I am trying to implement the speck cipher as specified here: Speck Cipher. On page 18 of the document you can find some speck pseudo-code I want to implement.
It seems that I got a problem on understanding the pseudo-code. As you can find there, x and y are plaintext words with length n. l[m-2],...l[0], k[0] are key words (as for words, they have length n right?). When you do the key expansion, we iterate for i from 0 to T-2, where T are the round numbers (for example 34). However I get an IndexOutofBoundsException, because the array with the l's has only m-2 positions and not T-2.
Can someone clarify what the key expansions does and how?
Ah, I get where the confusion lies:
l[m-2],...l[0], k[0]
these are the input key words, in other words, they represent the key. These are not declarations of the size of the arrays, as you might expect if you're a developer.
Then the subkey's in array k should be derived, using array l for intermediate values.
According to the formulas, taking the largest i, i.e. i_max = T - 2 you get a highest index for array l of i_max + m - 1 = T - 2 + m - 1 = T + m - 3 and therefore a size of the array of one more: T + m - 2. The size of a zero-based array is always the index of the last element - plus one, after all.
Similarly, for subkey array k you get a highest index of i_max + 1, which is T - 2 + 1 or T - 1. Again, the size of the array is one more, so there are T elements in k. This makes a lot of sense if you require T round keys :)
Note that it seems possible to simply redo the subkey derivation for each round if you require a minimum of RAM. The entire l array doesn't seem necessary either. For software implementations that doesn't matter a single iota of course.
Related
Say I am tossing a fair coin where 'tails' is assigned the value x = -1/2 and 'heads' is assigned x = 1/2.
I do this N times and I want to obtain the sum. This is what I have tried:
p = 0.5
N = 1e4
X(N,p)=(rand(N).<p)
I know this is incomplete but when I check (rand(N).<p) I see an array consisting of true, false. I interpret this as 'Tails' or 'Heads'. However, I don't know how to assign the values 1/2 and -1/2 to each of these elements in order for me to find the sum. If I simply use sum((rand(N).<p)) I do get an integer value, but I don't think this is the right way to do it because I haven't specified the values 1/2 and -1/2 anywhere.
Any help is greatly appreciated.
As indicated by the comments already, you want to do
sum(rand([-0.5, 0.5], N))
where N must be an integer (you wrote N=1e4, therefore typeof(N) == Float64 and rand won't work).
The documentation of rand (obtained by ?rand) describes what rand(S, N) does:
Pick a random element or array of random elements from the set of
values specified by S
Here, S can be an optional indexable collection, an array of values in your case (or a type like Int). So, above S = [-0.5, 0.5] and rand draws N random elements from this collection, which we can afterwards sum up.
Assigning specific values to a boolean array
Since this is the title of your question, and the answer above doesn't actually address this, let me comment on this as well.
You could do sum((rand(N).<p)-0.5), i.e. you shift all the ones to 0.5 and all the zeros to -0.5 to get the wanted result. Note that this is a general strategy: Let's say you want true to be a and false to be b, where a and b are numbers. You achieve this by (rand(N).<p)*(a-b) + b.
However, beyond being more "complicated", sum((rand(N).<p)-0.5) will allocate temporary arrays, first one of booleans, then one of numbers, the latter of which will eventually go into sum. Because of these unnecessary allocations this approach will be slower than the solution above.
I am looking for help with pseudo code (unless you are a user of Game Maker 8.0 by Mark Overmars and know the GML equivalent of what I need) for how to generate a list / array of unique combinations of a set of X number of integers which size is variable. It can be 1-5 or 1-1000.
For example:
IntegerList{1,2,3,4}
1,2
1,3
1,4
2,3
2,4
3,4
I feel like the math behind this is simple I just cant seem to wrap my head around it after checking multiple sources on how to do it in languages such as C++ and Java. Thanks everyone.
As there are not many details in the question, I assume:
Your input is a natural number n and the resulting array contains all natural numbers from 1 to n.
The expected output given by the combinations above, resembles a symmetric relation, i. e. in your case [1, 2] is considered the same as [2, 1].
Combinations [x, x] are excluded.
There are only combinations with 2 elements.
There is no List<> datatype or dynamic array, so the array length has to be known before creating the array.
The number of elements in your result is therefore the binomial coefficient m = n over 2 = n! / (2! * (n - 2)!) (which is 4! / (2! * (4 - 2)!) = 24 / 4 = 6 in your example) with ! being the factorial.
First, initializing the array with the first n natural numbers should be quite easy using the array element index. However, the index is a property of the array elements, so you don't need to initialize them in the first place.
You need 2 nested loops processing the array. The outer loop ranges i from 1 to n - 1, the inner loop ranges j from 2 to n. If your indexes start from 0 instead of 1, you have to take this into consideration for the loop limits. Now, you only need to fill your target array with the combinations [i, j]. To find the correct index in your target array, you should use a third counter variable, initialized with the first index and incremented at the end of the inner loop.
I agree, the math behind is not that hard and I think this explanation should suffice to develop the corresponding code yourself.
So in my text book there is this example of a recursive function using f#
let rec gcd = function
| (0,n) -> n
| (m,n) -> gcd(n % m,m);;
with this function my text book gives the example by executing:
gcd(36,116);;
and since the m = 36 and not 0 then it ofcourse goes for the second clause like this:
gcd(116 % 36,36)
gcd(8,36)
gcd(36 % 8,8)
gcd(4,8)
gcd(8 % 4,4)
gcd(0,4)
and now hits the first clause stating this entire thing is = 4.
What i don't get is this (%)percentage sign/operator or whatever it is called in this connection. for an instance i don't get how
116 % 36 = 8
I have turned this so many times in my head now and I can't figure how this can turn into 8?
I know this is probably a silly question for those of you who knows this but I would very much appreciate your help the same.
% is a questionable version of modulo, which is the remainder of an integer division.
In the positive, you can think of % as the remainder of the division. See for example Wikipedia on Euclidean Divison. Consider 9 % 4: 4 fits into 9 twice. But two times four is only eight. Thus, there is a remainder of one.
If there are negative operands, % effectively ignores the signs to calculate the remainder and then uses the sign of the dividend as the sign of the result. This corresponds to the remainder of an integer division that rounds to zero, i.e. -2 / 3 = 0.
This is a mathematically unusual definition of division and remainder that has some bad properties. Normally, when calculating modulo n, adding or subtracting n on the input has no effect. Not so for this operator: 2 % 3 is not equal to (2 - 3) % 3.
I usually have the following defined to get useful remainders when there are negative operands:
/// Euclidean remainder, the proper modulo operation
let inline (%!) a b = (a % b + b) % b
So far, this operator was valid for all cases I have encountered where a modulo was needed, while the raw % repeatedly wasn't. For example:
When filling rows and columns from a single index, you could calculate rowNumber = index / nCols and colNumber = index % nCols. But if index and colNumber can be negative, this mapping becomes invalid, while Euclidean division and remainder remain valid.
If you want to normalize an angle to (0, 2pi), angle %! (2. * System.Math.PI) does the job, while the "normal" % might give you a headache.
Because
116 / 36 = 3
116 - (3*36) = 8
Basically, the % operator, known as the modulo operator will divide a number by other and give the rest if it can't divide any longer. Usually, the first time you would use it to understand it would be if you want to see if a number is even or odd by doing something like this in f#
let firstUsageModulo = 55 %2 =0 // false because leaves 1 not 0
When it leaves 8 the first time means that it divided you 116 with 36 and the closest integer was 8 to give.
Just to help you in future with similar problems: in IDEs such as Xamarin Studio and Visual Studio, if you hover the mouse cursor over an operator such as % you should get a tooltip, thus:
Module operator tool tip
Even if you don't understand the tool tip directly, it'll give you something to google.
If I enter a value, for example
1234567 ^ 98787878
into Wolfram Alpha it can provide me with a number of details. This includes decimal approximation, total length, last digits etc. How do you evaluate such large numbers? As I understand it a programming language would have to have a special data type in order to store the number, let alone add it to something else. While I can see how one might approach the addition of two very large numbers, I can't see how huge numbers are evaluated.
10^2 could be calculated through repeated addition. However a number such as the example above would require a gigantic loop. Could someone explain how such large numbers are evaluated? Also, how could someone create a custom large datatype to support large numbers in C# for example?
Well it's quite easy and you can have done it yourself
Number of digits can be obtained via logarithm:
since `A^B = 10 ^ (B * log(A, 10))`
we can compute (A = 1234567; B = 98787878) in our case that
`B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...`
integer part + 1 (601767807 + 1 = 601767808) is the number of digits
First, say, five, digits can be gotten via logarithm as well;
now we should analyze fractional part of the
B * log(A, 10) = 98787878 * log(1234567, 10) = 601767807.4709646...
f = 0.4709646...
first digits are 10^f (decimal point removed) = 29577...
Last, say, five, digits can be obtained as a corresponding remainder:
last five digits = A^B rem 10^5
A rem 10^5 = 1234567 rem 10^5 = 34567
A^B rem 10^5 = ((A rem 10^5)^B) rem 10^5 = (34567^98787878) rem 10^5 = 45009
last five digits are 45009
You may find BigInteger.ModPow (C#) very useful here
Finally
1234567^98787878 = 29577...45009 (601767808 digits)
There are usually libraries providing a bignum datatype for arbitrarily large integers (eg. mapping digits k*n...(k+1)*n-1, k=0..<some m depending on n and number magnitude> to a machine word of size n redefining arithmetic operations). for c#, you might be interested in BigInteger.
exponentiation can be recursively broken down:
pow(a,2*b) = pow(a,b) * pow(a,b);
pow(a,2*b+1) = pow(a,b) * pow(a,b) * a;
there also are number-theoretic results that have engenedered special algorithms to determine properties of large numbers without actually computing them (to be precise: their full decimal expansion).
To compute how many digits there are, one uses the following expression:
decimal_digits(n) = 1 + floor(log_10(n))
This gives:
decimal_digits(1234567^98787878) = 1 + floor(log_10(1234567^98787878))
= 1 + floor(98787878 * log_10(1234567))
= 1 + floor(98787878 * 6.0915146640862625)
= 1 + floor(601767807.4709647)
= 601767808
The trailing k digits are computed by doing exponentiation mod 10^k, which keeps the intermediate results from ever getting too large.
The approximation will be computed using a (software) floating-point implementation that effectively evaluates a^(98787878 log_a(1234567)) to some fixed precision for some number a that makes the arithmetic work out nicely (typically 2 or e or 10). This also avoids the need to actually work with millions of digits at any point.
There are many libraries for this and the capability is built-in in the case of python. You seem primarily concerned with the size of such numbers and the time it may take to do computations like the exponent in your example. So I'll explain a bit.
Representation
You might use an array to hold all the digits of large numbers. A more efficient way would be to use an array of 32 bit unsigned integers and store "32 bit chunks" of the large number. You can think of these chunks as individual digits in a number system with 2^32 distinct digits or characters. I used an array of bytes to do this on an 8-bit Atari800 back in the day.
Doing math
You can obviously add two such numbers by looping over all the digits and adding elements of one array to the other and keeping track of carries. Once you know how to add, you can write code to do "manual" multiplication by multiplying digits and putting the results in the right place and a lot of addition - but software will do all this fairly quickly. There are faster multiplication algorithms than the one you would use manually on paper as well. Paper multiplication is O(n^2) where other methods are O(n*log(n)). As for the exponent, you can of course multiply by the same number millions of times but each of those multiplications would be using the previously mentioned function for doing multiplication. There are faster ways to do exponentiation that require far fewer multiplies. For example you can compute x^16 by computing (((x^2)^2)^2)^2 which involves only 4 actual (large integer) multiplications.
In practice
It's fun and educational to try writing these functions yourself, but in practice you will want to use an existing library that has been optimized and verified.
I think a part of the answer is in the question itself :) To store these expressions, you can store the base (or mantissa), and exponent separately, like scientific notation goes. Extending to that, you cannot possibly evaluate the expression completely and store such large numbers, although, you can theoretically predict certain properties of the consequent expression. I will take you through each of the properties you talked about:
Decimal approximation: Can be calculated by evaluating simple log values.
Total number of digits for expression a^b, can be calculated by the formula
Digits = floor function (1 + Log10(a^b)), where floor function is the closest integer smaller than the number. For e.g. the number of digits in 10^5 is 6.
Last digits: These can be calculated by the virtue of the fact that the expression of linearly increasing exponents form a arithmetic progression. For e.g. at the units place; 7, 9, 3, 1 is repeated for exponents of 7^x. So, you can calculate that if x%4 is 0, the last digit is 1.
Can someone create a custom datatype for large numbers, I can't say, but I am sure, the number won't be evaluated and stored.
EDIT
So it seems I "underestimated" what varying length numbers meant. I didn't even think about situations where the operands are 100 digits long. In that case, my proposed algorithm is definitely not efficient. I'd probably need an implementation who's complexity depends on the # of digits in each operands as opposed to its numerical value, right?
As suggested below, I will look into the Karatsuba algorithm...
Write the pseudocode of an algorithm that takes in two arbitrary length numbers (provided as strings), and computes the product of these numbers. Use an efficient procedure for multiplication of large numbers of arbitrary length. Analyze the efficiency of your algorithm.
I decided to take the (semi) easy way out and use the Russian Peasant Algorithm. It works like this:
a * b = a/2 * 2b if a is even
a * b = (a-1)/2 * 2b + a if a is odd
My pseudocode is:
rpa(x, y){
if x is 1
return y
if x is even
return rpa(x/2, 2y)
if x is odd
return rpa((x-1)/2, 2y) + y
}
I have 3 questions:
Is this efficient for arbitrary length numbers? I implemented it in C and tried varying length numbers. The run-time in was near-instant in all cases so it's hard to tell empirically...
Can I apply the Master's Theorem to understand the complexity...?
a = # subproblems in recursion = 1 (max 1 recursive call across all states)
n / b = size of each subproblem = n / 1 -> b = 1 (problem doesn't change size...?)
f(n^d) = work done outside recursive calls = 1 -> d = 0 (the addition when a is odd)
a = 1, b^d = 1, a = b^d -> complexity is in n^d*log(n) = log(n)
this makes sense logically since we are halving the problem at each step, right?
What might my professor mean by providing arbitrary length numbers "as strings". Why do that?
Many thanks in advance
What might my professor mean by providing arbitrary length numbers "as strings". Why do that?
This actually change everything about the problem (and make your algorithm incorrect).
It means than 1234 is provided as 1,2,3,4 and you cannot operate directly on the whole number. You need to analyze your algorithm in terms of #additions, #multiplications, #divisions.
You should expect a division to be a bit more expensive than a multiplication, and a multiplication to be lot more expensive than an addition. So a good algorithm try to reduce the number of divisions and multiplications.
Check out the Karatsuba algorithm, (ps don't copy it that's not what your teacher want) is one of the fastest for this specification.
Add 3): Native integers are limited in how large (or small) numbers they can represent (32- or 64-bit integers for example). To represent arbitrary length numbers you can choose strings, because then you are not really limited by this. The problem is then, of course, that your arithmetic units are not really made to add strings ;-)