I am new to cryptography kindly help to solve the following vigenere cipher problem with well defined steps
Assume you are given a 300 character encrypted message, encrypted in Vigenere cryptosystem, in which you know the plaintext word CRYPTOGRAPHY occurs exactly two times, and we know that the ciphertext sequence TICRMQUIRTJR is the encryption of CRYPTOGRAPHY. The first occurrence starts at character position 10 and second at character position 241 (we start counting from 1). What is the length of the key used for encryption
Answer is 7
Solution To estimate the period we use the Kasiski test. The distance between the two occurrences given is
241 − 10 = 231 = 3 · 7 · 11
positions.
Possible periods are thus 3, 7 and 11. If the guess is correct, we can immediately find the
corresponding shifts: at position 10 the shift is
T − c = 19 − 2 = 17 = r
. Similar computations for the other positions gives the shift keys
rrectcorrect
We now see that this is not periodic with periods 3 or 11, while period 7 is possible. The keyword
of length 7 starts at position 15; hence the keyword is
correct.
Related
I'm trying to understand the code that returns the size_in_bytes for a RSA key (I'm looking at PyCryptoDomex). They calculate it:
def size_in_bytes(self):
return (self.size_in_bits() - 1) // 8 + 1
I would have thought that it would simply be
self.size_in_bits()//8
The specific code is here: github_code_location
I'm sure there's a reason to subtract 1 from the bits and then add 1 to the integer after division but I'd like to understand why.
7 // 8 == 0
But you cannot store 7 bits in zero bytes.
I am studying cryptography from Cristof Paar's book. There is a question about LFSR's I have trouble with. I just can't understand one point here. Question is this:
We want to perform an attack on another LFSR-based stream cipher. In order
to process letters, each of the 26 uppercase letters and the numbers 0, 1, 2, 3, 4, 5
are represented by a 5-bit vector according to the following mapping:
A -> 0 = 00000
.
.
.
Z -> 25 = 11001
0 -> 26 = 11010
.
.
.
5 -> 31= 11111
(binary)
We happen to know the following facts about the system:
-The degree of the LFSR is m = 6.
-Every message starts with the header WPI
We observe now on the channel the following message (the fourth letter is a zero): j5a0edj2b
What are the feedback coefficients of the LFSR? (This one!)
Solution:
I can't understand the matrix in this solution where did these numbers come?
Using WPI, we have plaintext begins with
P=>(10110)(01111)(01000)
Using j5a0edj2b we have the ciphertext
C=>(01001)(11111)(00000)(11010)(00100)(00011)(01001)............
then by addition of P and C in mod 2, the key stream is
S=>(11111)(10000)(01000)....
we find the matrix from key stream
s0=1,s1=1,s2=1,s3=1,s4=1,s5=1,s6=0,s7=0,s8=0,s9=0,s10=0,s11=1 etc
For the matrix
first line.... (s0,s1,s2,s3,s4,s5)
second line....(s1,s2,s3,s4,s5,s6)
third line.....(s2,s3,s4,s5,s6,s7)
4th (s3,s4,s5,s6,s7,s8)
5th (s4,s5,s6,s7,s8,s9)
last line (s5,s6,s7,s8,s9,s10)
this calulations are given in LFSRs in details
I'm trying to assemble proper track data given a CVC3 and a bunch of positional parameters. But the EMV C-2 Kernel book is about as obtuse as you could imagine (would it kill somebody to include an example!?!). Can anyone help work this example:
9f62 - pcvc3(t1) - Position of CVC3 in track1: 0x38 (4-6?)
9f63 - punatc(t1) - Unpredictable Number Track1 Pos: 0x3C6 (2-3 7-10?)
9f64 - natc(t1) - Digits in track1 ATC: 4
9f65 - pcvc3(t2) - Position of CVC3 in track2: 0x38 (4-6)
9f66 - punatc(t2) - Unpredictable Number Track2 Pos: 0x3C6 (2-3 7-10?)
9f67 - Digits in track2 ATC: 4
After successful checksum generation:
9f61 - track2 CVC3 - 2EF4
9f60 - track1 CVC3 - 609B
9f36 - ATC - 1E47
assuming the discretionary data field starts out as all 0s, how does it end up? The spec says this:
Convert the binary encoded CVC3 (Track2) to the BCD encoding of the
corresponding number expressed in base 10. Copy the q least significant digits of the
BCD encoded CVC3 (Track2) in the eligible positions of the 'Discretionary Data' in
Track 2 Data. The eligible positions are indicated by the q non-zero bits in
PCVC3(Track2).
I read that as:
CVC3 = 0x609B = 24731 (so copy 731? What does BCD have to do with this? Or are they just saying "copy the 731 as bcd encoded to the byte array"?)
yes you are correct it is rather obtuse. you are correct that you would convert your p values (pCVC3, and PUNATC) to binary. (0011 1000, 0011 1100 0110 for your track1 p values) you then right align the proper values with the discretionary data. example
Bxxxxxxxxxxxxxxxx^ /^14111014010000000000
....000000000000000000CCC000
....00000000000000AAAA000UU0
so you say that you CVC3 for track 1 is 609B which is 24,731, since you PCVC3 is asking for only 3 characters youd set 731 in. your ATC is 1E47 which is 7,751. your PUNATC is asking for 4 digits so you'd use 7751. FYI... if the ATC is lower than the requested characters you'd pad with 0's. Your unpredictable number is even more tricky... so you make a 4 byte random number. convert it to a uint (base 10) and then mark the 8 most significant bytes as 0. example. lets say your random 4 bytes is 29A6 06AE. in base 10 that is 698,746,542. mark out the first 8 characters with 0. and you are left with 000,000,002. you'd place 02 in for you unpredictable number placment.. so.. all that said your track would look like this
Bxxxxxxxxxxxxxxxx^ /^14111014017751731020
the last character is equal to the length of the unpredicatable number (numeric) digits. which was 02.. so the last digit is 2 making your final track data
%Bxxxxxxxxxxxxxxxx^ /^1411014017751731022;
track2 is very similar. good luck with it. I understand your frustrations with this. :)
I'm wondering if it is possible to represent a number as a sequence of bits, each having approximately the same significance, such that if we flip one of the bits, the overall value does not change by much.
For example, we can use sequences of 4-bits, where each group represents a value from 0 to 15 and the overall value is the sum of all these values.
0110 0101 1101 1010 1011 → 6 + 5 + 13 + 10 + 11 = 45
and now flipping any bit can only incur in a maximum difference of 8 in the final value.
Some drawbacks obviously exist with this approach:
values have multiple representations, with some values having more representations than other ones (for example, there are 39280 distinct representations for the number 38, and only 1 for the number 0);
the amount of values that can be represented is greatly reduced (this representation allows for integers from 0 to 75, while 20 bits could normally represent 220 ~ 1 million different integers).
Are there any resources I can find concerning this problem? I can't seem to find anything online, but maybe I'm not searching with the right keywords. What other alternatives exist to my approach? Do they improve on its disadvantages?
I'm trying to emulate a function in SQL that a client has produced in Excel. In effect, they have a unique, 10-digit numeric value (VARCHAR) as the primary key in one of their enterprise database systems. Within another database, they require a unique, 5-digit alphanumeric identifier. They want that 5-digit alphanumeric value to be a representation of the 10-digit number. So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
The EXCEL equation is:
=IF(VALUE(MID(A2,1,4))>0,DEC2HEX(VALUE(MID(A2,3,2)))&DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX(VALUE(MID(A2,9,2))),DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX((VALUE(MID(A2,9,2)))))
I need the SQL equivalent of this. Of course, should someone out there know a better way to accomplish their goal of "a 5-digit alphanumeric identifier" based off the 10-digit number, I'm all ears.
ADDED 8/2/2011
First of all, thank you to everyone for the replies. Nice to see folks willing to help and even enjoying it! Based on all the responses, I'm apt to tell my client they're intent is sound, only their method is off kilter. I'd also like to recommend a solution. So the challenge remains, just modified slightly:
CHALLENGE: Within SQL, take a 10 digit, unique NUMERIC string and represent it ALPHANUMERICALLY in as few characters as possible. The resulting string must also be unique.
Note that the first 3-4 characters in the 10-digit string are likely to be zeros, and that they could be stripped to shorten the resulting alphanumeric string. Not required, but perhaps helpful.
This problem is inherently impossible. You have a 10 digit numeric value that you want to convert to a 5 digit alphanumeric value. Since there are 10 numeric characters, this means that there are 10^10 = 10 000 000 000 unique values for your 10 digit number. Since there are 36 alphanumeric characters (26 letters + 10 numbers), there are 36^5 = 60 466 176 unique values for your 5 digit number. You cannot map a set of 10 billion elements into a set with around 60 million.
Now, lets take a closer look at what your client's code is doing:
So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
This isn't 100% accurate. The excel code never uses the first 2 digits, but performs this operation on the remaining 8. There are two main problems with this algorithm which may not be intuitively obvious:
Two 10 digit numbers can map to the same 5 digit number. Consider the numbers 1000000117 and 1000001701. The last four digits of 1000000117 get mapped to 1 11, where the last four digits of 1000001701 get mapped to 11 1. This causes both to map to 00111.
The 5 digit number may not even end up being 5 digits! For example, 1000001616 gets mapped to 001010.
So, what is a possible solution? Well, if you don't care if that 5 digit number is unique or not, in MySQL you can use something like:
hex(<NUMERIC VALUE> % 0xFFFFF)
The log of 10^10 base 2 is 33.219280948874
> return math.log(10 ^ 10) / math.log(2)
33.219280948874
> = 2 ^ 33.21928
9999993422.9114
So, it takes 34 bits to represent this number. In hex this will take 34/4 = 8.5 characters, much more than 5.
> return math.log(10 ^ 10) / math.log(16)
8.3048202372184
The Excel macro is ignoring the first 4 (or 6) characters of the 10 character string.
You could try encoding in base 36 instead of 16. This will get you to 7 characters or less.
> return math.log(10 ^ 10) / math.log(36)
6.4254860446923
The popular base 64 encoding will get you to 6 characters
> return math.log(10 ^ 10) / math.log(64)
5.5365468248123
Even Ascii85 encoding won't get you down to 5.
> return math.log(10 ^ 10) / math.log(85)
5.1829075929158
You need base 100 to get to 5 characters
> return math.log(10 ^ 10) / math.log(100)
5
There aren't 100 printable ASCII characters, so this is not going to work, as zkhr explained as well, unless you're willing to go beyond ASCII.
I found your question interesting (although I don't claim to know the answer) - I googled a bit for you out of interest and found this which may help you http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html