Self-Assessment Quiz: Deutsch-Jozsa and Bernstein-Vazirani

Twenty questions on the first quantum algorithms, oracle models, promise problems, and how seriously to take their speedups. Aim for 16+.


Question 1

Deutsch–Jozsa decides whether $f:\{0,1\}^n\to\{0,1\}$ is:

A) Injective or surjective B) Constant or balanced, given a promise that it is one of the two C) Linear or nonlinear D) Computable

Question 2

Classically, deciding this with certainty requires in the worst case:

A) 1 query B) $n$ queries C) $2^{n-1}+1$ queries D) $2^n$ queries

Question 3

Deutsch–Jozsa requires:

A) 1 query B) $n$ queries C) $\sqrt{2^n}$ queries D) $2^n$ queries

Question 4

Classically, with a randomized algorithm accepting small error probability, you need:

A) $2^{n-1}+1$ queries B) $O(1)$ queries C) $O(n)$ queries D) $O(2^n)$

Question 5

That last fact means the Deutsch–Jozsa speedup is:

A) Exponential over any classical method B) Exponential only over deterministic classical algorithms C) Nonexistent D) Quadratic

Question 6

The final measurement in Deutsch–Jozsa gives all zeros iff:

A) $f$ is balanced B) $f$ is constant C) $f$ is linear D) Always

Question 7

Bernstein–Vazirani finds the hidden string $s$ in $f(x) = s\cdot x \bmod 2$ using:

A) 1 query B) $n$ queries C) $2^n$ queries D) $\log n$ queries

Question 8

Classically, finding $s$ requires:

A) 1 query B) $n$ queries C) $2^n$ queries D) $n^2$ queries

Question 9

So Bernstein–Vazirani's speedup is:

A) Exponential B) Linear ($n\to1$) C) Quadratic D) None

Question 10

The state after the oracle in Bernstein–Vazirani is:

A) $\frac{1}{\sqrt{2^n}}\sum_x(-1)^{s\cdot x}|x\rangle$ B) $|s\rangle$ C) Uniform with no phases D) Entangled with the ancilla

Question 11

Applying $H^{\otimes n}$ to that state yields:

A) Uniform superposition B) $|s\rangle$ exactly C) $|0\rangle$ D) A random state

Question 12

The oracle in these algorithms is:

A) A physical device you buy B) A black box counted per query — the resource being measured C) Always free D) Classical only

Question 13

A "promise problem" is one where:

A) Any input is allowed B) Inputs are guaranteed to satisfy a condition, and behavior is unconstrained otherwise C) The answer is probabilistic D) The oracle may lie

Question 14

Deutsch–Jozsa's practical importance is mainly:

A) Direct application to real problems B) Historical and conceptual — the first provable exponential separation in the query model C) Cryptanalysis D) Chemistry simulation

Question 15

Building the oracle circuit for a given $f$ typically costs:

A) Nothing B) Gates that may exceed the query savings C) Exactly one gate D) $2^n$ measurements

Question 16

True or false: Deutsch–Jozsa proves quantum computers beat classical ones on practical problems.

Question 17

True or false: Bernstein–Vazirani's output is deterministic on a noiseless device.

Question 18

True or false: These algorithms need entanglement between the input qubits.

Question 19

Short answer. Explain why the query-model speedup can be misleading when the oracle must be built from gates.

Question 20

Short answer. Why is Bernstein–Vazirani a good hardware benchmark despite being a modest speedup?


Answer Key

Q Ans Note
1 B The promise is essential — without it the problem is different and the algorithm's guarantee does not apply.
2 C You must check just over half the inputs: $2^{n-1}$ agreeing values still leaves both cases open.
3 A One oracle query, with certainty.
4 B A handful of random samples distinguishes constant from balanced with overwhelming probability — a few queries suffice for any fixed error.
5 B This is the honest framing, and it is frequently omitted. Against randomized classical algorithms the separation collapses.
6 B Constant $f$ gives amplitude $\pm1$ on all zeros; balanced gives exactly 0.
7 A One query.
8 B Query $e_i$ for each basis vector to read one bit of $s$ at a time.
9 B $n$ queries to 1 — real and provable, including against randomized classical algorithms, but linear rather than exponential.
10 A Phase kickback writes $s\cdot x$ into the sign.
11 B That state is the Hadamard transform of $|s\rangle$, and $H$ is its own inverse.
12 B The query model measures oracle calls, deliberately ignoring the cost of implementing the oracle.
13 B Behavior on promise-violating inputs is undefined — which is why these results do not translate directly into general-purpose algorithms.
14 B Nobody needs to know whether a function is constant or balanced. Its value is as the first proof that quantum query complexity can be exponentially smaller.
15 B Constructing $U_f$ for a specific $f$ can require many gates — the saved queries may cost more than they save.
16 False It is a promise problem in an artificial oracle model, with the separation holding only against deterministic classical algorithms.
17 True The output is exactly $|s\rangle$ with probability 1 in the noiseless case.
18 False Both algorithms work on product states throughout the input register; the oracle may entangle input and ancilla transiently, but no entanglement between input qubits is required. This is a useful reminder that entanglement is not the sole source of quantum advantage (Ch. 6).
19 — The query model charges one unit per oracle call and nothing for constructing the oracle. In practice $U_f$ must be compiled into gates, and for a complicated $f$ that circuit can be enormous — potentially exponential. An algorithm saving exponentially many queries while requiring an exponentially large oracle circuit has saved nothing. Query complexity is a lower-bound tool and a poor cost model for end-to-end applications.
20 — Its output is a single deterministic bitstring on an ideal device, so any deviation is directly attributable to hardware error — no statistical inference is needed to interpret the result. It scales to any qubit count, its circuit depth is shallow and constant in $n$, and success probability degrades smoothly with error rate. That makes it an excellent, cheap system-level check, which is why it appears in vendor benchmark suites far more often than its algorithmic importance would justify.