高级算法 (Fall 2018)/Hashing and Sketching

Distinct Elements

Consider the following problem of counting distinct elements: Suppose that [math]\displaystyle{ \Omega }[/math] is a sufficiently large universe.

• Input: a sequence of (not necessarily distinct) elements [math]\displaystyle{ x_1,x_2,\ldots,x_n\in\Omega }[/math];
• Output: an estimation of the total number of distinct elements [math]\displaystyle{ z=|\{x_1,x_2,\ldots,x_n\}| }[/math].

A straightforward way of solving this problem is to maintain a dictionary data structure, which costs at least linear ([math]\displaystyle{ O(n) }[/math]) space. For big data, where [math]\displaystyle{ n }[/math] is very large, this is still too expensive. However, due to an information-theoretical argument, linear space is necessary if you want to compute the exact value of [math]\displaystyle{ z }[/math].

Our goal is to relax the problem a little bit to significantly reduce the space cost by tolerating approximate answers. The form of approximation we consider is [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator.

[math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator

A random variable [math]\displaystyle{ \widehat{Z} }[/math] is an [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator of a quantity [math]\displaystyle{ z }[/math] if [math]\displaystyle{ \Pr[\,(1-\epsilon)z\le \widehat{Z}\le (1+\epsilon)z\,]\ge 1-\delta }[/math]. [math]\displaystyle{ \widehat{Z} }[/math] is said to be an unbiased estimator of [math]\displaystyle{ z }[/math] if [math]\displaystyle{ \mathbb{E}[\widehat{Z}]=z }[/math].

Usually [math]\displaystyle{ \epsilon }[/math] is called approximation error and [math]\displaystyle{ \delta }[/math] is called confidence error.

We now present an elegant algorithm introduced by Flajolet and Martin in 1984. The algorithm can be implemented in data stream model: The input elements [math]\displaystyle{ x_1,x_2,\ldots,x_n }[/math] is presented to the algorithm one at a time, where the size of data [math]\displaystyle{ n }[/math] is unknown to the algorithm. The algorithm maintains a value [math]\displaystyle{ \widehat{Z} }[/math] which is an [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator of the total number of distinct elements [math]\displaystyle{ z=|\{x_1,x_2,\ldots,x_n\}| }[/math], using only a small amount of memory space to memorize (with loss) the data set [math]\displaystyle{ \{x_1,x_2,\ldots,x_n\} }[/math].

A famous quotation of Flajolet describes the performance of this algorithm as:

"Using only memory equivalent to 5 lines of printed text, you can estimate with a typical accuracy of 5% and in a single pass the total vocabulary of Shakespeare."

An estimator by hashing

Suppose that we can access to an idealized random hash function [math]\displaystyle{ h:\Omega\to[0,1] }[/math] which is uniformly distributed over all mappings from the universe [math]\displaystyle{ \Omega }[/math] to unit interval [math]\displaystyle{ [0,1] }[/math].

Recall that the input sequence [math]\displaystyle{ x_1,x_2,\ldots,x_n\in\Omega }[/math] consists of [math]\displaystyle{ z=|\{x_1,x_2,\ldots,x_n\}| }[/math] distinct elements. These elements are mapped by the random function [math]\displaystyle{ h }[/math] to [math]\displaystyle{ z }[/math] hash values uniformly and independently distributed in [math]\displaystyle{ [0,1] }[/math]. We could maintain these hash values instead of the original elements, but this would still be too expensive because in the worst case we still have up to [math]\displaystyle{ n }[/math] distinct values to maintain. However, due to the idealized random hash function, the unit interval [math]\displaystyle{ [0,1] }[/math] will be partitioned into [math]\displaystyle{ z+1 }[/math] subintervals by these [math]\displaystyle{ z }[/math] uniform and independent hash values. The typical length of the subinterval gives an estimation of the number [math]\displaystyle{ z }[/math].

Proposition

[math]\displaystyle{ \mathbb{E}\left[\min_{1\le i\le n}h(x_i)\right]=\frac{1}{z+1} }[/math].

Proof. The input sequence [math]\displaystyle{ x_1,x_2,\ldots,x_n\in\Omega }[/math] consisting of [math]\displaystyle{ z }[/math] distinct elements are mapped to [math]\displaystyle{ z }[/math] random hash values uniformly and independently distributed in [math]\displaystyle{ [0,1] }[/math]. These [math]\displaystyle{ z }[/math] hash values partition the unit interval [math]\displaystyle{ [0,1] }[/math] into [math]\displaystyle{ z+1 }[/math] subintervals [math]\displaystyle{ [0,v_1],[v_1,v_2],[v_2,v_3]\ldots,[v_{z-1},v_z],[v_z,1] }[/math], where [math]\displaystyle{ v_i }[/math] denotes the [math]\displaystyle{ i }[/math]-th smallest value among all hash values [math]\displaystyle{ \{h(x_1),h(x_2),\ldots,h(x_n)\} }[/math]. Clearly we have [math]\displaystyle{ v_1=\min_{1\le i\le n}h(x_i) }[/math]. Meanwhile, since all hash values are uniformly and independently distributed in [math]\displaystyle{ [0,1] }[/math], the lengths of all subintervals [math]\displaystyle{ v_1, v_2-v_1, v_3-v_2,\ldots, v_z-v_{z-1}, 1-v_z }[/math] are identically distributed. By symmetry, they have the same expectation, therefore [math]\displaystyle{ (z+1)\mathbb{E}[v_1]= \mathbb{E}[v_1]+\sum_{i=1}^{z-1}\mathbb{E}[v_{i+1}-v_i]+\mathbb{E}[1-v_z] =\mathbb{E}\left[v_1+(v_2-v_1)+(v_3-v_2)+\cdots+(v_{z}-v_{z-1})+1-v_z\right] =1, }[/math] which implies that [math]\displaystyle{ \mathbb{E}\left[\min_{1\le i\le n}h(x_i)\right]=\mathbb{E}[v_1]=\frac{1}{z+1} }[/math]. [math]\displaystyle{ \square }[/math]

The quantity [math]\displaystyle{ \min_{1\le i\le n}h(x_i) }[/math] can be computed with small space cost (for storing the current smallest hash value) by scan the input sequence in a single pass. Because as we proved its expectation is [math]\displaystyle{ \frac{1}{z+1} }[/math], the smallest hash value [math]\displaystyle{ Y=\min_{1\le i\le n}h(x_i) }[/math] gives an unbiased estimator for [math]\displaystyle{ \frac{1}{z+1} }[/math]. However, [math]\displaystyle{ \frac{1}{Y-1} }[/math] is not necessarily a good estimator for [math]\displaystyle{ z }[/math]. Actually, it is a rather poor estimator. Consider for example when [math]\displaystyle{ z=1 }[/math], all input elements are the same. In this case, there is only one hash value and [math]\displaystyle{ Y=\min_{1\le i\le n}h(x_i) }[/math] is distributed uniformly over [math]\displaystyle{ [0,1] }[/math], thus [math]\displaystyle{ \frac{1}{Y-1} }[/math] fails to be close enough to the correct answer 1 with high probability.

Flajolet-Martin algorithm

The reason that the above estimator of a single hash function performs poorly is that the unbiased estimator [math]\displaystyle{ \min_{1\le i\le n}h(x_i) }[/math] has large variance. So a natural way to reduce this variance is to have multiple independent hash functions and take the average. This is precisely what Flajolet-Martin algorithm does.

Suppose that we can access to [math]\displaystyle{ k }[/math] independent random hash functions [math]\displaystyle{ h_1,h_2,\ldots,h_k }[/math], where each [math]\displaystyle{ h_j:\Omega\to[0,1] }[/math] is uniformly and independently distributed over all functions mapping [math]\displaystyle{ \Omega }[/math] to [math]\displaystyle{ [0,1] }[/math]. Here [math]\displaystyle{ k }[/math] is a parameter to be fixed by the desired approximation error [math]\displaystyle{ \epsilon }[/math] and confidence error [math]\displaystyle{ \delta }[/math]. The Flajolet-Martin algorithm is given by the following pseudocode.

Flajolet-Martin algorithm

Suppose that [math]\displaystyle{ h_1,h_2,\ldots,h_k:\Omega\to[0,1] }[/math] are [math]\displaystyle{ k }[/math] uniform and independent random hash functions, where [math]\displaystyle{ k }[/math] is a parameter to be fixed later. Scan the input sequence [math]\displaystyle{ x_1,x_2,\ldots,x_n\in\Omega }[/math] in a single pass to compute: [math]\displaystyle{ Y_j=\min_{1\le i\le n}h_j(x_i) }[/math] for every [math]\displaystyle{ j=1,2,\ldots,k }[/math]; average value [math]\displaystyle{ \overline{Y}=\frac{1}{k}\sum_{j=1}^kY_j }[/math]; return [math]\displaystyle{ \widehat{Z}=\frac{1}{\overline{Y}}-1 }[/math] as the estimator.

The algorithm is easy to implement in data stream model, with a space cost of storing [math]\displaystyle{ k }[/math] hash values. The following theorem guarantees that the algorithm returns an [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator of the total number of distinct elements for some big enough [math]\displaystyle{ k=O\left(\frac{1}{\epsilon^2\delta}\right) }[/math].

Theorem

For any [math]\displaystyle{ \epsilon,\delta\lt 1/2 }[/math], if the parameter [math]\displaystyle{ k }[/math] is set as [math]\displaystyle{ k\ge\left\lceil\frac{4}{\epsilon^2\delta}\right\rceil }[/math] then the output [math]\displaystyle{ \widehat{Z} }[/math] always gives an [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator of the correct answer [math]\displaystyle{ z }[/math].

In the following we prove this main theorem.

An obstacle to analyze the estimator [math]\displaystyle{ \widehat{Z}=\frac{1}{\overline{Y}}-1 }[/math] is that it is a nonlinear function of [math]\displaystyle{ \overline{Y} }[/math] who is easier to analyze. Nevertheless, we observe that [math]\displaystyle{ \widehat{Z} }[/math] is an [math]\displaystyle{ (\epsilon,\delta) }[/math]-estimator of [math]\displaystyle{ z }[/math] as long as [math]\displaystyle{ \overline{Y} }[/math] is an [math]\displaystyle{ (\epsilon/2,\delta) }[/math]-estimator of [math]\displaystyle{ \frac{1}{z+1} }[/math]. This can be deduced by just verifying the following:

[math]\displaystyle{ \frac{1-\epsilon/2}{z+1}\le \overline{Y}\le \frac{1+\epsilon/2}{z+1} \implies (1-\epsilon)z\le\frac{1}{\overline{Y}}-1\le (1+\epsilon)z }[/math],

for [math]\displaystyle{ \epsilon\lt \frac{1}{2} }[/math]. Therefore,

[math]\displaystyle{ \Pr\left[\,(1-\epsilon)z\le \widehat{Z} \le (1+\epsilon)z\,\right]\ge \Pr\left[\,\frac{1-\epsilon/2}{z+1}\le \overline{Y}\le \frac{1+\epsilon/2}{z+1}\,\right] =\Pr\left[\,\left|\overline{Y}-\frac{1}{z+1}\right|\le \frac{\epsilon/2}{z+1}\,\right] }[/math].

It is then sufficient to show that [math]\displaystyle{ \Pr\left[\,\left|\overline{Y}-\frac{1}{z+1}\right|\le \frac{\epsilon/2}{z+1}\,\right]\ge 1-\delta }[/math] for proving the main theorem above. We will see that this is equivalent to show the concentration inequality

[math]\displaystyle{ \Pr\left[\,\left|\overline{Y}-\mathbb{E}\left[\overline{Y}\right]\right|\le \frac{\epsilon/2}{z+1}\,\right]\ge 1-\delta\quad\qquad({\color{red}*}) }[/math].

Lemma

The followings hold for each [math]\displaystyle{ Y_j }[/math], [math]\displaystyle{ j=1,2\ldots,k }[/math], and [math]\displaystyle{ \overline{Y}=\frac{1}{k}\sum_{j=1}^kY_j }[/math]: [math]\displaystyle{ \mathbb{E}\left[\overline{Y}\right]=\mathbb{E}\left[Y_j\right]=\frac{1}{z+1} }[/math]; [math]\displaystyle{ \mathbf{Var}\left[Y_j\right]\le\frac{1}{(z+1)^2} }[/math], and consequently [math]\displaystyle{ \mathbf{Var}\left[\overline{Y}\right]\le\frac{1}{k(z+1)^2} }[/math].

Proof. As in the case of single hash function, by symmetry it holds that [math]\displaystyle{ \mathbb{E}[Y_j]=\frac{1}{z+1} }[/math] for every [math]\displaystyle{ j=1,2,\ldots,k }[/math]. Therefore, [math]\displaystyle{ \mathbb{E}\left[\overline{Y}\right]=\frac{1}{k}\sum_{j=1}^k\mathbb{E}[Y_j]=\frac{1}{z+1} }[/math]. Recall that each [math]\displaystyle{ Y_j }[/math] is the minimum of [math]\displaystyle{ z }[/math] random hash values uniformly and independently distributed over [math]\displaystyle{ [0,1] }[/math]. By geometry probability, it holds that for any [math]\displaystyle{ y\in[0,1] }[/math], [math]\displaystyle{ \Pr[Y_j\gt y]=(1-y)^z }[/math], which means [math]\displaystyle{ \Pr[Y_j\le y]=1-(1-y)^z }[/math]. Taking the derivative with respect to [math]\displaystyle{ y }[/math], we obtain the probability density function of random variable [math]\displaystyle{ Y_j }[/math], which is [math]\displaystyle{ z(1-y)^{z-1} }[/math]. We then compute the second moment. [math]\displaystyle{ \mathbb{E}[Y_j^2]=\int^{1}_0y^2z(1-y)^{z-1}\,\mathrm{d}y=\frac{2}{(z+1)(z+2)} }[/math]. The variance is bounded as [math]\displaystyle{ \mathbf{Var}\left[Y_j\right]=\mathbb{E}\left[Y_j^2\right]-\mathbb{E}\left[Y_j\right]^2=\frac{2}{(z+1)(z+2)}-\frac{1}{(z+1)^2}\le\frac{1}{(z+1)^2} }[/math]. Due to the (pairwise) independence between [math]\displaystyle{ Y_j }[/math]'s, [math]\displaystyle{ \mathbf{Var}\left[\overline{Y}\right]=\mathbf{Var}\left[\frac{1}{k}\sum_{j=1}^kY_j\right]=\frac{1}{k^2}\sum_{j=1}^k\mathbf{Var}\left[Y_j\right]\le \frac{1}{k(z+1)^2} }[/math]. [math]\displaystyle{ \square }[/math]

We resume to prove the inequality [math]\displaystyle{ ({\color{red}*}) }[/math]. By Chebyshev's inequality, it holds that

[math]\displaystyle{ \Pr\left[\,\left|\overline{Y}-\mathbb{E}\left[\overline{Y}\right]\right|\gt \frac{\epsilon/2}{z+1}\,\right] \le\frac{4}{\epsilon^2}(z+1)^2\mathbf{Var}\left[\overline{Y}\right] \le\frac{4}{\epsilon^2k} }[/math].

When [math]\displaystyle{ k\ge\left\lceil\frac{4}{\epsilon^2\delta}\right\rceil }[/math], this probability is at most [math]\displaystyle{ \delta }[/math]. The inequality [math]\displaystyle{ ({\color{red}*}) }[/math] is proved. As we discussed above, this proves the main theorem.

Set Membership

Perfect hashing

Perfect hashing is a data structure for storing a static dictionary. In a static dictionary, a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] items from the universe [math]\displaystyle{ [N] }[/math] are preprocessed and stored in a table. Once the table is constructed, it will nit be changed any more, but will only be used for search operations: a search for an item gives the location of the item in the table or returns that the item is not in the table. You may think of an application that we store an encyclopedia in a DVD, so that searches are very efficient but there will be no updates to the data.

This problem can be solved by binary search on a sorted table or balanced search trees in [math]\displaystyle{ O(\log n) }[/math] time for a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] elements. We show how to solve this problem with [math]\displaystyle{ O(1) }[/math] time by perfect hashing.

Perfect hashing using quadratic space

The idea of perfect hashing is that we use a hash function [math]\displaystyle{ h }[/math] to map the [math]\displaystyle{ n }[/math] items to distinct entries of the table; store every item [math]\displaystyle{ x\in S }[/math] in the entry [math]\displaystyle{ h(x) }[/math]; and also store the hash function [math]\displaystyle{ h }[/math] in a fixed location in the table (usually the beginning of the table). The algorithm for searching for an item is as follows:

search for [math]\displaystyle{ x }[/math] in table [math]\displaystyle{ T }[/math]:

1. retrieve [math]\displaystyle{ h }[/math] from a fixed location in the table;
2. if [math]\displaystyle{ x=T[h(x)] }[/math] return [math]\displaystyle{ h(x) }[/math]; else return NOT_FOUND;

This scheme works as long as that the hash function satisfies the following two conditions:

• The description of [math]\displaystyle{ h }[/math] is sufficiently short, so that [math]\displaystyle{ h }[/math] can be stored in one entry (or in constant many entries) of the table.
• [math]\displaystyle{ h }[/math] has no collisions on [math]\displaystyle{ S }[/math], i.e. there is no pair of items [math]\displaystyle{ x_1,x_2\in S }[/math] that are mapped to the same value by [math]\displaystyle{ h }[/math].

The first condition is easy to guarantee for 2-universal hash families. As shown by Carter-Wegman construction, a 2-universal hash function can be uniquely represented by two integers [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math], which can be stored in two entries (or just one, if the word length is sufficiently large) of the table.

Our discussion is now focused on the second condition. We find that it relies on the perfectness of the hash function for a data set [math]\displaystyle{ S }[/math].

A hash function [math]\displaystyle{ h }[/math] is perfect for a set [math]\displaystyle{ S }[/math] of items if [math]\displaystyle{ h }[/math] maps all items in [math]\displaystyle{ S }[/math] to different values, i.e. there is no collision.

We have shown by the birthday problem for 2-universal hashing that when [math]\displaystyle{ n }[/math] items are mapped to [math]\displaystyle{ n^2 }[/math] values, for an [math]\displaystyle{ h }[/math] chosen uniformly from a 2-universal family of hash functions, the probability that a collision occurs is at most 1/2. Thus

[math]\displaystyle{ \Pr[h\mbox{ is perfect for }S]\ge\frac{1}{2} }[/math]

for a table of [math]\displaystyle{ n^2 }[/math] entries.

The construction of perfect hashing is straightforward then:

For a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] elements:

1. uniformly choose an [math]\displaystyle{ h }[/math] from a 2-universal family [math]\displaystyle{ \mathcal{H} }[/math]; (for Carter-Wegman's construction, it means uniformly choose two integer [math]\displaystyle{ 1\le a\le p-1 }[/math] and [math]\displaystyle{ b\in[p] }[/math] for a sufficiently large prime [math]\displaystyle{ p }[/math].)
2. check whether [math]\displaystyle{ h }[/math] is perfect for [math]\displaystyle{ S }[/math];
3. if [math]\displaystyle{ h }[/math] is NOT perfect for [math]\displaystyle{ S }[/math], start over again; otherwise, construct the table;

This is a Las Vegas randomized algorithm, which construct a perfect hashing for a fixed set [math]\displaystyle{ S }[/math] with expectedly at most two trials (due to geometric distribution). The resulting data structure is a [math]\displaystyle{ O(n^2) }[/math]-size static dictionary of [math]\displaystyle{ n }[/math] elements which answers every search in deterministic [math]\displaystyle{ O(1) }[/math] time.

FKS perfect hashing

In the last section we see how to use [math]\displaystyle{ O(n^2) }[/math] space and constant time for answering search in a set. Now we see how to do it with linear space and constant time. This solves the problem of searching asymptotically optimal for both time and space.

This was once seemingly impossible, until Yao's seminal paper:

• Yao. Should tables be sorted? Journal of the ACM (JACM), 1981.

Yao's paper shows a possibility of achieving linear space and constant time at the same time by exploiting the power of hashing, but assumes an unrealistically large universe.

Inspired by Yao's work, Fredman, Komlós, and Szemerédi discover the first linear-space and constant-time static dictionary in a realistic setting:

• Fredman, Komlós, and Szemerédi. Storing a sparse table with O(1) worst case access time. Journal of the ACM (JACM), 1984.

The idea of FKS hashing is to arrange hash table in two levels:

• In the first level, [math]\displaystyle{ n }[/math] items are hashed to [math]\displaystyle{ n }[/math] buckets by a 2-universal hash function [math]\displaystyle{ h }[/math].

Let [math]\displaystyle{ B_i }[/math] be the set of items hashed to the [math]\displaystyle{ i }[/math]th bucket.

• In the second level, construct a [math]\displaystyle{ |B_i|^2 }[/math]-size perfect hashing for each bucket [math]\displaystyle{ B_i }[/math].

The data structure can be stored in a table. The first few entries are reserved to store the primary hash function [math]\displaystyle{ h }[/math]. To help the searching algorithm locate a bucket, we use the next [math]\displaystyle{ n }[/math] entries of the table as the "pointers" to the bucket: each entry stores the address of the first entry of the space to store a bucket. In the rest of table, the [math]\displaystyle{ n }[/math] buckets are stored in order, each using a [math]\displaystyle{ |B_i|^2 }[/math] space as required by perfect hashing.

It is easy to see that the search time is constant. To search for an item [math]\displaystyle{ x }[/math], the algorithm does the followings:

• Retrieve [math]\displaystyle{ h }[/math].
• Retrieve the address for bucket [math]\displaystyle{ h(x) }[/math].
• Search by perfect hashing within bucket [math]\displaystyle{ h(x) }[/math].

Each line takes constant time. So the worst-case search time is constant.

We then need to guarantee that the space is linear to [math]\displaystyle{ n }[/math]. At the first glance, this seems impossible because each instance of perfect hashing for a bucket costs a square-size of space. We will prove that although the individual buckets use square-sized spaces, the sum of the them is still linear.

For a fixed set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] items, for a hash function [math]\displaystyle{ h }[/math] chosen uniformly from a 2-universe family which maps the items to [math]\displaystyle{ [n] }[/math], called [math]\displaystyle{ n }[/math] buckets, let [math]\displaystyle{ Y_i=|B_i| }[/math] be the number of items in [math]\displaystyle{ S }[/math] mapped to the [math]\displaystyle{ i }[/math]th bucket. We are going to bound the following quantity:

[math]\displaystyle{ Y=\sum_{i=1}^n Y_i^2. }[/math]

Since each bucket [math]\displaystyle{ B_i }[/math] use a space of [math]\displaystyle{ Y_i^2 }[/math] for perfect hashing. [math]\displaystyle{ Y }[/math] gives the size of the space for storing the buckets.

We will show that [math]\displaystyle{ Y }[/math] is related to the total number of collision pairs. (Indeed, the number of collision pairs can be computed by a degree-2 polynomial, just like [math]\displaystyle{ Y }[/math].)

Note that a bucket of [math]\displaystyle{ Y_i }[/math] items contributes [math]\displaystyle{ {Y_i\choose 2} }[/math] collision pairs. Let [math]\displaystyle{ X }[/math] be the total number of collision pairs. [math]\displaystyle{ X }[/math] can be computed by summing over the collision pairs in every bucket:

[math]\displaystyle{ X=\sum_{i=1}^n{Y_i\choose 2}=\sum_{i=1}^n\frac{Y_i(Y_i-1)}{2}=\frac{1}{2}\left(\sum_{i=1}^nY_i^2-\sum_{i=1}^nY_i\right)=\frac{1}{2}\left(\sum_{i=1}^nY_i^2-n\right). }[/math]

Therefore, the sum of squares of the sizes of buckets is related to collision number by:

[math]\displaystyle{ \sum_{i=1}^nY_i^2=2X+n. }[/math]

By our analysis of the collision number, we know that for [math]\displaystyle{ n }[/math] items mapped to [math]\displaystyle{ n }[/math] buckets, the expected number of collision pairs is: [math]\displaystyle{ \mathbf{E}[X]\le \frac{n}{2} }[/math]. Thus,

[math]\displaystyle{ \mathbf{E}\left[\sum_{i=1}^nY_i^2\right]=\mathbf{E}[2X+n]\le 2n. }[/math]

Due to Markov's inequality, [math]\displaystyle{ \sum_{i=1}^nY_i^2=O(n) }[/math] with a constant probability. For any set [math]\displaystyle{ S }[/math], we can find a suitable [math]\displaystyle{ h }[/math] after expected constant number of trials, and FKS can be constructed with guaranteed (instead of expected) linear-size which answers each search in constant time.