People - Apple Puzzle [Inspired by client-puzzle protocol] - math

Question

I am learning a client-puzzle protocol and i have a question about finding the possibility of a solution. Instead of going into the dry protocol facts, here is a scenario:
Lets say i have x people and I have y apples:
Each person must have at least 1 apple
Each person can have at most z apples.
Is there a formula to calculate the number of scenarios?
Example:
4 people [x], 6 apples [y], 15 MAX apples [z]
No. of scenarios calculated by hand: 10.
If my number is very huge, I hope to calculate it using a formula.
Thank you for any help.

Answer 1

Your problem is equivalent to "finds the number of ways you can get x by adding together z numbers, each of which lies between min and max." Sample Python implementation:
def possible_sums(x, z, min, max):
if min*z > x or max*z < x:
return 0
if z == 1:
if x >= min and x <= max:
return 1
else:
return 0
total = 0
#iterate from min, up to and including max
for i in range(min, max+1):
total += possible_sums(x-i, z-1, min, max)
return total
print possible_sums(6, 4, 1, 15)
Result:
10
This function can become quite expensive when called with large numbers, but runtime can be improved with memoization. How this can be accomplished depends on the language, but the conventional Python approach is to store previously calculated values in a dictionary.
def memoize(fn):
results = {}
def f(*args):
if args not in results:
results[args] = fn(*args)
return results[args]
return f
#memoize
def possible_sums(x, z, min, max):
#rest of code goes here
Now print possible_sums(60, 40, 1, 150), which would have taken a very long time to calculate, returns 2794563003870330 in an instant.

Answer 2

There are ways to do this mathematically. It is similar to asking how many ways there are to roll a total of 10 on 3 6-sided dice (x=3, y=10, z=6). You can implement this in a few different ways.
One approach is to use inclusion-exclusion. The number of ways to write y as a sum of x positive numbers with no maximum is y-1 choose x-1 by the stars-and-bars argument. You can calculate the number of ways to write y as a sum of x positive numbers so that a particular set of s of them are at least z+1: 0 if y-x-sz is negative, and y-1-s z choose x-1 if it is nonnegative. Then you can use inclusion-exclusion to write the count as the sum over nonnegative values of s so that y-x-sz is nonnegative of (-1)^s (x choose s)(y-1-sz choose x-1).
You can use generating functions. You can let powers of some variable, say t, hold the total, and the coefficients say how many combinations there are with that total. Then you are asking for the coefficient of t^y in (t+t^2+...+t^z)^x. You can compute this in a few ways.
One approach is with dynamic programming, computing coefficients of (t+t^2+...+t^z)^k for k up to x. The naive approach is probably fast enough: You can compute this for k=1, 2, 3, ..., x. It is a bit faster to use something like repeated squaring, e.g., to compute the 87th power, you could expand 87 in binary as 64+16+4+2+1=0b1010111 (written as a binary literal). You could compute the 1st, 2nd, 4th, 16th, and 64th powers by squaring and multiply these, or you could compute the 0b1, 0b10, 0b101, 0b1010, 0b10101, 0b101011, and 0b1010111 powers by squaring and multiplying to save a little space.
Another approach is to use the binomial theorem twice.
(t+t^2+...+t^z)^x = t^x ((t^z-1)/(t-1))^x
= t^x (t^z-1)^x (t-1)^-x.
The binomial theorem with exponent x lets us rewrite (t^z-1)^x as a sum of (-1)^s t^(z(x-s))(x choose s) where s ranges from 0 to x. It also lets us rewrite (t-1)^-x as an infinite sum of (r+x-1 choose x-1)t^r over nonnegative r. Then we can pick out the finite set of terms which contribute to the coefficient of t^y (r = y-x-sz), and we get the same sum as by inclusion-exclusion above.
For example, suppose we have x=1000, y=1100, z=30. The value is
=1.29 x 10^144.