Discrete Mathematics: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Propositional logic & proofs

Rows of a truth table
2ⁿ
Implication
p → q ≡ ¬p ∨ q ≡ ¬q → ¬p
Negated implication
¬(p → q) ≡ p ∧ ¬q
Biconditional
p ↔ q ≡ (p → q) ∧ (q → p)
De Morgan
¬(p ∧ q) ≡ ¬p ∨ ¬q, ¬(p ∨ q) ≡ ¬p ∧ ¬q
Distributive laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r), p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Negated quantifiers
¬∀x P(x) ≡ ∃x ¬P(x), ¬∃x P(x) ≡ ∀x ¬P(x)
Modus ponens, modus tollens
p → q, p ⊢ q; p → q, ¬q ⊢ ¬p

Method

  1. Count the different variables: the truth table has 2ⁿ rows, starting with all of them 1.
  2. Give every subformula its own column and fill the columns from the inside out.
  3. Apply the rule of each connective: → is 0 only for 1 → 0, ↔ is 1 when both sides agree.
  4. For equivalences, compare the final columns row by row; for an argument, look only at the rows in which every premise is 1.
  5. To negate a quantified statement, move ¬ inwards, swap ∀ and ∃, and negate the inner formula.

Sets, relations & functions

Two sets
|A ∪ B| = |A| + |B| − |A ∩ B|
Three sets
|A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|
Difference and complement
A − B = A ∩ B′, |A′| = |U| − |A|
Power set and product
|𝒫(A)| = 2|A|, |A × B| = |A|·|B|
Subsets of size k
C(n, k) = n!/(k!(n − k)!)
Functions A → B
nᵐ; injective n!/(n − m)!; surjective Σ (−1)ᵏ C(n, k)(n − k)ᵐ
Relations on n elements
2n²; reflexive 2n² − n; symmetric 2n(n+1)/2
Composition
(g ∘ f)(x) = g(f(x))

Method

  1. Draw a Venn diagram and decide region by region what belongs to the set expression.
  2. For counting problems, fill in the regions from the centre outwards: all sets, then pairs, then single sets, then the outside.
  3. Use inclusion–exclusion as a check: |A ∪ B| = |A| + |B| − |A ∩ B|.
  4. For a relation, test each property on every pair and give one counterexample when it fails.
  5. For a function, count the arrows arriving at each element of B: at most one means injective, at least one surjective.

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

Number theory & modular arithmetic

Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n

Method

  1. Reduce every number modulo n first; remainders lie between 0 and n − 1.
  2. For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
  3. For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
  4. Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
  5. For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).

2Practice problems

1.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (p → r) → (r → p) a tautology, a contradiction or a contingency? Click your answer.

(p → r) → (r → p)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
2.Propositional logic & proofsHard
pqp ∨ qp¬q11100100

The argument with premises p ∨ q; p and conclusion ¬q is invalid. Click the row of the truth table that is a counterexample.

p ∨ qp∴ ¬q
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
3.Sets, relations & functionsMedium
UABC101411113361

Each number in the Venn diagram is the number of elements in that region. Find |(A ∩ C) ∩ (C′ ∪ A)|.

|U| = 59
|(A ∩ C) ∩ (C′ ∪ A)| =
4.Sets, relations & functionsHard

How many different relations are there on a set with 4 elements? Call the number N.

|X| = 4
N =
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
7103110214875ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

start: Around 5
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 4 (mod 7) and x ≡ 7 (mod 10).

x ≡ 4 (mod 7)x ≡ 7 (mod 10)
x =
8.Number theory & modular arithmeticHard
442c = me​ mod n?m = cd​ mod nm(n, e) = (1763, 19)d

Encrypt the message m = 442 with the RSA public key (n, e) = (1763, 19). Find the ciphertext c.

n = 1763e = 19m = 442
c =

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

p = 1, r = 1: p → r = 1, r → p = 1, (p → r) → (r → p) = 1
p = 1, r = 0: p → r = 0, r → p = 1, (p → r) → (r → p) = 1
p = 0, r = 1: p → r = 1, r → p = 0, (p → r) → (r → p) = 0
p = 0, r = 0: p → r = 1, r → p = 1, (p → r) → (r → p) = 1

Classify

(p → r) → (r → p): 1, 1, 0, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
2.Propositional logic & proofs

Answer: p = 1, q = 1

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

p = 1, q = 1: 1 1 | 0
p = 1, q = 0: 1 1 | 1
p = 0, q = 1: 1 0 | 0
p = 0, q = 0: 0 0 | 1

Counterexample

p = 1, q = 1

Answer

p = 1, q = 1

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: |(A ∩ C) ∩ (C′ ∪ A)| = 4

Sets and Venn diagrams: A ∪ B holds the elements in A or B (or both), A ∩ B the elements in both, A − B the elements of A that are not in B, and the complement A′ everything in U that is not in A. In a Venn diagram every set expression is a union of regions, so count it region by region. Inclusion–exclusion: |A ∪ B| = |A| + |B| − |A ∩ B|, and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.

Regions of the expression

A ∩ B′ ∩ C: 3
A ∩ B ∩ C: 1

Add

|(A ∩ C) ∩ (C′ ∪ A)| = 3 + 1 = 4

Answer

|(A ∩ C) ∩ (C′ ∪ A)| = 4
4.Sets, relations & functions

Answer: N = 65536

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.

Count the free choices

N = 216 = 65536

Answer

N = 65536
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: G

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

1. A (0): B: 7
2. B (7): C: 17, D: 10, G: 15
3. D (10): E: 20, F: 12
4. F (12)
5. G (15)
6. C (17): E: 18
7. E (18)

round 5

G

Answer

G
7.Number theory & modular arithmetic

Answer: x = 67

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 = 4 + 7k

Second congruence

4 + 7k ≡ 7 (mod 10) → 7k ≡ 3 (mod 10)
k ≡ 3·3 ≡ 9 (mod 10)

Result

x = 4 + 7·9 = 67
x = 67
check: 67 mod 7 = 4, 67 mod 10 = 7

Answer

x = 67
8.Number theory & modular arithmetic

Answer: c = 665

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.

Repeated squaring

19 = 16 + 2 + 1
4421 ≡ 442 (mod 1763)
4422 ≡ 1434 (mod 1763)
4424 ≡ 698 (mod 1763)
4428 ≡ 616 (mod 1763)
44216 ≡ 411 (mod 1763)
44219 ≡ 442·1434·411 ≡ 665 (mod 1763)

Ciphertext

c = 665

Answer

c = 665

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