Complete Search with Recursion

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Solution - Apple Division

Since $n\le 20$, we can try all possible divisions of $n$ apples into two sets and find the one with the minimum difference in weights. Here are two ways to do this.

Permutations

A permutation is a reordering of a list of elements.

Lexicographical Order

This term is mentioned quite frequently, ex. in USACO Bronze - Photoshoot.

Think about how are words ordered in a dictionary. (In fact, this is where the term "lexicographical" comes from.)

In dictionaries, you will see that words beginning with the letter a appears at the very beginning, followed by words beginning with b, and so on. If two words have the same starting letter, the second letter is used to compare them; if both the first and second letters are the same, then use the third letter to compare them, and so on until we either reach a letter that is different, or we reach the end of some word (in this case, the shorter word goes first).

Permutations can be placed into lexicographical order in almost the same way. We first group permutations by their first element; if the first element of two permutations are equal, then we compare them by the second element; if the second element is also equal, then we compare by the third element, and so on.

For example, the permutations of 3 elements, in lexicographical order, are

Notice that the list starts with permutations beginning with 1 (just like a dictionary that starts with words beginning with a), followed by those beginning with 2 and those beginning with 3. Within the same starting element, the second element is used to make comparisions.

Generally, unless you are specifically asked to find the lexicographically smallest/largest solution, you do not need to worry about whether permutations are being generated in lexicographical order. However, the idea of lexicographical order does appear quite often in programming contest problems, and in a variety of contexts, so it is strongly recommended that you familiarize yourself with its definition.

Generating Permutations

Some problems will ask for an ordering of elements that satisfies certain conditions. In these problems, if $N \leq 10$, we can probably iterate through all permutations and check each permutation for validity. For a list of $N$ elements, there are $N!$ ways to permute them, and generally we'll need to read through each permutation once to check its validity, for a time complexity of $O(N \cdot N!)$.

do {
    check(v); // process or check the current permutation for validity
} while(next_permutation(v.begin(), v.end()));

Solution - Creating Strings I

Backtracking

Resources

Solution - Chessboard & Queens

Problems