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 −76 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −768 bit
U =
2.Number systemsHard
2566 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 2566 items its own binary code of the same length. What is the smallest number of bits b needed?

2566
b = bit
3.Propositional logic & proofsMedium
(q → p) ∨ (r → p)q ∨ (r → p)(q ∧ r) → p(q → p) ∧ (r → p)

Which formula is logically equivalent to (q ∨ r) → p? Click it.

(q ∨ r) → p
  1. 1(q → p) ∨ (r → p)
  2. 2q ∨ (r → p)
  3. 3(q ∧ r) → p
  4. 4(q → p) ∧ (r → p)
4.Propositional logic & proofsHard
∀x (F(x) ∧ ¬G(x))∃x (F(x) ∧ ¬G(x))∀x (¬F(x) → ¬G(x))∀x (F(x) → ¬G(x))

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

¬∃x (F(x) → G(x))
  1. 1∀x (F(x) ∧ ¬G(x))
  2. 2∃x (F(x) ∧ ¬G(x))
  3. 3∀x (¬F(x) → ¬G(x))
  4. 4∀x (F(x) → ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = B′C + A′B′C + A′BC + AB′C to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = B′C + A′B′C + A′BC + AB′C
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Solve 36x ≡ 33 (mod 89). Give the solution x with 0 < x < 89.

36x ≡ 33 (mod 89)
x =
8.Number theory & modular arithmeticHard

How many solutions N does 36x ≡ 24 (mod 40) have in {0, 1, …, 39}?

36x ≡ 24 (mod 40)
N =
9.Algorithms & complexityMedium

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

  1. ALinear
  2. BExponential
  3. CConstant
  4. DLogarithmic
10.Algorithms & complexityHard

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. AGeneral SAT on formulas in conjunctive normal form
  2. B2-SAT, where each clause has at most two literals
  3. C3-SAT, where each clause has exactly three literals
  4. DCircuit satisfiability
11.Data structuresMedium

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(V²)
  2. BO(V + E)
  3. CO(V × E)
  4. DO(E²)
12.Data structuresHard

How does a radix tree (compressed trie) improve on a plain trie?

  1. AIt keeps only the first letter of each word
  2. BIt stores each word in a hash table at the root instead of in separate nodes
  3. CIt sorts the children of each node by frequency
  4. DIt merges chains of single-child nodes into one edge labelled with a string
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 4 vertices of degree 4, 3 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 34 × deg 43 × deg 5
L =
14.Graph theoryHard
843613135ABCDEF

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

start: Aend: D
d(D) =
15.Databases & SQLMedium

Second normal form (2NF) removes:

  1. ATransitive dependencies between non-key columns via another non-key column
  2. BForeign keys
  3. CPartial dependencies of non-key columns on part of a composite key
  4. DAll NULL values
16.Databases & SQLHard

After customers LEFT JOIN orders, why use COUNT(o.id) rather than COUNT(*) to count orders per customer?

  1. ACOUNT(*) ignores rows with NULL values
  2. BCOUNT(*) is not allowed after a join
  3. CCOUNT(*) counts the NULL-filled row as 1 for customers without orders
  4. DCOUNT(o.id) counts only distinct customers
17.Computer networksMedium

What is the usual order of messages when a client obtains an address from DHCP?

  1. ARequest, Offer, Discover, Acknowledge
  2. BDiscover, Offer, Request, Acknowledge
  3. COffer, Discover, Acknowledge, Request
  4. DDiscover, Request, Offer, Acknowledge
18.Computer networksHard

DNS queries normally use UDP. When does DNS typically use TCP instead?

  1. AFor every query sent from a web browser
  2. BOnly for reverse lookups of IP addresses
  3. CFor responses too large for UDP and for zone transfers
  4. DOnly when the domain name is longer than ten characters
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1990 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1990 hosts
p =
20.IPv4 subnettingHard
last octet = 254112816413211618141201networkhost

A host has the address 10.20.110.254/28. What is the fourth octet x of its network address?

10.20.110.254/28
x =

3Solutions

1.Number systems

Answer: U = 180

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

76 = 0100 1100
inverted: 1011 0011
+1 → 1011 0100

Read it as unsigned

1011 0100₂ = 256 − 76 = 180

Answer

U = 180
2.Number systems

Answer: b = 12 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

211 = 2048 < 2566
212 = 4096 ≥ 2566

Answer

b = 12
3.Propositional logic & proofs

Answer: (q → p) ∧ (r → p)

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).

The law

(a ∨ b) → c needs c whenever either a or b holds: (a → c) ∧ (b → c).
(q ∨ r) → p ≡ (q → p) ∧ (r → p)

Check with the truth table

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

Answer

(q → p) ∧ (r → p)

The columns after the bar are the options in the order shown.

4.Propositional logic & proofs

Answer: ∀x (F(x) ∧ ¬G(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) → G(x)) ≡ ∀x (F(x) ∧ ¬G(x))

Answer

∀x (F(x) ∧ ¬G(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: L = 4

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

B′C → m(1, 5)
A′B′C → m(1)
A′BC → m(3)
AB′C → m(5)
ABC00011110010011130204150706

Group and read off

F = B′C + A′C

Count the literals

L = 4

Answer

L = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 1 → Q = 0
2: J = 1, K = 0 → Q = 1
3: J = 1, K = 1 → Q = 0
4: J = 0, K = 1 → Q = 0
5: J = 0, K = 1 → Q = 0
6: J = 1, K = 1 → Q = 1
7: J = 0, K = 1 → Q = 0

Count the changes

0 → 0 → 1 → 0 → 0 → 0 → 1 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 38

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 36

r = 89, t = 0
r = 36, q = 2, t = 1
r = 17, q = 2, t = −2
r = 2, q = 8, t = 5
r = 1, q = 2, t = −42
36−1 ≡ 47 (mod 89)

Solve

x ≡ 47·33 = 1551 ≡ 38 (mod 89)
check: 36·38 = 1368 ≡ 33 (mod 89)

Answer

x = 38
8.Number theory & modular arithmetic

Answer: N = 4

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(36, 40) = 4
9x ≡ 6 (mod 10)

Solve the reduced congruence

9−1 ≡ 9 (mod 10) → x ≡ 4 (mod 10)

Solutions modulo 40

x ∈ {4, 14, 24, 34}
N = 4

Answer

N = 4
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: B · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

B · 2-SAT, where each clause has at most two literals
11.Data structures

Answer: B · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

B · O(V + E)
12.Data structures

Answer: D · It merges chains of single-child nodes into one edge labelled with a string

Why this is the answer

Collapsing paths with no branching reduces the number of nodes and the memory used while keeping prefix search.

Answer

D · It merges chains of single-child nodes into one edge labelled with a string
13.Graph theory

Answer: L = 23

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.

Handshake lemma for a tree

L + 43 = 2(L + 11) − 2
L = 23

Answer

L = 23

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: d(D) = 13

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: 8, C: 4
C (4): B: 7, E: 5
E (5): F: 18
B (7): D: 13
D (13): F: 16
F (16)

Shortest path

A → C → B → D
d(D) = 13

Answer

d(D) = 13
15.Databases & SQL

Answer: C · Partial dependencies of non-key columns on part of a composite key

Why this is the answer

2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.

Answer

C · Partial dependencies of non-key columns on part of a composite key
16.Databases & SQL

Answer: C · COUNT(*) counts the NULL-filled row as 1 for customers without orders

Why this is the answer

The left join gives one row with NULL order columns for such customers; COUNT(o.id) skips the NULL and correctly reports 0.

Answer

C · COUNT(*) counts the NULL-filled row as 1 for customers without orders
17.Computer networks

Answer: B · Discover, Offer, Request, Acknowledge

Why this is the answer

The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.

Answer

B · Discover, Offer, Request, Acknowledge
18.Computer networks

Answer: C · For responses too large for UDP and for zone transfers

Why this is the answer

If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.

Answer

C · For responses too large for UDP and for zone transfers
19.IPv4 subnetting

Answer: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1990
211 − 2 = 2046 ≥ 1990 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 240

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

24 = 16

Subnet of this host

network: ⌊254/16⌋·16 = 240
broadcast: 240 + 16 − 1 = 255
usable hosts: 241 … 254

Answer

x = 240

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