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

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

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

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

r ∨ qr → pq → p
  1. 1r ∧ q
  2. 2r
  3. 3p
  4. 4¬p
3.Sets, relations & functionsMedium
UMPC

Of 61 first-year students, 26 passed Mathematics (M), 32 passed Physics (P) and 24 passed Chemistry (C). 17 passed M and P, 11 passed M and C, 8 passed P and C, and 3 passed all three. How many are in exactly two of the three sets? Call the number n.

|U| = 61
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, b), (2, d), (3, a), (4, c)}
  1. 1injective, not surjective
  2. 2surjective, not injective
  3. 3bijective
  4. 4neither
5.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
1516183114617ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 3.

start: Ck = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF
7.Number theory & modular arithmeticMedium

Find the inverse of 11 modulo 73: the number x with 0 < x < 73 and 11·x ≡ 1 (mod 73).

a = 11n = 73
11−1 mod 73 =
8.Number theory & modular arithmeticHard

11 is prime. Use Fermat’s little theorem to find 2574727 mod 11.

p = 11k = 74727
2574727 mod 11 =

3Solutions

1.Propositional logic & proofs

Answer: N = 6

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

Count the rows with value 1

(q → ¬r) ∨ (p ↔ (¬r ∨ ¬p)): 0, 1, 1, 1, 0, 1, 1, 1
N = 6

Answer

N = 6
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: r ∧ q = 1, r = 1, p = 1, ¬p = 0
p = 1, q = 1, r = 0: r ∧ q = 0, r = 0, p = 1, ¬p = 0
p = 1, q = 0, r = 1: r ∧ q = 0, r = 1, p = 1, ¬p = 0

Conclusion

p (proof by cases)

Answer

p
3.Sets, relations & functions

Answer: n = 27

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: 3
M ∩ P only: 17 − 3 = 14
M ∩ C only: 11 − 3 = 8
P ∩ C only: 8 − 3 = 5

Single sets

M only: 26 − 14 − 8 − 3 = 1
P only: 32 − 14 − 5 − 3 = 10
C only: 24 − 8 − 5 − 3 = 8

Answer

14 + 8 + 5 = 27

Answer

n = 27
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: neither

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.

Degrees

deg(A) = 2, deg(B) = 2, deg(C) = 3, deg(D) = 1, deg(E) = 3, deg(F) = 1

Odd vertices

C, D, E, F → 4
neither

Answer

neither
6.Graph theory

Answer: edge BD

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.

Prim’s algorithm

1. CE (11)
2. DE (4)
3. BD (3)
4. DF (6)
5. AB (15)

edge number 3

BD

Answer

edge BD
7.Number theory & modular arithmetic

Answer: 11−1 mod 73 = 20

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·11 (mod 73).

r = 73, t = 0
r = 11, q = 6, t = 1
r = 7, q = 1, t = −6
r = 4, q = 1, t = 7
r = 3, q = 1, t = −13
r = 1, q = 3, t = 20

Inverse

1 ≡ 20·11 (mod 73)
x = 20
check: 11·20 = 220 = 3·73 + 1

Answer

11−1 mod 73 = 20
8.Number theory & modular arithmetic

Answer: 2574727 mod 11 = 9

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.

Reduce the exponent

74727 mod 10 = 7
2574727 ≡ 257 (mod 11)

Small power

257 mod 11 = 9

Answer

2574727 mod 11 = 9

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