Complete Search with Recursion

Subsets

Resources

Solution - Apple Division

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

Generating Subsets Recursively

The first method would be to write a recursive function which searches over all possibilities.

At some index, we either add $weigh{t}_i$ to the first set or the second set, storing two sums $s_1$ and $s_2$ with the sum of values in each set.

Then, we return the difference between the two sums once we've reached the end of the array.

1#include <bits/stdc++.h>
2using namespace std;
3using ll = long long;
4using vi = vector<int>;
5#define pb push_back
6#define rsz resize
7#define all(x) begin(x), end(x)
8#define sz(x) (int)(x).size()
9using pi = pair<int,int>;
10#define f first

1import java.util.*;
2import java.io.*;
3
4public class Main {
5
6    static int N;
7    static int weights[];
8
9    public static void main(String[] args) throws Exception {
10        FastIO sc = new FastIO(System.in);  //See "Intro - FastIO" - this class acts similarly to Scanner, but faster

1n = int(input())
2p = list(map(int, input().split()))
3
4def solve(i, s1, s2):
5    if i == n:
6        return abs(s2-s1)
7    return min(solve(i+1, s1+p[i], s2), 
8               solve(i+1, s1, s2+p[i]))
9
10print(solve(0, 0, 0))

Generating Subsets with Bitmasks

A bitmask is an integer whose binary representation is used to represent a subset. For a particular bitmask, if the $i$'th bit is turned on (equal to $1$), we say the $i$'th apple is in $s_1$. Then, the rest of the apples are in $s_2$. We can iterate through all subsets $s_1$ if we check all bitmasks ranging from $0$ to $2^N-1$. For each bitmask, find the sum of $s_1$ and $s_2$ and find the minimum difference between their sums.

Note the fancy bitwise operations:

• 1 << x for an integer $x$ is another way of writing $2^x$, which, in binary, has only the $x$'th bit turned on.
• The & (and) operator will take two integers and return a new integer. a & b for integers $a$ and $b$ will return a new integer whose $i$'th bit is turned on if and only if the $i$'th bit is turned on for both $a$ and $b$. Thus, mask & (1 << x) will return a positive value only if the $x$'th bit is turned on in $mask$.

1#include <bits/stdc++.h>
2using namespace std;
3using ll = long long;
4using vi = vector<int>;
5#define pb push_back
6#define rsz resize
7#define all(x) begin(x), end(x)
8#define sz(x) (int)(x).size()
9using pi = pair<int,int>;
10#define f first

1import java.util.*;
2import java.io.*;
3
4public class Main {
5    public static void main(String[] args) throws Exception {
6        FastIO sc = new FastIO(System.in); //see "Intro - FastIO" - this class acts similarly to Scanner
7        PrintWriter pw = new PrintWriter(System.out);
8
9        int N = sc.nextInt();
10        int weights[] = new int[N];

1n = int(input())
2p = list(map(int, input().split()))
3
4ans = float('inf')
5for mask in range(1<<n):
6    s1,s2 = 0,0
7    for i in range(n):
8        if mask & (1<<i):
9            s1 += p[i]
10        else:

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.

Some problems will ask for an ordering of elements that satisfies certain conditions. In these problems, if $N\le 10$, we can just iterate through all $N!=N\cdot (N-1)\cdot (N-2)\cdots 1$ permutations and check each permutation for validity.

Solution - Creating Strings I

Generating Permutations Recursively

Just a slight modification of method 1 from CPH.

1int N, cnt[26];
2str S, perm;
3vs perms;
4
5void search() {
6    if (sz(perm) == sz(S)) {
7        perms.pb(perm);
8        return;
9    }
10    F0R(i,26) if (cnt[i]) {

1import java.util.*;
2import java.io.*;
3
4public class Main {
5    static String s, perm;
6    static ArrayList<String> perms;
7    static int N, cnt[];
8        
9    static void search() {
10        if (perm.length() == s.length()) {

Generating Permutations Using next_permutation

Alternatively, we can just use the next_permutation() function. This function takes in a range and modifies it to the next greater permutation. If there is no greater permutation, it returns false. To iterate through all permutations, place it inside a do-while loop. We are using a do-while loop here instead of a typical while loop because a while loop would modify the smallest permutation before we got a chance to process it.

What's going to be in the check function depends on the problem, but it should verify whether the current permutation satisfies the constraints given in the problem.

1do {
2    check(v); // process or check the current permutation for validity
3} while(next_permutation(v.begin(), v.end()));

Each call to next_permutation makes a constant number of swaps on average if we go through all $N!$ permutations of size $N$.

1#include <bits/stdc++.h>
2using namespace std;
3using ll = long long;
4using vi = vector<int>;
5#define pb push_back
6#define rsz resize
7#define all(x) begin(x), end(x)
8#define sz(x) (int)(x).size()
9using pi = pair<int,int>;
10#define f first

Backtracking

Resources

Solution - Chessboard & Queens

Using next_permutation

Since no two queens can be in the same column, let's generate a permutation of length $8$. Then, the $p_i$ represents the column that the $i$-th queen goes on.

By generating all permutations, we can quickly test all possible placements, and count how many are valid.

To make the implementation easier, we can observe that some bottom-left to top-right diagonal can be represented as all squares $i,j$ such that $i+j=S$ for some $S$. Similarly, some bottom-right to top-left diagonal can be represented as $i+7-j$ if $i,j$ are zero-indexed.

1#include <bits/stdc++.h>
2using namespace std;
3using ll = long long;
4using vi = vector<int>;
5#define pb push_back
6#define rsz resize
7#define all(x) begin(x), end(x)
8#define sz(x) (int)(x).size()
9using pi = pair<int,int>;
10#define f first

1import java.io.*;
2import java.util.*;
3
4public class Chessboard {
5    public static boolean ok[][] = new boolean[8][8];
6    public static ArrayList < Integer > perm = new ArrayList < Integer > ();
7    public static boolean[] chosen = new boolean[8];
8    public static int ans = 0;
9    public static void main(String[] args) throws IOException {
10        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

Using Backtracking

According to CPH:

A backtracking algorithm begins with an empty solution and extends the solution step by step. The search recursively goes through all different ways how a solution can be constructed.

Since the bounds are small, we can recursively backtrack over all ways to place the queens, storing the current state of the board.

Then, we can try to place a queen at all squares $x,y$ if it isn't attacked by a queen or blocked and recurse, before removing this queen and backtracking.

Finally, when we have placed all the queens and the board's state is valid, then increment the answer.

1string g[8];
2bool sum[15], dif[15], c[8];
3int ans = 0;
4 
5void dfs(int r) { // place queen in r-th row
6    if (r == 8) {
7        ++ ans; // found valid placement
8        return;
9    }
10    F0R(i,8) if (g[r][i] == '.' && !c[i] && !sum[i+r] && !dif[i-r+7]) {

1import java.util.*;
2import java.io.*;
3
4public class Main {
5    static String g[];
6    static boolean sum[], dif[], c[];
7    static int ans = 0;
8    static void dfs(int r) { //place queen in row r
9        if (r == 8) {
10            ++ ans; 

Pruning

Both of the resources below describe this well so I won't repeat it here.

Problems

None of these require pruning.

You can find more problems at the CP2 link given above or at USACO Training. However, these sorts of problems appear much less frequently then they once did.