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 −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
What is the most negative integer N that can be stored in 20-bit two's complement?
You want to prove (r ∧ p) → q by contradiction. Which statement do you assume at the start? Click it.
Which statement is equivalent to ¬∀x ∃y R(x, y), 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, D) = Σm(0, 3, 5, 6, 7, 8, 9, 11, 14, 15). How many sum terms (factors) N does it have?
A 4-bit binary down counter starts at 3 (decimal, Q = 0011). Which count Q (in decimal) does it show after 41 clock pulses?
Find the smallest positive integer x with x ≡ 1 (mod 3) and x ≡ 6 (mod 8).
An RSA key uses the primes p = 7 and q = 17 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).
What is the worst-case time complexity of quicksort?
What does the recurrence T(n) = 2T(n/2) + n log n solve to?
A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?
In the BST with root 50, left subtree 30 (children 20, 40) and right subtree 70 (children 60, 80), the root is deleted using the in-order successor. Which key becomes the new root?
How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.
How many walks of length 3 go from E to D? A walk may repeat vertices and edges. Call the number N.
Which statement about HAVING is correct?
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?
Why does TCP choose an unpredictable initial sequence number for each connection?
The network 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A host has the address 10.20.187.19/30. What is the fourth octet x of its network address?
Answer: U = 138
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: N = −524288
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 ∧ p) ∧ ¬q
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).
If the assumption leads to a contradiction, the implication must be true.
Answer: ∃x ∀y ¬R(x, y)
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: Q = 10
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.
Every 16 pulses bring the counter back to the same state.
Answer: x = 22
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 = 37
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: B · O(n²)
With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).
Answer: C · Θ(n log² n)
Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.
Answer: D · Adjacency list
A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.
Answer: A · 60
The leftmost node of the right subtree is 60, so 60 replaces 50 and its old position is removed; using the predecessor would give 40.
Answer: H = 360
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: N = 5
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Answer: A · It can use aggregate functions to filter groups
HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.
Answer: B · 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: D · 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: D · To avoid confusion with old segments and make spoofed segments harder to inject
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer: H = 14
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 = 16
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).
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