# Tag Info

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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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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 ...

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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 ...

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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.

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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 ...

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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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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 & ...

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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 ...

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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: ...

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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 ...

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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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