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 −38 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 24-bit two's complement?
Which formula is logically equivalent to ¬(q ∧ p)? Click it.
The argument with premises q → (p ∨ r); q and conclusion p is invalid. Click the row of the truth table that is a counterexample.
A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C, I₁ = C, I₂ = 0, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.
A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.
Find the inverse of 47 modulo 53: the number x with 0 < x < 53 and 47·x ≡ 1 (mod 53).
Use repeated squaring to find 53164 mod 86.
Dynamic programming is useful when a problem has:
What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?
How much memory does an adjacency list use for a graph with V vertices and E edges?
Why are B-trees used in databases and file systems?
A tree has 3 vertices of degree 3, 4 vertices of degree 4, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
Find the total weight W of a minimum spanning tree of the weighted graph.
Second normal form (2NF) removes:
What does the write-ahead logging (WAL) rule require?
What is the IPv6 loopback address?
Why does TCP choose an unpredictable initial sequence number for each connection?
A department needs 818 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
A host has the address 10.20.28.156/25. What is the fourth octet x of the last usable host address in its subnet?
Answer: U = 218
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 = −8.389 × 10⁶
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: ¬q ∨ ¬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: p = 0, q = 1, r = 1
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.
An argument with this form looks like a valid rule, but one row is enough to show that it is not.
Answer: N = 2
Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.
Answer: clock edge 3
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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.
Answer: 47−1 mod 53 = 44
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·47 (mod 53).
Answer: 53164 mod 86 = 9
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 · Overlapping subproblems and optimal substructure
Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.
Answer: B · O(n log n)
The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).
Answer: D · O(V + E)
Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).
Answer: B · Their high branching factor keeps the tree shallow, reducing slow disk reads
B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.
Answer: L = 19
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.
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
Answer: W = 54
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 · Partial dependencies of non-key columns on part of a composite key
2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.
Answer: A · A change's log record must reach stable storage before the changed data page does
Writing the log first lets the DBMS redo committed work and undo uncommitted work after a crash, supporting atomicity and durability.
Answer: B · ::1
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
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: 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 = 254
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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