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
place values in two's complement0−12816403201618041211

The 8-bit pattern 0100 1011 stores a signed integer in two's complement. What is its value N in decimal?

0100 10118 bit
N =
2.Number systemsHard
962 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 962 items its own binary code of the same length. What is the smallest number of bits b needed?

962
b = bit
3.Propositional logic & proofsMedium
(r ∧ q) ∨ (r ∧ p)r ∨ (q ∧ p)(r ∨ q) ∧ (r ∨ p)(r ∧ q) ∨ p

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

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

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

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∀x ∀y (R(x, y) ∧ ¬R(y, x))
  2. 2∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
  3. 3∃x ∃y (R(x, y) → ¬R(y, x))
  4. 4∃x ∃y (R(x, y) ∧ ¬R(y, x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302041507160121130151140809111110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 5, 6, 10, 11, 13, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 5, 6, 10, 11, 13, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​CLKMOD-4

A 12 kHz clock drives a mod-4 counter. Find the frequency fout of its most significant output, which completes one cycle per counting cycle.

fclk = 12 kHzN = 4
fout = kHz
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(968), the number of integers from 1 to 968 that are coprime to 968.

n = 968 = 23·112
φ(968) =
8.Number theory & modular arithmeticHard
313c = me​ mod n?m = cd​ mod nm(n, e) = (527, 13)d

Encrypt the message m = 313 with the RSA public key (n, e) = (527, 13). Find the ciphertext c.

n = 527e = 13m = 313
c =
9.Algorithms & complexityMedium

What is the amortised time complexity of appending to a dynamic array that doubles its capacity when full?

  1. AO(1)
  2. BO(n)
  3. CO(n²)
  4. DO(log n)
10.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n² 2ⁿ)
  2. BO(n³)
  3. CO(2n²)
  4. DO(n!)
11.Data structuresMedium

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

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

In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?

  1. AStoring each vertex's current distance from the source during the search
  2. BChoosing the starting vertex
  3. CChecking whether an edge's two endpoints are already connected
  4. DSorting the edges by weight
13.Graph theoryMedium

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

3 × deg 4
L =
14.Graph theoryHard
6161418417191012ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 9
W =
15.Databases & SQLMedium

A column holds the values 10, 20 and NULL. What does AVG(col) return?

  1. ANULL
  2. B15
  3. C30
  4. D10
16.Databases & SQLHard

A correlated subquery is one that:

  1. AReferences columns of the outer query and is evaluated per outer row
  2. BJoins two tables with a foreign key
  3. CUses only aggregate functions
  4. DRuns only once before the outer query and caches its result for every row
17.Computer networksMedium

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

  1. AOffer, Discover, Acknowledge, Request
  2. BDiscover, Offer, Request, Acknowledge
  3. CDiscover, Request, Offer, Acknowledge
  4. DRequest, Offer, Discover, Acknowledge
18.Computer networksHard

With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?

  1. AGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
  2. BNeither ever resends lost frames
  3. CGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
  4. DBoth resend the whole window from the start
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/243 subnets
H =
20.IPv4 subnettingHard
last octet = 7012806403201608141211networkhost

A host has the address 10.20.63.7/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.63.7/28
x =

3Solutions

1.Number systems

Answer: N = 75

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.

Add the place values

The other bits keep their normal positive values.

64 + 8 + 2 + 1 = 75

Answer

N = 75
2.Number systems

Answer: b = 10 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

29 = 512 < 962
210 = 1024 ≥ 962

Answer

b = 10
3.Propositional logic & proofs

Answer: (r ∧ q) ∨ (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

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
r ∧ (q ∨ p) ≡ (r ∧ q) ∨ (r ∧ p)

Check with the truth table

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

Answer

(r ∧ q) ∨ (r ∧ p)

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

4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, 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 ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, 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 = 5

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.

ABCD000111100001111010010302041507160121130151140809111110

Group the 0s

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

(B + C + D′)
(B′ + C + D)
(B′ + C′ + D′)
(A′ + C + D)
(A + B + C′)

Minimal POS form

F = (B + C + D′)(B′ + C + D)(B′ + C′ + D′)(A′ + C + D)(A + B + C′)
N = 5

Answer

N = 5

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: fout = 3 kHz

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.

Frequency division

fout = 12 kHz / 4 = 3 kHz

Answer

fout = 3 kHz
7.Number theory & modular arithmetic

Answer: φ(968) = 440

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

968 = 23·112

Totient

φ(968) = φ(8)·φ(121) = 4·110 = 440

Answer

φ(968) = 440
8.Number theory & modular arithmetic

Answer: c = 210

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

13 = 8 + 4 + 1
3131 ≡ 313 (mod 527)
3132 ≡ 474 (mod 527)
3134 ≡ 174 (mod 527)
3138 ≡ 237 (mod 527)
31313 ≡ 313·174·237 ≡ 210 (mod 527)

Ciphertext

c = 210

Answer

c = 210
9.Algorithms & complexity

Answer: A · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

A · O(1)
10.Algorithms & complexity

Answer: A · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

A · O(n² 2ⁿ)
11.Data structures

Answer: A · 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

A · O(V + E)
12.Data structures

Answer: C · Checking whether an edge's two endpoints are already connected

Why this is the answer

Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.

Answer

C · Checking whether an edge's two endpoints are already connected
13.Graph theory

Answer: L = 8

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 + 12 = 2(L + 3) − 2
L = 8

Answer

L = 8

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

14.Graph theory

Answer: W = 63

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

DF (4): take
AC (6): take
GC (10): take
GD (12): take
BD (14): take
BC (16): skip, closes a cycle
EF (17): take

Total weight

W = 4 + 6 + 10 + 12 + 14 + 17 = 63

Answer

W = 63
15.Databases & SQL

Answer: B · 15

Why this is the answer

AVG ignores NULL, so it averages 10 and 20 to get 15, not 30 / 3 = 10.

Answer

B · 15
16.Databases & SQL

Answer: A · References columns of the outer query and is evaluated per outer row

Why this is the answer

Correlated subqueries depend on the current outer row, so they are logically evaluated per row (optimisers may rewrite them as joins).

Answer

A · References columns of the outer query and is evaluated per outer row
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: A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one

Why this is the answer

A Go-Back-N receiver discards out-of-order frames, so the sender must repeat everything after the loss. Selective Repeat buffers them and asks only for the missing one.

Answer

A · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
19.IPv4 subnetting

Answer: H = 62

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).

Bits to borrow

22 = 4 ≥ 3 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: x = 15

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: ⌊7/16⌋·16 = 0
broadcast: 0 + 16 − 1 = 15
usable hosts: 1 … 14

Answer

x = 15

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