Lecture 6 on 02/11/2026 - FKS Hashing Analysis and Preprocessing

Scribes: Olivia Xu and Laura Torres

Summary of Lecture

• Comparison of FKS Hashing vs. Chaining (Worst-case vs. Expected)
• Mathematical Intuition: Minimizing Sum of Squares for Equality
• Analysis of Step 2: Probability of Second-level Collisions
• Analysis of Step 1: Expected Collisions and Markov Bound

FKS Hashing vs. Hashing with Chaining

Professor Goswami began by distinguishing the guarantees:

• Hashing with Chaining: Query time is $O(1)$ expected. Preprocessing is always $O(n)$.
• FKS Hashing: Query time is $O(1)$ worst-case. Preprocessing is $O(n)$ expected.

First a Brief Summary of FKS Hashing Preprocessing Phase

Before analyzing the FKS hashing preprocessing, let’s review its steps.

• Step 2: We take each $b_i$ row from the first step and find a hash function for each row that can distribute the keys into a row that is $2b_i^2$ in length such that there are no collisions.
• By the end, we will have used $n+1$ hash functions: 1 in step 1 and $n$ in step 2.

Intuition: Why $\sum b_i^2$?

To understand why Step 1 limits the sum of squares, we consider a calculus problem:

Problem: Given $x + y = 1$ and $x, y \geq 0$, minimize $f(x,y) = x^2 + y^2$.

• Substituting $y = 1-x$, we get $f(x) = x^2 + (1-x)^2 = 2x^2 - 2x + 1$.
• To minimize it, we differentiate it once and set it equal to zero:

$$f'(x) = 4x - 2 = 0 \implies x = 1/2, y = 1/2$$

• Since the second derivative is positive, $f''(x) = 4 > 0$, this means that $x = 1/2$ and $y = 1/2$ is a minimum.
• Insight: Minimizing the sum of squares is a mathematical way to enforce an equal distribution.
• In FKS, $\sum b_i = n$. The sum of squares $\sum b_i^2$ ranges from $n$ (perfectly equal, $b_i=1$) to $n^2$ (all in one bucket). FKS accepts any $h$ where $\sum b_i^2 \leq 4n$.
  • The most equal distribution will be when there are no collisions and each $b_i$ is equal to 1. Then $\sum b_i^2 = n$:

$$b_1^2+b_2^2+\dots+b_n^2 = 1^2+1^2+\dots+1^2=n$$

• The most unequal distribution will occur when all keys go into the same bucket, so all the other buckets have 0 keys:

$$\sum b_i^2 = n^2$$ $$b_1^2+b_2^2+\dots+b_n^2 = n^2+0^2+\dots+0^2=n^2$$

Step 2 Analysis: Second-level Collision Probability

We need to prove that Step 2 terminates quickly.

Scenario: We hash $b_i$ keys into $m_i = 2b_i^2$ cells.

• Let $C$ be the total number of collisions in a bucket.
• Using a Universal Hash Family, for any pair of keys, $\Pr(\text{collision}) \leq 1/m_i = 1/(2b_i^2)$.
• Total number of pairs is $\binom{b_i}{2} = \frac{b_i(b_i-1)}{2}$.

$$E[C] = \sum_{\text{pairs}} \Pr(\text{collision}) = \frac{b_i(b_i-1)}{2} \cdot \frac{1}{2b_i^2} < \frac{b_i^2}{4b_i^2} = 1/4$$

• By Markov’s Inequality: $\Pr(C \geq 1) \leq \frac{E[C]}{1} < 1/4$.
• Conclusion: Since the failure probability is $< 1/2$, the number of tries follows a Geometric Random Variable with success $p \geq 1/2$. Expected tries $\leq 2$.

Step 1 Analysis: Preprocessing Time

We prove that we don’t need to resample $h$ too many times. Step 1 succeeds with probability at least $1/2$, so expected tries is at most 2.

From Collisions to Sum of Squares

Key observation: The total number of collisions is related to how keys distribute across buckets.

When $b_i$ keys hash to bucket $i$, how many collision pairs are there? If we count ordered pairs (each unordered pair counted twice), we get $b_i(b_i-1)$ ordered collision pairs from bucket $i$. Summing over all buckets:

$$C = \sum_{i=1}^{m} b_i(b_i-1) = \sum_{i=1}^{m} (b_i^2 - b_i) = \sum_{i=1}^{m} b_i^2 - \sum_{i=1}^{m} b_i = \sum b_i^2 - n$$

Since $\sum b_i = n$ (all keys must go somewhere). Therefore:

$$\sum b_i^2 = C + n$$

This connects the sum of squares directly to collision count, tying Step 1’s success condition to a probability argument.

Expected Number of Collisions

When hashing $n$ keys into $n$ buckets using a universal hash function:

What’s the expected number of keys that collide with one particular key $x_j$?

• There are $n-1$ other keys
• Each hashes to the same bucket as $x_j$ with probability $1/n$
• Expected collisions with $x_j$: $(n-1) \cdot \frac{1}{n} < 1$

Since this holds for every key:

$$E[C] = \sum_{j=1}^{n} E[\text{collisions with } x_j] < n$$

Important fact: When hashing $n$ keys into $n$ buckets, $E[C] < n$.

Markov Bound for Step 1

Using the result above where $E[C] < n$, and $C = \sum b_i^2 - n$:

$$E[\sum b_i^2 - n] = E[C] < n \implies E[\sum b_i^2] < 2n$$

Step 1 fails when $\sum b_i^2 > 4n$. By Markov’s inequality:

$$\Pr\left(\sum b_i^2 > 4n\right) \leq \frac{E[\sum b_i^2]}{4n} < \frac{2n}{4n} = \frac{1}{2}$$

Conclusion:

• Probability Step 1 succeeds: $\geq 1/2$
• Expected number of tries: $\leq 2$ (geometric with success probability $\geq 1/2$)
• Total expected work in Step 1: $2 \cdot O(n) = O(n)$

Summary of Guarantees

• Query: $O(1)$ worst-case (two hash function evaluations).
• Space: $O(n)$ (since $\sum 2b_i^2 \leq 8n$ from Step 1’s condition).
• Preprocessing: $O(n)$ expected (Geometric trials for both levels).