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    <p>You may be able to simply brute force all options.</p> <p>Here is some example Python code:</p> <pre><code>from itertools import combinations F=['apple','banana','clementine'] T=['compass','telephone'] C=['socks','gloves','hat'] weights={'apple':1,'banana':2,'clementine':3,'compass':4,'telephone':5, 'socks':6,'gloves':7,'hat':8} values={'apple':10,'banana':9,'clementine':8,'compass':7,'telephone':6, 'socks':5,'gloves':4,'hat':3} choices=[] W=45 # Maximum allowed weight n=5 # Number of choices to display for M in range(3): # M represents which category gets an extra item num_food = 3 if M==0 else 2 num_toys = 2 if M==1 else 1 num_clothes = 3 if M==2 else 2 for food_to_take in combinations(F,num_food): for toys_to_take in combinations(T,num_toys): for clothes_to_take in combinations(C,num_clothes): things_to_take = food_to_take+toys_to_take+clothes_to_take weight = sum(weights[a] for a in things_to_take) value = sum(values[a] for a in things_to_take) if weight&lt;=W: choices.append([value,weight,things_to_take]) for value,weight,things_to_take in sorted(choices,reverse=True)[:n]: print value,weight,things_to_take </code></pre> <p>prints</p> <pre><code>43 23 ('apple', 'banana', 'clementine', 'compass', 'socks', 'gloves') 42 24 ('apple', 'banana', 'clementine', 'telephone', 'socks', 'gloves') 42 24 ('apple', 'banana', 'clementine', 'compass', 'socks', 'hat') 41 25 ('apple', 'banana', 'compass', 'telephone', 'socks', 'gloves') 41 25 ('apple', 'banana', 'clementine', 'telephone', 'socks', 'hat') </code></pre>
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    1. PO01 Knapsack specialization
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    1. USPeter de Rivaz
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    1. COI don't think that brute forcing is an option. Ultimately I want to go on longer trips, so I'm looking at taking 8 or 9 items and some of the categories have 200 or so items to choose from.
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      1. USFuriousGeorge
 

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