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 −16 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
A system must give every one of 1598 items its own binary code of the same length. What is the smallest number of bits b needed?
You want to prove (b ∨ a) → ¬c by contradiction. Which statement do you assume at the start? Click it.
Which statement is equivalent to ¬∃x (P(x) ∧ Q(x)), with the negation moved all the way inside? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(7, 8, 11, 12, 13, 14). How many literals L does it contain in total (a primed variable counts as one literal)?
For F(A, B, C, D) = Σm(4, 6, 7, 10, 12, 13, 15) + d(0, 3), where d marks the don't-care conditions, find the number N of essential prime implicants.
Find the inverse of 38 modulo 67: the number x with 0 < x < 67 and 38·x ≡ 1 (mod 67).
An RSA key uses p = 5, q = 17 and e = 23. Decrypt the ciphertext c = 77: find the message m.
Why is the base of a logarithm usually left out in Big O, as in O(log n)?
Taking both endpoints of every edge in a maximal matching gives a vertex cover. How good is this cover?
Which data structure is used to check whether the brackets in "{[()]}" are balanced?
An object is stored as a key in a hash map, and then a field used by its hash function is changed. What is the likely result?
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.
Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 2; skipped edges do not count.
What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?
The query FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.status = 'paid' unexpectedly drops customers with no orders. Why?
How many usable host addresses does a /26 IPv4 subnet have?
By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?
The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A host has the address 10.20.203.213/25. What is the fourth octet x of its network address?
Answer: U = 240
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: b = 11 bit
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: (b ∨ a) ∧ ¬¬c
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 (P(x) → ¬Q(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: L = 17
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 = 1
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.
The largest groups that can be formed from 1s and Xs.
Answer: 38−1 mod 67 = 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).
Each remainder r is written as r ≡ t·38 (mod 67).
Answer: m = 53
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 · Logarithms with different bases differ only by a constant factor
Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.
Answer: B · At most twice the size of a minimum vertex cover
The matching edges share no vertices, so the optimum needs at least one vertex per edge, while this method takes two: a 2-approximation.
Answer: C · A stack
Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.
Answer: A · Looking up the object may fail because it sits in the bucket for its old hash
The key's location was chosen from its old hash; with a new hash, lookups search the wrong bucket, so keys should be immutable.
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: edge DE
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: B · Each name with the label 'pass' or 'fail' depending on the grade
A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.
Answer: A · The WHERE condition on o rejects the NULL-filled rows, turning it into an inner join
Unmatched rows have o.status NULL, so the WHERE test is not TRUE; moving the condition into the ON clause keeps every customer.
Answer: C · 62
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer: A · 12 kbit/s
2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.
Answer: H = 30
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 = 128
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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