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The Next Hope

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This was my first hope conference (The Next HOPE Conference)despite being in New York City for more than half a decade. Always it seemed that work would send me out of town just before the con. However, this time around I had the good fortune of being in the city during the conference.
There were a few good talks some of which were not so technical but kindled the questions for privacy fanatics.
The talks I attended included Alessio Pennasilico’s talk about DDoS attack on Bakeca.it, Modern Crimeware and Tools talk by Alexander Heid, Steven Rambam’s talk on Privacy is Dead, Blaze Mouse Cheswick et. al’s talk which was abstract but awesome. I did attend a few more talks and it was fun. All in all a great conference.

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PCI SSC Forbids SSL and “Early TLS”

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On April 15, 2015 the PCI SSC released the PCI DSS v3.1.  The main cause for concern for most merchants and other entities (called “entities” hereonforth) that store, transmit and process cardholder data is the prohibition of using SSL and “Early TLS”.  The PCI SSC also released a supplement to assist entities in mitigating the issue.   The supplement references the NIST guideline SP800-52 rev1 for determining which are good ciphers and which are not.

The key point being what does “Early TLS” mean?  Does it mean TLSv1.0 and TLSv1.1 OR does it mean only TLSv1.0?  Are the entities supposed to disable all ciphers except anything that’s TLSv1.2?

Answer is (in consultant speak) “it depends”. 🙂

TLSv1.1 does theoretically have ciphers that are not ideal.  Example: CBC mode ciphers that are TLSv1.1 but there may be a potential for attacks on them given that in the past couple of years CBC has fallen multiple times (BEAST, POODLE).

Google Chrome lists the use of CBC-based ciphers (despite the fact that they’re TLSv1.1) to be obsolete.  Google Chrome essentially makes “obsolete cryptography” a function of using TLS v1.2-based ciphers.

Untitled2

Firefox allows the configuration of disabling TLSv1.0 and that can be done by typing “about:config” in the address bar.  The security.tls.version.min = 0 (means SSLv3), 1 (means TLSv1.0), 2 (means TLSv1.1) and 3 (means TLSv1.2).  The following screenshot shows the configuration snapshot (here the lowest allowed version is TLSv1.0).

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Let’s start with what is definitely ok for PCI:

https://www.openssl.org/docs/apps/ciphers.html#TLS-v1.2-cipher-suites

 TLS_RSA_WITH_NULL_SHA256                  NULL-SHA256
 TLS_RSA_WITH_AES_128_CBC_SHA256           AES128-SHA256
 TLS_RSA_WITH_AES_256_CBC_SHA256           AES256-SHA256
 TLS_RSA_WITH_AES_128_GCM_SHA256           AES128-GCM-SHA256
 TLS_RSA_WITH_AES_256_GCM_SHA384           AES256-GCM-SHA384

 TLS_DH_RSA_WITH_AES_128_CBC_SHA256        DH-RSA-AES128-SHA256
 TLS_DH_RSA_WITH_AES_256_CBC_SHA256        DH-RSA-AES256-SHA256
 TLS_DH_RSA_WITH_AES_128_GCM_SHA256        DH-RSA-AES128-GCM-SHA256
 TLS_DH_RSA_WITH_AES_256_GCM_SHA384        DH-RSA-AES256-GCM-SHA384

 TLS_DH_DSS_WITH_AES_128_CBC_SHA256        DH-DSS-AES128-SHA256
 TLS_DH_DSS_WITH_AES_256_CBC_SHA256        DH-DSS-AES256-SHA256
 TLS_DH_DSS_WITH_AES_128_GCM_SHA256        DH-DSS-AES128-GCM-SHA256
 TLS_DH_DSS_WITH_AES_256_GCM_SHA384        DH-DSS-AES256-GCM-SHA384

 TLS_DHE_RSA_WITH_AES_128_CBC_SHA256       DHE-RSA-AES128-SHA256
 TLS_DHE_RSA_WITH_AES_256_CBC_SHA256       DHE-RSA-AES256-SHA256
 TLS_DHE_RSA_WITH_AES_128_GCM_SHA256       DHE-RSA-AES128-GCM-SHA256
 TLS_DHE_RSA_WITH_AES_256_GCM_SHA384       DHE-RSA-AES256-GCM-SHA384

 TLS_DHE_DSS_WITH_AES_128_CBC_SHA256       DHE-DSS-AES128-SHA256
 TLS_DHE_DSS_WITH_AES_256_CBC_SHA256       DHE-DSS-AES256-SHA256
 TLS_DHE_DSS_WITH_AES_128_GCM_SHA256       DHE-DSS-AES128-GCM-SHA256
 TLS_DHE_DSS_WITH_AES_256_GCM_SHA384       DHE-DSS-AES256-GCM-SHA384

 TLS_ECDH_RSA_WITH_AES_128_CBC_SHA256      ECDH-RSA-AES128-SHA256
 TLS_ECDH_RSA_WITH_AES_256_CBC_SHA384      ECDH-RSA-AES256-SHA384
 TLS_ECDH_RSA_WITH_AES_128_GCM_SHA256      ECDH-RSA-AES128-GCM-SHA256
 TLS_ECDH_RSA_WITH_AES_256_GCM_SHA384      ECDH-RSA-AES256-GCM-SHA384

 TLS_ECDH_ECDSA_WITH_AES_128_CBC_SHA256    ECDH-ECDSA-AES128-SHA256
 TLS_ECDH_ECDSA_WITH_AES_256_CBC_SHA384    ECDH-ECDSA-AES256-SHA384
 TLS_ECDH_ECDSA_WITH_AES_128_GCM_SHA256    ECDH-ECDSA-AES128-GCM-SHA256
 TLS_ECDH_ECDSA_WITH_AES_256_GCM_SHA384    ECDH-ECDSA-AES256-GCM-SHA384

 TLS_ECDHE_RSA_WITH_AES_128_CBC_SHA256     ECDHE-RSA-AES128-SHA256
 TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA384     ECDHE-RSA-AES256-SHA384
 TLS_ECDHE_RSA_WITH_AES_128_GCM_SHA256     ECDHE-RSA-AES128-GCM-SHA256
 TLS_ECDHE_RSA_WITH_AES_256_GCM_SHA384     ECDHE-RSA-AES256-GCM-SHA384

 TLS_ECDHE_ECDSA_WITH_AES_128_CBC_SHA256   ECDHE-ECDSA-AES128-SHA256
 TLS_ECDHE_ECDSA_WITH_AES_256_CBC_SHA384   ECDHE-ECDSA-AES256-SHA384
 TLS_ECDHE_ECDSA_WITH_AES_128_GCM_SHA256   ECDHE-ECDSA-AES128-GCM-SHA256
 TLS_ECDHE_ECDSA_WITH_AES_256_GCM_SHA384   ECDHE-ECDSA-AES256-GCM-SHA384

 TLS_DH_anon_WITH_AES_128_CBC_SHA256       ADH-AES128-SHA256
 TLS_DH_anon_WITH_AES_256_CBC_SHA256       ADH-AES256-SHA256
 TLS_DH_anon_WITH_AES_128_GCM_SHA256       ADH-AES128-GCM-SHA256
 TLS_DH_anon_WITH_AES_256_GCM_SHA384       ADH-AES256-GCM-SHA384
 TLS_ECDHE_ECDSA_WITH_CAMELLIA_128_CBC_SHA256 ECDHE-ECDSA-CAMELLIA128-SHA256
 TLS_ECDHE_ECDSA_WITH_CAMELLIA_256_CBC_SHA384 ECDHE-ECDSA-CAMELLIA256-SHA384
 TLS_ECDH_ECDSA_WITH_CAMELLIA_128_CBC_SHA256  ECDH-ECDSA-CAMELLIA128-SHA256
 TLS_ECDH_ECDSA_WITH_CAMELLIA_256_CBC_SHA384  ECDH-ECDSA-CAMELLIA256-SHA384
 TLS_ECDHE_RSA_WITH_CAMELLIA_128_CBC_SHA256   ECDHE-RSA-CAMELLIA128-SHA256
 TLS_ECDHE_RSA_WITH_CAMELLIA_256_CBC_SHA384   ECDHE-RSA-CAMELLIA256-SHA384
 TLS_ECDH_RSA_WITH_CAMELLIA_128_CBC_SHA256    ECDH-RSA-CAMELLIA128-SHA256
 TLS_ECDH_RSA_WITH_CAMELLIA_256_CBC_SHA384    ECDH-RSA-CAMELLIA256-SHA384

Now let’s see what may potentially be good from TLSv1.1 perspective (from NIST SP8000-52 rev1):

TLS_RSA_WITH_3DES_EDE_CBC_SHA           DES-CBC3-SHA
TLS_RSA_WITH_AES_128_CBC_SHA            AES128-SHA

Here’s a problem though per OpenSSL man page:
Untitled

If you’re using OpenSSL, how do you ensure that the browser is not negotiating the vulnerable TLSv1.0 ciphers? The only real answer seems to be by providing a cipher order for negotiation and hoping the client doesn’t cheat.  Most likely, the browser will negotiate a better cipher when it exists in the server and on the client and you’d avert the possibility of negotiation of a bad cipher.

According to experts, anything that uses CBC is inherently broken.  But disabling TLSv1.0 may make the server inaccessible to various older Android devices.  Also, if you’re using older Java Development Kits (JDK7 and below), do remember that the default ciphers may not hit the spot for PCI.

There’s an excellent site to help you configure each type of the server so you could become PCI compliant. This is an excellent site by Ivan Ristic to test your Internet-facing servers for configuration of SSL/TLS encryption.

In conclusion, configure browsers to minimally allow TLSv1.1 and configure servers to use TLSv1.2 to be PCI DSS compliant.  The road to TLSv1.1 compatibility and PCI DSS is filled with potholes and death-falls so do it at your own risk.

8

Certified Reverse Engineering Analyst Certified

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This week I got the CREA certification to add to my list of CISSP, CEPT, Visa QSA. This certification required a good practical and conceptual knowledge of reverse engineering. The certification requires a good working knowledge of components such as IA-32 assembly language, malware reversing, expert level knowledge of IDA Pro, OllyDbg, HiEW, Dumpbin etc., PE File header, repairing packed and compacted binaries, using system level reversing etc. The exam was good and tested on the concepts of the reverse engineer.

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VMWare snapshots issue

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VMWare is excellent for malware analysts because it lets you keep snapshots of pristine Virtual machine states and you can revert back to them when you want to.
I encountered a weired error this time around on my Windows XP Pro VM. Whenever I would try to take any snapshots I would get an error: “Error taking snapshot: Windows XP Professional.VMX-Snapshot1.vmsn file already exists”. When I looked into the folder there was no .vmsn file with that name. I deleted all the files .lck and .lock files and still to no avail. Then I saw the files named as
Windows XP Professional-000001-s00?.VMDK.
The regex for these files was:
Windows XP Professional-00000?-s00?.VMDK
where ? is one character replaced by 0-9. Upon deleting these files, my snapshots started working properly.

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GooScan compilation errors

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I was just browsing away when I stumbled upon Johnny Long’s GooScan. He says that this is a Linux only tool but it seems to compile (not without problems though) on cygwin.
I kept getting the following errors:


L:\tools\gooscan-v1.0.9>gcc gooscan.c
gooscan.c: In function `inet_send':
gooscan.c:575: error: `MSG_WAITALL' undeclared (first
use in this function)
gooscan.c:575: error: (Each undeclared identifier is
reported only once
gooscan.c:575: error: for each function it appears in.)

Then I read somewhere that MSG_WAITALL is not defined for Cygwin and that instead of that zero would work. There are many neater solutions to this…but I’m a hacker and I’ll do the stuff that’s easiest and hassle-free.
Some people say that the following will work:
#ifdef __CYGWIN__
#define MSG_WAITALL 0

So in order to compile this bad boy, you need to goto line 574 in your favorite editor.
It looks like this:
recv(sock, recvbuf, sizeof(recvbuf), MSG_WAITALL);

You need to make it look like this:
recv(sock, recvbuf, sizeof(recvbuf), 0);//MSG_WAITALL);

You are all set:
gcc gooscan.c -o gooscan.exe

Compilation works! But then I observed that the results were not coming well. However, if you run it through a local proxy such as burp it still works…I bet it has something to do with socket establishment and receiving and being incompatible with the MSG_WAITALL flag.
But as long as you can get the results … who cares? If someone figures out exactly how to make this work, please post it as a comment.

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DefCon CTF Quals GrabBag 300 Writeup

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The question was:
Question: This is semi-real. 🙁
140.197.217.85:10435
Password: 5fd78efc6620f6

When you would connect using netcat you would see a 9 numbers and a user PIN. This would repeat thrice and then you would have to choose the right pin for the fourth pair 6×6 matrix of numbers. My first reaction was either the PINS were constant or they were following a pattern. So I wrote up this quick python script to solve this puzzle which helped me understand the problem also.

#!/usr/bin/python
import socket, re, threading, time
 
lookupdict = []

def process_array_pin(fs,s):
	i = 6
	temp = ''
	pin = ''
	while i > 0:
		line = fs.readline()
		#print line
		#re.match(".{11}(.).{12}(.).{12}(.)", line).group(1)
		test = re.split(' ',line)
		#print test[1],' ',test[3],' ',test[5],' ',test[7],' ',test[9],' ',test[11]
		i = i - 1
		try:
			temp += test[1]+test[3]+test[5]+test[7]+test[9]+test[11]
		except IndexError:
			pass
			#i = 15
			#while i > 0:
			#	print fs.readline()
			#	i = i - 1
			#s.send('2\n')
			#i = 15
			#while i > 0:
			#	print fs.readline()
			#	i = i - 1
	line = fs.readline()
	try:
		pin = re.match("..........User entered: (.*)", line).group(1)
	except:
		pass
	#pin = fs.readline()
	#print 'Line: '+line
	#print 'Pin is : '+pin
	strpin = re.sub(' ','',pin)
	#strpin = re.split(' ',pin)
	#lookupdict[temp] = strpin
	print 'Pin for : ' + temp+' is '+strpin+'\n'
	return temp,strpin
def play():
	global fs, s
	s = socket.create_connection(('140.197.217.85', 10435))
	fs = s.makefile()
	s.send('5fd78efc6620f6\n')
	print fs.readline()
	print fs.readline()
	print fs.readline()
	answer = []
	numTimes = 0
	while numTimes < 5:
		j = 3
		while j > 0:
			test = process_array_pin(fs,s)
			lookupdict.append(test[0])
			lookupdict.append(test[1])
			j = j - 1
			if j > 0:
				numlines = 3
				while numlines > 0:
					fs.readline()
					numlines = numlines - 1
		fs.readline()
		pindigits = list(lookupdict[1])
		#print pindigits
		pinpos = 0
		for num in pindigits:
			i = 0
			start = 0
			end = len(lookupdict[0])
			while i < lookupdict[0].count(num):
				indofinterest = lookupdict[0].find(num,start,end)
				#print 'index of interest '+str(indofinterest)
				if lookupdict[2][indofinterest] == lookupdict[3][pinpos]:
					if lookupdict[4][indofinterest] == lookupdict[5][pinpos]:
						answer.append(indofinterest)
						break
				i = i + 1
				start = indofinterest+1
			pinpos = pinpos + 1
		#print answer
		# Get question
		i = 6
		temp1 = ""
		while i > 0:
			line = fs.readline()
			#print line
			#re.match(".{11}(.).{12}(.).{12}(.)", line).group(1)
			test = re.split(' ',line)
			#print test[1],' ',test[3],' ',test[5],' ',test[7],' ',test[9],' ',test[11]
			temp1 += test[1]+test[3]+test[5]+test[7]+test[9]+test[11]
			i = i - 1
		#fs.read(14)
		#fs.flush()
		print "Question : " +temp1+'\n'
		answerstr = ''
		count = 0
		for i in answer:
			answerstr += temp1[i]
			#print temp1[i],
			count = count + 1
			if count < 4:
				answerstr += ' '
			else:
				answerstr += '\n'
		print "Answer : "+answerstr
		s.send(answerstr)
		output = fs.readline()
		#output = fs.readline()
		print output
		if output.find('Sun') > -1:
			output = fs.readline()
		else:
			a = 10
			while a > 0:
				print fs.readline()
				a = a - 1
			#output = fs.readline()
			#print 'Inside else\n'
			#if output.find('NOVA') > -1:
			#	print 'NOVAFOUND!!!!!\n'
			s.send('2\n')
			print 'Sent last\n'
			a = 100
			while a > 0:
				print fs.readline()
				s.send('%d%n\n')
				a = a - 1
			#print fs.readline()
			break
		del answer[:]
		del lookupdict[:]
		del pindigits[:]
		numTimes += 1
	
	s.close()
#for i in range(2000):
#threading.Thread(target=play).start()
play()

The above file reads the numbers, filters out the formatting that adds color to the digits and picks out the indices that would be chosen as the key.

So to solve this, each pattern of digits had fixed matrix positions that would be chosen as the pin. Once you successfully solve the puzzle four time you are presented with an ATM screen as follows:

 ***NOVABANK ATM menu***

 Balance: $9238740982570237012935.32

 1) withdraw
 2) deposit
 3) transfer
 4) exit

 <disconnected>

The real part is the balance i.e., 9238740982570237012935.32 is the answer. It took me various attempts to solve this one because the answer was for some reason not being accepted by the scoreboard until my teammate submitted it at which time it worked.

This was a really cool problem. Thanks DDTEK.

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ERROR 1045 (28000): Access denied for user ‘root’@’localhost’ (using password: NO)

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If this is the error you are getting then one of the solutions is to reset your root password on the MySQL database server.

$ pkill mysql
$ sudo mysqld --skip-grant-privileges
$ mysql

At this point you get the mysql command shell. You will need to update the root password and flush the table when you reset the password.

mysql> set UPDATE mysql.user SET Password=PASSWORD('YOUR_NEW_PASSWORD') WHERE User='root';
mysql> FLUSH PRIVILEGES;

Now that you’ve flushed your passwords, just restart your mysql daemon.

$ sudo pkill mysqld
$ sudo /etc/init.d/mysqld start
$ mysql -u root -p
Enter Password: YOUR_NEW_PASSWORD
mysql>

You should be all set now!