Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
Set the bits so they store −81 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
A system must give every one of 4060 items its own binary code of the same length. What is the smallest number of bits b needed?
Which formula is logically equivalent to c ↔ b? Click it.
The premises q ∨ r; q → p; r → p are given. Which conclusion follows from them? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(5, 10, 11, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?
A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?
Find the inverse of 34 modulo 37: the number x with 0 < x < 37 and 34·x ≡ 1 (mod 37).
Encrypt the message m = 96 with the RSA public key (n, e) = (161, 7). Find the ciphertext c.
What does the recurrence T(n) = T(n/2) + O(1) solve to?
A problem is NP-complete if it is:
A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?
Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.
Which statement about HAVING is correct?
A reporting database copies customer names into the orders table to avoid joins. What is the main cost of this denormalisation?
Using even parity, which parity bit is added to the data bits 1011001?
What is the maximum transmission unit (MTU) of standard Ethernet payload?
A department needs 1151 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
A host has the address 172.20.56.235/23. What is the third octet x of its network address?
Answer: U = 175
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Answer: b = 12 bit
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Answer: (c → b) ∧ (b → c)
Logical equivalence: two formulas are equivalent (≡) when they have the same value in every row of the truth table. A tautology is 1 in every row, a contradiction is 0 in every row, and a contingency is neither. Useful laws: De Morgan, ¬(a ∧ b) ≡ ¬a ∨ ¬b and ¬(a ∨ b) ≡ ¬a ∧ ¬b; a → b ≡ ¬a ∨ b ≡ ¬b → ¬a (contrapositive); ¬(a → b) ≡ a ∧ ¬b; and the distributive laws, a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
The columns after the bar are the options in the order shown.
Answer: p
Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.
Answer: L = 5
Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Answer: N = 3
Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.
A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.
Answer: 34−1 mod 37 = 12
Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).
Each remainder r is written as r ≡ t·34 (mod 37).
Answer: c = 54
Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.
Answer: D · O(log n)
The problem halves at each step with constant work per step, so there are about log₂ n steps.
Answer: A · In NP and every NP problem reduces to it in polynomial time
NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.
Answer: B · Index 4
10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.
Answer: B · O(n)
Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.
Answer: Euler circuit
Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.
Answer: d(F) = 25
Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
Answer: D · It can use aggregate functions to filter groups
HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.
Answer: D · Redundant copies must be kept consistent whenever customer data changes
Denormalisation speeds up reads but brings back redundancy, so updates must change every copy or anomalies appear.
Answer: C · 0
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer: A · 1500 bytes
Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.
Answer: p = 21
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Answer: x = 56
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
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