The arrow diagram shows a function f: A → B with A = {1, 2, 3} and B = {a, b}. Click the description that fits f.
f = {(1, a), (2, b), (3, a)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
A tree has 2 vertices of degree 3, 2 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?
2 × deg 32 × deg 5
L =
6.Graph theoryHard
Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.
start: Aend: F
d(F) =
7.Number theory & modular arithmeticMedium
Find the inverse of 51 modulo 67: the number x with 0 < x < 67 and 51·x ≡ 1 (mod 67).
a = 51n = 67
51−1 mod 67 =
8.Number theory & modular arithmeticHard
Solve 6x ≡ 8 (mod 26). Give the smallest positive solution x.
6x ≡ 8 (mod 26)
x =
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
b = 1, c = 1: c → b = 1, b → c = 1, (c → b) → (b → c) = 1
b = 1, c = 0: c → b = 1, b → c = 0, (c → b) → (b → c) = 0
b = 0, c = 1: c → b = 0, b → c = 1, (c → b) → (b → c) = 1
b = 0, c = 0: c → b = 1, b → c = 1, (c → b) → (b → c) = 1
Classify
(c → b) → (b → c): 1, 0, 1, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.
Answer
contingency
2.Propositional logic & proofs
Answer: ¬p
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.
Rows with all premises true
p = 0, q = 0: q = 0, ¬p = 1, ¬p ∧ q = 0, p = 0
Conclusion
¬p (modus tollens)
Answer
¬p
3.Sets, relations & functions
Answer: transitive
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: (2, 1) ∈ R, (1, 2) ∉ R
antisymmetric
fails: (2, 4), (4, 2) ∈ R
transitive
holds
Answer
transitive
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(1) = f(3) = a → not injective
Surjective?
every element of B is hit → surjective
Answer
surjective, not injective
Answer
surjective, not injective
5.Graph theory
Answer: L = 10
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 + 16 = 2(L + 4) − 2
L = 10
Answer
L = 10
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
6.Graph theory
Answer: d(F) = 32
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
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
A (0): B: 13
B (13): C: 27, D: 25, G: 24
G (24): E: 30
D (25): F: 36
C (27)
E (30): F: 32
F (32)
Shortest path
A → B → G → E → F
d(F) = 32
Answer
d(F) = 32
7.Number theory & modular arithmetic
Answer: 51−1 mod 67 = 46
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·51 (mod 67).
r = 67, t = 0
r = 51, q = 1, t = 1
r = 16, q = 3, t = −1
r = 3, q = 5, t = 4
r = 1, q = 3, t = −21
Inverse
1 ≡ −21·51 (mod 67)
x = 46
check: 51·46 = 2346 = 35·67 + 1
Answer
51−1 mod 67 = 46
8.Number theory & modular arithmetic
Answer: x = 10
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).