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.
Related
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.
How do I make a biased random number generator (RNG) in VB.NET?
I know I could make it by fiddling with the output of the Randomize()/Rnd methods, but is there a built-in way of doing this?
I want the biased RNG to give me either a 2 or 4 (though using 1 or 2 as a substitute is also OK by me), with 2 occurring on average 90% of the time and 4 occurring on average 10% of the time.
Create a random number generator to return values from 1-10, if the value from the random number generator is between 1 and 9 send a 2 if the value is 10 send a 4.
You might want to look at this
http://msdn.microsoft.com/en-us/library/vstudio/ctssatww(v=vs.100).aspx?cs-save-lang=1&cs-lang=vb#code-snippet-2
If you want to come out with a mask to generate your values
Here is what I think you can do.
Dim numbers() as integer = {2,2,2,2,4,2,2,2,2,2} ' set 10% for 4, 90% for 2
Dim r as new Random()
Return numbers(r.Next(0, 10))
I'm trying to create a binary to decimal converter, and have got stuck on the code. I have researched forums for any help, but they all seam to use functions, which can not be used within a private sub. Please can anyone give me help on a solution to this problem?
I would use the positional notation method:
http://en.wikipedia.org/wiki/Positional_notation
http://www.wikihow.com/Convert-from-Binary-to-Decimal
So basically, without giving you the answer, you want to loop through binary place holders, filling up a variable as you go along. You would use an index to move from the least significant placeholder to the most.
For example : 10011011 in binary is 155 decimal.
So every place holder is a power with a base of two. Then you add the value for each one until your finished, like so:
placeholder 1 is: 2 pow 0 equals 1.
placeholder 2 is: 2 pow 1 equals 2.
placeholder 3 is: 2 pow 2 equals 4.
placeholder 4 is: 2 pow 3 equals 8.
placeholder 5 is: 2 pow 4 equals 16.
placeholder 6 is: 2 pow 5 equals 32.
placeholder 7 is: 2 pow 6 equals 64.
placeholder 8 is: 2 pow 7 equals 128.
Now we only add for the placeholders that have 1s.
128+16+8+2+1 = 155
What you will need:
A loop looping through indexes, and incrementing the exponent value as you go along, only adding the value if the index equals 1 in the binary number.
Hope my explanation makes sense. Good luck.
I understand the Modulus operator in terms of the following expression:
7 % 5
This would return 2 due to the fact that 5 goes into 7 once and then gives the 2 that is left over, however my confusion comes when you reverse this statement to read:
5 % 7
This gives me the value of 5 which confuses me slightly. Although the whole of 7 doesn't go into 5, part of it does so why is there either no remainder or a remainder of positive or negative 2?
If it is calculating the value of 5 based on the fact that 7 doesn't go into 5 at all why is the remainder then not 7 instead of 5?
I feel like there is something I'm missing here in my understanding of the modulus operator.
(This explanation is only for positive numbers since it depends on the language otherwise)
Definition
The Modulus is the remainder of the euclidean division of one number by another. % is called the modulo operation.
For instance, 9 divided by 4 equals 2 but it remains 1. Here, 9 / 4 = 2 and 9 % 4 = 1.
In your example: 5 divided by 7 gives 0 but it remains 5 (5 % 7 == 5).
Calculation
The modulo operation can be calculated using this equation:
a % b = a - floor(a / b) * b
floor(a / b) represents the number of times you can divide a by b
floor(a / b) * b is the amount that was successfully shared entirely
The total (a) minus what was shared equals the remainder of the division
Applied to the last example, this gives:
5 % 7 = 5 - floor(5 / 7) * 7 = 5
Modular Arithmetic
That said, your intuition was that it could be -2 and not 5. Actually, in modular arithmetic, -2 = 5 (mod 7) because it exists k in Z such that 7k - 2 = 5.
You may not have learned modular arithmetic, but you have probably used angles and know that -90° is the same as 270° because it is modulo 360. It's similar, it wraps! So take a circle, and say that its perimeter is 7. Then you read where is 5. And if you try with 10, it should be at 3 because 10 % 7 is 3.
Two Steps Solution.
Some of the answers here are complicated for me to understand. I will try to add one more answer in an attempt to simplify the way how to look at this.
Short Answer:
Example 1:
7 % 5 = 2
Each person should get one pizza slice.
Divide 7 slices on 5 people and every one of the 5 people will get one pizza slice and we will end up with 2 slices (remaining). 7 % 5 equals 2 is because 7 is larger than 5.
Example 2:
5 % 7 = 5
Each person should get one pizza slice
It gives 5 because 5 is less than 7. So by definition, you cannot divide whole 5items on 7 people. So the division doesn't take place at all and you end up with the same amount you started with which is 5.
Programmatic Answer:
The process is basically to ask two questions:
Example A: (7 % 5)
(Q.1) What number to multiply 5 in order to get 7?
Two Conditions: Multiplier starts from `0`. Output result should not exceed `7`.
Let's try:
Multiplier is zero 0 so, 0 x 5 = 0
Still, we are short so we add one (+1) to multiplier.
1 so, 1 x 5 = 5
We did not get 7 yet, so we add one (+1).
2 so, 2 x 5 = 10
Now we exceeded 7. So 2 is not the correct multiplier.
Let's go back one step (where we used 1) and hold in mind the result which is5. Number 5 is the key here.
(Q.2) How much do we need to add to the 5 (the number we just got from step 1) to get 7?
We deduct the two numbers: 7-5 = 2.
So the answer for: 7 % 5 is 2;
Example B: (5 % 7)
1- What number we use to multiply 7 in order to get 5?
Two Conditions: Multiplier starts from `0`. Output result and should not exceed `5`.
Let's try:
0 so, 0 x 7 = 0
We did not get 5 yet, let's try a higher number.
1 so, 1 x 7 = 7
Oh no, we exceeded 5, let's get back to the previous step where we used 0 and got the result 0.
2- How much we need to add to 0 (the number we just got from step 1) in order to reach the value of the number on the left 5?
It's clear that the number is 5. 5-0 = 5
5 % 7 = 5
Hope that helps.
As others have pointed out modulus is based on remainder system.
I think an easier way to think about modulus is what remains after a dividend (number to be divided) has been fully divided by a divisor. So if we think about 5%7, when you divide 5 by 7, 7 can go into 5 only 0 times and when you subtract 0 (7*0) from 5 (just like we learnt back in elementary school), then the remainder would be 5 ( the mod). See the illustration below.
0
______
7) 5
__-0____
5
With the same logic, -5 mod 7 will be -5 ( only 0 7s can go in -5 and -5-0*7 = -5). With the same token -5 mod -7 will also be -5.
A few more interesting cases:
5 mod (-3) = 2 i.e. 5 - (-3*-1)
(-5) mod (-3) = -2 i.e. -5 - (-3*1) = -5+3
It's just about the remainders. Let me show you how
10 % 5=0
9 % 5=4 (because the remainder of 9 when divided by 5 is 4)
8 % 5=3
7 % 5=2
6 % 5=1
5 % 5=0 (because it is fully divisible by 5)
Now we should remember one thing, mod means remainder so
4 % 5=4
but why 4?
because 5 X 0 = 0
so 0 is the nearest multiple which is less than 4
hence 4-0=4
modulus is remainders system.
So 7 % 5 = 2.
5 % 7 = 5
3 % 7 = 3
2 % 7 = 2
1 % 7 = 1
When used inside a function to determine the array index. Is it safe programming ? That is a different question. I guess.
Step 1 : 5/7 = 0.71
Step 2 : Take the left side of the decimal , so we take 0 from 0.71 and multiply by 7
0*7 = 0;
Step # : 5-0 = 5 ; Therefore , 5%7 =5
Modulus operator gives you the result in 'reduced residue system'. For example for mod 5 there are 5 integers counted: 0,1,2,3,4. In fact 19=12=5=-2=-9 (mod 7). The main difference that the answer is given by programming languages by 'reduced residue system'.
lets put it in this way:
actually Modulus operator does the same division but it does not care about the answer , it DOES CARE ABOUT reminder for example if you divide 7 to 5 ,
so , lets me take you through a simple example:
think 5 is a block, then for example we going to have 3 blocks in 15 (WITH Nothing Left) , but when that loginc comes to this kinda numbers {1,3,5,7,9,11,...} , here is where the Modulus comes out , so take that logic that i said before and apply it for 7 , so the answer gonna be that we have 1 block of 5 in 7 => with 2 reminds in our hand! that is the modulus!!!
but you were asking about 5 % 7 , right ?
so take the logic that i said , how many 7 blocks do we have in 5 ???? 0
so the modulus returns 0...
that's it ...
A novel way to find out the remainder is given below
Statement : Remainder is always constant
ex : 26 divided by 7 gives R : 5
This can be found out easily by finding the number that completely divides 26 which is closer to the
divisor and taking the difference of the both
13 is the next number after 7 that completely divides 26 because after 7 comes 8, 9, 10, 11, 12 where none of them divides 26 completely and give remainder 0.
So 13 is the closest number to 7 which divides to give remainder 0.
Now take the difference (13 ~ 7) = 5 which is the temainder.
Note: for this to work divisor should be reduced to its simplest form ex: if 14 is the divisor, 7 has to be chosen to find the closest number dividing the dividend.
As you say, the % sign is used to take the modulus (division remainder).
In w3schools' JavaScript Arithmetic page we can read in the Remainder section what I think to be a great explanation
In arithmetic, the division of two integers produces a quotient and a
remainder.
In mathematics, the result of a modulo operation is the
remainder of an arithmetic division.
So, in your specific case, when you try to divide 7 bananas into a group of 5 bananas, you're able to create 1 group of 5 (quotient) and you'll be left with 2 bananas (remainder).
If 5 bananas into a group of 7, you won't be able to and so you're left with again the 5 bananas (remainder).
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
Hi, I am new to Visual Basic, I have a project where I need to be able to manipulate individual bits in a value.
I need to be able to switch these bits between 1 and 0 and combine multiple occurrences of bits into one variable in my code.
Each bit will represent a single TRUE / FALSE value, so I'm not looking for how to do a single TRUE / FALSE value in one variable, but rather multiple TRUE / FALSE values in one variable.
Can someone please explain to me how I can achieve this please.
Many thanks in advance.
Does it have to be exactly one bit?
Why don't you just use the actual built in VB data type of Boolean for this.
http://msdn.microsoft.com/en-us/library/wts33hb3(v=vs.80).aspx
It's sole reason for existence is so you can define variables that have 2 states, true or false.
Dim myVar As Boolean
myVar = True
myVar = Flase
if myVar = False Then
myVar = True
End If
UPDATE (1)
After reading through the various answers and comments from the OP I now understand what it is the OP is trying to achieve.
As others have said the smallest unit one can use in any of these languages is an 8 bit byte. There is simply no order of data type with a smaller bit size than this.
However, with a bit of creative thinking and a smattering of binary operations, you can refer to the contents of that byte as individual bits.
First however you need to understand the binary number system:
ALL numbers in binary are to the power of two, from right to left.
Each column is the double of it's predecessor, so:
1 becomes 2, 2 becomes 4, 4 becomes 8 and so on
looking at this purely in a binary number your columns would be labelled thus:
128 64 32 16 8 4 2 1 (Remember it's right to left)
this gives us the following:
The bit at position 1 = 1;
The bit at position 2 = 2;
The bit at position 3 = 4;
The bit at position 4 = 8;
and so on.
Using this method on the smallest data type you have (The byte) you can pack 8 bit's into one value. That is you could use one variable to hold 8 separate values of 1 or 0
So while you cannot go any smaller than a byte, you can still reduce memory consumption by packing 8 values into 1 variable.
How do you read and write the values?
Remember the column positions? well you can use something called Bit Shifting and Bit masks.
Bit Shifting is the process of using the
<<
and
>>
operators
A shifting operation takes as a parameter the number of columns to shift.
EG:
Dim byte myByte
myByte = 1 << 4
In this case the variable 'myByte' would become equal to 16, but you would have actually set bit position 5 to a 1, if we illustrate this, it will make better sense:
mybyte = 0 = 00000000 = 0
mybyte = 1 = 00000001 = 1
mybyte = 2 = 00000010 = (1 << 1)
mybyte = 4 = 00000100 = (1 << 2)
mybyte = 8 = 00001000 = (1 << 3)
mybyte = 16 = 00010000 = (1 << 4)
the 0 through to 16 if you note is equal to the right to left column values I mentioned above.
given what Iv'e just explained then, if you wanted to set bits 5, 4 and 1 to be equal to 1 and the rest to be 0, you could simply use:
mybyte = 25(16 + 8 + 1) = 00011001 = (1 << 4) + (1 << 3) + 1
to get your bits back out, into a singleton you just bit shift the other way
retrieved bit = mybyte >> 4 = 00000001
Now there is unfortunately however one small flaw with the bit shifting method.
by shifting back and forth you are highly likely to LOOSE information from any bits you might already have set, in order to prevent this from happening, it's better to combine your bit shifting operations with bit masks and boolean operations such as 'AND' & 'OR'
To understand what these do you first need to understand simple logic principles as follows:
AND
Output is one if both the A and B inputs are 1
Illustrating this graphically
A B | Output
-------------
0 0 | 0
0 1 | 0
1 0 | 0
1 1 | 1
As you can see if a bit position in our input number is a 1 and the same position in our input number B is 1, then we will keep that position in our output number, otherwise we will discard the bit and set it to a 0, take the following example:
00011001 = Bits 5,4 and 1 are set
00010000 = Our mask ONLY has bit 5 set
if we perform
00011001 AND 0010000
we will get a result of
00010000
which we can then shift down by 5
00010000 >> 5 = 00000001 = 1
so by using AND we now have a way of checking an individual bit in our byte for a value of 1:
if ((mybyte AND 16) >> 1) = 1 then
'Bit one is set
else
'Bit one is NOT set
end if
by using different masks, with the different values of 2 in the right to left columns as shown previously, we can easily extract different singular values from our byte and treat them as a simple bit value.
Setting a byte is just as easy, except you perform the operation the opposite way using an 'OR'
OR
Output is one if either the A or B inputs are 1
Illustrating this graphically
A B | Output
-------------
0 0 | 0
0 1 | 1
1 0 | 1
1 1 | 1
eg:
00011001 OR 00000100 = 00011101
as you can see the bit at position 4 has been set.
To answer the fundamental question that started all this off however, you cannot use a data type in VB that has any resolution less than 1 byte, I suspect if you need absolute bit wise accuracy I'm guessing you must be writing either a compression algorithm or some kind of encryption system. :-)
01010100 01110010 01110101 01100101, is the string value of the word "TRUE"
What you want is to store the information in a boolean
Dim v As Boolean
v = True
v = False
or
If number = 84 Then ' 84 = 01010100 = T
v = True
End If
Other info
Technicaly you can't store anything in a bit, the smallest value is a char which is 8 bit. You'll need to learn how to do bitwise operation. Or use the BitArray class.
VB.NET (nor any other .NET language that I know of) has a "bit" data type. The smallest that you can use is a Byte. (Not a Char, they are two-bytes in size). So while you can read and convert a byte of value 84 into a byte with value 1 for true, and convert a byte of value 101 into a byte of value 0 for false, you are not saving any memory.
Now, if you have a small and fixed number of these flags, you CAN store several of them in one of the integer data types (in .NET the largest integer data type is 64 bits). Or if you have a large number of these flags you can use the BitArray class (which uses the same technique but backs it with an array so storage capacity is greater).