The A5 Cracking Project
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Contents
1. LICENSE
GSM Software Project License Version 1, January 2007 All code, information or data [from now on "data"] available from the GSM Software Project or any other project linked from this or other pages is owned by the creator who created the data. The copyright, license right, distribution right and any other rights lies with the creator. It is prohibitied to use the data without the written agreement of the creator. This included using ideas in other projects (commercial or not commercial). Where data was created by more than 1 creator a written agreement from each of the creators has to be obtained. Please contact steve [at] segfault.net for any questions.
2. About
We are security enthusiasts. Our goal is to implement a system that can crack A5/1. Our results will be used with the GSM Software Project to demonstrate weaknesses in GSM. The A5 algorithm has been broken (in theory) in 1998 but it's still widely used. The mobile operators still insist that the GSM customers (that's you and me!) are protected and that our data is safe.
We want to bring together all the folks who worked on the theory of cracking A5/1.
Subscribe to our mailinglist by sending an email to a5subscribe [at] lists.segfault.net
3. How you can help
 Add links and information to this page or send them to steve at segfault.net
 Sponsor us! We need hardware, books and coffee!
 Come up with smart ideas.
4. TODO
 Come up with example data (e.g. first encrypted burst from BTS to MS and first burst from MS to BTS).
 Enhance the attack on A5/1
 Implement a A5/2 crack.
5. Requirements
The project comes in stages.
 Understand current state of A5/1 cracking (THAT'S WHERE WE ARE IN NOW!)
 Implement A5/2 crack (the weaker of both algorithms)
 Implement one of the many A5/1 cracks from the academic papers
 Research and Implement new ways to crack A5/1
Our ultimate goal is to crack A5/1:
 by only intercepting data (passiv)
 require less than 4Terabyte HD.
 able to decrypt short encrypted bursts (like SMS, last less than 0.1 seconds).
 Cracking time less than 1 day.
6. A5 weakness
A5 is weak. That's A5/1 and A5/2. When you look at the algorithm it just gives you a bad feeling.
 The registers are to small
 The trap registers are all on one side
 The 3 LSFRs do not mix results amoung each other
 Protocol implementation is faulty: An attacker can record all encrypted traffic. If the attacker gains access to the sim at any point in the future he can decrypt all traffic sniffed in the past. This works by putting the sim card into a sim reader and running the gsm_runalgorithm() function on the sim. The sim will decode any traffic without us knowing the Ki. This attack requires access to the sim for 30 seconds and can decode any GSM converstation that happened in the past.
 etc etc etc
I did a quick example to visualize the entroypy. Crypto people love entropy. An easy way to visualize the entropy is to generate a picture of the relationship between two, three or four successive numbers generated by the algorithm. Ideally we should not see any structure. All pixels should be distributed randomly. lcamtufs ISN analyzsis explains more details about this method.
I use a matlab script to generate the graphics. x.txt contains the output of the a5/1 key initialization algorithm.
a = 0; b = 0; c = 0; d = 0; XD = 256; YD = 256; ZD = 256; M = dlmread('x.txt', ' '); V = M(2,2) I(1:((XD  1) * 2), 1:((YD  1) * 2)) = 0; for i=1:25600 x = b  a; % 255 .. 255 y = c  b; % 255 .. 255 z = d  c; % 255 .. 255 I(x + XD, y + YD) = cast(z + (ZD  1), 'double') / ((ZD1) * 2); a = b; b = c; c = d; d = cast(mod(M(i, 1),256), 'int16'); % val between 0..255 %d = cast(rand(1,1) * XD, 'int16'); % val between 0..255 end imshow(I);
Figure 1: Key set to 0. FrameNumber runs from 025600. We can see a structure. There is a relationship between the key state with FrameNumber N and the key state with FrameNumber N  1.
TODO: add more.
7. A5/GSM encryption example
TODO: write down how a5 works and how the data looks that is encrypted and what the first encrypted message from/to basestation is and which bits are static/known/guessable.
The Frame Number (FN) wrapps around every 3h 28min 53 sec and 750ms.
A layer 1 GSM message is 23 octet long. It is padded with 0x2b if less than 23 octet content data are to be send.
How to encode 1 GSM message (after padding):
23 * 8 = 184 bit content data per GSM message. [Output: 184 bit]
Add 40 bit fire code (crc) and 4 bit tail (0x00). [Output: 228 bit]
Convolutional encode the 228 bit. This duplicates the number of (known) bits. [Output: 456 bit]
Interleave the 456 bit. [Output: 456 bit]
Chop the 456 bit into 8 packs, each 57 bit long. Take the first two 57 bit chunks and send them in the first GSM burst. The 3rd and 4th are send in the second GSM burst and so on and so on.[Output: 4x114 bit]
 The frame number is known and incremented for each GSM burst. A5 is reinitialized for _each_ burst. This means each burst is encoded under the same Kc but under a different frame number. The A5 state is thus different for each GSM burst.
First encrypted message send from MS to BTS is 'Ciphering Mode Complete':
000: ?? ?? ?? 06 32 2b 2b 2b  2b 2b 2b 2b 2b 2b 2b 2b 001: 2b 2b 2b 2b 2b 2b 2b 0: ?? 1 Extended Address: 1 octet long 0: ?? 0 C/R: Response 0: ?? 000 SAPI: RR, MM and CC 0: ?? 00 Link Protocol Disciminator: GSM (not Cell Broadcasting) 1: ?? 01 Supvervisory Frame 1: ?? 00 RR Frame (Receive ready) 1: ?? 0 Poll/Final bit (P/F) 1: ?? 000 N(R), Retransmission counter: 0 2: ?? 0 EL, Extended Length: n 2: ?? 0 M, segmentation: N 2: ?? 000010 Length: 2 3: 06 0 Direction: From originating site 3: 06 000 0 TransactionID 3: 06 0110 Radio Resouce Management 4: 32 00110010 RR Cipher Mode Complete
This message tells the BTS to start ciphering. The first encrypted message send from the BTS to the MS is either a MMIdentityRequest followed by a empty GSM message or a empty GSM message. Both of them contain plenty known plaintext: The 0x2b GSM message padding octet.
000: 03 42 0d 05 18 03 2b 2b  2b 2b 2b 2b 2b 2b 2b 2b 001: 2b 2b 2b 2b 2b 2b 2b 0: 03 1 Extended Address: 1 octet long 0: 03 1 C/R: Command 0: 03 000 SAPI: RR, MM and CC 0: 03 00 Link Protocol Disciminator: GSM (not Cell Broadcasting) 1: 42 0 Information Frame 1: 42 001 N(S), Sequence counter: 1 1: 42 0 P 1: 42 010 N(R), Retransmission counter: 2 2: 0d 1 EL, Extended Length: y 2: 0d 0 M, segmentation: N 2: 0d 000011 Length: 3 3: 05 0 Direction: From originating site 3: 05 000 0 TransactionID 3: 05 0101 Mobile Management Message (non GPRS) 4: 18 00 SendSequenceNumber: 0 4: 18 011000 MMIdentidyRequest 5: 03 011 IMEISV
or
000: 03 03 01 2b 2b 2b 2b 2b  2b 2b 2b 2b 2b 2b 2b 2b 001: 2b 2b 2b 2b 2b 2b 2b 0: 03 1 Extended Address: 1 octet long 0: 03 1 C/R: Command 0: 03 000 SAPI: RR, MM and CC 0: 03 00 Link Protocol Disciminator: GSM (not Cell Broadcasting) 1: 03 11 Unnumbered Frame 1: 03 0 P 1: 03 00000 UI frame (Unnumbered information) 2: 01 1 EL, Extended Length: y 2: 01 0 M, segmentation: N 2: 01 000000 Length: 0
8. Misc Ideas
 Shall we do a brute force with FPGA or do a smart attack as outlined in the 2001 paper?
 Can we use the weakness in A8/A3 to calculate Kc for A5/1?
 What happened to the cypherpunks mailinglist? The LNE links seem to be down! Anyone?
 I'm not concerned if we need 50 FPGA's or 4TB or harddrives. Some people say that it's not practical to carry 4TB of harddrives in a rucksack. We can always host the solution and when on a cracking mission the challenge can be send (via sms?) to the hosted Cracking Server which sends the results back after a couple of seconds.
 Can we devide the A5/1 cracking problems into smaller problems and solve each on its own? This means finding a new attack against A5/1.
8.1. FPGA Ideas
8.1.1. Brute Force
Some initial thoughts on A5/1 and FPGA. All this needs to be calculated more precisely.
Each clock cycle the A5 implementation should output 64 bit of streamcipher. We can put multiple A5 implementations on the same FPGA chip. The calculation is based on a pipelined implementation of A5.
The three LSFR registers are in total 19 + 22 + 23 = 64bit long. The first LSFR requires 5 Logical Units (LU's, e.g xor). The second requires 3 LU's and the last one requires 5 LU's. All together 13 LU's and 64 bit. The Trap register add's 1 LU per LSFR. Makes 16 LU's and 64bit.
Generating the state (with key and FrameNumber (FN)) requires 64 + 22 = 88 steps. This is followed by another 100 cycles. Each of the 100 cycles requires 1 LU less per LSFR. After these 100 cycles we want to generate about 64 bit of output (e.g. enother 64 cycles).
 LU's: 16 * 88 + 13 * 100 + 13 * 64 = 3540
 Registers: 64 * 88 + 64 * 100 + 64 * 64 = 16128
After 88 + 100 + 64 cycles we will start seeing 64 bit of stream cipher output for each cycle.
This is all not optimized. We do not need the first 9 steps because the Tap register only start at bit 8. we also do not need all the LU's or registers for the first 18 steps because the first LSFR is not fully used until step 18. Same for the last 64 steps. For each of the last 64 steps we only need 2 LU's and 1 register less for each step.
We decided to use Xilinx. Altera is a good choice as well but at the moment most of us worked with xilinx before.
The Virtex5 from Xilinx LX330 has 330.000 LU's and runs at 500 Mhz. That brings us down to 4 days per development board?! But the boards and chips are to expensive. Better to stick with LX50.
8.1.2. Brute Force II
Some more precise calculation by David Hulton:[[BR]] The LX50 can run at 200300Mhz and cost $300 each (just the chip, without dev board). I pipelined my version of A5/1 and came up with some rough numbers on the Virtex5 LX50. This is purely just computing the 186 clock cycles for setup and only computing a single bit of output from the pipeline on each clock cycle. I'm sure we could optimize it a little bit but once we factor in the overhead of doing the key compares and other bridge code it probably won't be much less than the numbers here..
With this design, we will probably only be able to fit 4 fully pipelined instances of A5/1 on here unless we can handoptimize the placement better than the Xilinx tools and code in some of the shortcuts that you mentioned on the a5 cracking page. I'll work on this a bit more and see if I can reduce the logic down.
Slice Logic Utilization: Number of Slice Registers: 7,289 out of 28,800 25% Number used as Flip Flops: 7,289 Number of Slice LUTs: 6,968 out of 28,800 24% Number used as logic: 6,566 out of 28,800 22% Number using O6 output only: 6,566 Number used as Memory: 402 out of 7,680 5% Number used as Shift Register: 402 Number using O6 output only: 402 Slice Logic Distribution: Number of occupied Slices: 2,670 out of 7,200 37% Number of LUT Flip Flop pairs used: 7,292 Number with an unused Flip Flop: 3 out of 7,292 1% Number with an unused LUT: 324 out of 7,292 4% Number of fully used LUTFF pairs: 6,965 out of 7,292 95% Number of unique control sets: 2 A LUT Flip Flop pair for this architecture represents one LUT paired with one Flip Flop within a slice. A control set is a unique combination of clock, reset, set, and enable signals for a registered element. The Slice Logic Distribution report is not meaningful if the design is overmapped for a nonslice resource or if Placement fails. IO Utilization: Number of bonded IOBs: 88 out of 220 40% Specific Feature Utilization: Number of BUFG/BUFGCTRLs: 1 out of 32 3% Number used as BUFGs: 1 Total equivalent gate count for design: 155,730 Additional JTAG gate count for IOBs: 4,224
8.1.3. possible boards
ML501 Xlinix LX50 ($955)
PicoComputing E16 LX50 ($2.000)
The LX330 boards cost $5.000. Because we can put 4x more a5/1 implementations on them and they run 6.6x faster it might be worth it.
8.2. Rainbow Table
Traditional rainbow tables take the key as input. Our key is 88 bit (of which the last 22 bit are the known Frame Number). We can not generate a rainbox table for 2^88 key combinations.
8.2.1. Idea I
The state table of all 3 LSFR's combined is just 64 bit. The A5 initialization process (e.g. seeding in key + FN and mixing it 100 cycles) is reverseable. Thus once we know the key state we can compute the key easily. Generating rainbow tables for 64 bit keys is difficult (TODO: calculate how difficult and how many FPGA's required).
This attack would work regardless of the frame number and regardless of the key length (54, 64 or 128 bit). It also uses less LU's than the normal key brute force implementation.
All 3 LSFR can be stuck together to get one 64bit register:  R1 19bit  R2 22bit  R3 23bit 
Rought idea of generating rainbow table with 2^36 tables:
 Start with key state bit 35..0 is set to 0000..001. Bit 63..36 is set to 0.
RainbowtableNumber++; Entries = 0;
 Calculate 64bit output from this keystate. Entries++;
 If output's bit 63..36 are all 0 then stop this rainbow table. Otherwise take 64 bit usefull output and use this output as state. Repeat 3.
 Increment value in bit 35..0 by 1 (e.g. start next rainbow table). Repeast 2.
Problems:
 What happens if we never hit a state that has bit 63..36 to all 0s (e.g. if we are stuck in a loop)? Break loop after a maximum number of iterations and call it an 'unlucky' rainbow table which is handles specially?
 Using bit 63..36 is just an example. In fact any number of bits (in sequence or not in sequence) can be used.
8.2.2. Idea II
(This Idea is now obsolete)
Maybe it's enough to generate a rainbow table for FrameNumber 0. Calculating all 2**54 keys with an FPGA and generating a rainbow tables is a matter of days (e.g. possible). Can a rainbow table generated with FrameNumber == 0 be used to decrypt packets that do not have Frame Number set to 0?
8.2.3. Idea III
Is it possible to reduce a LSFR register? By this i mean exist there a shorter LSFR register that would produce the same output (for a certain class of keys)?
8.2.4. Idea IV
We do not need to generate rainbow tables for all possible keystates. Let's assume we generate rainbow tables for 1/4 of all keystates (e.g 62bit). If we sniff 64 bit known plaintext our chances that we can crack it with the rainbow table is 25%.
A5 is reversable: Let N be the index of current working bit of the A5 algorithm (e.g. after N bits of output have been produced and N bit of plaintext have been encrypted). Let keystate(N) be the state of the keystate after N bits have been produced. Let plaintext(N) be the Nth bit of the plaintext. It is possible to calculate keystate(N1) if keystate(N) and plaintext(0..N) is known.
Let's assume we know 65 bit of plaintext. We first try to find a match in the rainbow table for bit 0..63 and then we try to find a match for bit 1..64. The probability for 65 bit known plaintext it is already 1  (3/4)**(65  64 + 1) = 43.75%. For 80 bit known plaintext it is 1  (3/4)**(80  64 + 1) = 98.997%.
Let's get this further down: Generate 1/64 of all rainbow tables (which makes it a 58bit problem): If we get 128 bit of known plaintext our chances of decoding it are 1  (63/64)**(128  64 + 1) == 64% or 95% if 256 bit of plaintext are known.
The maximum number of bits that are encrypted under the same keystate is 114. There are 4 bursts of 114 bit and the plaintext of each of the bursts is known. For each burst the propability of cracking it with only 1/64th of the rainbow table is:
1(63/64)^(114  64 + 1) = 55.2%
Considering that we have a 55.2% chance for each of the 4 burst:
1  (1  0.552)**4 = 95.97%
Limitation: It is obivous that this is working if we are dealing with successive bits of plaintext. It is less obvious that this also works as long as the 65 bit plaintext as distributed equaly (FIXME: can we optimize this?).
 Does NOT work: bit 0..63 in one sequence followed by some unknown plaintext followed by bit 64 of known plaintext.
 DOES work: plaintext bit 0 followed by 1 unknown plaintext bit followed by known plaintext bit 1, followed by unknown plaintext bit followed by known plaintext bit 2, ... until 64.
Further optimization:
 Do this over multiple messages (e.g. if we know 128 bit in the first packet and another 128 bit in the second message it dramaticaly increases our chances of finding the key state in one of our rainbow tables).
 Remember that for each message the BTS sends the MS also sends a message. Again, increasing our chances.
8.2.5. Idea V
We have known plaintext. The first encrypted message send from the BTS to the MS is amost all 0x2b (except for the first three octets). This means we can implement the attack by Anderson and Roe: Guessing the 41 bit in the shorter R1 and R2 registers, and deriving the 23bit of the longer R3 register from the output.
Anderson and Roe's attack is further described in A5/1 FPGA cracking.
Calculating Rainbow tables for this is the next challenge. Combing this with Idea IV makes it a 416 = 35 bit problem.
8.2.6. Idea VI
Are there 'useless' bits in R2? It only has two trap registers. Does this help us calculating the value of others?
9. Resources
CS200607crackinga5.pdf Barkan, Biham and Keller. Most recent research paper about cracking A5/1.
PHD200604.pdf Elad Pinhas Barkan, Cryptoanalyzis of Ciphers (A5, Rainbow tables)
GsmSecurity.pdf 15 Dec 2006, Stausholm, Dahl. Explaining A5 and different attack vectors.
2000, Biryukov, Shamir, Wagner (WWW). (PDF) Real Time Cryptanalysis of A5/1 on a PC.
ekdahl03a51a.pdf Different Attack. Requires 25 mins of data. Not practical but good A5 explanation.
Ross Anderson original email posting.
a512.c Most recent A5/1 and A5/2 implementation by Marc Briceno.
a3a8.txt A3 and A8 implementation by Briceno, Goldberg and Wagner.
9.1. List of used encryption around the World
Known GSM Netowrk Encryption usage
Version 1.12 8th December 2005
gsm_network_encryption_list.csv
If you have updates (what about France??) please send an email to steve at segfault.net.
MCC 
Country 
MNC 
Network 
Crypto 
Date & City 
Comments 
204 
Netherlands 
4 
Vodafone 
A5/1 


204 
Netherlands 
8 
KPN 
A5/1 


204 
Netherlands 
16 
TMobile 
A5/1 


204 
Netherlands 
12 
O2 
A5/1 


204 
Netherlands 
20 
Orange 
A5/1 


206 
Belgium 
1 
Proximus 
A5/1 


206 
Belgium 
10 
Mobilstar 
A5/1 


206 
Belgium 
20 
Base 
A5/1 


208 
France 
10F 
SFR 
A5/1, A5/0 
20070525 Grenoble 
A5/1 for TCH, A5/0 for SMS 
214 
Spain 
1 
Vodafone 
A5/1 


214 
Spain 
3 
Amena 
A5/1 


214 
Spain 
7 
Movistar 
A5/1 


222 
Italy 
1 
TIM 
A5/1 


222 
Italy 
10 
Vodafone 
A5/1 


234 
United Kingdom 
10 
O2 
A5/1 


234 
United Kingdom 
15 
Vodafone 
A5/1 


234 
United Kingdom 
30 
TMobile 
A5/1 


234 
United Kingdom 
33 
Orange 
A5/1 


238 
Denmark 
1 
TDC 
A5/1 


242 
Norway 
1 
Telenor Mobil 
A5/1 


242 
Norway 
2 
Netcom 
A5/1 


250 
Russia 
1 
MTS 
A5/1 


250 
Russia 
2 
Megafon 
A5/1 


250 
Russia 
99 
Beeline 
A5/1 


262 
Germany 
2 
Vodafone 
A5/1 


262 
Germany 
3 
Eplus 
A5/1 


262 
Germany 
7 
O2 
A5/1 


272 
Ireland 
2 
O2 
A5/1 


293 
Slovenia 
40 
SI Mobil Vodafone 
A5/2 


293 
Slovenia 
41 
SI Mobitel GSM 
A5/1 


293 
Slovenia 
70 
Vega 
A5/1 


404 
India 
4 
IDEA 
A5/0 


404 
India 
10 
A5/0 



404 
India 
11 
Essar 
A5/0 


404 
India 
20 
Orange 
A5/0 


404 
India 
68 
Dolphin 
A5/0 


424 
United Arab Emirates 
1 
Etisalat 
A5/1 


505 
Australia 
1 
Telstra 
A5/1 


505 
Australia 
2 
Optus 
A5/1 


505 
Australia 
3 
Vodafone 
A5/1 


515 
Philippines 
2 
Globe 
A5/1 


515 
Philippines 
3 
Smart 
A5/1 


515 
Philippines 
5 
Sun 
A5/1 


639 
Kenya 
2 
Safaricom 
A5/2 


639 
Kenya 
3 
Celtel 
A5/2 


Converting the CSV to wiki table:
cat gsm_network_encryption_list.csv  sed 's/"//g'  while read x; do echo "`echo "$x"  sed 's/,//g'`"; done
History: When A5/1 came out mostly germany (as the bordering country to the soviet block) wanted to implement strong encryption. Other Nato members (led by france) were worried that the middle east would use strong encryption. Thus they cut a deal to come up with a weaker version, A5/2. These days both (A5/1 and A5/2) have been broken. A5/3 has not been seen in the wild yet.
Other comments:
 No encryption in Russia/Ukraine, during emergencies (which can last weeks!)
 No encryption if BTS is under load (can somebody confirm??)
 No encryption in germany during HLR/VLR outages
 In some arab countries without reason some areas without encryption.
 SMS are sometimes unencrypted even when TCH is encrypted.
9.2. How to check if A5/1 is used
There are two ways. You can either use Nokia's Netmonitor (aka Field Tester) or you can use any dct3 mobile (like the nokia 3310) and gammu + PC to find out. The netmonitor is the easier way because you do not need a PC. The netmonitor software runs on many famous mobiles phones (nokia 6630, 6680, n70, sony erricson, ..)
Make sure your phone is using GSM (and not 3G/UMTS or DUAL). Go to Menu > Tools > Settings > Network > Network mode and switch to GSM.
 Install the netmonitor by connecting your phone to the PC (via usb cable).
 Launch netmonitor
 Go to screen 1.10. Send a SMS to the phone. See if the 'Ciphering val' changes from OFF to something else.
 Go to screen 1.10. Call the mobile phone. See if the 'Ciphering val' changes from OFF to something else.
 Send an email to steve [at] segfault dot net including the country, mobile operator and cipher used (See example results below).
Example how it looks like:
Results of this example:
 Date: 2007/05/25 09:32
 Country Code: 234
 Network Code: 10F
 Location area: 12124 (central london)
 A51 when receiving SMS
 A51 when receiving voice call
 Hopping: On
The other method is by using gammu and a dct3 trace mobile (like the nokia 3310) connected to the PC. Start a trace, make a phonecall and send in the out.xml file that gammu produces. See our main project page on how to use gammu and dct3 trace mobiles. Check the GSMSP Project for more infos on how to use gammu.