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1

Depending on your specific inner product, I will assume that the inner product is given by $\langle u,v\rangle = \sum(u)_i(v)_i$ and that we are working in $\mathbb{F}_2^7$ as our vector space (i.e. the space of binary sequences of length seven). So, the inner product between the two is given by multiplying each position together and then adding: ...


0

Consider the following question: $\dagger$ What is the number of 0-1 sequences of length $Y$ with exactly $S$ many '1's but no subsequence of $X+1$ consecutive '1's? I will only address the case that $X+1$ divides $Y$ and leave the general case to you. Let us split the sequence of length $Y$ into chunks of length $X+1$ and then put exactly one '0' into ...


1

Hint: By parity, Player B can guarantee that he can duplicate Player A's move. The only concern is if Player A drew from the 1 pile. Deal with that.


0

Hint: the binary representation of $\frac{a}{2^k}$ is obtained by shifting the digits (or bits, if you prefer) in the binary representation of $a$ by $k$ places to the right.


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I do not think you will get a satisfactory answer here. In fact, after thinking about the most appropriate stackexchange site for your question, I concluded that it is the philosophy stackexchange. All joking aside, the question you are asking is very hard from multiple perspectives. To show causal relationships ideally you want randomized experiments and ...


1

To expand on my comment: Let's say we want to convert $409$ to binary. We first need to know how many bits we'll need; we need to know the largest power of $2$ that's less than $409$. We'll also need all smaller powers of $2$, so let's write those down (they are all very easy to calculate by hand). \begin{array}{c|c|c|c|c|c} k & 8 & 7 & 6 & ...


1

You can convert to a different power-of-two base, e.g. octal or hexadecimal, first. That's $3$ or $4$ times fewer divisions, and then you can very quickly convert to binary using a lookup table for the octal or hexadecimal digits.


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You have a bit string with a point, $(b_{N-1}b_{N-2}\ldots b_P\color{red}.b_{P-1}\ldots b_0)_{\text{two}}$, with $P$ bits to the right of the point (which is easy to miss, so I’ve colored it red). Let’s ignore the leftmost bit for a moment. The remaining $N-1$ bits are first interpreted in the usual way. The binary number $(b_{N-2}b_{N-3}\ldots ...


1

This is why: 0000 -> 0111 = 0 -> 7 1111 -> 1000 = -1 -> -8 both series can represent 8 numbers, the positive series already include the 0 so the negative serie doesn't have to do this.


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Trying to catch the direction of the question, probably with something like taking advantage of the fact that 10=5*2 y making use of modular arithmetic to calculate the decimal overflow, i could'n t imagine other way without entry in a kind of complex "iterating" stuff... and then transform to binary. pd: Like a detour... ...


0

If your $x$ is between $0$ and $1$, you can write $x=\sum_{i=1}^\infty a_i2^{-i}$ where $a_i \in \{0,1\}$ are binary digits of the expansion. If it is not, you can add the integral part of $x$ converted to binary to this expression. You can't have an infinite binary string to the left of the fraction point as the value would be infinite.


1

Suppose that the root is on level $n$. If $n=0$ or $n=1$, none of the nodes pays or receives anything, so each node has a balance of $0$. (To avoid clutter I omit the dollar sign.) If $n=2$, the $4$ nodes on level $0$ pay $7$ each and receive nothing, so they end up with a balance of $-7$; the $2$ nodes on level $1$ neither pay nor receive and have a balance ...


2

Recall that for a parity check matrix $H$ and codeword $c$, $Hc^T = 0$ (by definition). When you send a codeword $c$ over the channel and receive $r$ on the other end, it effectively gets a noise vector $n$ added to it: $r = c + n$. Thus the syndrome is $s = Hr^T = H (c + n)^T = Hc^T + Hn^T = Hn^T$. The most likely noise vector is the one with the lowest ...


1

Flipping all possible bits until the errors are fixed is very similar to minimum distance decoding, and that's what you probably have to do if there are more errors than can be corrected by the code. Syndrome decoding also works with a lookup-table, but with fewer entries. I'm not sure, but it seems that "interpreting the syndrome as an integer" might only ...


0

Let the decimal be $0.d_1d_2d_3...$. When we are trying to express it in binary, we want to express it as $\frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + ...$ Now, if the original fraction is more than $\frac{1}{2}$, we want $a_1 = 1$, otherwise $a_1 = 0$. The best way to check that is to as to multiply LHS by $2$. If you get an integer part 1, put $a_1 ...


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It has to do with the invention of the method of calculus. The slope y/x us making x to be the thickness of 1. Then del x was used as 0. Also in number theory, what are the numbers that has meaning beyond just to "count". Then 0 and 1 are the only numbers. Also the truth of Yin and Yang really are every where by as big as the heliocentric world and the ...


1

The simplest thing to do, especially from an algorithmic standpoint, is to do successive doubling: if $n$ is the decimal integer to be converted to binary, compute powers of $2$ using a recursive loop, at each step checking the condition that you have not exceeded $n$: So if $n = 237$, then we calculate: $$1, 2, 4, 8, 16, 32, 64, 128, 256.$$ The final ...


2

I'll give you a method and then I'll explain it. I'll use 237 to demonstrate the method. Calculate $\log_2{(237)} = 7.88\ldots$ we only care about the first digit as it tells us that $2^7$ is the largest power of 2 smaller than 237. An explanation of this is that is that the logarithm tells us what we need to take 2 to the power of to get 237. As the ...



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