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    Breaking VISA PIN

    Below is an article I found recently. This one of the most comprehensive descriptions of PIN Verification Value (PVV) hacking.

    I thought I would replicate it here for my local reference.

    As comments have been made regarding the grammar used in the original text, I have corrected some of the obvious errors whilst maintaining the context of the original material.

    ——– Original Text ———-

    Have you ever wonder what would happen if you lose your credit or debit card and someone finds it. Would this person be able to withdraw cash from an ATM guessing, somehow, your PIN? Moreover, if you were who finds someone’s card would you try to guess the PIN and take the chance to get some easy money? Of course the answer to both questions should be “no”. This work does not deal with the second question, it is a matter of personal ethics. Herewith I try to answer the first question.

    All the information used for this work is public and can be freely found in Internet. The rest is a matter of mathematics and programming, thus we can learn something and have some fun. I reveal no secrets. Furthermore, the aim (and final conclusion) of this work is to demonstrate that PIN algorithms are still strong enough to provide sufficient security. We all know technology is not the weak point.

    This work analyses one of the most common PIN algorithms, VISA PVV, used by many ATM cards (credit and debit cards) and tries to find out how resistant is to PIN guessing attacks. By “guessing” I do not mean choosing a random PIN and trying it in an ATM. It is well known that generally we are given three consecutive trials to enter the right PIN, if we fail ATM keeps the card. As VISA PIN is four digit long it’s easy to deduce that the chance for a random PIN guessing is 3/10000 = 0.0003, it seems low enough to be safe; it means you need to lose your card more than three thousand times (or losing more than three thousand cards at the same time 🙂 until there is a reasonable chance of losing money.

    What I really meant by “guessing” was breaking the PIN algorithm so that given any card you can immediately know the associated PIN. Therefore this document studies that possibility, analyzing the algorithm and proposing a method for the attack. Finally we give a tool which implements the attack and present results about the estimated chance to break the system. Note that as long as other banking security related algorithms (other PIN formats such as IBM PIN or card validation signatures such as CVV or CVC) are similar to VISA PIN, the same analysis can be done yielding nearly the same results and conclusions.

    VISA PVV algorithm

    One of the most common PIN algorithms is the VISA PIN Verification Value (PVV). The customer is given a PIN and a magnetic stripe card. Encoded in the magnetic stripe is a four digit number, called PVV. This number is a cryptographic signature of the PIN and other data related to the card. When a user enters his/her PIN the ATM reads the magnetic stripe, encrypts and sends all this information to a central computer. There a trial PVV is computed using the customer entered PIN and the card information with a cryptographic algorithm. The trial PVV is compared with the PVV stored in the card, if they match the central computer returns to the ATM authorization for the transaction. See in more detail.

    The description of the PVV algorithm can be found in two documents linked in the previous page. In summary it consists in the encryption of a 8 byte (64 bit) string of data, called Transformed Security Parameter (TSP), with DES algorithm (DEA) in Electronic Code Book mode (ECB) using a secret 64 bit key. The PVV is derived from the output of the encryption process, which is a 8 byte string. The four digits of the PVV (from left to right) correspond to the first four decimal digits (from left to right) of the output from DES when considered as a 16 hexadecimal character (16 x 4 bit = 64 bit) string. If there are no four decimal digits among the 16 hexadecimal characters then the PVV is completed taken (from left to right) non decimal characters and decimalizing them by using the conversion A->0, B->1, C->2, D->3, E->4, F->5. Here is an example:

    Output from DES: 0FAB9CDEFFE7DCBA

    PVV: 0975

    The strategy of avoiding decimalization by skipping characters until four decimal digits are found (which happens to be nearly all the times as we will see below) is very clever because it avoids an important bias in the distribution of digits which has been proven to be fatal for other systems, although the impact on this system would be much lower. See also a related problem not applying to VISA PVV.

    The TSP, seen as a 16 hexadecimal character (64 bit) string, is formed (from left to right) with the 11 rightmost digits of the PAN (card number) excluding the last digit (check digit), one digit from 1 to 6 which selects the secret encrypting key and finally the four digits of the PIN. Here is an example:

    PAN: 1234 5678 9012 3445
    Key selector: 1
    PIN: 2468

    TSP: 5678901234412468

    Obviously the problem of breaking VISA PIN consists in finding the secret encrypting key for DES. The method for that is to do a brute force search of the key space. Note that this is not the only method, one could try to find a weakness in DEA, many tried, but this old standard is still in wide use (now been replaced by AES and RSA, though). This demonstrates it is robust enough so that brute force is the only viable method (there are some better attacks but not practical in our case, for a summary see LASEC memo and for the dirty details see Biham & Shamir 1990, Biham & Shamir 1991, Matsui 1993, Biham & Biryukov 1994 and Heys 2001).

    The key selector digit was very likely introduced to cover the possibility of a key compromise. In that case they just have to issue new cards using another key selector. Older cards can be substituted with new ones or simply the ATM can transparently write a new PVV (corresponding to the new key and keeping the same PIN) next time the customer uses his/her card. For the shake of security all users should be asked to change their PINs, however it would be embarrassing for the bank to explain the reason, so very likely they would not make such request.

    Preparing the attack

    A brute force attack consists in encrypting a TSP with known PVV using all possible encrypting keys and compare each obtained PVV with the known PVV. When a match is found we have a candidate key. But how many keys we have to try? As we said above the key is 64 bit long, this would mean we have to try 2^64 keys. However this is not true. Actually only 56 bits are effective in DES keys because one bit (the least significant) out of each octet was historically reserved as a checksum for the others; in practice those 8 bits (one for each of the 8 octets) are ignored.

    Therefore the DES key space consists of 2^56 keys. If we try all these keys will we find one and only one match, corresponding to the bank secret key? Certainly not. We will obtain many matching keys. This is because the PVV is only a small part (one fourth) of the DES output. Furthermore the PVV is degenerated because some of the digits (those between 0 and 5 after the last, seen from left to right, digit between 6 and 9) may come from a decimal digit or from a decimalized hexadecimal digit of the DES output. Thus many keys will produce a DES output which yields to the same matching PVV.

    Then what can we do to find the real key among those other false positive keys? Simply we have to encrypt a second different TSP, also with known PVV, but using only the candidate keys which gave a positive matching with the first TSP-PVV pair. However there is no guarantee we won’t get again many false positives along with the true key. If so, we will need a third TSP-PVV pair, repeat the process and so on.

    Before we start our attack we have to know how many TSP-PVV pairs we will need. For that we have to calculate the probability for a random DES output to yield a matching PVV just by chance. There are several ways to calculate this number and here I will use a simple approach easy to understand but which requires some background in mathematics of probability.

    A probability can always be seen as the ratio of favorable cases to possible cases. In our problem the number of possible cases is given by the permutation of 16 elements (the 0 to F hexadecimal digits) in a group of 16 of them (the 16 hexadecimal digits of the DES output). This is given by 16^16 ~ 1.8 * 10^19 which of course coincides with 2^64 (different numbers of 64 bits). This set of numbers can be separated into five categories:

    Those with at least four decimal digits (0 to 9) among the 16 hexadecimal digits (0 to F) of the DES output.

    Those with exactly only three decimal digits.

    Those with exactly only two decimal digits.

    Those with exactly only one decimal digit.

    Those with no decimal digits (all between A and F).

    Let’s calculate how many numbers fall in each category. If we label the 16 hexadecimal digits of the DES output as X1 to X16 then we can label the first four decimal digits of any given number of the first category as Xi, Xj, Xk and Xl. The number of different combinations with this profile is given by the product 6 i-1 * 10 * 6j-i-1 * 10 * 6k-j-1 * 10 * 6 l-k-1 * 10 * 1616-l where the 6’s come from the number of possibilities for an A to F digit, the 10’s come from the possibilities for a 0 to 9 digit, and the 16 comes from the possibilities for a 0 to F digit. Now the total numbers in the first category is simply given by the summation of this product over i, j, k, l from 1 to 16 but with i < j < k < l. If you do some math work you will see this equals to the product of 104/6 with the summation over i from 4 to 16 of (i-1) * (i-2) * (i-3) * 6i-4 * 16 16-i ~ 1.8 * 1019.

    Analogously the number of cases in the second category is given by the summation over i, j, k from 1 to 16 with i < j < k of the product 6i-1 * 10 * 6j-i-1 * 10 * 6k-j-1 * 10 * 616-k which you can work it out to be 16!/(3! * (16-13)!) * 103 * 6 13 = 16 * 15 * 14/(3 * 2) * 103 * 613 = 56 * 104 * 613 ~ 7.3 * 1015. Similarly for the third category we have the summation over i, j from 1 to 16 with i < j of 6 i-1 * 10 * 6j-i-1 * 10 * 616-j which equals to 16!/(2! * (16-14)!) * 102 * 614 = 2 * 103 * 615 ~ 9.4 * 1014. Again, for the fourth category we have the summation over i from 1 to 16 of 6i-1 * 10 * 616-i = 160 * 615 ~ 7.5 * 1013. And finally the amount of cases in the fifth category is given by the permutation of six elements (A to F digits) in a group of 16, that is, 616 ~ 2.8 * 1012.

    I hope you followed the calculations up to this point, the hard part is done. Now as a proof that everything is right you can sum the number of cases in the 5 categories and see it equals the total number of possible cases we calculated before. Do the operations using 64 bit numbers or rounding (for floats) or overflow (for integers) errors won’t let you get the exact result.

    Up to now we have calculated the number of possible cases in each of the five categories, but we are interested in obtaining the number of favorable cases instead. It is very easy to derive the latter from the former as this is just fixing the combination of the four decimal digits (or the required hexadecimal digits if there are no four decimal digits) of the PVV instead of letting them free. In practice this means turning the 10’s in the formula above into 1’s and the required amount of 6’s into 1’s if there are no four decimal digits. That is, we have to divide the first result by 104, the second one by 103 * 6, the third one by 102 * 62 , the fourth one by 10 * 63 and the fifth one by 64 . Then the number of favorable cases in the five categories are approximately 1.8 * 1015, 1.2 * 1012, 2.6 * 1011 , 3.5 * 1010, 2.2 * 109 respectively.

    Now we are able to obtain what is the probability for a DES output to match a PVV by chance. We just have to add the five numbers of favorable cases and divide it by the total number of possible cases. Doing this we obtain that the probability is very approximately 0.0001 or one out of ten thousand. Is it strange this well rounded result? Not at all, just have a look at the numbers we calculated above. The first category dominates by several orders of magnitude the number of favorable and possible cases. This is rather intuitive as it seems clear that it is very unlikely not having four decimal digits (10 chances out of 16 per digit) among 16 hexadecimal digits. We saw previously that the relationship between the number of possible and favorable cases in the first category was a division by 10^4, that’s where our result p = 0.0001 comes from.

    Our aim for all these calculations was to find out how many TSP-PVV pairs we need to carry a successful brute force attack. Now we are able to calculate the expected number of false positives in a first search: it will be the number of trials times the probability for a single random false positive, i.e. t * p where t = 2^56, the size of the key space. This amounts to approximately 7.2 * 10^12, a rather big number. The expected number of false positives in the second search (restricted to the positive keys found in the first search) will be (t * p) * p, for a third search will be ((t * p) * p) * p and so on. Thus for n searches the expected number of false positives will be t * p^n.

    We can obtain the number of searches required to expect just one false positive by expressing the equation t * p^n = 1 and solving for n. So n equals to the logarithm in base p of 1/t, which by properties of logarithms it yields n = log(1/t)/log(p) ~ 4.2. Since we cannot do a fractional search it is convenient to round up this number. Therefore what is the expected number of false positives if we perform five searches? It is t * p^5 ~ 0.0007 or approximately 1 out of 1400. Thus using five TSP-PVV pairs is safe to obtain the true secret key with no false positives.

    The attack

    Once we know we need five TSP-PVV pairs, how do we get them? Of course we need at least one card with known PIN, and due to the nature of the PVV algorithm, that’s the only thing we need. With other PIN systems, such as IBM, we would need five cards, however this is not necessary with VISA PVV algorithm. We just have to read the magnetic stripe and then change the PIN four times but reading the card after each change.

    It is necessary to read the magnetic stripe of the card to get the PVV and the encrypting key selector. You can buy a commercial magnetic stripe reader or make one yourself following the instructions you can find in the previous page and links therein. Once you have a reader see this description of standard magnetic tracks to find out how to get the PVV from the data read. In that document the PVV field in tracks 1 and 2 is said to be five character long, but actually the true PVV consists of the last four digits. The first of the five digits is the key selector. I have only seen cards with a value of 1 in this digit, which is consistent with the standard and with the secret key never being compromised (and therefore they did not need to move to another key changing the selector).

    I did a simple C program, getpvvkey.c, to perform the attack. It consists of a loop to try all possible keys to encrypt the first TSP, if the derived PVV matches the true PVV a new TSP is tried, and so on until there is a mismatch, in which case the key is discarded and a new one is tried, or the five derived PVVs match the corresponding true PVVs, in which case we can assume we got the bank secret key, however the loop goes on until it exhausts the key space. This is done to assure we find the true key because there is a chance (although very low) the first key found is a false positive.

    It is expected the program would take a very long time to finish and to minimize the risks of a power cut, computer hang out, etc. it does checkpoints into the file getpvvkey.dat from time to time (the exact time depends on the speed of the computer, it’s around one hour for the fastest computers now in use). For the same reason if a positive key is found it is written on the file getpvvkey.key. The program only displays one message at the beginning, the starting position taken from the checkpoint file if any, after that nothing more is displayed.

    The DES algorithm is a key point in the program, it is therefore very important to optimize its speed. I tested several implementations: libdes, SSLeay, openssl, cryptlib, nss, libgcrypt, catacomb, libtomcrypt, cryptopp, ufc-crypt. The DES functions of the first four are based on the same code by Eric Young and is the one which performed best (includes optimized C and x86 assembler code). Thus I chose libdes which was the original implementation and condensed all relevant code in the files encrypt.c (C version) and x86encrypt.s (x86 assembler version). The code is slightly modified to achieve some enhancements in a brute force attack: the initial permutation is a fixed common steep in each TSP encryption and therefore can be made just one time at the beginning. Another improvement is that I wrote a completely new setkey function (I called it nextkey) which is optimum for a brute force loop.

    To get the program working you just have to type in the corresponding place five TSPs and their PVVs and then compile it. I have tested it only in UNIX platforms, using the makefile Makegetpvvkey to compile (use the command “make -f Makegetpvvkey”). It may compile on other systems but you may need to fix some things. Be sure that the definition of the type long64 corresponds to a 64 bit integer. In principle there is no dependence on the endianness of the processor. I have successfully compiled and run it on Pentium-Linux, Alpha-Tru64, Mips-Irix and Sparc-Solaris. If you do not have and do not want to install Linux (you don’t know what you are missing 😉 you still have the choice to run Linux on CD and use my program, see my page running Linux without installing it.

    Once you have found the secret bank key if you want to find the PIN of an arbitrary card you just have to write a similar program (sorry I have not written it, I’m too lazy 🙂 that would try all 10^4 PINs by generating the corresponding TSP, encrypting it with the (no longer) secret key, deriving the PVV and comparing it with the PVV in the magnetic stripe of the card. You will get one match for the true PIN. Only one match? Remember what we saw above, we have a chance of 0.0001 that a random encryption matches the PVV. We are trying 10000 PINs (and therefore TSPs) thus we expect 10000 * 0.0001 = 1 false positive on average.

    This is a very interesting result, it means that, on average, each card has two valid PINs: the customer PIN and the expected false positive. I call it “false” but note that as long as it generates the true PVV it is a PIN as valid as the customer’s one. Furthermore, there is no way to know which is which, even for the ATM; only customer knows. Even if the false positive were not valid as PIN, you still have three trials at the ATM anyway, enough on average. Therefore the probability we calculated at the beginning of this document about random guessing of the PIN has to be corrected. Actually it is twice that value, i.e., it is 0.0006 or one out of more than 1600, still safely low.


    It is important to optimize the compilation of the program and to run it in the fastest possible processor due to the long expected run time. I found that the compiler optimization flag -O gets the better performance, thought some improvement is achieved adding the -fomit-frame-pointer flag on Pentium-Linux, the -spike flag on Alpha-Tru64, the -IPA flag on Mips-Irix and the -fast flag on Sparc-Solaris. Special flags (-DDES_PTR -DDES_RISC1 -DDES_RISC2 -DDES_UNROLL -DASM) for the DES code have generally benefits as well. All these flags have already been tested and I chose the best combination for each processor (see makefile) but you can try to fine tune other flags.

    According to my tests the best performance is achieved with the AMD Athlon 1600 MHz processor, exceeding 3.4 million keys per second. Interestingly it gets better results than Intel Pentium IV 1800 MHz and 2000 MHz (see figures below, click on them to enlarge). I believe this is due to some I/O saturation, surely cache or memory access, that the AMD processor (which has half the cache of the Pentium) or the motherboard in which it is running, manages to avoid. In the first figure below you can see that the DES breaking speed of all processors has more or less a linear relationship with the processor speed, except for the two Intel Pentium I mentioned before. This is logical, it means that for a double processor speed you’ll get double breaking speed, but watch out for saturation effects, in this case it is better the AMD Athlon 1600 MHz, which will be even cheaper than the Intel Pentium 1800 MHz or 2000 MHz.

    In the second figure we can see in more detail what we would call intrinsic DES break power of the processor. I get this value simply dividing the break speed by the processor speed, that is, we get the number of DES keys tried per second and per MHz. This is a measure of the performance of the processor type independently of its speed. The results show that the best processor for this task is the AMD Athlon, then comes the Alpha and very close after it is the Intel Pentium (except for the higher speed ones which perform very poor due to the saturation effect). Next is the Mips processor and in the last place is the Sparc. Some Alpha and Mips processors are located at bottom of scale because they are early releases not including enhancements of late versions. Note that I included the performance of x86 processors for C and assembler code as there is a big difference. It seems that gcc is not a good generator of optimized machine code, but of course we don’t know whether a manual optimization of assembler code for the other processors (Alpha, Mips, Sparc) would boost their results compared to the native C compilers (I did not use gcc for these other platforms) as it happens with the x86 processor.


    Here is an article where these techniques may have been used.

    Technology is always being challenged

    I read a very interesting paper created by the University of Massachusetts, RSA Laboratories and Innealta, Inc.<<

    This paper primarily relates to the compromise of contact less payment technologies (RFID) if the RFID and/or reader have not been implemented correctly or the solution provider has used an inappropriate type of RFID and discusses the challenges around Chip and Pin with respect to financial transactions e.g. EMV standards and compliance.

    Additionally, the paper describes a RFID relay method which is being discussed within many forums around the world and we have now begun to see equipment being produced for the RFID skimmers/clonners to use for malicious means.

    The overarching point of this paper is to use an appropriate RFID & Chip solutions which supports the security/privacy of the user and purpose of the transaction (financial or non financial)<<

    The paper can be found at

    In modern payment RFID & Chip solutions, newer devices can be used which possess a high degree of processing power and are therefore able to execute strong cryptographic methods (such as digital signatures) to protect the identification and payment information whilst the transaction is occurring.

    These systems often utilise bidirectional authentication between the RFID/Chip scanner and the RFID tag/Chip prior to performing the transaction. These methods and cryptographic algorithms are accepted and proven to work within the traditional payment markets.

    As mentioned in the paper, some solution store static digitally signed and/or encrypted data which is provided to the RFID/Chip reader when queried, but this data never changes from one transaction to another. This may allow a malicious individual to capture and re-inject the data into the reader at a later stage. The alternative to storing static digitally signed and/or encrypted data is to negotiate a key exchange at the time of the transaction in which the card/value information is encrypted and subsequently transmitted. With this method the transmitted data
    changes on every transaction and therefore even if a malicious individual was to capture the encrypted transaction data from one transaction, this would not be accepted by the reader if re-injected at a later stage.

    Although this is the case today, older RFID/Chip solutions often use technologies which are not appropriate for financial transactions and therefore may be compromised easily and in some cases without the knowledge of the card holder, merchant or acquirer.

    I find this interesting how some of these less secure solution have been approved for use by acquiring banks and the card schemes around the world (if they were told) in recent years, where it has been seen that these solutions have utilised techniques or deployment methods which can be compromised. These technologies and techniques would never be approved within the Point of Sale (PoS) or traditional banking markets.

    It can only be assumed that the need to get product to market quickly at the expense of proper testing, understanding and with due consideration to industry lessons learnt has succeeded again.

    How to Build a Low-Cost, Extended-Range RFID Skimmer Filed under: RFID </a>

    Check it Out…

    here is a local copy.

    How to Build a Low-Cost, Extended-Range RFID Skimmer

    Also some of the supporting documents.

    A Practical Relay Attack on ISO 14443 Proximity Cards

    S4100 Multi-Function Reader Module Data Sheet

    Security Analysis of a Cryptographically-Enabled RFID’s

    Antenna Circuit Design for RFID Applications

    ISO 14443

    FAQ Interoperability

    Are MIFARE and ISO/IEC 14443 Type A the same?

    MIFARE and ISO/IEC 14443 Type A are not the same. While MIFARE is often viewed as an extension to or subset of ISO/IEC 14443 Type A, it is a proprietary encryption/conditional access protocol owned and licensed by Philips Semiconductors to multiple vendors of card ICs and reader ICs.

    Because MIFARE has been so predominantly used with products employing ISO/IEC 14443 Type A technology, it has mistakenly become synonymous with the standard. However, ISO/IEC 14443 Type A is a completely open standard when used independently of the MIFARE encryption/conditional access scheme.

    What changes to contactless standards and technology are expected in the future?

    Many vendors are actively developing new technologies to address the increasing market need for secure contactless technologies for a wide variety of applications. Changes in government regulations will also provide opportunities for enhancing contactless technology performance. It is important to note, however, that standards development is a lengthy process so it takes time for new technology developments to be reflected in standards that help to drive the availability of interoperable solutions. A few examples of new technologies that are expected include:

    • Changes to technology based on the ISO/IEC 15693 standard. Contactless cards supporting the ISO/IEC 15693 standard currently operate at 1.65 Kb/sec to meet FCC limits on sideband power in this frequency range. The FCC is expected to lift its restriction in late 2002, which would allow cards based on the ISO/IEC 15693 standard to improve their data rates.
    • Changes for higher speed operation. ISO working groups plan to add higher speed modes of operation to ISO/IEC 14443. This will increase the speed supported by this standard from 106 Kb/sec to the 848 Kb/sec that has already been demonstrated by IC manufacturers.
    • Alternative access control reader networking solutions. Wireless readers offer a significant advantage in lower costs of installation, particularly in older facilities. New security approaches can ensure strong authenticated channels between hosts or panels and new wireless readers. IP readers also permit direct connectivity to LANbased management and control applications.
    • The ability for a single contactless chip in a card to operate in full ISO/IEC 14443 and ISO/IEC 15693 modes.

    Is there a risk of someone listening or stealing the information from a contactless card?

    One risk with contactless cards is the ability for the card to be activated when it enters a reader’s RF range without the owner being aware of it. To prevent a contactless card activation without the card owner being aware of it, the application can be configured to always ask for the owner’s authorisation (password, PIN or biometric) before providing any user information or processing on the user’s behalf.


    e level of security of communication required between the contactless card and the reader must be defined as part of the system design and security controls must put in place so that un-invited listeners cannot intercept the data in any meaningful way. For example, all of the contactless technologies can use data encryption to protect data on the card and during transmission; this helps to ensure that, if information is intercepted, the information cannot be used by the recipient. It is important that all of the application’s requirements be understood and defined prior to any technology selection and implementation so that the appropriate security features are designed into the system.
    Additionally, the contactless chip is designed to self destruct if anyone tries to hack into it.

    What do you mean by three technologies on one card?

    There is confusing terminology used in the market to refer to cards that can support a combination of technologies. Cards are described as multiple technology when multiple, independent technologies share a common plastic card and do not communicate or interact with each other (e.g., magnetic stripe and contactless or contact chip). Cards are described as having a “dual-interface” when the card has a single integrated circuit (IC) that can communicate with a smart card reader/terminal via either contact or contactless.