Question

// enumerate the different combinations
// for select k objects from the set of objects given in list L
public static List<List<String>> comb(List<String> L, int k) {
if (k == 0) return new ArrayList<List<String>>(){{add(new ArrayList<String>( ));}};
if (L.size() == k) return new ArrayList<List<String>>(){{add(new ArrayList<String>(L));}};
List<List<String>> L1 = comb(L.subList(1,L.size()),k);
List<List<String>> L2 = comb(L.subList(1,L.size()),k-1);
for (List<String> list : L2) list.add(L.get(0));
L1.addAll(L2);
return L1;
}

Q.Study the given Java codes. Explain in plain English the logic for enumerating combinations of a certain size for a given set.

Answer 1

The logic is:

While choosing K objects from a list of say N objects in the list L, we are excluding the first object and from the remaining N-1 objects, we are choosing K objects in each list of L1, and K-1 objects in each list of L2, and then we are including the first object which we excluded from L into each list of L2 (think of when L2 will be empty), which completes the K objects in each list of L2.
After this, we are concatenating all the objects from L1 and L2 to form a single List<List<String> > Object in which each List<String> consists of K objects.

Let's take a small example for this:

Let L = {"a","b","c"} and K = 2, then our answer should be {{"a","b"},{"b","c"},{"a","c"}} right?
Let's work on the code to find out what the answer is:

Case 1:

L={"a","b","c"}, K=2

K!=0

L.size()=3!=K

L1=comb({"b","c"},2)

Case 1.a:
L={"b","c"},K=2

K!=0

L.size()=2=K therefore this function will return {{"b","c"}}

Therefore, L1={{"b","c"}}

L2=comb({"b","c"},1)

Case 1.b

L={"b","c"},K=1

K!=0
L.size()=2!=K

L1=comb{"c",1)

Case 1.b.a

L={"c"}, K =1

K!=0

L.size()=1=K

Therefore this will return {{"c"}}

Therefore L1={{"c"}}

L2=comb({"c"},0)

Case 1.b.b

K==0, therefore this will return {{}}

Therefore L2={{}}

for(List<String> list: L2) list.add(L.get(0))

Therefore this for loop will modify L2 to {{"b"}}

L1.addAll(L2)

Therefore L1 will become {{"c","b"}}

Therefore L1={{"c","b"}}

Now L2=comb({"b","c"},1)

By similar logic L2 would be {{"b"},{"c"}}

Now we will add L[0] so L2 will become {{"a","b"},{"a","c"}}

Then L1.addAll(L2)

Therefore L1 will return {{"a","b"},{"a","c"},{"b","c"}} which is our required answer.
Hence done.