### Re: rosettacode 500 milestone passed

- Posted by petelomax in March
- 393 views

I translated the C dynamic programming version:

sequence items = { {"map", 9, 150, 1}, {"compass", 13, 35, 1}, {"water", 153, 200, 2}, {"sandwich", 50, 60, 2}, {"glucose", 15, 60, 2}, {"tin", 68, 45, 3}, {"banana", 27, 60, 3}, {"apple", 39, 40, 3}, {"cheese", 23, 30, 1}, {"beer", 52, 10, 3}, {"suntan cream", 11, 70, 1}, {"camera", 32, 30, 1}, {"T-shirt", 24, 15, 2}, {"trousers", 48, 10, 2}, {"umbrella", 73, 40, 1}, {"waterproof trousers", 42, 70, 1}, {"waterproof overclothes",43, 75, 1}, {"note-case", 22, 80, 1}, {"sunglasses", 7, 20, 1}, {"towel", 18, 12, 2}, {"socks", 4, 50, 1}, {"book", 30, 10, 2}, }; sequence {names,weights,points,counts} = columnize(items) constant n = length(items) function knapsack(int w) int v -- m is the achievable points matrix: -- Note that Phix uses 1-based indexes, so m[1][1] -- actually holds points for 0 items of weight 0, -- and m[n+1][w+1] is for n items at weight w. seq m = repeat(repeat(0,w+1),n+1) for i=1 to n do for j=1 to w+1 do -- (0 to w really) m[i+1][j] = m[i][j] for k=1 to counts[i] do if k*weights[i]>j-1 then exit end if v = m[i][j-k*weights[i]]+k*points[i] if v>m[i+1][j] then m[i+1][j] = v end if end for end for end for seq s = repeat(0,n) int j = w+1 -- (w -> 0 really) for i=n+1 to 2 by -1 do -- (n to 1 really) v = m[i][j] int k = 0 while v!=m[i-1][j]+k*points[i-1] do s[i-1] += 1 j -= weights[i-1] k += 1 end while end for return s end function int tc = 0, tw = 0, tv = 0 seq s = knapsack(400) for i=1 to n do int si = s[i] if si then printf(1,"%-22s %5d %5d %5d\n", {names[i], si, si*weights[i], si*points[i]}) tc += si tw += si*weights[i] tv += si*points[i] end if end for printf(1,"%-22s %5d %5d %5d\n", {"count, weight, points:", tc, tw, tv})

Simple and elegant, yet quite difficult to get your head around - there are some youtube vids (search for knapsack problem) that might help.

Pete