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 r → q? Click it.
r → q
1¬r ∨ q
2q → r
3¬r → ¬q
4r ∨ ¬q
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∀x (A(x) → B(x)), with the negation moved all the way inside? Click it.
¬∀x (A(x) → B(x))
1∃x (A(x) ∧ ¬B(x))
2∃x (A(x) → ¬B(x))
3∀x (A(x) ∧ ¬B(x))
4∃x (¬A(x) ∧ B(x))
3.Sets, relations & functionsMedium
In a group of 56 students, 13 speak French (F), 31 speak German (G) and 2 speak both. How many are in exactly one of the two sets?
|U| = 56|F| = 13|G| = 31|F ∩ G| = 2
|(F − G) ∪ (G − F)| =
4.Sets, relations & functionsHard
The arrow diagram shows a function f: A → B with A = {1, 2, 3, 4} and B = {a, b, c, d}. Click the description that fits f.
f = {(1, c), (2, b), (3, a), (4, d)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
A tree has 4 vertices of degree 3, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
4 × deg 32 × deg 53 × deg 2
L =
6.Graph theoryHard
Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 3.
start: Ck = 3
1edge AB
2edge AC
3edge BC
4edge CE
5edge DE
6edge DF
7edge EF
8edge BE
7.Number theory & modular arithmeticMedium
Find the inverse of 43 modulo 79: the number x with 0 < x < 79 and 43·x ≡ 1 (mod 79).
a = 43n = 79
43−1 mod 79 =
8.Number theory & modular arithmeticHard
Use repeated squaring to find 4258 mod 89.
a = 42k = 58n = 89
4258 mod 89 =
3Solutions
1.Propositional logic & proofs
Answer: ¬r ∨ q
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
a → b is 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
r → q ≡ ¬r ∨ q
Check with the truth table
q = 1, r = 1: 1 | 1 1 1 1
q = 1, r = 0: 1 | 1 0 0 0
q = 0, r = 1: 0 | 0 1 1 1
q = 0, r = 0: 1 | 1 1 1 1
Answer
¬r ∨ q
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∃x (A(x) ∧ ¬B(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 (A(x) → B(x)) ≡ ∃x (A(x) ∧ ¬B(x))
Answer
∃x (A(x) ∧ ¬B(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: |(F − G) ∪ (G − F)| = 40
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|.
Regions
F ∩ G: 2
F − G: 13 − 2 = 11
G − F: 31 − 2 = 29
(F ∪ G)′: 56 − (11 + 29 + 2) = 14
Answer
|(F − G) ∪ (G − F)| = 40
Answer
|(F − G) ∪ (G − F)| = 40
4.Sets, relations & functions
Answer: bijective
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?
every element of B is hit at most once → injective
Surjective?
every element of B is hit → surjective
Answer
bijective
Answer
bijective
5.Graph theory
Answer: L = 12
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 + 28 = 2(L + 9) − 2
L = 12
Answer
L = 12
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: edge BE
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. AC (7)
2. CE (16)
3. BE (4)
4. EF (10)
5. DF (9)
edge number 3
BE
Answer
edge BE
7.Number theory & modular arithmetic
Answer: 43−1 mod 79 = 68
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·43 (mod 79).
r = 79, t = 0
r = 43, q = 1, t = 1
r = 36, q = 1, t = −1
r = 7, q = 5, t = 2
r = 1, q = 7, t = −11
Inverse
1 ≡ −11·43 (mod 79)
x = 68
check: 43·68 = 2924 = 37·79 + 1
Answer
43−1 mod 79 = 68
8.Number theory & modular arithmetic
Answer: 4258 mod 89 = 25
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.