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 ¬b → (c ∨ a) by contradiction. Which statement do you assume at the start? Click it.
¬b → (c ∨ a)
1¬¬b → ¬(c ∨ a)
2¬b ∧ ¬(c ∨ a)
3¬¬b ∧ (c ∨ a)
4¬¬b ∧ ¬(c ∨ a)
2.Propositional logic & proofsHard
How many of the 16 rows of the truth table make ((d → a) ↔ ¬(d ∨ c)) → (c ↔ b) true? Call this number N.
((d → a) ↔ ¬(d ∨ c)) → (c ↔ b)
N =
3.Sets, relations & functionsMedium
R = {(1, 3), (2, 3), (3, 1), (3, 3)} is a relation on {1, 2, 3}. What is the smallest number k of ordered pairs that must be added to R to make it transitive (the size of the transitive closure minus |R|)?
R = {(1, 3), (2, 3), (3, 1), (3, 3)}
k =
4.Sets, relations & functionsHard
The arrow diagram shows a function f: A → B with A = {1, 2, 3, 4} and B = {a, b}. Click the description that fits f.
f = {(1, a), (2, b), (3, b), (4, b)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
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| = 5
1Euler circuit
2Euler path, no circuit
3neither
6.Graph theoryHard
How many walks of length 2 go from A to B? A walk may repeat vertices and edges. Call the number N.
M = adjacency matrixL = 2
N =
7.Number theory & modular arithmeticMedium
Find Euler’s totient φ(242), the number of integers from 1 to 242 that are coprime to 242.
n = 242 = 2·112
φ(242) =
8.Number theory & modular arithmeticHard
An RSA key uses the primes p = 17 and q = 43 and the public exponent e = 19. Find the private exponent d (0 < d < φ(n)).
p = 17q = 43e = 19
d =
3Solutions
1.Propositional logic & proofs
Answer: ¬b ∧ ¬(c ∨ a)
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 = ¬b, b = c ∨ a
Negation of the implication
¬(a → b) ≡ a ∧ ¬b = ¬b ∧ ¬(c ∨ a)
Answer
¬b ∧ ¬(c ∨ a)
If the assumption leads to a contradiction, the implication must be true.
2.Propositional logic & proofs
Answer: N = 12
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.
Columns
d → a; d ∨ c; ¬(d ∨ c); (d → a) ↔ ¬(d ∨ c); c ↔ b; ((d → a) ↔ ¬(d ∨ c)) → (c ↔ b)
Split on a
a = 1: 7
a = 0: 5
N = 7 + 5 = 12
Answer
N = 12
3.Sets, relations & functions
Answer: k = 2
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, 1), (2, 1)}
Count
k = 2
Answer
k = 2
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(2) = f(3) = b → not injective
Surjective?
every element of B is hit → surjective
Answer
surjective, not injective
Answer
surjective, not injective
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.
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Adjacency matrix
A: 0 1 1 0 0 0
B: 1 0 1 1 0 0
C: 1 1 0 0 1 0
D: 0 1 0 0 1 0
E: 0 0 1 1 0 1
F: 0 0 0 0 1 0
Common neighbours
(M²)AB = 0·1 + 1·0 + 1·1 + 0·1 + 0·0 + 0·0 = 1
Answer
N = 1
7.Number theory & modular arithmetic
Answer: φ(242) = 110
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).
Prime factors
242 = 2·112
Totient
φ(242) = φ(121)·φ(2) = 110·1 = 110
Answer
φ(242) = 110
8.Number theory & modular arithmetic
Answer: d = 283
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.