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 −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
Set the bits to the 8-bit result of 0101 0111 XOR 0010 1100. What is the result's value R in decimal?
Which formula is logically equivalent to a ∧ (c ∨ b)? Click it.
Which statement is equivalent to ¬∀x (F(x) → G(x)), with the negation moved all the way inside? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 3, 10, 11, 13). How many product terms N does it have?
Find the minimal SOP form of F(A, B, C, D) = Σm(3, 4, 5, 6, 15) + d(0, 1, 2, 7, 9), where d marks the don't-care conditions. How many literals L does it contain?
Solve 14x ≡ 63 (mod 83). Give the solution x with 0 < x < 83.
Solve 18x ≡ 48 (mod 75). Give the smallest positive solution x.
In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?
To show that a new problem X is NP-hard, what must you do?
A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?
Which problem of linear probing does double hashing mainly reduce?
A connected graph has 8 vertices and 11 edges. How many edges k must be removed to leave a spanning tree?
Find the total weight W of a minimum spanning tree of the weighted graph.
What does the "A" in ACID stand for?
There is a composite index on (last_name, first_name). Which condition can use it most effectively?
Which IPv4 header field stops packets from circulating forever in a routing loop?
What does TCP fast retransmit do?
A department needs 1784 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
A router has routes to these networks: 10.87.42.0/24, 10.87.43.0/24, 10.87.44.0/24, 10.87.45.0/24, 10.87.46.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: U = 208
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: R = 123
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.
XOR gives 1 where the two bits differ.
Answer: (a ∧ c) ∨ (a ∧ b)
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: ∃x (F(x) ∧ ¬G(x))
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.
The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.
Answer: N = 4
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: L = 4
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.
The largest groups that can be formed from 1s and Xs.
Answer: x = 46
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: x = 11
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: D · Pick the activity that finishes earliest, then repeat with the compatible ones
Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.
Answer: D · Reduce a known NP-hard problem to X in polynomial time
If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.
Answer: C · O(1)
Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).
Answer: A · Primary clustering, where long runs of occupied slots form
Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.
Answer: k = 4
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: W = 40
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 · Atomicity
ACID: atomicity, consistency, isolation, durability. Atomicity means a transaction is applied completely or not at all.
Answer: C · WHERE last_name = 'Jansen'
A composite B-tree index is ordered by last_name and then first_name, so a search on the leading column works, but first_name alone cannot use the order.
Answer: D · Time to live (TTL)
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer: D · Resends a segment after three duplicate ACKs, without waiting for the timeout
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
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: p = 21
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: 10.87.40.0/21 (it also covers 3 network(s) that are not in the list).
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