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 −94 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
A system must give every one of 1421 items its own binary code of the same length. What is the smallest number of bits b needed?
You want to prove c → (a ∨ b) by contrapositive. Which implication do you prove instead? Click it.
The argument with premises a → b; ¬a and conclusion ¬b 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 = 1, I₁ = C′, I₂ = C, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.
Find the minimal SOP form of F(A, B, C, D) = Σm(0, 4, 8, 10, 13, 14) + d(7, 12), where d marks the don't-care conditions. How many literals L does it contain?
Find the smallest positive integer x with x ≡ 6 (mod 10) and x ≡ 7 (mod 13).
An RSA key uses p = 23, q = 31 and e = 23. Decrypt the ciphertext c = 96: find the message m.
What does the recurrence T(n) = T(n − 1) + O(1) solve to?
What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?
What is the purpose of a sentinel (dummy) node in a linked list?
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?
A connected graph has 18 vertices and 25 edges. How many edges k must be removed to leave a spanning tree?
Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.
What does INTERSECT return when used between two SELECT queries?
What is a covering index for a query?
Using even parity, which parity bit is added to the data bits 1011001?
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: 10.85.196.0/24, 10.85.197.0/24, 10.85.198.0/24, 10.85.199.0/24, 10.85.200.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: U = 162
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: 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: ¬(a ∨ b) → ¬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).
The contrapositive is equivalent to the original statement, so proving it proves the statement.
Answer: a = 0, b = 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 = 5
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: L = 7
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: x = 46
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: m = 487
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(n)
There are n levels with constant work each, as in a recursive sum of an array.
Answer: C · 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: B · To simplify code by removing special cases at the head or tail
With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.
Answer: B · 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: k = 8
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: d(F) = 20
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.
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
Answer: D · Rows that appear in both results
INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.
Answer: B · An index that contains all columns the query needs, so the table itself need not be read
If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.
Answer: D · 0
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer: B · 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 = 20
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: 10.85.192.0/20 (it also covers 11 network(s) that are not in the list).
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