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 −61 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 862 items its own binary code of the same length. What is the smallest number of bits b needed?
Which formula is logically equivalent to (q ∧ p) → r? Click it.
Which statement is equivalent to ¬∀x F(x), 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.
Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 6, 7, 8, 12, 14, 15) + d(9, 10), where d marks the don't-care conditions. How many product terms N does it have?
Find the inverse of 26 modulo 29: the number x with 0 < x < 29 and 26·x ≡ 1 (mod 29).
Use repeated squaring to find 7137 mod 34.
What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?
Which bound is correct for log(n!)?
How is a new element inserted into a binary heap?
In a hash table using open addressing, why is deletion often done with a "tombstone" marker?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
How many walks of length 3 go from A to F? A walk may repeat vertices and edges. Call the number N.
A table has rows where col is NULL. Which rows does WHERE col = NULL return?
A table has a million rows and a Boolean column is_active that is TRUE for 95% of them. Why might the optimiser ignore an index on is_active for WHERE is_active = TRUE?
What is the main purpose of a VLAN?
TCP congestion control's "slow start" phase:
A subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?
A router has routes to these networks: 10.4.66.0/24, 10.4.67.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: U = 195
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 = 10 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: q → (p → r)
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: ∃x ¬F(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: AB = 00, CD = 00
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 = 3
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: 26−1 mod 29 = 19
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·26 (mod 29).
Answer: 7137 mod 34 = 27
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: D · O(n log n)
The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).
Answer: C · Θ(n log n)
log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).
Answer: A · Place it at the end of the array and sift it up while it violates the heap property
Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.
Answer: A · Emptying the slot could break the probe sequence for other keys
If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.
Answer: neither
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: N = 2
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Answer: B · No rows, because the comparison is unknown
Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.
Answer: C · Most rows match, so a full table scan is cheaper than many index lookups
An index helps when it narrows the search to few rows; when most rows qualify, random index lookups cost more than reading the table sequentially.
Answer: C · To split one physical switched network into separate broadcast domains
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer: D · Increases the congestion window exponentially until a threshold or loss
Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.
Answer: H = 62
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 = 23
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: 10.4.66.0/23
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