Now I would like to know the maximum amount of E (mE) I can make with the contents of the chest.

Derived just from logic: The maximum is the minimum of the "maximums". The "maximums" are the amounts calculated, when looking at each one required item type separately.

Eg

Code: Select all

```
r1= 2, r2= 3
c1=19, c2=23
q1 = 19/2 = 9
q2 = 23/3 = 7
mE = min(9,7) = 7
```

- Step 1: For each pair (rx,cx) I can set up a circuit that performs a division and returns a quotient qx. qx is the whole multiple of rx in cx. The growth rate of combinators for this step is O(n).

- Step 2: I would need to perform a calculation mE = min(q1, q2, q3, ... qn). And I believe the growth rate of combinators in this step would also be O(n). As I would perform m12 = min(q1,q2), m13= min(m12,q3), ... mE = min(m1n-1, qn)

Can you get to mE with growth rates smaller than O(n)?

Edit 2937: Added an example. Added step numbers for my approach.