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 −91 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 3324 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.
The argument with premises r → q; p → q and conclusion r → p is invalid. Click the row of the truth table that is a counterexample.
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.
For F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 11, 15) + d(4, 10), where d marks the don't-care conditions, find the number N of essential prime implicants.
Find the inverse of 43 modulo 99: the number x with 0 < x < 99 and 43·x ≡ 1 (mod 99).
Solve 9x ≡ 3 (mod 12). Give the smallest positive solution x.
What does the recurrence T(n) = T(n/2) + O(1) solve to?
The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?
AVL trees and red-black trees guarantee:
In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?
A tree has 3 vertices of degree 4, 4 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 4.
A table stores the customer's address on every order row. Changing the address in one row but not in the others is an example of:
Two transactions both read stock = 10, each subtracts 1 and writes the result, giving 9 instead of 8. Which anomaly is this?
Which IPv4 header field stops packets from circulating forever in a routing loop?
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
The network 192.168.63.0/24 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 3?
A host has the address 10.20.62.240/29. What is the fourth octet x of the last usable host address in its subnet?
Answer: U = 165
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 = 12 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) ∨ (q ∧ 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: p = 0, q = 1, r = 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: AB = 00, CD = 11
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: 43−1 mod 99 = 76
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·43 (mod 99).
Answer: x = 3
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: D · O(log n)
The problem halves at each step with constant work per step, so there are about log₂ n steps.
Answer: D · W is a number, so the time is exponential in the number of bits used to write W
The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.
Answer: C · O(log n) height through rebalancing
Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.
Answer: B · Checking whether an edge's two endpoints are already connected
Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.
Answer: L = 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.
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
Answer: edge EF
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: C · An update anomaly
Redundant storage means one change must be made in many rows; missing some leaves the data inconsistent, which normalisation prevents.
Answer: D · Lost update
In a lost update, both transactions work from the same old value and the second write erases the first; locking (SELECT ... FOR UPDATE) or atomic updates prevent it.
Answer: D · Time to live (TTL)
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer: C · /27
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
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).
Answer: x = 246
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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