Hello! In this tutorial, we will cover one of the frequently asked problems in coding tests, the Traveling Salesman Problem (TSP). This problem helps to understand important concepts in graphs and dynamic programming.
1. Problem Description
The Traveling Salesman Problem is the problem of finding the minimum cost path that visits all given cities once and returns to the starting city. This problem is typically represented by a distance matrix that provides the cost of moving between cities, and the salesman must visit all cities exactly once.
2. Mathematical Representation of the Problem
Let the number of cities be represented as N. The distance matrix between cities is defined in the form of cost[i][j]. Here, cost[i][j] represents the cost of moving from city i to city j. The path that the salesman must take can be expressed as follows:
min(cost[0][1] + cost[1][2] + ... + cost[N-1][0])
3. Approach to Solve the Problem
Since the Traveling Salesman Problem is NP-hard, the time complexity is very high for solving it using brute force. Therefore, we can solve the problem more efficiently using dynamic programming.
To solve this problem using a dynamic programming approach, we will use bit masking. Bit masking helps to easily check the visiting status by representing each city as a bit. Let’s approach the problem using the following algorithmic steps.
3.1 State Representation through Bit Masking
We represent the state of visited cities as a bitmask. For example, if there are 4 cities:
- 0000: No city visited
- 0001: City 0 visited
- 0010: City 1 visited
- 0011: Cities 0 and 1 visited
- …All combinations can be expressed in this way.
3.2 Definition of the Dynamic Programming Table
The DP table dp[mask][i] stores the minimum cost of visiting the cities corresponding to mask and starting from city i. In the initial state, mask is set to 1, and all other states are initialized to infinity (INFINITY).
4. Algorithm Implementation
Now, let’s implement the algorithm we wrote in JavaScript.
function tsp(cost) {
const N = cost.length;
const INF = Number.MAX_SAFE_INTEGER;
const dp = new Array(1 << N).fill(null).map(() => new Array(N).fill(INF));
// Starting point
dp[1][0] = 0;
for (let mask = 0; mask < (1 << N); mask++) {
for (let u = 0; u < N; u++) {
if (dp[mask][u] === INF) continue; // Skip if u is not visited
// Visit all other cities not in the current mask
for (let v = 0; v < N; v++) {
if (mask & (1 << v)) continue; // Skip if v is already visited
const nextMask = mask | (1 << v);
dp[nextMask][v] = Math.min(dp[nextMask][v], dp[mask][u] + cost[u][v]);
}
}
}
let ans = INF;
for (let i = 1; i < N; i++) {
ans = Math.min(ans, dp[(1 << N) - 1][i] + cost[i][0]);
}
return ans;
}
5. Time Complexity
The time complexity of this algorithm is O(N^2 * 2^N). Since the method of representing states through bit masking and the process of updating the DP table are combined, significant computation time is required as the number of cities increases. Therefore, this algorithm is only practical when N is less than or equal to 20.
6. Testing and Examples
Now, let's provide some examples to test the algorithm. Below is the cost matrix between cities:
const costMatrix = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
];
console.log(tsp(costMatrix)); // Expected output: 80
7. Conclusion
By dealing with the Traveling Salesman Problem, we gained an understanding of the pathfinding techniques using dynamic programming and bit masking. This problem may seem like a simple route-finding problem, but it is an important algorithmic problem that can be applied in various fields, such as route optimization for internet companies.
I hope this tutorial has been helpful in solving algorithmic problems using JavaScript, and I encourage you to solve more problems through additional practice. Thank you!