I have been doing more research on the topic of DWT Steganography. I have came across the code below on the web. This is the first time I have came across subbands coefficients being specified. I have an idea what the code does but I would like someone to verify it!
steg_coeffs = [4, 4.75, 5.5, 6.25, 7];
for jj=1:size(message,2)+1
if jj > size(message,2)
charbits = [0,0,0,0,0,0,0,0];
else
charbits = dec2bin(message(jj),8)';
charbits = charbits(:)'-'0';
end
for ii=1:8
bit_count = bit_count + 1;
if charbits(ii) == 1
if HH(bit_count) <= 0
HH(bit_count) = steg_coeffs(randi(numel(steg_coeffs)));
end
else
if HH(bit_count) >= 0
HH(bit_count) = -1 * steg_coeffs(randi(numel(steg_coeffs)));
end
end
end
I think the steg_coeffs are selected coeffiecnt of the HH subband, where bits will be embedded in these selected coefficients. I have googled randi and believe that it will randomise these specified coeffs on each iteration of the loop and embed in random selection coeffs. I am correct?? Thank you
Typing help randi, you find out that randi(IMAX) will return a scalar, which will be an integer uniformly distributed (based on a prng) in the range 1:IMAX. To put simply, it chooses a random integer between 1 and IMAX.
numel(matrix) returns the total number of elements in the matrix.
So, steg_coeffs(randi(numel(steg_coeffs))) chooses a random element from steg_coeffs, by choosing a random index between 1 and 5.
The embedding algorithm is implemented in the following block.
if charbits(ii) == 1
...
else
...
end
Basically, if you're embedding a 1, the HH coefficient has to be positive. If it isn't, substitute it with one from steg_coeffs. Similarly, if you're embedding a 0, the HH coefficient has to be negative. If it isn't, substitute it with the negative of one from steg_coeffs.
The idea is that when you extract the secret, all you have to check is whether the HH coefficient is positive or negative, to know whether the bit has to be 1 or 0.
Related
I'm new to Scilab (and programming in general). I'm trying to implement a Scilab code to solve the cutting stock problem aka 'bin packing'.
The problem: given 'n' items of sizes[] (a vector from s1 to sn), and same capacity for all bins (c=1000), I need to minimize the number of bins required to fit all items.
I'm trying the 'next item algorithm', i.e., pick the first item from the vector, put it in the bin, then pick the next item and try to put in the same bin, in case there is no enough space, then create another bin.
Actually I don't need help in improving the algorithm, but rather in implement the code for this specific one.
Here is what I've tried so far:
// 'n' is the number of items to be packed
// 'c' is the capacity (how much items fit in the bin)
// 'sizes' a vector containing the size of n items
// 'nbins' number of bins used so far
// 'bin_rem' space left in current bin
sizes=[400,401,402,403,404,405,406,408,409,411,428,450,482]
c=1000
n=length(sizes)
nbins = 0
bin_rem = c
function y = bins(sizes,c,n)
for i=0; i<n; i=i+1
if sizes[i] > bin_rem
nbins=nbins+1
bin_rem = c - sizes(i)
bin_rem = bin_rem - sizes(i)
end
endfunction
disp ("Number of bins needed "+string(bins([sizes,c,n])))
end
I'm stuck with this error below and have no idea on how to solve it.
at line 20 of executed file
endfunction
^~~~~~~~~~~^
Error: syntax error, unexpected endfunction, expecting end
Any help?
First, seems like you still don't quite understand Scilab's syntax, since I see you using sizes[i], instead of sizes(i), and calling bins([sizes,c,n]). So, for now, try not to use functions. As for the error you get, it happens because you forgot one end. The way you wrote your code, only the if statement is closed, and for loop is still open.
Secondly, when you correct that, you'll notice that your program does not work properly, and that is because you defined the loop wrong. In Scilab, the for loop is actually a "for each" loop, therefore, you need to provide a full range of values for each iteration, instead of starting value (i=0), condition (i<n) and increment function (i=i+1).
Thirdly, seems like you understand the algorithm you're trying to use, but you implemented it wrong: the last line inside the loop should be the else statement.
Solving all those, you have the following piece of code:
sizes=[400,401,402,403,404,405,406,408,409,411,428,450,482]
c=1000
n=length(sizes)
nbins = 0 //should start as 0
bin_rem = 0 //should start as 0
for i = 1:n
if sizes(i) > bin_rem
nbins = nbins + 1;
bin_rem = c - sizes(i);
else
bin_rem = bin_rem - sizes(i);
end
end
disp ("Number of bins needed "+string(nbins))
Just to clarify, 1:n means a vector from 1 to n with pace of 1. You could have also written for i = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].
I am working with a legacy code base written in VB and have run into a conditional operator that I don't understand and cannot figure out what to search for to resolve it.
What I am dealing with is the following code and the variables that result as true. The specific parts that I do not understand are (1) the relationship between the first X and the first parens (-2 and (2) the role of X < 2
If X is a value below -2 it evaluates as false.
If X is a value above 2 it evaluates as true.
If Y is below 5 it evaluates to true as expected.
X = 3
Y = 10
If X > (-2 And X < 2) Or Y < 5 Then
'True
Else
'False
End If
I'm gonna leave off the Or Y < 5 part of the expression for this post as uninteresting, and limit myself to the X > (-2 And X < 2) side of the expression.
I haven't done much VB over the last several years, so I started out with some digging into Operator Precedence rules in VB, to be sure I have things right. I found definitive info on VBA and VB.Net, and some MSDN stuff that might have been VB6 but could also have been the 2013 version of VB.Net. All of them, though, gave the < and > comparison operators higher precedence over the And operator, regardless of whether you see And as logical or bitwise.
With that info, and also knowing that we must look inside parentheses first, I'm now confident the very first part of the expression to be evaluated is X < 2 (rather than -2 And X). Further, we know this will produce a Boolean result, and this Boolean result must be then converted to an Integer to do a bitwise (not logical) And with -2. That result (I'll call it n), which is still an Integer, can at last be compared to see if X > n, which will yield the final result of the expression as a Boolean.
I did some more digging and found this Stack Overflow answer about converting VB Booleans to Integers. While not definitive documentation, I was once privileged to meet the author (Hi #JaredPar) and know he worked on the VB compiler team at Microsoft, so he should know what he's talking about. It indicates that VB Boolean True has the surprising value of -1 as an integer! False becomes the more-normal 0.
At this point we need to talk about the binary representation of negative numbers. Using this reference as a guide (I do vaguely remember learning about this in college, but it's not the kind of thing I need every day), I'm going to provide a conversion table for integers from -3 to +3 in an imaginary integer size of only 4 bits (short version: invert the bit pattern and add one to get the negative representation):
-3 1101
-2 1110
-1 1111 --this helps explain **why** -1 was used for True
0 0000
1 0001
2 0010
3 0010
Stepping back, let's now consider the original -2 And X < 2 parenthetical and look at the results from the True (-1) and False (0) possible outcomes for X < 2 after a bitwise And with -2:
-2 (1110) And True (1111) = 1110 = -2
-2 (1110) And False (0000) = 0000 = 0
Really the whole point here from using the -1 bit pattern is anything And True produces that same thing you started with, whereas anything And False produces all zeros.
So if X < 2 you get True, which results in -2; otherwise you end up with 0. It's interesting to note here that if our language used positive one for True, you'd end up with the same 0000 value doing a bitwise And with -2 that you get from False (1110 And 0001).
Now we know enough to look at some values for X and determine the result of the entire original expression. Any positive integer is greater than both -2 and 0, so the expression should result in True. Zero and -1 are similar: they are less than two, and so will be compared again only as greater than -2 and thus always result in True. Negative two, though, and anything below, should be False.
Unfortunately, this means you could simplify the entire expression down to X > -2. Either I'm wrong about operator precedence, my reference for negative integer bit patterns is wrong, you're using a version of VB that converts True to something other than -1, or this code is just way over-complicated from the get-go.
I believe the sequence is as follows:
r = -2 and X 'outputs X with the least significant bit set to 0 (meaning the first lesser even number)
r = (r < 2) 'outputs -1 or 0
r = (X > r) 'outputs -1 or 0
r = r Or (Y < 5)
Use AndAlso and OrElse.
I'm sorry the title is so confusingly worded, but it's hard to condense this problem down to a few words.
I'm trying to find the minimum value of a specific equation. At first I'm looping through the equation, which for our purposes here can be something like y = .245x^3-.67x^2+5x+12. I want to design a loop where the "steps" through the loop get smaller and smaller.
For example, the first time it loops through, it uses a step of 1. I will get about 30 values. What I need help on is how do I Use the three smallest values I receive from this first loop?
Here's an example of the values I might get from the first loop: (I should note this isn't supposed to be actual code at all. It's just a brief description of what's happening)
loop from x = 1 to 8 with step 1
results:
x = 1 -> y = 30
x = 2 -> y = 28
x = 3 -> y = 25
x = 4 -> y = 21
x = 5 -> y = 18
x = 6 -> y = 22
x = 7 -> y = 27
x = 8 -> y = 33
I want something that can detect the lowest three values and create a loop. From theses results, the values of x that get the smallest three results for y are x = 4, 5, and 6.
So my "guess" at this point would be x = 5. To get a better "guess" I'd like a loop that now does:
loop from x = 4 to x = 6 with step .5
I could keep this pattern going until I get an absurdly accurate guess for the minimum value of x.
Does anybody know of a way I can do this? I know the values I'm going to get are going to be able to be modeled by a parabola opening up, so this format will definitely work. I was thinking that the values could be put into a column. It wouldn't be hard to make something that returns the smallest value for y in that column, and the corresponding x-value.
If I'm being too vague, just let me know, and I can answer any questions you might have.
nice question. Here's at least a start for what I think you should do for this:
Sub findMin()
Dim lowest As Integer
Dim middle As Integer
Dim highest As Integer
lowest = 999
middle = 999
hightest = 999
Dim i As Integer
i = 1
Do While i < 9
If (retVal(i) < retVal(lowest)) Then
highest = middle
middle = lowest
lowest = i
Else
If (retVal(i) < retVal(middle)) Then
highest = middle
middle = i
Else
If (retVal(i) < retVal(highest)) Then
highest = i
End If
End If
End If
i = i + 1
Loop
End Sub
Function retVal(num As Integer) As Double
retVal = 0.245 * Math.Sqr(num) * num - 0.67 * Math.Sqr(num) + 5 * num + 12
End Function
What I've done here is set three Integers as your three Min values: lowest, middle, and highest. You loop through the values you're plugging into the formula (here, the retVal function) and comparing the return value of retVal (hence the name) to the values of retVal(lowest), retVal(middle), and retVal(highest), replacing them as necessary. I'm just beginning with VBA so what I've done likely isn't very elegant, but it does at least identify the Integers that result in the lowest values of the function. You may have to play around with the values of lowest, middle, and highest a bit to make it work. I know this isn't EXACTLY what you're looking for, but it's something along the lines of what I think you should do.
There is no trivial way to approach this unless the problem domain is narrowed.
The example polynomial given in fact has no minimum, which is readily determined by observing y'>0 (hence, y is always increasing WRT x).
Given the wide interpretation of
[an] equation, which for our purposes here can be something like y =
.245x^3-.67x^2+5x+12
many conditions need to be checked, even assuming the domain is limited to polynomials.
The polynomial order is significant, and the order determines what conditions are necessary to check for how many solutions are possible, or whether any solution is possible at all.
Without taking this complexity into account, an iterative approach could yield an incorrect solution due to underflow error, or an unfortunate choice of iteration steps or bounds.
I'm not trying to be hard here, I think your idea is neat. In practice it is more complicated than you think.
I have this piece of code which checks whether a given number is prime:
If x Mod 2 = 0 Then
Return False
End If
For i = 3 To x / 2 + 1 Step 2
If x Mod i = 0 Then
Return False
End If
Next
Return True
I only use it for numbers 1E7 <= x <= 2E7. However, it is extremely slow - I can hardly check 300 numbers a second, so checking all x's would take more than 23 days...
Could someone give some improvements tips or say what I might be doing redundantly this way?
That is general algorithm for checking prime number.
If you want to check prime number in bulk use algorithm http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
Look up the term "Sieve of Eratosthenes". It's a two thousand years old algorithm which is way better than yours. It's often taught in school.
You should definitely change your algorithm! You can try Sieve of Eratosthenes or a more advanced Fermat primality test. Beware that your code will be more complicated, as you would need to implement modular arithmetics. Look here for the list of some even more mathematically advanced methods.
You can also look for AKS primality test.
This is a good algorithm for checking primality.
Since x/2 + 1 is a constant through out the looping operation, keep it in a separate variable before the For loop. Thus, saving a division & addition operation every time you loop. Though this might slightly increase the performance.
Use the Sieve of Eratosthenes to create a Set that contains all the prime numbers up to the largest number you need to check. It will take a while to set up the Set, but then checking if a number exists in it will be very fast.
Split range in some chunks, and do checks in two or more threads, if you have multicore cpu. Or use Parallel.For.
To check if the number is prime you have only check if it can't be divided by primes less then it.
Please check following snippet:
Sub Main()
Dim primes As New List(Of Integer)
primes.Add(1)
For x As Integer = 1 To 1000
If IsPrime(x, primes) Then
primes.Add(x)
Console.WriteLine(x)
End If
Next
End Sub
Private Function IsPrime(x As Integer, primes As IEnumerable(Of Integer)) As Boolean
For Each prime In primes
If prime <> 1 AndAlso prime <> x Then
If x Mod prime = 0 Then
Return False
End If
End If
Next
Return True
End Function
it slow because you use the x/2. I modified your code a little bit. (I don't know about syntax of VB, Maybe you have to change my syntax.)
If x < 2 Then
Return False
IF x == 2 Then
Return True
If x Mod 2 = 0 Then
Return False
End If
For i = 3 To (i*i)<=x Step 2
If x Mod i = 0 Then
Return False
End If
Next
Return True
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.