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
a ∧ (b ∨ c)(a ∧ b) ∨ (a ∧ c)(a ∨ b) ∧ c(a ∨ b) ∧ (a ∨ c)

Which formula is logically equivalent to a ∨ (b ∧ c)? Click it.

a ∨ (b ∧ c)
  1. 1a ∧ (b ∨ c)
  2. 2(a ∧ b) ∨ (a ∧ c)
  3. 3(a ∨ b) ∧ c
  4. 4(a ∨ b) ∧ (a ∨ c)
2.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((q ∨ p) → ¬r) ∨ ((¬p ∨ r) → ¬(s ∧ ¬q)) true? Call this number N.

((q ∨ p) → ¬r) ∨ ((¬p ∨ r) → ¬(s ∧ ¬q))
N =
3.Sets, relations & functionsMedium
UABC6571112355

Each number in the Venn diagram is the number of elements in that region. Find |C ∩ B|.

|U| = 54
|C ∩ B| =
4.Sets, relations & functionsHard
AB1234abcdinjective, not surjectivesurjective, not injectivebijectiveneither

The arrow diagram shows a function f: A → B with A = {1, 2, 3, 4} and B = {a, b, c, d}. Click the description that fits f.

f = {(1, b), (2, a), (3, d), (4, c)}
  1. 1injective, not surjective
  2. 2surjective, not injective
  3. 3bijective
  4. 4neither
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
6.Graph theoryHard
ABCDE

How many walks of length 3 go from E to A? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 5 (mod 8) and x ≡ 3 (mod 9).

x ≡ 5 (mod 8)x ≡ 3 (mod 9)
x =
8.Number theory & modular arithmeticHard
133c = me​ mod n?m = cd​ mod nm(n, e) = (143, 23)d

Encrypt the message m = 133 with the RSA public key (n, e) = (143, 23). Find the ciphertext c.

n = 143e = 23m = 133
c =

3Solutions

1.Propositional logic & proofs

Answer: (a ∨ b) ∧ (a ∨ c)

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

∨ distributes over ∧: a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c)

Check with the truth table

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

Answer

(a ∨ b) ∧ (a ∨ c)

The columns after the bar are the options in the order shown.

2.Propositional logic & proofs

Answer: N = 15

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.

Columns

q ∨ p; ¬r; (q ∨ p) → ¬r; ¬p; ¬p ∨ r; ¬q; s ∧ ¬q; ¬(s ∧ ¬q); (¬p ∨ r) → ¬(s ∧ ¬q); ((q ∨ p) → ¬r) ∨ ((¬p ∨ r) → ¬(s ∧ ¬q))

Split on p

p = 1: 7
p = 0: 8
N = 7 + 8 = 15

Answer

N = 15
3.Sets, relations & functions

Answer: |C ∩ B| = 10

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: 5
A ∩ B ∩ C: 5

Add

|C ∩ B| = 5 + 5 = 10

Answer

|C ∩ B| = 10
4.Sets, relations & functions

Answer: bijective

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?

every element of B is hit → surjective

Answer

bijective

Answer

bijective
5.Graph theory

Answer: H = 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.

Count

H = (5 − 1)!/2 = 24/2 = 12

Answer

H = 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
B: 1 0 0 1 1
C: 1 0 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)EA = 2·0 + 1·1 + 0·1 + 1·0 + 3·0 = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: x = 21

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 = 5 + 8k

Second congruence

5 + 8k ≡ 3 (mod 9) → 8k ≡ 7 (mod 9)
k ≡ 8·7 ≡ 2 (mod 9)

Result

x = 5 + 8·2 = 21
x = 21
check: 21 mod 8 = 5, 21 mod 9 = 3

Answer

x = 21
8.Number theory & modular arithmetic

Answer: c = 100

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

23 = 16 + 4 + 2 + 1
1331 ≡ 133 (mod 143)
1332 ≡ 100 (mod 143)
1334 ≡ 133 (mod 143)
1338 ≡ 100 (mod 143)
13316 ≡ 133 (mod 143)
13323 ≡ 133·100·133·133 ≡ 100 (mod 143)

Ciphertext

c = 100

Answer

c = 100

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