Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

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.

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

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).

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).

IPv4 subnetting

Hosts per subnet
2(32 − prefix) − 2 (/31: 2, /32: 1)
Block size
256 − (mask octet)
Network address
IP AND mask
Broadcast address
IP OR (NOT mask): all host bits set to 1
Subnets from n borrowed bits
2n

2Practice problems

1.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1128 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1010 0011 >> 2. What is the result's value R in decimal?

1010 0011>> 2
R =
3.Propositional logic & proofsMedium
abc¬c ↔ ¬b(¬c ↔ ¬b) ∨ a(b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a)111?110?101?100?011?010?001?000?

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

(b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a)
N =
4.Propositional logic & proofsHard
pqp → q¬p¬q11100100

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

p → q¬p∴ ¬q
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​CI1​0I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 0, I₁ = C, I₂ = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = CI₂ = 0I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLK

A 4-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 5 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 4tpd = 5 ns
fmax = MHz
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 3 (mod 10) and x ≡ 4 (mod 13).

x ≡ 3 (mod 10)x ≡ 4 (mod 13)
x =
8.Number theory & modular arithmeticHard
89c = me​ mod n?m = cd​ mod nm(n, e) = (119, 7)d

Encrypt the message m = 89 with the RSA public key (n, e) = (119, 7). Find the ciphertext c.

n = 119e = 7m = 89
c =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms with different bases differ only by a constant factor
  2. BAll logarithms in computer science have base 10, so it is implied
  3. CThe base only matters for small n
  4. DLogarithms are always rounded to integers
10.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is NP-hard
  2. BIt can be solved by Dijkstra's algorithm with negated weights
  3. CIt is undecidable
  4. DIt is always solvable in O(V + E)
11.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. AAscending sorted order
  2. BInsertion order
  3. CLevel order
  4. DDescending order always
12.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. AThe need for a hash function
  2. BThe need to store keys in the table
  3. CPrimary clustering, where long runs of occupied slots form
  4. DThe O(n) cost of reading a key's value
13.Graph theoryMedium

A connected graph has 11 vertices and 21 edges. How many edges k must be removed to leave a spanning tree?

|V| = 11|E| = 21
k =
14.Graph theoryHard
21589184ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.

k = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

What does INTERSECT return when used between two SELECT queries?

  1. ARows that appear in both results
  2. BThe Cartesian product of the results
  3. CRows in the first result but not the second
  4. DAll rows from both results
16.Databases & SQLHard

What is a phantom read?

  1. AA repeated query returns new rows inserted by another committed transaction
  2. BReading uncommitted changes
  3. CA query that returns no rows
  4. DReading data that was never written
17.Computer networksMedium

What does an MX record in DNS specify?

  1. AThe name servers for a zone
  2. BAn alias name for another host
  3. CThe mail servers that accept email for a domain
  4. DThe IPv4 address of a web server
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIPv6 only works on local networks, so addresses never need to be shared globally
  2. BIPv6 does not use addresses
  3. CIts 128-bit address space is large enough to give every device a global address
  4. DIPv6 encrypts all addresses
19.IPv4 subnettingMedium
255.255.255.24811111111111111111111111111111000network bitshost bits

A subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?

255.255.255.248
H =
20.IPv4 subnettingHard
last octet = 111012816413201618141211networkhost

A host has the address 10.20.156.111/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.156.111/27
x =

3Solutions

1.Number systems

Answer: U = 144

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000

Read it as unsigned

1001 0000₂ = 256 − 112 = 144

Answer

U = 144
2.Number systems

Answer: R = 40

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Bit by bit

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1010 0011 >> 2
= 0010 1000

Value of the result

0010 1000₂ = 40

Answer

R = 40
3.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

a = 1, b = 1, c = 1: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 0, ¬c = 0, ¬b = 0, ¬c ↔ ¬b = 1, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1
a = 1, b = 1, c = 0: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 0, ¬c = 1, ¬b = 0, ¬c ↔ ¬b = 0, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1
a = 1, b = 0, c = 1: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 1, ¬c = 0, ¬b = 1, ¬c ↔ ¬b = 0, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1
a = 1, b = 0, c = 0: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 1, ¬c = 1, ¬b = 1, ¬c ↔ ¬b = 1, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1
a = 0, b = 1, c = 1: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 0, ¬c = 0, ¬b = 0, ¬c ↔ ¬b = 1, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1
a = 0, b = 1, c = 0: c ∨ a = 0, ¬(c ∨ a) = 1, b → ¬(c ∨ a) = 1, ¬c = 1, ¬b = 0, ¬c ↔ ¬b = 0, (¬c ↔ ¬b) ∨ a = 0, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 0
a = 0, b = 0, c = 1: c ∨ a = 1, ¬(c ∨ a) = 0, b → ¬(c ∨ a) = 1, ¬c = 0, ¬b = 1, ¬c ↔ ¬b = 0, (¬c ↔ ¬b) ∨ a = 0, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 0
a = 0, b = 0, c = 0: c ∨ a = 0, ¬(c ∨ a) = 1, b → ¬(c ∨ a) = 1, ¬c = 1, ¬b = 1, ¬c ↔ ¬b = 1, (¬c ↔ ¬b) ∨ a = 1, (b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a) = 1

Count the rows with value 1

(b → ¬(c ∨ a)) → ((¬c ↔ ¬b) ∨ a): 1, 1, 1, 1, 1, 0, 0, 1
N = 6

Answer

N = 6
4.Propositional logic & proofs

Answer: p = 0, q = 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

p = 1, q = 1: 1 0 | 0
p = 1, q = 0: 0 0 | 1
p = 0, q = 1: 1 1 | 0
p = 0, q = 0: 1 1 | 1

Counterexample

p = 0, q = 1

Answer

p = 0, q = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

5.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′BC + ABC′

Rows with F = 1

ABC = 011, 110
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: fmax = 50 MHz

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Worst-case settling time

T = 4·5 ns = 20 ns

Maximum frequency

fmax = 1/20 ns = 50 MHz

Answer

fmax = 50 MHz
7.Number theory & modular arithmetic

Answer: x = 43

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 = 3 + 10k

Second congruence

3 + 10k ≡ 4 (mod 13) → 10k ≡ 1 (mod 13)
k ≡ 4·1 ≡ 4 (mod 13)

Result

x = 3 + 10·4 = 43
x = 43
check: 43 mod 10 = 3, 43 mod 13 = 4

Answer

x = 43
8.Number theory & modular arithmetic

Answer: c = 47

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

7 = 4 + 2 + 1
891 ≡ 89 (mod 119)
892 ≡ 67 (mod 119)
894 ≡ 86 (mod 119)
897 ≡ 89·67·86 ≡ 47 (mod 119)

Ciphertext

c = 47

Answer

c = 47
9.Algorithms & complexity

Answer: A · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

A · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: A · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

A · It is NP-hard
11.Data structures

Answer: A · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

A · Ascending sorted order
12.Data structures

Answer: C · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

C · Primary clustering, where long runs of occupied slots form
13.Graph theory

Answer: k = 11

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 = 21 − 10 = 11

Answer

k = 11
14.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.

Kruskal’s algorithm

1. AC (1)
2. AB (2)
3. BE (4)
4. BD (8)

edge number 4

BD

Answer

edge BD
15.Databases & SQL

Answer: A · Rows that appear in both results

Why this is the answer

INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.

Answer

A · Rows that appear in both results
16.Databases & SQL

Answer: A · A repeated query returns new rows inserted by another committed transaction

Why this is the answer

Phantoms occur when another transaction inserts rows that match a range condition; serializable isolation prevents them.

Answer

A · A repeated query returns new rows inserted by another committed transaction
17.Computer networks

Answer: C · The mail servers that accept email for a domain

Why this is the answer

MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.

Answer

C · The mail servers that accept email for a domain
18.Computer networks

Answer: C · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

C · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: H = 6

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Prefix length

255.255.255.248 = /29

Usable hosts

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 127

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Block size

25 = 32

Subnet of this host

network: ⌊111/32⌋·32 = 96
broadcast: 96 + 32 − 1 = 127
usable hosts: 97 … 126

Answer

x = 127

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