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 −20 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 1001 0001 XOR 0010 1101. What is the result's value R in decimal?
In how many of the 8 rows of its truth table is (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) true? Call this number N.
The premises a → c; a are given. Which conclusion follows from them? Click it.
Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 4, 9, 10, 12). How many sum terms (factors) N does it have?
A positive-edge-triggered D 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 31 modulo 75: the number x with 0 < x < 75 and 31·x ≡ 1 (mod 75).
An RSA key uses the primes p = 43 and q = 47 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).
What does the recurrence T(n) = T(n − 1) + O(n) solve to?
Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?
Which abstract data type does Dijkstra's algorithm use to choose the next vertex to process?
Which data structure supports an LRU (least recently used) cache with O(1) get and put?
A connected graph has 8 vertices and 10 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.
Third normal form (3NF) removes:
Which window function assigns 1, 2, 3, ... to rows within each partition without ties?
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
The network 192.168.98.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
The network 192.168.54.0/24 is split with VLSM for these LANs: A: 3 hosts, B: 43 hosts, C: 102 hosts, D: 9 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 D?
Answer: U = 236
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 = 188
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: N = 2
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: c
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: 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: N = 5
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: 31−1 mod 75 = 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).
Each remainder r is written as r ≡ t·31 (mod 75).
Answer: d = 527
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: C · O(n²)
The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.
Answer: D · A decision tree with n possible outcomes must have depth at least log₂ n
Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.
Answer: D · A min-priority queue
A min-priority queue, usually a binary heap, returns the unvisited vertex with the smallest distance efficiently.
Answer: C · A hash map combined with a doubly linked list
The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).
Answer: k = 3
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 · Transitive dependencies between non-key attributes
In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.
Answer: A · ROW_NUMBER()
ROW_NUMBER gives unique sequential numbers; RANK leaves gaps after ties; DENSE_RANK does not.
Answer: B · Detecting transmission errors
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer: C · Name resolution (DNS)
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer: x = 32
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 = 192
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 D: 192.168.54.192/28.
ExamProblems · examproblems.com/courses/computer-science-fundamentals