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 (q ∧ p) → ¬r by contradiction. Which statement do you assume at the start? Click it.
(q ∧ p) → ¬r
1(q ∧ p) ∧ ¬¬r
2¬(q ∧ p) ∧ ¬¬r
3¬(q ∧ p) → ¬¬r
4¬(q ∧ p) ∧ ¬r
2.Propositional logic & proofsHard
Which statement is equivalent to ¬∀x P(x), with the negation moved all the way inside? Click it.
¬∀x P(x)
1∀x ¬P(x)
2∀x P(x)
3∃x ¬P(x)
4∃x P(x)
3.Sets, relations & functionsMedium
R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 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 transitive (the size of the transitive closure minus |R|)?
On X = {1, 2, …, 9} the relation x ~ y holds when x ≡ y (mod 5). How many elements are in the equivalence class [9]?
X = {1, …, 9}x ~ y ⟺ 5 | (x − y)
|[9]| =
5.Graph theoryMedium
How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.
K4
H =
6.Graph theoryHard
Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.
k = 4
1edge AB
2edge AC
3edge BC
4edge BD
5edge DE
6edge DF
7edge EF
8edge BE
7.Number theory & modular arithmeticMedium
Solve 28x ≡ 52 (mod 73). Give the solution x with 0 < x < 73.
28x ≡ 52 (mod 73)
x =
8.Number theory & modular arithmeticHard
31 is prime. Use Fermat’s little theorem to find 2088311 mod 31.
p = 31k = 88311
2088311 mod 31 =
3Solutions
1.Propositional logic & proofs
Answer: (q ∧ 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).
Parts of the implication
a = q ∧ p, b = ¬r
Negation of the implication
¬(a → b) ≡ a ∧ ¬b = (q ∧ p) ∧ ¬¬r
Answer
(q ∧ p) ∧ ¬¬r
If the assumption leads to a contradiction, the implication must be true.
2.Propositional logic & proofs
Answer: ∃x ¬P(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 P(x) ≡ ∃x ¬P(x)
Answer
∃x ¬P(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: k = 3
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
{(3, 2), (3, 3), (3, 4)}
Count
k = 3
Answer
k = 3
4.Sets, relations & functions
Answer: |[9]| = 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.
The class
[9] = {4, 9}
|[9]| = 2
Answer
|[9]| = 2
5.Graph theory
Answer: H = 3
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 = (4 − 1)!/2 = 6/2 = 3
Answer
H = 3
6.Graph theory
Answer: edge BC
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.
Kruskal’s algorithm
1. BE (1)
2. AB (2)
3. DE (6)
4. BC (8)
5. EF (15)
edge number 4
BC
Answer
edge BC
7.Number theory & modular arithmetic
Answer: x = 54
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 28
r = 73, t = 0
r = 28, q = 2, t = 1
r = 17, q = 1, t = −2
r = 11, q = 1, t = 3
r = 6, q = 1, t = −5
r = 5, q = 1, t = 8
r = 1, q = 5, t = −13
28−1 ≡ 60 (mod 73)
Solve
x ≡ 60·52 = 3120 ≡ 54 (mod 73)
check: 28·54 = 1512 ≡ 52 (mod 73)
Answer
x = 54
8.Number theory & modular arithmetic
Answer: 2088311 mod 31 = 4
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.