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
pqr¬q ∨ p(r ∧ ¬p) → (¬q ∨ p)(q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) true? Call this number N.

(q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p))
N =
2.Propositional logic & proofsHard
bcb → ccb11100100

The argument with premises b → c; c and conclusion b is invalid. Click the row of the truth table that is a counterexample.

b → cc∴ b
  1. 1b = 1, c = 1
  2. 2b = 1, c = 0
  3. 3b = 0, c = 1
  4. 4b = 0, c = 0
3.Sets, relations & functionsMedium
123

R = {(1, 3), (3, 2), (3, 3)} is a relation on {1, 2, 3}. What is the smallest number k of ordered pairs that must be added to R to make it reflexive?

R = {(1, 3), (3, 2), (3, 3)}
k =
4.Sets, relations & functionsHard

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

|X| = 4
N =
5.Graph theoryMedium

A connected graph has 10 vertices and 21 edges. How many edges k must be removed to leave a spanning tree?

|V| = 10|E| = 21
k =
6.Graph theoryHard
ABCDEF

How many walks of length 2 go from A to D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 8 modulo 61: the number x with 0 < x < 61 and 8·x ≡ 1 (mod 61).

a = 8n = 61
8−1 mod 61 =
8.Number theory & modular arithmeticHard
mc = me​ mod n333m = cd​ mod n?(n, e) = (341, 13)d

An RSA key uses p = 11, q = 31 and e = 13. Decrypt the ciphertext c = 333: find the message m.

p = 11q = 31e = 13c = 333
m =

3Solutions

1.Propositional logic & proofs

Answer: N = 7

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: q ↔ r = 1, ¬p = 0, r ∧ ¬p = 0, ¬q = 0, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 1, q = 1, r = 0: q ↔ r = 0, ¬p = 0, r ∧ ¬p = 0, ¬q = 0, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 1, q = 0, r = 1: q ↔ r = 0, ¬p = 0, r ∧ ¬p = 0, ¬q = 1, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 1, q = 0, r = 0: q ↔ r = 1, ¬p = 0, r ∧ ¬p = 0, ¬q = 1, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 0, q = 1, r = 1: q ↔ r = 1, ¬p = 1, r ∧ ¬p = 1, ¬q = 0, ¬q ∨ p = 0, (r ∧ ¬p) → (¬q ∨ p) = 0, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 0
p = 0, q = 1, r = 0: q ↔ r = 0, ¬p = 1, r ∧ ¬p = 0, ¬q = 0, ¬q ∨ p = 0, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 0, q = 0, r = 1: q ↔ r = 0, ¬p = 1, r ∧ ¬p = 1, ¬q = 1, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1
p = 0, q = 0, r = 0: q ↔ r = 1, ¬p = 1, r ∧ ¬p = 0, ¬q = 1, ¬q ∨ p = 1, (r ∧ ¬p) → (¬q ∨ p) = 1, (q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)) = 1

Count the rows with value 1

(q ↔ r) → ((r ∧ ¬p) → (¬q ∨ p)): 1, 1, 1, 1, 0, 1, 1, 1
N = 7

Answer

N = 7
2.Propositional logic & proofs

Answer: b = 0, c = 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

b = 1, c = 1: 1 1 | 1
b = 1, c = 0: 0 0 | 1
b = 0, c = 1: 1 1 | 0
b = 0, c = 0: 1 0 | 0

Counterexample

b = 0, c = 1

Answer

b = 0, c = 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: k = 2

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.

Pairs to add

{(1, 1), (2, 2)}

Count

k = 2

Answer

k = 2
4.Sets, relations & functions

Answer: N = 1024

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 = 24 + 6 = 1024

Answer

N = 1024
5.Graph theory

Answer: k = 12

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

10 − 1 = 9

Edges to remove

k = 21 − 9 = 12

Answer

k = 12
6.Graph theory

Answer: N = 1

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 1 1 0 0 0
B: 1 0 1 1 1 0
C: 1 1 0 0 1 0
D: 0 1 0 0 0 1
E: 0 1 1 0 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)AD = 0·0 + 1·1 + 1·0 + 0·0 + 0·0 + 0·1 = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: 8−1 mod 61 = 23

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·8 (mod 61).

r = 61, t = 0
r = 8, q = 7, t = 1
r = 5, q = 1, t = −7
r = 3, q = 1, t = 8
r = 2, q = 1, t = −15
r = 1, q = 2, t = 23

Inverse

1 ≡ 23·8 (mod 61)
x = 23
check: 8·23 = 184 = 3·61 + 1

Answer

8−1 mod 61 = 23
8.Number theory & modular arithmetic

Answer: m = 339

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.

Private exponent

n = 341, φ(n) = 300
d = 13−1 mod 300 = 277

Decrypt by repeated squaring

277 = 256 + 16 + 4 + 1
3331 ≡ 333 (mod 341)
3332 ≡ 64 (mod 341)
3334 ≡ 4 (mod 341)
3338 ≡ 16 (mod 341)
33316 ≡ 256 (mod 341)
33332 ≡ 64 (mod 341)
33364 ≡ 4 (mod 341)
333128 ≡ 16 (mod 341)
333256 ≡ 256 (mod 341)
333277 ≡ 333·4·256·256 ≡ 339 (mod 341)

Message

m = 339
check: 33913 mod 341 = 333

Answer

m = 339

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