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, b), (3, c)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
A tree has 3 vertices of degree 3, and all its other vertices are leaves (degree 1). How many leaves L does it have?
3 × deg 3
L =
6.Graph theoryHard
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.
start: Around 4
1A
2B
3C
4D
5E
6F
7.Number theory & modular arithmeticMedium
Find the inverse of 9 modulo 53: the number x with 0 < x < 53 and 9·x ≡ 1 (mod 53).
a = 9n = 53
9−1 mod 53 =
8.Number theory & modular arithmeticHard
Encrypt the message m = 37 with the RSA public key (n, e) = (235, 3). Find the ciphertext c.
n = 235e = 3m = 37
c =
3Solutions
1.Propositional logic & proofs
Answer: contingency
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).
Truth table row by row
p = 1, q = 1: q → p = 1, p → q = 1, (q → p) → (p → q) = 1
p = 1, q = 0: q → p = 1, p → q = 0, (q → p) → (p → q) = 0
p = 0, q = 1: q → p = 0, p → q = 1, (q → p) → (p → q) = 1
p = 0, q = 0: q → p = 1, p → q = 1, (q → p) → (p → q) = 1
Classify
(q → p) → (p → q): 1, 0, 1, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.
Answer
contingency
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.
Columns
a ∨ b; (a ∨ b) ∨ d; c → b; a ↔ c; (c → b) ∧ (a ↔ c); ((a ∨ b) ∨ d) ↔ ((c → b) ∧ (a ↔ c))
Split on a
a = 1: 2
a = 0: 4
N = 2 + 4 = 6
Answer
N = 6
3.Sets, relations & functions
Answer: symmetric
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
holds
antisymmetric
fails: (1, 3), (3, 1) ∈ R
transitive
fails: (1, 3), (3, 1) ∈ R, (1, 1) ∉ R
Answer
symmetric
4.Sets, relations & functions
Answer: bijective
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?
every element of B is hit → surjective
Answer
bijective
Answer
bijective
5.Graph theory
Answer: L = 5
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.
Handshake lemma for a tree
L + 9 = 2(L + 3) − 2
L = 5
Answer
L = 5
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
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: 14, C: 9
2. C (9): E: 15
3. B (14): D: 27
4. E (15): D: 19, F: 28
5. D (19)
6. F (28)
round 4
E
Answer
E
7.Number theory & modular arithmetic
Answer: 9−1 mod 53 = 6
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).
Extended Euclidean algorithm
Each remainder r is written as r ≡ t·9 (mod 53).
r = 53, t = 0
r = 9, q = 5, t = 1
r = 8, q = 1, t = −5
r = 1, q = 8, t = 6
Inverse
1 ≡ 6·9 (mod 53)
x = 6
check: 9·6 = 54 = 1·53 + 1
Answer
9−1 mod 53 = 6
8.Number theory & modular arithmetic
Answer: c = 128
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.