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.
Write −76 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
Set the bits to the 8-bit result of 1110 0100 >> 3. What is the result's value R in decimal?
In how many of the 8 rows of its truth table is ¬((p ∨ (q ↔ r)) → (¬r ∨ ¬(q ∧ p))) true? Call this number N.
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 product-of-sums (POS) form of F(A, B, C) = Σm(1, 3, 4, 5, 6). How many sum terms (factors) N does it have?
A mod-12 counter has to count through 12 different states. What is the minimum number of flip-flops n it needs?
Find the inverse of 16 modulo 19: the number x with 0 < x < 19 and 16·x ≡ 1 (mod 19).
An RSA key uses the primes p = 41 and q = 43 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).
Why do many library sorting routines switch to insertion sort for very small subarrays?
Topological sorting is possible only for:
What is the purpose of a sentinel (dummy) node in a linked list?
In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?
A connected graph has 17 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.
A view in SQL is:
When is decomposing R into R1 and R2 guaranteed to be lossless?
What is the role of a default gateway on a host?
TCP congestion control's "slow start" phase:
A department needs 373 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: 10.117.3.0/24, 10.117.4.0/24, 10.117.5.0/24, 10.117.6.0/24, 10.117.7.0/24, 10.117.8.0/24, 10.117.9.0/24, 10.117.10.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: U = 180
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: R = 28
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.
Shifting right by 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.
Answer: N = 1
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: ∃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 = 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
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.
Answer: 16−1 mod 19 = 6
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·16 (mod 19).
Answer: d = 593
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 · Insertion sort has little overhead and is fast on small or nearly sorted inputs
For small n the simple loop of insertion sort beats the recursion overhead of quicksort or merge sort, even though it is O(n²).
Answer: C · Directed acyclic graphs
A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.
Answer: B · To simplify code by removing special cases at the head or tail
With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.
Answer: B · Checking whether an edge's two endpoints are already connected
Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.
Answer: k = 2
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: A · A stored query that behaves like a virtual table
A view stores a SELECT statement; querying it runs the underlying query (materialised views store the result).
Answer: D · When the common attributes of R1 and R2 form a superkey of R1 or of R2
If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.
Answer: A · It is the router used for traffic to other networks
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer: B · Increases the congestion window exponentially until a threshold or loss
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer: p = 23
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 = 20
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: 10.117.0.0/20 (it also covers 8 network(s) that are not in the list).
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