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In the picture below from CLRS, I fail to understand why exactly $h(k)$ = the $p$ highest-order bits of the lower w-bit half of the product.

For context, this is supposed to compute $h(k) = \lfloor m (k A \; \text{mod} 1) \rfloor $

enter image description here

For further context, CLRS mentions the following, but I still don't quite get why those $p$ highest-order bits are the ones we are looking for.

enter image description here

Josh
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1 Answers1

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$(kA \bmod 1)$ is in the range $[0,1)$. So multiplying that by $2^p$ gives a number in the range $[0,2^p)$. That is:

$$\left\lfloor 2^p (kA \bmod 1) \right\rfloor = \left\lfloor 2^p (kA \bmod 1) \right \rfloor \bmod 2^p$$

Once you've worked that out, it's not too hard to see:

$$\left\lfloor 2^p (kA \bmod 1) \right \rfloor \bmod 2^p = \left\lfloor kA 2^p \right\rfloor \bmod 2^p = \left\lfloor \frac{ks}{2^{w-p}} \right\rfloor \bmod 2^p$$

So you can implement this by taking $ks$, shifting it $w-p$ bits to the right, then taking the lowest order $p$ bits. Which is exactly the same as taking the highest order $p$ bits of the low word.

Pseudonym
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    In case it helps others who reads this: The result of $x ; \text{mod} ; b^p$ is the last $p$ digits of the number $x$ represented in base $b$. For example: $x ; \text{mod} ; 10$ gives you the last digit of $x$ in base 10. Similarly, $x ; \text{mod} ; 100$ gives you the last two digits of $x$ in base 10, etc. – Josh Apr 09 '20 at 16:43
  • Maybe I read this too quickly, sorry. I fail to see the first equivalence: $$\left\lfloor 2^p (kA \bmod 1) \right\rfloor = \left\lfloor 2^p (kA \bmod 1) \right \rfloor \bmod 2^p$$ – Josh Apr 09 '20 at 19:37
  • Do you $(kA see that \bmod 1) \in [0,1)$? Therefore $\left\lfloor 2^p (kA \bmod 1) \right\rfloor < 2^p$. – Pseudonym Apr 10 '20 at 07:41
  • Thanks - the first formula that you wrote in your comment just above didn't come out well, but I think I follow you. I see that $(k A \bmod 1) \in [0,1)$, i.e. this formula takes the fractional part of the product $kA$ so the result must be $\in [0, 1)$. I also see that if $n$ is a positive integer and $x \in [0,1)$, we must have $n x \leq n$, and thus we would also have $n x \bmod n \leq n$. Is that the argument behind the first Eq. in your answer, i.e. $\left\lfloor 2^p (kA \bmod 1) \right\rfloor = \left\lfloor 2^p (kA \bmod 1) \right \rfloor \bmod 2^p$ ? – Josh Apr 11 '20 at 01:44
  • Yes, that's it. – Pseudonym Apr 11 '20 at 01:49
  • Thanks! And the last bit, why is it that $\left\lfloor 2^p (kA \bmod 1) \right \rfloor \bmod 2^p = \left\lfloor kA 2^p \right\rfloor \bmod 2^p$ ? (i.e. why can we "drop" the $\text{mod} ;1$ part? – Josh Apr 11 '20 at 01:52
  • Because the $\bmod 2^p$ part does the job. – Pseudonym Apr 12 '20 at 02:27
  • @Pseudonym Can you please elaborate on the last part a bit further? – Yueor Jul 17 '21 at 15:12