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 −17 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −178 bit
U =
2.Number systemsHard
1822 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 1822 items its own binary code of the same length. What is the smallest number of bits b needed?

1822
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is r ∧ ¬r a tautology, a contradiction or a contingency? Click your answer.

r ∧ ¬r
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∃x ¬F(x)∀x F(x)∀x ¬F(x)∃x F(x)

Which statement is equivalent to ¬∀x F(x), with the negation moved all the way inside? Click it.

¬∀x F(x)
  1. 1∃x ¬F(x)
  2. 2∀x F(x)
  3. 3∀x ¬F(x)
  4. 4∃x F(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302040517060120130150141819111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 7, 8, 9, 10, 11). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 7, 8, 9, 10, 11)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X00113020405X70601211311501418X9111110

For F(A, B, C, D) = Σm(3, 8, 10, 11, 13, 15) + d(0, 7, 9), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(3, 8, 10, 11, 13, 15) + d(0, 7, 9)
N =
7.Number theory & modular arithmeticMedium

Solve 41x ≡ 4 (mod 43). Give the solution x with 0 < x < 43.

41x ≡ 4 (mod 43)
x =
8.Number theory & modular arithmeticHard

Solve 20x ≡ 6 (mod 38). Give the smallest positive solution x.

20x ≡ 6 (mod 38)
x =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALogarithmic
  2. BExponential
  3. CConstant
  4. DLinear
10.Algorithms & complexityHard

Which algorithm finds a minimum spanning tree by repeatedly adding the cheapest edge that does not form a cycle?

  1. AKruskal's algorithm
  2. BFloyd–Warshall
  3. CPrim's algorithm starting from one node
  4. DDijkstra's algorithm
11.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(V)
  2. BO(E)
  3. CO(log V)
  4. DO(1)
12.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(n) time and O(n) extra space
  2. BO(n) time and O(1) extra space
  3. CO(1) time and O(1) extra space
  4. DO(n log n) time and O(1) extra space
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDEF

How many closed walks of length 2 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Which ACID property guarantees that committed changes survive a system crash?

  1. AAtomicity
  2. BDurability
  3. CIsolation
  4. DConsistency
16.Databases & SQLHard

x is NULL and y = 1. What is the value of the condition (x > 5 OR y = 1)?

  1. AFALSE
  2. BNULL, so the row is always rejected
  3. CUNKNOWN
  4. DTRUE
17.Computer networksMedium

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.2.2.3
  2. B10.0.2.3
  3. C10.1.200.5
  4. D11.1.2.3
18.Computer networksHard

What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?

  1. AThe server can push updates over UDP
  2. BResponses can arrive in any order on one stream, so slow replies do not block others
  3. CThe connection stays encrypted without TLS
  4. DSeveral requests reuse one TCP connection, avoiding repeated handshakes
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
third octet = 168112806413201618040201networkhost

A host has the address 172.20.168.69/17. What is the third octet x of its subnet's broadcast address?

172.20.168.69/17
x =

3Solutions

1.Number systems

Answer: U = 239

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

17 = 0001 0001
inverted: 1110 1110
+1 → 1110 1111

Read it as unsigned

1110 1111₂ = 256 − 17 = 239

Answer

U = 239
2.Number systems

Answer: b = 11 bit

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Powers of two

210 = 1024 < 1822
211 = 2048 ≥ 1822

Answer

b = 11
3.Propositional logic & proofs

Answer: contradiction

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).

Truth table row by row

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

Classify

r ∧ ¬r: 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ∃x ¬F(x)

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.

Move the negation inwards

¬∀x F(x) ≡ ∃x ¬F(x)

Answer

∃x ¬F(x)

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: N = 3

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

ABCD000111100001111010010302040517060120130150141819111110

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

B′C′D′
AB′
A′BCD

Minimal SOP form

F = B′C′D′ + AB′ + A′BCD
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

Prime implicants

The largest groups that can be formed from 1s and Xs.

CD, B′C′D′, AD, AB′
ABCD0001111000011110X00113020405X70601211311501418X9111110

Essential prime implicants

CD, AD, AB′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 41

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).

Inverse of 41

r = 43, t = 0
r = 41, q = 1, t = 1
r = 2, q = 20, t = −1
r = 1, q = 2, t = 21
41−1 ≡ 21 (mod 43)

Solve

x ≡ 21·4 = 84 ≡ 41 (mod 43)
check: 41·41 = 1681 ≡ 4 (mod 43)

Answer

x = 41
8.Number theory & modular arithmetic

Answer: x = 6

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).

Divide by the gcd

gcd(20, 38) = 2
10x ≡ 3 (mod 19)

Solve the reduced congruence

10−1 ≡ 2 (mod 19) → x ≡ 6 (mod 19)

Solutions modulo 38

x ∈ {6, 25}
x = 6

Answer

x = 6
9.Algorithms & complexity

Answer: B · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

B · Exponential
10.Algorithms & complexity

Answer: A · Kruskal's algorithm

Why this is the answer

Kruskal sorts edges by weight and adds them if they connect different components; Prim grows one tree from a start vertex.

Answer

A · Kruskal's algorithm
11.Data structures

Answer: D · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

D · O(1)
12.Data structures

Answer: B · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

B · O(n) time and O(1) extra space
13.Graph theory

Answer: Euler circuit

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) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: N = 3

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

Common neighbours

(M²)CC = 1·1 + 1·1 + 0·0 + 0·0 + 1·1 + 0·0 = 3

Answer

N = 3
15.Databases & SQL

Answer: B · Durability

Why this is the answer

Durability means that once a transaction commits, its effects are kept, usually by writing a log to stable storage.

Answer

B · Durability
16.Databases & SQL

Answer: D · TRUE

Why this is the answer

x > 5 is UNKNOWN, but OR with TRUE is TRUE regardless, so the row is kept; UNKNOWN AND FALSE would likewise be FALSE.

Answer

D · TRUE
17.Computer networks

Answer: C · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

C · 10.1.200.5
18.Computer networks

Answer: D · Several requests reuse one TCP connection, avoiding repeated handshakes

Why this is the answer

Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.

Answer

D · Several requests reuse one TCP connection, avoiding repeated handshakes
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 = 255

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 in the third octet

27 = 128

Subnet of this host

network: 172.20.128.0
broadcast: 172.20.255.255

Answer

x = 255

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