比特幣挖礦算法簡化模擬及耗時估算

簡化模擬：

0.假設挖礦難度爲整數d(d>0)
1.隨機生成一個整數x
2.對整數x進行摘要計算hash(x),獲得s，即s=hash(x)
3.對s前導字符串斷定:如爲d個'0'，則礦已挖到，結束；不然跳到步驟1.

一句話:不斷隨機生成整數，直到對生成的整數被進行摘要計算獲得的字符串以某個數目的0開頭爲止。

摘要算法選用sha256

代碼實現:

 1 package cn.ubingo.block_chain;
 2 
 3 import java.util.Collections;
 4 import java.util.Date;
 5 import java.util.Random;
 6 
 7 public class Demo {
 8     static Random r = new Random();
 9 
10     public static void main(String[] args) {
11 
12         int dificulty = 3;
13         String zeros = String.join("", Collections.nCopies(dificulty, "0"));
14         System.out.println(zeros);
15 
16         Date start = new Date();
17         long times = 0;
18         while (true) { 19             String sha = Sha.getSHA256StrJava(r.nextInt() + ""); 20             // System.out.println(sha);
21             times++; 22             if (sha.startsWith(zeros)) 23                 break; 24  } 25         Date end = new Date();
26         long use = end.getTime() - start.getTime();
27 
28         System.out.println(times + " in " + use + "ms with dificulty " + dificulty);
29     }
30 
31 }

輸出結果：

000

4973 in 106ms with dificulty 3

 

耗時計算：

讓難度從1遞增，每一個難度挖10次礦，取耗時最久的那次做爲參考。

package cn.ubingo.block_chain;

import java.util.Collections;
import java.util.Date;
import java.util.Random;

public class App {
    static Random r = new Random();

    public static void main(String[] args) {

        int dificulty = 0;

        while (true) {
            dificulty++;
            String zeros = String.join("", Collections.nCopies(dificulty, "0"));
            System.out.println(zeros);
            long maxUse = 0;
            long maxTimes = 0;
            for (int i = 0; i < 10; i++) {
                Date start = new Date();
                long times = 0;
                while (true) { String sha = Sha.getSHA256StrJava(r.nextInt() + ""); //System.out.println(sha);
                    times++; if (sha.startsWith(zeros)) break; }
                Date end = new Date();
                long use = end.getTime() - start.getTime();
                if (use > maxUse) {
                    maxUse = use;
                    maxTimes = times;
                }
            }
            System.out.println(maxTimes + " in " + maxUse + "ms with dificulty " + dificulty);
        }

    }
}

輸出結果（代碼還在跑）：

0
19 in 32ms with dificulty 1
00
366 in 11ms with dificulty 2
000
11377 in 73ms with dificulty 3
0000
153888 in 377ms with dificulty 4
00000
3160837 in 4568ms with dificulty 5
000000
55559940 in 81454ms with dificulty 6
0000000
637485941 in 977748ms with dificulty 7
00000000
8540898666 in 12283075ms with dificulty 8
000000000

===========================================================

從上面結果能夠看出，當難度爲8時，一我的挖，耗時204分鐘(12283075ms)左右。

因爲挖礦算法具備的隨機性，如要求在10分鐘內挖到礦，須要204/10=20個同等算力的礦工同時挖。

若是難度爲72（比特幣當前的難度），可想而知須要的礦工數量得是什麼級別。

因爲挖礦算法具備的隨機性，誰的算力強，誰的礦工多，誰挖到礦的概率就大，誰就能發大財。（因此礦機和礦場登場了）

===========================================================

代碼中用到的工具類以下（網上找的）：

package cn.ubingo.block_chain;

import java.io.UnsupportedEncodingException;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;

public class Sha {

    /**
     *  利用java原生的摘要實現SHA256摘要計算
     * @param content 原文
     * @return
     */
    public static String getSHA256StrJava(String content){
        MessageDigest messageDigest;
        String encodeStr = "";
        try {
            messageDigest = MessageDigest.getInstance("SHA-256");
            messageDigest.update(content.getBytes("UTF-8"));
            encodeStr = byte2Hex(messageDigest.digest());
        } catch (NoSuchAlgorithmException e) {
            e.printStackTrace();
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }
        return encodeStr;
    }

    /**
     * 將byte轉爲16進制
     * @param bytes
     * @return
     */
    private static String byte2Hex(byte[] bytes){
        StringBuffer stringBuffer = new StringBuffer();
        String temp = null;
        for (int i=0;i<bytes.length;i++){
            temp = Integer.toHexString(bytes[i] & 0xFF);
            if (temp.length()==1){
                //1獲得一位的進行補0操做
                stringBuffer.append("0");
            }
            stringBuffer.append(temp);
        }
        return stringBuffer.toString();
    }
    
}