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¬p ∨ r
2¬p ∧ ¬r
3¬p ∧ r
4¬p ∨ ¬r
2.Propositional logic & proofsHard
The premises p ∨ q; p → r; q → r are given. Which conclusion follows from them? Click it.
p ∨ qp → rq → r
1r
2p
3p ∧ q
4¬r
3.Sets, relations & functionsMedium
R = {(1, 4), (2, 2), (3, 1), (3, 3), (4, 1), (4, 4)} is a relation on {1, 2, 3, 4}. What is the smallest number k of ordered pairs that must be added to R to make it symmetric?
The arrow diagram shows a function f: A → B with A = {1, 2, 3, 4, 5} and B = {a, b}. Click the description that fits f.
f = {(1, b), (2, b), (3, a), (4, b), (5, b)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
A tree has 2 vertices of degree 5, 2 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
2 × deg 52 × deg 2
L =
6.Graph theoryHard
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.
start: Around 5
1A
2B
3C
4D
5E
6F
7G
7.Number theory & modular arithmeticMedium
Solve 7x ≡ 5 (mod 20). Give the solution x with 0 < x < 20.
7x ≡ 5 (mod 20)
x =
8.Number theory & modular arithmeticHard
Use repeated squaring to find 14170 mod 43.
a = 14k = 170n = 43
14170 mod 43 =
3Solutions
1.Propositional logic & proofs
Answer: ¬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 law
De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(p ∧ r) ≡ ¬p ∨ ¬r
Check with the truth table
p = 1, r = 1: 0 | 1 0 0 0
p = 1, r = 0: 1 | 0 0 0 1
p = 0, r = 1: 1 | 1 0 1 1
p = 0, r = 0: 1 | 1 1 0 1
Answer
¬p ∨ ¬r
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: r
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.
Rows with all premises true
p = 1, q = 1, r = 1: r = 1, p = 1, p ∧ q = 1, ¬r = 0
p = 1, q = 0, r = 1: r = 1, p = 1, p ∧ q = 0, ¬r = 0
p = 0, q = 1, r = 1: r = 1, p = 0, p ∧ q = 0, ¬r = 0
Conclusion
r (proof by cases)
Answer
r
3.Sets, relations & functions
Answer: k = 1
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 to add
{(1, 3)}
Count
k = 1
Answer
k = 1
4.Sets, relations & functions
Answer: surjective, not injective
Functions and counting: a function f: A → B gives every element of A exactly one image in B. f is injective (one-to-one) when different elements have different images, surjective (onto) when every element of B is an image, and bijective when both hold. With |A| = m and |B| = n there are nᵐ functions, n!/(n − m)! injective ones (m ≤ n), and by inclusion–exclusion Σ (−1)ᵏ·C(n, k)·(n − k)ᵐ surjective ones. |𝒫(A)| = 2ᵐ and |A × B| = m·n.
Injective?
f(1) = f(2) = b → not injective
Surjective?
every element of B is hit → surjective
Answer
surjective, not injective
Answer
surjective, not injective
5.Graph theory
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.
Handshake lemma for a tree
L + 14 = 2(L + 4) − 2
L = 8
Answer
L = 8
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: E
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.
Dijkstra round by round
1. A (0): B: 11
2. B (11): D: 16, G: 21
3. D (16): E: 27, G: 20
4. G (20): C: 29
5. E (27): F: 34
6. C (29)
7. F (34)
round 5
E
Answer
E
7.Number theory & modular arithmetic
Answer: x = 15
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).
Inverse of 7
r = 20, t = 0
r = 7, q = 2, t = 1
r = 6, q = 1, t = −2
r = 1, q = 6, t = 3
7−1 ≡ 3 (mod 20)
Solve
x ≡ 3·5 = 15 ≡ 15 (mod 20)
check: 7·15 = 105 ≡ 5 (mod 20)
Answer
x = 15
8.Number theory & modular arithmetic
Answer: 14170 mod 43 = 24
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.