I was going through the text Introduction to Algorithms by Cormen et. al. where I came across the following statement:

If the keys are known to be random real numbers $k$ independently and uniformly distributed in the range $0 \leq k < 1$ , the hash function

$$h(k) = \lfloor km \rfloor$$ satisfies the condition of simple uniform hashing.

Now what I can understand that they are probably considering uniform disturbution in the "continuous" sense and not in the discrete sense. Had it been in the discrete sense then suppose for $n$ keys the probabity mass function (p.m.f) shall be constant and equal to $1/n$ and so it shall be equally likely for each key to be used in the hashing there-by yeilding the desired result.

But we seem sort of in trouble if the distribution being referred to is continuous (I feel so because of the line: "uniformly distributed in the range $0 \leq k < 1$")

Let $f(x)$ be the associated probability density function (p.d.f) and from the given information we have $f(x)=1$,(which is quite easily found, integrating $f(x)$ in the range $0$ to $1$ and equating it with $1$ and noting that in uniform distribution the p.d.f is a constant).

Now though the p.d.f is a constant but p.d.f is not the probability. Rather probability at a spectrum point is $0$. Now how to use this result to get to the claim of the authors.

Or am I entirely at fault considering the distribution to be continuous?

(There is an answer here, but it does not go into this detail as the question there is different after all).

  • 1
    $\begingroup$ $k$ is a continuous random variable with a constant density function, nothing wrong with that. $\endgroup$
    – Ariel
    Commented Jun 25, 2020 at 20:48

1 Answer 1


$h\in [m]^U$ satisfies the simple uniform hashing assumption if when $x\in U$ is chosen uniformly at random, then $h(x)$ is uniformly distributed over $[m]$, or equivalently $\forall i\in[m]: \Pr\limits_{x\in U}[h(x)=i]=\frac{1}{m}$. In our case we have:

$\Pr[h(x)=i]=\Pr\big[\lfloor mx \rfloor=i\big]=\Pr[i\le mx < i+1]=\frac{i+1}{m}-\frac{i}{m}=\frac{1}{m}$.

We used the fact that if $x$ is uniformly distributed over $[0,1]$ then $\Pr[a\le x\le b]=b-a$ (the equality holds with all four combinations of $\le$ and $<$).

  • $\begingroup$ can you please say the meaning of the notation $h\in [m]^U$ $\endgroup$ Commented Jun 25, 2020 at 20:52
  • 2
    $\begingroup$ In this context $[m]=\{0,1,...,,m-1\}$ and $[m]^U$ is the set of functions from $U$ to $[m]$. $\endgroup$
    – Ariel
    Commented Jun 25, 2020 at 20:53
  • $\begingroup$ thanks for the help. $\endgroup$ Commented Jun 25, 2020 at 20:55

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