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 (p ∧ q) → r by contradiction. Which statement do you assume at the start? Click it.
(p ∧ q) → r
1¬(p ∧ q) → ¬r
2¬(p ∧ q) ∧ r
3¬(p ∧ q) ∧ ¬r
4(p ∧ q) ∧ ¬r
2.Propositional logic & proofsHard
How many of the 16 rows of the truth table make (b ∨ a) → (¬d ∧ (¬c ∧ b)) true? Call this number N.
(b ∨ a) → (¬d ∧ (¬c ∧ b))
N =
3.Sets, relations & functionsMedium
The relation R on {1, 2, 3, 4} is drawn as a directed graph (a loop means (x, x) ∈ R). Exactly one of the four properties holds. Click it.
R = {(1, 2), (2, 3), (3, 1), (4, 2)}
1reflexive
2symmetric
3antisymmetric
4transitive
4.Sets, relations & functionsHard
The bijection f on {1, 2, 3, 4, 5} is given by its table. Find f−1(4).
f: 1 ↦ 2, 2 ↦ 3, 3 ↦ 4, 4 ↦ 1, 5 ↦ 5
f−1(4) =
5.Graph theoryMedium
How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.
K7
H =
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 ≡ 1 (mod 17). Give the solution x with 0 < x < 17.
7x ≡ 1 (mod 17)
x =
8.Number theory & modular arithmeticHard
An RSA key uses p = 5, q = 17 and e = 23. Decrypt the ciphertext c = 8: find the message m.
p = 5q = 17e = 23c = 8
m =
3Solutions
1.Propositional logic & proofs
Answer: (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).
Parts of the implication
a = p ∧ q, b = r
Negation of the implication
¬(a → b) ≡ a ∧ ¬b = (p ∧ q) ∧ ¬r
Answer
(p ∧ q) ∧ ¬r
If the assumption leads to a contradiction, the implication must be true.
2.Propositional logic & proofs
Answer: N = 6
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.
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.
reflexive
fails: (1, 1) ∉ R
symmetric
fails: (1, 2) ∈ R, (2, 1) ∉ R
antisymmetric
holds
transitive
fails: (1, 2), (2, 3) ∈ R, (1, 3) ∉ R
Answer
antisymmetric
4.Sets, relations & functions
Answer: f−1(4) = 3
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.
Read the table backwards
f(3) = 4 → f−1(4) = 3
Answer
f−1(4) = 3
5.Graph theory
Answer: H = 360
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 = (7 − 1)!/2 = 720/2 = 360
Answer
H = 360
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: 6, C: 8
2. B (6): D: 7, G: 11
3. D (7): E: 10, F: 17
4. C (8)
5. E (10)
6. G (11)
7. F (17)
round 5
E
Answer
E
7.Number theory & modular arithmetic
Answer: x = 5
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 = 17, t = 0
r = 7, q = 2, t = 1
r = 3, q = 2, t = −2
r = 1, q = 3, t = 5
7−1 ≡ 5 (mod 17)
Solve
x ≡ 5·1 = 5 ≡ 5 (mod 17)
check: 7·5 = 35 ≡ 1 (mod 17)
Answer
x = 5
8.Number theory & modular arithmetic
Answer: m = 32
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.