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

You want to prove q → (p ∨ r) by contradiction. Which statement do you assume at the start? Click it.

q → (p ∨ r)
  1. 1q ∧ ¬(p ∨ r)
  2. 2¬q → ¬(p ∨ r)
  3. 3¬q ∧ ¬(p ∨ r)
  4. 4¬q ∧ (p ∨ r)
2.Propositional logic & proofsHard
q ∧ r¬pqp

The premises q ∨ r; q → p; r → p are given. Which conclusion follows from them? Click it.

q ∨ rq → pr → p
  1. 1q ∧ r
  2. 2¬p
  3. 3q
  4. 4p
3.Sets, relations & functionsMedium
123reflexivesymmetricantisymmetrictransitive

The relation R on {1, 2, 3} is drawn as a directed graph (a loop means (x, x) ∈ R). Exactly one of the four properties holds. Click it.

R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
  1. 1reflexive
  2. 2symmetric
  3. 3antisymmetric
  4. 4transitive
4.Sets, relations & functionsHard
AB12abcinjective, not surjectivesurjective, not injectivebijectiveneither

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

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

A connected graph has 17 vertices and 23 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 23
k =
6.Graph theoryHard
913610133111393ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(147), the number of integers from 1 to 147 that are coprime to 147.

n = 147 = 3·72
φ(147) =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 34171 mod 73.

a = 34k = 171n = 73
34171 mod 73 =

3Solutions

1.Propositional logic & proofs

Answer: q ∧ ¬(p ∨ r)

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).

Parts of the implication

a = q, b = p ∨ r

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = q ∧ ¬(p ∨ r)

Answer

q ∧ ¬(p ∨ r)

If the assumption leads to a contradiction, the implication must be true.

2.Propositional logic & proofs

Answer: p

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.

Rows with all premises true

p = 1, q = 1, r = 1: q ∧ r = 1, ¬p = 0, q = 1, p = 1
p = 1, q = 1, r = 0: q ∧ r = 0, ¬p = 0, q = 1, p = 1
p = 1, q = 0, r = 1: q ∧ r = 0, ¬p = 0, q = 0, p = 1

Conclusion

p (proof by cases)

Answer

p
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: (3, 3) ∉ R

symmetric

fails: (3, 1) ∈ R, (1, 3) ∉ R

antisymmetric

fails: (1, 2), (2, 1) ∈ R

transitive

holds

Answer

transitive
4.Sets, relations & functions

Answer: injective, not surjective

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?

b receives no arrow → not surjective

Answer

injective, not surjective

Answer

injective, not surjective
5.Graph theory

Answer: k = 7

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

17 − 1 = 16

Edges to remove

k = 23 − 16 = 7

Answer

k = 7
6.Graph theory

Answer: d(F) = 26

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 9, C: 13
B (9): D: 15, G: 20
C (13): E: 23
D (15)
G (20)
E (23): F: 26
F (26)

Shortest path

A → C → E → F
d(F) = 26

Answer

d(F) = 26
7.Number theory & modular arithmetic

Answer: φ(147) = 84

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

147 = 3·72

Totient

φ(147) = φ(49)·φ(3) = 42·2 = 84

Answer

φ(147) = 84
8.Number theory & modular arithmetic

Answer: 34171 mod 73 = 22

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.

Binary exponent

171 = 101010112 = 128 + 32 + 8 + 2 + 1

Repeated squaring

341 ≡ 34 (mod 73)
342 ≡ 61 (mod 73)
344 ≡ 71 (mod 73)
348 ≡ 4 (mod 73)
3416 ≡ 16 (mod 73)
3432 ≡ 37 (mod 73)
3464 ≡ 55 (mod 73)
34128 ≡ 32 (mod 73)

Multiply the needed powers

34171 ≡ 34·61·4·37·32 ≡ 22 (mod 73)

Answer

34171 mod 73 = 22

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