0/1 Knapsack with irrational weights

Consider the 0/1 knapsack problem. The standard Dynamic Programming algorithm applies only when the capacity as well as the weights to fill the knapsack with are integers/ rational numbers. What do you do when the capacity/weights are irrational?

The issue is that we can't memoize like we do for integer weights because we may need potentially infinite decimal places for irrational weights - leading to an infinitely large number开发者_高级运维 of columns for the Dynamic Programming Table .

Is there any standard method for solving this? Any comments on the complexity of this problem? Any heuristics?

What about associated recurrences like (for example):

f(x)=1, for x< sqrt(2)

f(x)=f(x-sqrt(2))+sqrt(3),otherwise ?

Or the Pibonacci number problem here: http://www.spoj.pl/problems/PIB/ ?

I don't know of any general method which will solve problems of the kind you stated. Perhaps the memoization technique used in Pibonacci (see second section below) can be used.

In any case, sometimes, we can give really fast algorithms by exploiting the problem (see the sqrt(2) & sqrt(3)) solution below.

Reducing such problems to knapsack might not be such a good idea as I expect there will be other ways which will be much faster.

So to answer your questions:

Problem involving sqrt(2) and sqrt(3)

I will answer your second question first.

f(x) = 1 for x < sqrt(2). (x >= 0 also, I presume)
f(x) = f(x-sqrt(2)) + sqrt(3)

This can be solved very fast (in O(log logn) time!), using only integer arithmetic (which is assumed O(1)), expect for one last step which requires multiplying by sqrt(3) and adding 1.

Given an n we need to find the smallest m such that

n - m sqrt(2) < sqrt(2)

i.e.

n - m sqrt(2) < sqrt(2) => n < (m+1)*sqrt(2) => n * sqrt(2) < m+1

and

n - (m-1)sqrt(2) > sqrt(2) => n > m sqrt(2) => n*sqrt(2) > m.

Thus m is the integer part of n*sqrt(2)

and we have that f(n) = (m-1)*sqrt(3) + 1.

Thus we only need to calculate [n *sqrt(2)] the integer part of n*sqrt(2).

This can be quickly calculated by using the Continued Fractions of sqrt(2) which are rational approximations to sqrt(2) and they are in some sense the 'best' approximations with given denominator sizes.

The continued fraction a(i)/b(i) of sqrt(2) can be formed using the recurrence:

a0 = 1
b0 = 1
a(i+1) = a(i) +2*b(i)
b(i+1) = a(i) + b(i)

It can be shown that in order to approximate [n*sqrt(2)] it is enough to consider some odd i for which b(i) > 10*n^2 (using Liouville's approximation Theorem and theorems on continued fractions) and that [n*sqrt(2)] = [n*a(i)/b(i)] for that i.

Now a(i), b(i) satisfies the matrix equation

[1 2] [a(i)]    [a(i+1)]
[1 1] [b(i)]  = [b(i+1)]

Thus we need to compute powers of the matrix

[1 2]
[1 1]

So that the entries get bigger than 10*n^2.

It can be shown that the required power of the matrix is O(logn) and thus can be calculated in O(log log n) time using only integer arithmetic (assuming that is O(1)).

So the value of your function f at n can be calculated in O(log logn) time using only integer arithmetic (except for the last step, where you need to multiply an integer by sqrt(3)).

Pibonacci Number

From your comment, this is the problem

g(x) = 1 if 0 <= x < 4
g(x) = g(x-1) + g(x-pi) x >= 4

This can be solved using memoization:

Let h(m,n) = g(m - n*pi)

Then we have that

h(m,n) = h(m-1, n) + h(m, n+1)

And so we have that

g(m) = g(m-1) + h(m, 1)

You can now use memoization by maintaining two tables, one for g(m) and other for h(m,n). Note that even though you need to calculate h(m,n+1), increasing n only reduces m -n*pi and will become between 0 and 4 within a reasonable time (O(m) I suppose), thus you won't keep going on forever.

This is not as nice (or fast) as the sqrt(2) and sqrt(3) solution, but I believe it does give a way to do the calculation.

0-1 Knapsack with irrational coefficients

Perhaps taking better and better rational approximations to the irrationals and then solving the 0-1 knapsack problem for approximation will ultimately converge to the right solution.

My guess is, the fixed point in this iteration will give you the solution.

Of course, as the approximations get better the W in O(nW) of they dynamic programming algorithm might become exponential soon and you might be better off just consider all possibilities.

As a quick comment on the complexity, the knapsack problem is NP-Complete (and so no correct polynomial time algorithm is known). It would seem, at least intuitively, that this would also be NP-Complete (if you could solve it correctly and fast with infinite decimal places, you could also solve it when the infinite decimal digits were all zeros...i.e. an integer knapsack).