marklin-latte

![https://ithelp.ithome.com.tw/upload/images/20251005/20089358Eui0deSxc3.png](https://ithelp.ithome.com.tw/upload/images/20251005/20089358Eui0deSxc3.png)

在上一篇文中，我們有從這篇論文中有提到兩種 RAG Post-Retrieval 的主要方向分別為 : 

- Reranking
- Context Compressing

[Retrieval-Augmented Generation for Large Language Models: A Survey](https://arxiv.org/pdf/2312.10997)

然後接下來我們就要說說 `Context Compressing` 的手法。

marklin-latte

// Bug Version
function createArrayOfFunctions(y) {
    var arr = [];
    for(var i = 0; i<y; i++) {
      arr[i] = function(x) { return x + i; } // <=== the i is bug ! the i will always be y+1;
    }
    return arr;
 }

marklin-latte

#include <iostream>
#include <string>
#include <iostream>
#include <algorithm>
#include <math.h> 
#include <limits.h>
#include <vector>
#include <unordered_map>
using namespace std;

marklin-latte

#include <iostream>
#include <string>
#include <iostream>
#include <algorithm>
#include <math.h> 
#include <limits.h>
#include <vector>
using namespace std;

/**

marklin-latte

function g(n){
    let res = 0;
    for (let i=1; i <= n; i++){
        let temp = i;
        while(temp != 0){
            let digit = temp % 10;
            if(digit == 7){
                res++;
                break;
            }

marklin-latte

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:

marklin-latte

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> neighbors;

    Node() {}

    Node(int _val, vector<Node*> _neighbors) {

marklin-latte

public class Solution {
    public int result;
    /**
     * @param amount: a total amount of money amount
     * @param coins: the denomination of each coin
     * @return: the number of combinations that make up the amount
     */
    public int change(int amount, int[] coins) {
        // write your code here
        if(coins == null || coins.length == 0 ){

marklin-latte

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function(coins, amount) {
    
    // Using the greedy method. (Finding the maximum benefit.)
    // 1. First, we should select the maximum denomination.
    // 2. then, we should select the second largest coin.

marklin-latte

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {string[]}