Draw a Venn diagram and decide region by region what belongs to the set expression.
For counting problems, fill in the regions from the centre outwards: all sets, then pairs, then single sets, then the outside.
Use inclusion–exclusion as a check: |A ∪ B| = |A| + |B| − |A ∩ B|.
For a relation, test each property on every pair and give one counterexample when it fails.
For a function, count the arrows arriving at each element of B: at most one means injective, at least one surjective.
Graph theory
Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij
Method
Write down the degree of every vertex; the degrees add up to twice the number of edges.
For Euler questions, check that the graph is connected and count the vertices with odd degree.
For trees, use n − 1 edges and the handshake lemma.
For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).
Number theory & modular arithmetic
Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n
Method
Reduce every number modulo n first; remainders lie between 0 and n − 1.
For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).
2Practice problems
1.Propositional logic & proofsMedium
Which formula is logically equivalent to p → r? Click it.
p → r
1¬r → ¬p
2¬r ∧ ¬p
3r → p
4¬p → ¬r
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∀x (F(x) → G(x)), with the negation moved all the way inside? Click it.
¬∀x (F(x) → G(x))
1∃x (¬F(x) ∧ G(x))
2∃x (F(x) → ¬G(x))
3∀x (F(x) → ¬G(x))
4∃x (F(x) ∧ ¬G(x))
3.Sets, relations & functionsMedium
Exactly one region of the Venn diagram is the set (B ∩ A) − (A ∩ C). Click its number.
(B ∩ A) − (A ∩ C)
1region 1
2region 2
3region 3
4region 4
5region 5
6region 6
7region 7
8region 8
4.Sets, relations & functionsHard
An equivalence relation R on {1, …, 9} has the classes {5}, {2, 9}, {1, 3, 4, 6}, {7, 8}. How many ordered pairs does R contain? Call the number |R|.
{5}{2, 9}{1, 3, 4, 6}{7, 8}
|R| =
5.Graph theoryMedium
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
|V| = 5|E| = 6
1Euler circuit
2Euler path, no circuit
3neither
6.Graph theoryHard
Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 2.
start: Bk = 2
1edge AB
2edge AC
3edge BC
4edge BD
5edge CE
6edge DE
7edge BE
7.Number theory & modular arithmeticMedium
Find the inverse of 49 modulo 83: the number x with 0 < x < 83 and 49·x ≡ 1 (mod 83).
a = 49n = 83
49−1 mod 83 =
8.Number theory & modular arithmeticHard
Solve 10x ≡ 70 (mod 85). Give the smallest positive solution x.
10x ≡ 70 (mod 85)
x =
3Solutions
1.Propositional logic & proofs
Answer: ¬r → ¬p
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 law
An implication is equivalent to its contrapositive ¬b → ¬a, not to its converse b → a or its inverse ¬a → ¬b.
p → r ≡ ¬r → ¬p
Check with the truth table
p = 1, r = 1: 1 | 1 0 1 1
p = 1, r = 0: 0 | 0 0 1 1
p = 0, r = 1: 1 | 1 0 0 0
p = 0, r = 0: 1 | 1 1 1 1
Answer
¬r → ¬p
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∃x (F(x) ∧ ¬G(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.
Move the negation inwards
¬∀x (F(x) → G(x)) ≡ ∃x (F(x) ∧ ¬G(x))
Answer
∃x (F(x) ∧ ¬G(x))
The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.
3.Sets, relations & functions
Answer: region 3
Sets and Venn diagrams: A ∪ B holds the elements in A or B (or both), A ∩ B the elements in both, A − B the elements of A that are not in B, and the complement A′ everything in U that is not in A. In a Venn diagram every set expression is a union of regions, so count it region by region. Inclusion–exclusion: |A ∪ B| = |A| + |B| − |A ∩ B|, and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.
Inside or outside each set
A ∩ B ∩ C′
The region
region 3
Answer
region 3
4.Sets, relations & functions
Answer: |R| = 25
Relations: a relation R on a set X is a set of ordered pairs (x, y). R is reflexive when (x, x) ∈ R for every x in X; symmetric when (x, y) ∈ R always comes with (y, x) ∈ R; antisymmetric when (x, y) and (y, x) are both in R only if x = y; and transitive when (x, y) and (y, z) in R force (x, z) ∈ R. An equivalence relation is reflexive, symmetric and transitive, and splits X into disjoint classes.
Pairs per class
|R| = 1² + 2² + 4² + 2² = 25
Answer
|R| = 25
5.Graph theory
Answer: Euler circuit
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.
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.
Prim’s algorithm
1. BC (5)
2. AB (7)
3. CE (8)
4. DE (6)
edge number 2
AB
Answer
edge AB
7.Number theory & modular arithmetic
Answer: 49−1 mod 83 = 61
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).
Extended Euclidean algorithm
Each remainder r is written as r ≡ t·49 (mod 83).
r = 83, t = 0
r = 49, q = 1, t = 1
r = 34, q = 1, t = −1
r = 15, q = 2, t = 2
r = 4, q = 3, t = −5
r = 3, q = 1, t = 17
r = 1, q = 3, t = −22
Inverse
1 ≡ −22·49 (mod 83)
x = 61
check: 49·61 = 2989 = 36·83 + 1
Answer
49−1 mod 83 = 61
8.Number theory & modular arithmetic
Answer: x = 7
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).