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
You want to prove (r ∨ p) → q by contrapositive. Which implication do you prove instead? Click it.
(r ∨ p) → q
1(r ∨ p) ∧ ¬q
2q → (r ∨ p)
3¬q → ¬(r ∨ p)
4¬(r ∨ p) → ¬q
2.Propositional logic & proofsHard
How many of the 16 rows of the truth table make ¬((q → s) → r) → ((¬q ∧ ¬p) → (¬s ↔ r)) true? Call this number N.
¬((q → s) → r) → ((¬q ∧ ¬p) → (¬s ↔ r))
N =
3.Sets, relations & functionsMedium
Exactly one region of the Venn diagram is the set (A ∪ B) ∩ (C − A). Click its number.
(A ∪ B) ∩ (C − A)
1region 1
2region 2
3region 3
4region 4
5region 5
6region 6
7region 7
8region 8
4.Sets, relations & functionsHard
The arrow diagram shows a function f: A → B with A = {1, 2, 3} and B = {a, b, c}. Click the description that fits f.
f = {(1, a), (2, c), (3, c)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.
K6
H =
6.Graph theoryHard
Use Dijkstra’s algorithm to find the length d(G) of the shortest path from A to G in the weighted graph.
start: Aend: G
d(G) =
7.Number theory & modular arithmeticMedium
Find the smallest positive integer x with x ≡ 5 (mod 7) and x ≡ 1 (mod 11).
x ≡ 5 (mod 7)x ≡ 1 (mod 11)
x =
8.Number theory & modular arithmeticHard
Encrypt the message m = 3 with the RSA public key (n, e) = (133, 13). Find the ciphertext c.
n = 133e = 13m = 3
c =
3Solutions
1.Propositional logic & proofs
Answer: ¬q → ¬(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).
Parts of the implication
a = r ∨ p, b = q
Contrapositive
¬b → ¬a = ¬q → ¬(r ∨ p)
Answer
¬q → ¬(r ∨ p)
The contrapositive is equivalent to the original statement, so proving it proves the statement.
2.Propositional logic & proofs
Answer: N = 15
Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.
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 4
Answer
region 4
4.Sets, relations & functions
Answer: neither
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(2) = f(3) = c → not injective
Surjective?
b receives no arrow → not surjective
Answer
neither
Answer
neither
5.Graph theory
Answer: H = 60
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.
Count
H = (6 − 1)!/2 = 120/2 = 60
Answer
H = 60
6.Graph theory
Answer: d(G) = 15
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
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
A (0): B: 6, C: 4
C (4): B: 5, G: 15
B (5): D: 7
D (7): E: 11, F: 9
F (9)
E (11)
G (15)
Shortest path
A → C → G
d(G) = 15
Answer
d(G) = 15
7.Number theory & modular arithmetic
Answer: x = 12
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).
First congruence
x = 5 + 7k
Second congruence
5 + 7k ≡ 1 (mod 11) → 7k ≡ 7 (mod 11)
k ≡ 8·7 ≡ 1 (mod 11)
Result
x = 5 + 7·1 = 12
x = 12
check: 12 mod 7 = 5, 12 mod 11 = 1
Answer
x = 12
8.Number theory & modular arithmetic
Answer: c = 52
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.