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This would make sense if we interpret $A \cap B$ as the bitwise "and" ($\land$) (or intersection, if you look at sets interpretation), $A \cup B$ as the bitwise "or" ($\lor$) of the numbers (which is their union in the set interpretation). E.g. $7 \cap 11 =$ 0b0111 $\land$ 0b1011 $=$ 0b0011 $= 3$, $7 \cup 11 =$ 0b0111 $\lor$ 0b1011 $=$ 0b1111 $=15$ , so ...

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By inclusion-exclusion, the probability that neither counter remained zero is $$1-\left(1-\frac1{l_a}\right)^k-\left(1-\frac1{l_b}\right)^k+\left(1-\frac1{l_a}-\frac1{l_b}+\frac{2^c}{l_al_b}\right)^k\;,$$ where the second term is the probability that the first counter remained zero, the third term is the probability that the second counter remained zero ...

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I will assume we are dealing with unsigned integers in binary representation or an boolean algebra which behave like that. i.e. algebra whose elements can be viewed as a sequence of binary digits and multiplication by $2$ corresponds to a shift of the binary digits to the right. i.e. $$z = (z_0, z_1, z_2, \ldots )\quad\implies\quad 2z = ( 0, z_0, z_1, \... 7 Think about the procedure that you had in mind. If the binary representation of n ends in 1, then n is odd; say n=2k+1. Then a_n=a_k+1, and the binary representation of k is simply what’s left when you drop the last digit of n. If the binary representation of n ends in 0, then n is even; say n=2k. Then a_n=a_k-1, and again the binary ... 4 Actually, you made a mistake breaking down a_{2015}. It should be a_{2015}=a_{2\times 1007+1}=a_{1007}+1. And a_{1007}=a_{2\times503+1}=a_{503}+1 a_{503}=a_{2\times 251+1}=a_{251}+1 a_{251}=a_{2\times 125+1}=a_{125}+1 a_{125}=a_{2\times 62+1}=a_{62}+1 a_{62}=a_{2\times 31+1}=a_{31}-1 a_{31}=a_{2\times 15+1}=a_{15}+1 a_{15}=a_{2\times ... 0 Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex. f(x) = \left \lceil \frac{1}{2\Gamma \left ( \left | x \right | \... 0 I would use something very simple if a piecewise function does not suffice:$$f\left( x\right) =\left( \dfrac {x}{x}\right) ^{-1}$$This works, because any number divided by intself is 1, whose inverse is still one—except at x=0. Using limits, most people agree that \dfrac {n}{0}=\pm \infty if n\in \mathbb{R} (although it could hypothetically hold ... 10 You've already defined your function (assuming you've also chosen its domain). One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so. Typical notation for a function created by the sort of description you give is a definition by cases:$$ f(x) := \begin{cases} 0 & x = 0 \\ 1 &...

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How about $f(x)= 1-\delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $\delta_{x,0}$ as kroneckerdelta(x,0) )

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How about $f(x)=\left\lceil\frac{x^2}{x^2+1}\right\rceil$ *Works for real numbers, with imaginary numbers you may divide by 0.

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If you have an $n$-bit word $w$, flipping all of the bits is the same as subtracting it from $\underbrace{11\cdots 11}_n{}_2$ -- imagine subtracting binary digit for digit; there's never any borrows! Therefore your first procedure computes $$\underbrace{11\cdots 11}_n{}_2 - (w-1)$$ By a simple algebraic rearrangement, this is the same as $$\underbrace{11\... 1 I'll answer your last question first since it will give the answer to your other questions. The basic method for establishing that 2 sets have a 1:1 correspondence is constructing a bijection i.e. a one to one and onto function between them. For example, the set of integers \mathbb{Z} is in one to one correspondence with the set of all rationals \mathbb{Q}... 2 C is the set of all infinite binary sequences - that is, an element of C looks like (b_1, b_2, b_3, . . .), where each b_i is either 0 or 1. So for instance (0, 1, 0, 1, 0, 1, . . . ) is an element of C, and in this case b_1=0, b_2=1, b_3=0, . . . Re: your first question, you're trying a bit too hard - there's a simpler way. Suppose I have ... 0 The question about a vector of length one, i.e. the scalar case, made the answer clear: the comparison (x \succ y) \ne (y \succ x) so the distance is not symmetric. It is not a distance metric. 1 It might be simpler than that, but without knowing exactly what pictures you have in your head, I can't be 100% sure. I think it is a question of remembering that concatenation is essentially multiplication by x^n. You take$$R(x) = M(x)\cdot x^n\mod G(x)$$You say that you "append" R(x) to M(x), which is to say that you calculate$$RM(x) = (M(x)\...

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I believe your doubt is not really related to CRC but with cyclic codes (on which CRC are based), more specifically with the construction of systematic cyclic codes. This is explained in any textboox. Here's a summary. A binary $(n,k)$ cyclic code has $2^k$ codewords, they correspond to a binary polynomial $c(X)$ of degree less than $n$. A particular code ...

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1101 + 0100 = 0001 is an overflow if it is a wrong answer and not an overflow if it is a correct answer. If these are unsigned binary numbers then 13+4=1 is wrong, so there is an overflow. In fact, with unsigned binary, a carry out is always an overflow. But you have specified 2s-complement binary. In that case, (-3)+4=1 is right, so there is no overflow. ...

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I do not quite get the question. If the question is "Does there occur an overflow", the answer is yes only if the frame size is 4 bits. If the frame is 4 bits only, you cannot add these numbers without the loss of the most significant bit. Addition starts at the least significant bit (the right one). 0+1 = 1, No carry 0+0 = 0, No carry 1+1 = 0, Carry value ...

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In the paper "On nonrepetitive sequences" by Entringer and Jackson, in J. Combin. Theory Ser. A 11 (1974), 159–164. The link is http://www.sciencedirect.com/science/article/pii/0097316574900417, the authors show that any binary sequence of length more than 18 must have two identical consecutive blocks, the length of which is at least 2. The proof is by case ...

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You can deliberately define an irrational number to have this non-repeating quality, if required. For example, for an irrational number that avoids 3-blocks of repeating digits, take the binary definition of $\pi$ and generate a new number such that each digit $x$ in $\pi$ is replaced by $11x00$. (This also avoids 4-blocks). To avoid any repetitions of ...

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