How many surjective (onto) functions f: A → B are there if |A| = 5 and |B| = 2? Call the number N.
|A| = 5|B| = 2
N =
5.Graph theoryMedium
A connected graph has 19 vertices and 26 edges. How many edges k must be removed to leave a spanning tree?
|V| = 19|E| = 26
k =
6.Graph theoryHard
Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 4.
start: Dk = 4
1edge AB
2edge AC
3edge BC
4edge BD
5edge CE
6edge DE
7edge BE
7.Number theory & modular arithmeticMedium
Find the smallest positive integer x with x ≡ 1 (mod 9) and x ≡ 8 (mod 13).
x ≡ 1 (mod 9)x ≡ 8 (mod 13)
x =
8.Number theory & modular arithmeticHard
An RSA key uses p = 37, q = 43 and e = 13. Decrypt the ciphertext c = 373: find the message m.
p = 37q = 43e = 13c = 373
m =
3Solutions
1.Propositional logic & proofs
Answer: ¬q ∧ ¬p
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).
The law
De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(q ∨ p) ≡ ¬q ∧ ¬p
Check with the truth table
p = 1, q = 1: 0 | 0 0 0 1
p = 1, q = 0: 0 | 1 0 1 0
p = 0, q = 1: 0 | 1 0 0 1
p = 0, q = 0: 1 | 1 1 0 1
Answer
¬q ∧ ¬p
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: ∃x (¬F(x) ∧ ¬G(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 (F(x) ∨ G(x)) ≡ ∃x (¬F(x) ∧ ¬G(x))
Answer
∃x (¬F(x) ∧ ¬G(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: 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: (1, 3) ∈ R, (3, 1) ∉ R
antisymmetric
fails: (3, 4), (4, 3) ∈ R
transitive
holds
Answer
transitive
4.Sets, relations & functions
Answer: N = 30
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.
Count
N = C(2, 0)·25 − C(2, 1)·15 = 30
Answer
N = 30
5.Graph theory
Answer: k = 8
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.
Edges of a spanning tree
19 − 1 = 18
Edges to remove
k = 26 − 18 = 8
Answer
k = 8
6.Graph theory
Answer: edge AC
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.
Prim’s algorithm
1. DE (13)
2. BE (9)
3. BC (7)
4. AC (5)
edge number 4
AC
Answer
edge AC
7.Number theory & modular arithmetic
Answer: x = 73
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).
First congruence
x = 1 + 9k
Second congruence
1 + 9k ≡ 8 (mod 13) → 9k ≡ 7 (mod 13)
k ≡ 3·7 ≡ 8 (mod 13)
Result
x = 1 + 9·8 = 73
x = 73
check: 73 mod 9 = 1, 73 mod 13 = 8
Answer
x = 73
8.Number theory & modular arithmetic
Answer: m = 448
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.