The arrow diagram shows a function f: A → B with A = {1, 2} and B = {a, b, c}. Click the description that fits f.
f = {(1, b), (2, c)}
1injective, not surjective
2surjective, not injective
3bijective
4neither
5.Graph theoryMedium
How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.
K7
H =
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 3 modulo 17: the number x with 0 < x < 17 and 3·x ≡ 1 (mod 17).
a = 3n = 17
3−1 mod 17 =
8.Number theory & modular arithmeticHard
Use repeated squaring to find 20121 mod 61.
a = 20k = 121n = 61
20121 mod 61 =
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).
Parts of the implication
a = q, b = ¬p
Negation of the implication
¬(a → b) ≡ a ∧ ¬b = q ∧ ¬¬p
Answer
q ∧ ¬¬p
If the assumption leads to a contradiction, the implication must be true.
2.Propositional logic & proofs
Answer: ∃x (P(x) ∧ ∀y ¬R(x, y))
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 (P(x) → ∃y R(x, y)) ≡ ∃x (P(x) ∧ ∀y ¬R(x, y))
Answer
∃x (P(x) ∧ ∀y ¬R(x, y))
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: reflexive
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
holds
symmetric
fails: (2, 4) ∈ R, (4, 2) ∉ R
antisymmetric
fails: (1, 3), (3, 1) ∈ R
transitive
fails: (1, 3), (3, 2) ∈ R, (1, 2) ∉ R
Answer
reflexive
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: H = 360
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.
Count
H = (7 − 1)!/2 = 720/2 = 360
Answer
H = 360
6.Graph theory
Answer: d(F) = 24
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: 10
C (10)
B (14): D: 23, G: 24
D (23): E: 26, F: 24
F (24)
G (24)
E (26)
Shortest path
A → B → D → F
d(F) = 24
Answer
d(F) = 24
7.Number theory & modular arithmetic
Answer: 3−1 mod 17 = 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·3 (mod 17).
r = 17, t = 0
r = 3, q = 5, t = 1
r = 2, q = 1, t = −5
r = 1, q = 2, t = 6
Inverse
1 ≡ 6·3 (mod 17)
x = 6
check: 3·6 = 18 = 1·17 + 1
Answer
3−1 mod 17 = 6
8.Number theory & modular arithmetic
Answer: 20121 mod 61 = 20
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.