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 → r by contrapositive. Which implication do you prove instead? Click it.
q → r
1¬q → ¬r
2¬r → ¬q
3r → q
4q ∧ ¬r
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 71 students, 39 speak French (F), 36 speak German (G) and 11 speak both. How many are in exactly one of the two sets?
|U| = 71|F| = 39|G| = 36|F ∩ G| = 11
|(F − G) ∪ (G − F)| =
4.Sets, relations & functionsHard
The arrow diagram shows a function f: A → B with A = {1, 2, 3} and B = {a, b, c, d}. Click the description that fits f.
f = {(1, d), (2, b), (3, c)}
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 closed walks of length 3 start and end at D? A walk may repeat vertices and edges. Call the number N.
M = adjacency matrixL = 3
N =
7.Number theory & modular arithmeticMedium
Find Euler’s totient φ(33), the number of integers from 1 to 33 that are coprime to 33.
n = 33 = 3·11
φ(33) =
8.Number theory & modular arithmeticHard
71 is prime. Use Fermat’s little theorem to find 773718 mod 71.
p = 71k = 73718
773718 mod 71 =
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).
Parts of the implication
a = q, b = r
Contrapositive
¬b → ¬a = ¬r → ¬q
Answer
¬r → ¬q
The contrapositive is equivalent to the original statement, so proving it proves the statement.
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)| = 53
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: 11
F − G: 39 − 11 = 28
G − F: 36 − 11 = 25
(F ∪ G)′: 71 − (28 + 25 + 11) = 7
Answer
|(F − G) ∪ (G − F)| = 53
Answer
|(F − G) ∪ (G − F)| = 53
4.Sets, relations & functions
Answer: injective, not surjective
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?
a receives no arrow → not surjective
Answer
injective, not surjective
Answer
injective, not surjective
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
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0
One more step
(M³)DD = 1·0 + 1·1 + 2·0 + 2·0 + 1·1 = 2
Answer
N = 2
7.Number theory & modular arithmetic
Answer: φ(33) = 20
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
33 = 3·11
Totient
φ(33) = φ(3)·φ(11) = 2·10 = 20
Answer
φ(33) = 20
8.Number theory & modular arithmetic
Answer: 773718 mod 71 = 27
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.