A day, B day… H day? The Flabbergasting Math behind the Rotating Block Schedule
Why do C days start with Block 3? Why, on C days, is Block 1 after Block 4? Why do C days even exist? After lunch, things might start to make sense—Block 7 followed by, lo and behold, Block 8—but Schrödinger’s schedule can’t stay in one place for long, and the day ends with Block 5 (naturally). A and B days, admittedly, are not nearly as maddening, but D days make up for it—Block 2 to 8? Block 8 to 5?
One is left to believe that Leonia High School’s rotating block schedule (RBS) is either the work of a depraved and sadistic mind or a sad display of our country’s declining numeracy skills, but a closer analysis reveals a beautiful method to the madness.
By writing out the RBS, we can begin to make sense of what at first appears to be random:
A: 1, 2, 3, 5, 6, 7
B: 2, 3, 4, 6, 7, 8
C: 3, 4, 1, 7, 8, 5
D: 4, 1, 2, 8, 5, 6
From this, we can derive three constraints that a hypothetical RBS must satisfy:
- No block should be repeated on any school day.
- A normal school day can accommodate exactly six (55 minute) blocks.
- By the end of a full RBS cycle (A to [letter]), every block (1 to 8) must be seen the same number of times.
A visualization of how Leonia High School’s current RBS accomplishes this is shown below:
(The highlighted numbers represent the missing blocks on that letter day.)
B: 1, 2, 3, 4, 5, 6, 7, 8
C: 1, 2, 3, 4, 5, 6, 7, 8
D: 1, 2, 3, 4, 5, 6, 7, 8
A: 1, 2, 3, 4, 5, 6, 7, 8
Now that it is apparent that Leonia High School’s current RBS properly satisfies these constraints, we must ask if it can be simplified. Based on the pattern above, it would appear that renaming B, C, D, and A days to A, B, C, and D days respectively would semantically simplify the RBS, but for our purposes I will define simplification as reducing the number of letter days. In other words, how can we find the minimum number of letter days that a hypothetical RBS requires to satisfy all three constraints? Is four (A, B, C, and D) the minimum?
To answer this, imagine a Tetris game, eight blocks wide. Imagine filling the game board with individual blocks, six at a time. After the first letter day is added, our game board would look like this:
| A | A | A | A | A | A | N/A | N/A |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
After the second letter day is added, our game board would like like this:
| B | B | B | B | N/A | N/A | N/A | N/A |
| A | A | A | A | A | A | B | B |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Clearly, this RBS generation system is thus far satisfying constraints (1) and (2). No blocks are repeated during any letter day, and every letter day is exactly six blocks. But constraint (3), that every block must be seen the same number of times by the end of the full RBS cycle, remains unsatisfied; by the end of the full RBS cycle depicted above, blocks 1 to 4 are seen twice as much as blocks 5 to 8.
Let’s try adding a third letter day:
| C | C | N/A | N/A | N/A | N/A | N/A | N/A |
| B | B | B | B | C | C | C | C |
| A | A | A | A | A | A | B | B |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Constraint (3) is still unsatisfied. What if we add a fourth letter day?
| C | C | D | D | D | D | D | D |
| B | B | B | B | C | C | C | C |
| A | A | A | A | A | A | B | B |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Constraint (3), the last constraint, is satisfied! By the end of the full RBS cycle depicted above, every block is seen exactly three times.
In fact, the visualization above represents a valid, minimum-letter-day RBS under our current constraints:
(The highlighted numbers represent the missing blocks on that letter day.)
Red (A): 1, 2, 3, 4, 5, 6, 7, 8
Orange (B): 1, 2, 3, 4, 5, 6, 7, 8
Yellow (C): 1, 2, 3, 4, 5, 6, 7, 8
Green (D): 1, 2, 3, 4, 5, 6, 7, 8
The minimum-letter-day rule we deciphered through Tetris can also be expressed as the following equation:
LCM(a, b) ÷ a =
the minimum number of letter days that a hypothetical RBS must contain to satisfy all three constraints, where a = the number of blocks per school day and b = the number of blocks total
(LCM stands for least common multiple, “the smallest positive integer that is divisible by both a and b.”)
In our current RBS,
a = 6 (there are six blocks per school day); and
b = 8 (there are eight blocks total); so
LCM(6, 8) ÷ 6 = 4 letter days minimum
But what if we modified constraint (1) and increased or decreased the number of blocks per school day? Of course, if we increased the number of blocks per school day from six to eight, the RBS would be dramatically simplified to just S:
A: 1, 2, 3, 4 , 5, 6, 7, 8
If we decreased the number of blocks per school day from six to four, the RBS would contain just two letter days:
(The highlighted numbers represent the missing blocks on that letter day.)
A: 1, 2, 3, 4, 5, 6, 7, 8
B: 1, 2, 3, 4, 5, 6, 7, 8
But modifying the school day by two full blocks might be challenging—what if we try modifying it by just one? In answering this question, we find a fascinating equilibrium:
LCM(7, 8) ÷ 7 = LCM(5, 8) ÷ 5 = 8 letter days minimum
Increasing or decreasing the number of blocks per school day by just one would be catastrophic! If either of these slight modifications were made, we would require an H day!
But why?
Because seven and five are both prime numbers (“a natural number that has exactly two divisors: one and itself”), their LCM with a given number is just their product with said number. So, in the examples above:
LCM(7, 8) = 7 × 8 = 56
LCM(5, 8) = 5 × 8 = 40
and
56 ÷ 7 = 40 ÷ 5 = 8
Memorizing four different letter day combinations of six blocks can be annoying, but it is certainly more bearable than would be memorizing eight different letter day combinations of seven (or five) blocks. Leonia High School’s RBS is not, in fact, the device of a wicked man or a dullard; rather, it is mathematically maximally simplified. Realistically, the constraints described could change. For a time during virtual learning, if I remember correctly, there were four longer blocks per school day and only two letter days—and I liked that better. What do you think?