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.
The 8-bit pattern 1101 0001 stores a signed integer in two's complement. What is its value N in decimal?
Set the bits to the 8-bit result of 0110 0111 AND 1101 0110. What is the result's value R in decimal?
Which formula is logically equivalent to r ∧ (q ∨ p)? Click it.
How many of the 16 rows of the truth table make ¬((s → q) ∨ r) ↔ ((r ∨ p) ∧ p) true? Call this number N.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 4, 7). How many product terms N does it have?
Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 4, 8, 13, 15) + d(10, 12), where d marks the don't-care conditions. How many product terms N does it have?
Find the smallest positive integer x with x ≡ 9 (mod 11) and x ≡ 6 (mod 13).
61 is prime. Use Fermat’s little theorem to find 530087 mod 61.
The lower bound for comparison-based sorting in the worst case is:
Is 22n in O(2ⁿ)?
An adjacency matrix for a graph with V vertices uses how much memory?
The time complexity of decrease-key in a binary heap (given the element's position) is:
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Prim’s algorithm builds a minimum spanning tree starting from vertex G. Click the edge it adds as edge number 3.
Which query returns the number of orders per customer?
What is a known drawback of decomposing tables into BCNF compared with 3NF?
What characterises a distributed denial-of-service (DDoS) attack?
Why does a browser warn when a site presents a self-signed certificate?
A department needs 634 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
The network 192.168.249.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 24 hosts, C: 6 hosts, D: 13 hosts, E: 30 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN B?
Answer: N = −47
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.
The other bits keep their normal positive values.
Answer: R = 70
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.
AND gives 1 only where both bits are 1.
Answer: (r ∧ q) ∨ (r ∧ p)
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: N = 8
Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.
Answer: N = 2
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 = 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.
Another grouping of the same size exists; it has the same number of terms and literals.
Answer: x = 97
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: 530087 mod 61 = 41
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: A · Ω(n log n)
Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.
Answer: A · No, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound
A constant added to the exponent gives a constant factor, but a constant multiplying the exponent changes the base and the growth rate.
Answer: D · O(V²)
An adjacency matrix stores V × V entries, efficient for dense graphs; adjacency lists use O(V + E), better for sparse graphs.
Answer: A · O(log n)
After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.
Answer: Euler path, no 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.
The Euler path starts at one odd vertex and ends at the other.
Answer: edge BC
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: B · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
GROUP BY customer_id makes one group per customer, and COUNT(*) counts the rows in each group.
Answer: D · Some functional dependencies may no longer be checkable within a single table
A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.
Answer: C · Many compromised machines flood a target with traffic
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer: A · No trusted certificate authority vouches that the key belongs to that site
Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.
Answer: p = 22
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 = 32
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
LAN B: 192.168.249.32/27.
ExamProblems · examproblems.com/courses/computer-science-fundamentals