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 1000 1001 stores a signed integer in two's complement. What is its value N in decimal?
Set the bits to the 8-bit result of 0101 0011 OR 1101 1111. What is the result's value R in decimal?
Which formula is logically equivalent to ¬(p ↔ q)? Click it.
The premises p ∨ q; ¬p ∨ r are given. Which conclusion follows from them? Click it.
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
A Johnson (twisted-ring) counter is made of 6 D flip-flops, with the inverted last output fed back to the first input, and starts at all zeros. How many different states N does it go through?
Solve 24x ≡ 7 (mod 31). Give the solution x with 0 < x < 31.
Encrypt the message m = 589 with the RSA public key (n, e) = (943, 13). Find the ciphertext c.
Why do many library sorting routines switch to insertion sort for very small subarrays?
Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?
A trie (prefix tree) is especially efficient for:
A segment tree is typically used to:
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.
In relational algebra, the projection operator (π) corresponds to:
What does the following return: SELECT name FROM students WHERE grade > (SELECT AVG(grade) FROM students)?
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
Why does IPv6 not need NAT for address shortage?
A subnet uses the mask 255.255.248.0. How many usable host addresses H does it have?
A router has routes to these networks: 10.55.19.0/24, 10.55.20.0/24, 10.55.21.0/24, 10.55.22.0/24, 10.55.23.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: N = −119
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 = 223
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.
OR gives 1 where at least one bit is 1.
Answer: 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).
The columns after the bar are the options in the order shown.
Answer: q ∨ r
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: AB = 01, CD = 10
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.
A 1 with a neighbouring 1 can join a group of two or more.
Answer: N = 12
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: x = 30
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: c = 908
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 · 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 · 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 · Prefix searches such as autocomplete
Tries store strings by characters along paths, so lookups and prefix queries take time proportional to the key length.
Answer: B · Answer range queries (such as range sums) and updates in O(log n)
A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.
Answer: Euler 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.
Answer: edge CE
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 · Choosing which columns appear in the result
Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).
Answer: C · Students with a grade above the average grade
The scalar subquery computes the average, and the outer query keeps students above it.
Answer: A · /24
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer: A · Its 128-bit address space is large enough to give every device a global address
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer: H = 2046
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 = 21
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: 10.55.16.0/21 (it also covers 3 network(s) that are not in the list).
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