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
pqrr ∧ q(r ∧ q) ∨ q(r ∨ p) ∨ ((r ∧ q) ∨ q)111?110?101?100?011?010?001?000?

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

(r ∨ p) ∨ ((r ∧ q) ∨ q)
N =
2.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((c → d) ↔ (¬c ∧ a)) ∨ ¬((a ∧ b) ↔ d) true? Call this number N.

((c → d) ↔ (¬c ∧ a)) ∨ ¬((a ∧ b) ↔ d)
N =
3.Sets, relations & functionsMedium
UMPC

Of 49 first-year students, 22 passed Mathematics (M), 14 passed Physics (P) and 28 passed Chemistry (C). 5 passed M and P, 12 passed M and C, 8 passed P and C, and 4 passed all three. How many are in exactly one of the three sets? Call the number n.

|U| = 49
n =
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, d), (2, a), (3, d), (4, d)}
  1. 1injective, not surjective
  2. 2surjective, not injective
  3. 3bijective
  4. 4neither
5.Graph theoryMedium

A connected graph has 11 vertices and 14 edges. How many edges k must be removed to leave a spanning tree?

|V| = 11|E| = 14
k =
6.Graph theoryHard
ABCDEF

How many closed walks of length 3 start and end at A? A walk may repeat vertices and edges. Call the number N.

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

Find the inverse of 30 modulo 43: the number x with 0 < x < 43 and 30·x ≡ 1 (mod 43).

a = 30n = 43
30−1 mod 43 =
8.Number theory & modular arithmeticHard
mc = me​ mod n204m = cd​ mod n?(n, e) = (893, 5)d

An RSA key uses p = 19, q = 47 and e = 5. Decrypt the ciphertext c = 204: find the message m.

p = 19q = 47e = 5c = 204
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: r ∨ p = 1, r ∧ q = 1, (r ∧ q) ∨ q = 1, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 1, q = 1, r = 0: r ∨ p = 1, r ∧ q = 0, (r ∧ q) ∨ q = 1, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 1, q = 0, r = 1: r ∨ p = 1, r ∧ q = 0, (r ∧ q) ∨ q = 0, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 1, q = 0, r = 0: r ∨ p = 1, r ∧ q = 0, (r ∧ q) ∨ q = 0, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 0, q = 1, r = 1: r ∨ p = 1, r ∧ q = 1, (r ∧ q) ∨ q = 1, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 0, q = 1, r = 0: r ∨ p = 0, r ∧ q = 0, (r ∧ q) ∨ q = 1, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 0, q = 0, r = 1: r ∨ p = 1, r ∧ q = 0, (r ∧ q) ∨ q = 0, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 1
p = 0, q = 0, r = 0: r ∨ p = 0, r ∧ q = 0, (r ∧ q) ∨ q = 0, (r ∨ p) ∨ ((r ∧ q) ∨ q) = 0

Count the rows with value 1

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

Answer

N = 7
2.Propositional logic & proofs

Answer: N = 13

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

c → d; ¬c; ¬c ∧ a; (c → d) ↔ (¬c ∧ a); a ∧ b; (a ∧ b) ↔ d; ¬((a ∧ b) ↔ d); ((c → d) ↔ (¬c ∧ a)) ∨ ¬((a ∧ b) ↔ d)

Split on a

a = 1: 7
a = 0: 6
N = 7 + 6 = 13

Answer

N = 13
3.Sets, relations & functions

Answer: n = 26

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|.

Centre and pairs

M ∩ P ∩ C: 4
M ∩ P only: 5 − 4 = 1
M ∩ C only: 12 − 4 = 8
P ∩ C only: 8 − 4 = 4

Single sets

M only: 22 − 1 − 8 − 4 = 9
P only: 14 − 1 − 4 − 4 = 5
C only: 28 − 8 − 4 − 4 = 12

Answer

9 + 5 + 12 = 26

Answer

n = 26
4.Sets, relations & functions

Answer: neither

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?

f(1) = f(3) = d → not injective

Surjective?

b, c receive no arrow → not surjective

Answer

neither

Answer

neither
5.Graph theory

Answer: k = 4

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

11 − 1 = 10

Edges to remove

k = 14 − 10 = 4

Answer

k = 4
6.Graph theory

Answer: N = 2

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

One more step

(M³)AA = 2·0 + 1·1 + 1·1 + 0·0 + 2·0 + 0·0 = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: 30−1 mod 43 = 33

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·30 (mod 43).

r = 43, t = 0
r = 30, q = 1, t = 1
r = 13, q = 2, t = −1
r = 4, q = 3, t = 3
r = 1, q = 4, t = −10

Inverse

1 ≡ −10·30 (mod 43)
x = 33
check: 30·33 = 990 = 23·43 + 1

Answer

30−1 mod 43 = 33
8.Number theory & modular arithmetic

Answer: m = 507

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 = 893, φ(n) = 828
d = 5−1 mod 828 = 497

Decrypt by repeated squaring

497 = 256 + 128 + 64 + 32 + 16 + 1
2041 ≡ 204 (mod 893)
2042 ≡ 538 (mod 893)
2044 ≡ 112 (mod 893)
2048 ≡ 42 (mod 893)
20416 ≡ 871 (mod 893)
20432 ≡ 484 (mod 893)
20464 ≡ 290 (mod 893)
204128 ≡ 158 (mod 893)
204256 ≡ 853 (mod 893)
204497 ≡ 204·871·484·290·158·853 ≡ 507 (mod 893)

Message

m = 507
check: 5075 mod 893 = 204

Answer

m = 507

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