
48 (MM)
15
3 (GS)
6 (GS)
9 (GS)
18 (GS)
21 (GS)
25 (GS)
28 (GS)
36 (GS)
39 (GS)
43 (GS)
46 (MM)
54 (MM)

Larger Squares
n=13

47 (MM) 		n=14

57 		n=15

64 (MM) 		n=16

75 (MM)
n=17

82 (MM) 		n=18

94 (MM) 		n=19

106 (MM)
n=20

117 (MM) 		n=21

130 (MM) 		n=22

144 (MM)

Triangles
n=5

1 		n=6

4 		n=7

5 		n=8

6 		n=9

8 		n=10

12 		n=11

14
n=12

17 (GS) 		n=13

21 (GS) 		n=14

24 (MM) 		n=15

28 (MM) 		n=16

34 (MM)
n=17

38 (GS) 		n=18

42 (MM) 		n=19

47 (MM) 		n=20

53 (GS)

Pyramids
n=3

1 		n=4

2 		n=5

4 (GS) 		n=6

6 (GS) 		n=7

9 		n=8

13 (GS) 		n=9

17 (GS)
n=10

22 (GS) 		n=11

27 (MM) 		n=12

34 (GS) 		n=13

40 (GS)
n=14

48 (MM) 		n=15

55 (GS) 		n=16

63 (GS)

Diamonds
n=3

1 (GS) 		n=4

3 (GS) 		n=5

7 (GS) 		n=6

11 (GS) 		n=7

17 (GS)
n=8

24 (MM) 		n=9

32 (GS) 		n=10

41 (MM)
n=11

52 (MM) 		n=12

63 (MM) 		n=13

76 (GS)

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 4/26/18.