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 −113 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 4613 items its own binary code of the same length. What is the smallest number of bits b needed?
Is (q ∧ (q → p)) → p a tautology, a contradiction or a contingency? Click your answer.
The premises r → p; p → q are given. Which conclusion follows from them? Click it.
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?
Find Euler’s totient φ(225), the number of integers from 1 to 225 that are coprime to 225.
An RSA key uses p = 31, q = 37 and e = 23. Decrypt the ciphertext c = 55: find the message m.
Is it correct to say that an O(n²) algorithm is also O(n³)?
What did the Cook–Levin theorem establish?
A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the postorder traversal?
Which structure lets you compute the maximum of every window of size k in an array in O(n) total?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.
What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?
Transaction T1 locks row A and waits for row B, while T2 locks row B and waits for row A. What does a typical DBMS do?
What does an MX record in DNS specify?
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
The network 10.202.64.0/20 is divided into equal subnets with prefix /23. How many subnets N are there?
A router has routes to these networks: 10.155.138.0/24, 10.155.139.0/24, 10.155.140.0/24, 10.155.141.0/24, 10.155.142.0/24, 10.155.143.0/24, 10.155.144.0/24, 10.155.145.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: U = 143
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 = 13 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: tautology
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).
Answer: r → q
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: AB = 10, CD = 00
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.
A 1 with a neighbouring 1 can join a group of two or more.
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 D flip-flop copies D at each rising edge: Q⁺ = D.
Answer: φ(225) = 120
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).
Answer: m = 757
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 · Yes, because Big O is an upper bound and n² grows no faster than n³
Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.
Answer: D · The Boolean satisfiability problem (SAT) is NP-complete
Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.
Answer: B · 4 5 2 3 1
Postorder processes 4 and 5 before 2, then 3, and finally the root 1.
Answer: B · A deque holding indices in decreasing order of value
The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.
Answer: neither
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: B
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.
Answer: D · Each name with the label 'pass' or 'fail' depending on the grade
A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.
Answer: D · Detects the deadlock and aborts one transaction as the victim
A cycle in the wait-for graph is a deadlock; the DBMS rolls back one transaction so the other can continue, and the victim may be retried.
Answer: A · The mail servers that accept email for a domain
MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.
Answer: B · 8 Mbit/s
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer: N = 8
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).
Each /23 subnet has 29 − 2 = 510 usable hosts.
Answer: p = 19
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Summary route: 10.155.128.0/19 (it also covers 24 network(s) that are not in the list).
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