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 −85 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 1101 1010 << 3. What is the result's value R in decimal?
In how many of the 8 rows of its truth table is ¬((¬q → r) → p) → (¬(r ∧ ¬p) ↔ q) true? Call this number N.
Which statement is equivalent to ¬∀x ∃y R(x, y), with the negation moved all the way inside? 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 ring counter is made of 6 D flip-flops with the last output fed back to the first input, and starts with a single 1. How many different states N does it go through?
Solve 27x ≡ 5 (mod 73). Give the solution x with 0 < x < 73.
71 is prime. Use Fermat’s little theorem to find 1094716 mod 71.
Dynamic programming is useful when a problem has:
What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?
What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?
The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:
A tree has 4 vertices of degree 3, 4 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?
Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 2.
Why are indexes created on columns?
Given the functional dependencies A → B, B → C and C → D on R(A, B, C, D), what is the closure A+?
Why is UDP often preferred for live video calls?
By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?
The network 192.168.181.0/24 is divided into /26 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 4?
A host has the address 10.20.236.128/28. What is the last number (fourth octet) x of its subnet's broadcast address?
Answer: U = 171
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 = 208
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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
Answer: N = 7
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 ∀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: AB = 00, CD = 01
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 = 6
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 = 11
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: 1094716 mod 71 = 36
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 · 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) time and O(1) space
Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.
Answer: B · n (it becomes a linked list)
Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.
Answer: B · O(α(n)), the inverse Ackermann function, which is practically constant
With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.
Answer: L = 18
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: edge AC
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 · To speed up searches and joins on those columns
Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.
Answer: C · {A, B, C, D}
Start with A, add B (A → B), then C (B → C), then D (C → D); since A+ contains all attributes, A is a key.
Answer: B · Low delay matters more than retransmitting every lost packet
Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.
Answer: D · 8 Mbit/s
C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.
Answer: x = 192
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 = 143
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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