Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
|V| = 5|E| = 6
1Euler circuit
2Euler path, no circuit
3neither
6.Graph theoryHard
How many closed walks of length 2 start and end at E? A walk may repeat vertices and edges. Call the number N.
M = adjacency matrixL = 2
N =
7.Number theory & modular arithmeticMedium
Find Euler’s totient φ(1083), the number of integers from 1 to 1083 that are coprime to 1083.
n = 1083 = 3·192
φ(1083) =
8.Number theory & modular arithmeticHard
An RSA key uses the primes p = 5 and q = 11 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).
p = 5q = 11e = 17
d =
3Solutions
1.Propositional logic & proofs
Answer: tautology
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
a = 1, c = 1: c → a = 1, a → c = 1, (c → a) ∨ (a → c) = 1
a = 1, c = 0: c → a = 1, a → c = 0, (c → a) ∨ (a → c) = 1
a = 0, c = 1: c → a = 0, a → c = 1, (c → a) ∨ (a → c) = 1
a = 0, c = 0: c → a = 1, a → c = 1, (c → a) ∨ (a → c) = 1
Classify
(c → a) ∨ (a → c): 1, 1, 1, 1
1 in every row: a tautology.
Answer
tautology
2.Propositional logic & proofs
Answer: ∃x (F(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 (F(x) → ∃y R(x, y)) ≡ ∃x (F(x) ∧ ∀y ¬R(x, y))
Answer
∃x (F(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: 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: (2, 2) ∉ R
symmetric
fails: (4, 1) ∈ R, (1, 4) ∉ R
antisymmetric
fails: (1, 3), (3, 1) ∈ R
transitive
holds
Answer
transitive
4.Sets, relations & functions
Answer: (f ∘ g)(2) = 3
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.
Inner function
g(2) = 2
Outer function
f(2) = 3
(f ∘ g)(2) = 3
Answer
(f ∘ g)(2) = 3
5.Graph theory
Answer: Euler path, no 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.
The Euler path starts at one odd vertex and ends at the other.
6.Graph theory
Answer: N = 3
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Adjacency matrix
A: 0 0 1 0 0
B: 0 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0
Common neighbours
(M²)EE = 0·0 + 1·1 + 1·1 + 1·1 + 0·0 = 3
Answer
N = 3
7.Number theory & modular arithmetic
Answer: φ(1083) = 684
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).
Prime factors
1083 = 3·192
Totient
φ(1083) = φ(3)·φ(361) = 2·342 = 684
Answer
φ(1083) = 684
8.Number theory & modular arithmetic
Answer: d = 33
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.