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, b)}
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| = 6|E| = 7
1Euler circuit
2Euler path, no circuit
3neither
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 smallest positive integer x with x ≡ 1 (mod 4) and x ≡ 6 (mod 13).
x ≡ 1 (mod 4)x ≡ 6 (mod 13)
x =
8.Number theory & modular arithmeticHard
89 is prime. Use Fermat’s little theorem to find 2357424 mod 89.
p = 89k = 57424
2357424 mod 89 =
3Solutions
1.Propositional logic & proofs
Answer: c → (a → b)
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
Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(c ∧ a) → b ≡ c → (a → b)
Check with the truth table
a = 1, b = 1, c = 1: 1 | 1 1 1 1
a = 1, b = 1, c = 0: 1 | 1 0 1 1
a = 1, b = 0, c = 1: 0 | 0 0 0 0
a = 1, b = 0, c = 0: 1 | 0 0 1 0
a = 0, b = 1, c = 1: 1 | 1 1 1 1
a = 0, b = 1, c = 0: 1 | 1 0 1 1
a = 0, b = 0, c = 1: 1 | 1 1 1 0
a = 0, b = 0, c = 0: 1 | 0 0 1 1
Answer
c → (a → b)
The columns after the bar are the options in the order shown.
2.Propositional logic & proofs
Answer: q = 1, r = 0
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.
Premises and conclusion row by row
q = 1, r = 1: 1 0 | 0
q = 1, r = 0: 1 1 | 0
q = 0, r = 1: 0 0 | 1
q = 0, r = 0: 1 1 | 1
Counterexample
q = 1, r = 0
Answer
q = 1, r = 0
An argument with this form looks like a valid rule, but one row is enough to show that it is not.
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, 2), (2, 1) ∈ R
transitive
fails: (1, 2), (2, 1) ∈ R, (1, 1) ∉ R
Answer
symmetric
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(2) = f(3) = b → not injective
Surjective?
every element of B is hit → surjective
Answer
surjective, not injective
Answer
surjective, not injective
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.
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: 14, C: 14
B (14): D: 22, E: 27
C (14): E: 19
E (19): F: 26
D (22)
F (26)
Shortest path
A → C → E → F
d(F) = 26
Answer
d(F) = 26
7.Number theory & modular arithmetic
Answer: x = 45
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 + 4k
Second congruence
1 + 4k ≡ 6 (mod 13) → 4k ≡ 5 (mod 13)
k ≡ 10·5 ≡ 11 (mod 13)
Result
x = 1 + 4·11 = 45
x = 45
check: 45 mod 4 = 1, 45 mod 13 = 6
Answer
x = 45
8.Number theory & modular arithmetic
Answer: 2357424 mod 89 = 64
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.