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 −72 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −728 bit
U =
2.Number systemsHard
4635 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4635 items its own binary code of the same length. What is the smallest number of bits b needed?

4635
b = bit
3.Propositional logic & proofsMedium
¬(q ∧ p) → ¬¬r(q ∧ p) ∧ ¬¬r¬(q ∧ p) ∧ ¬r¬(q ∧ p) ∧ ¬¬r

You want to prove (q ∧ p) → ¬r by contradiction. Which statement do you assume at the start? Click it.

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

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

¬∃x P(x)
  1. 1∃x P(x)
  2. 2∃x ¬P(x)
  3. 3∀x ¬P(x)
  4. 4∀x P(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110312141517060120131150140819111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 2, 4, 5, 7, 9, 10, 11, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(1, 2, 4, 5, 7, 9, 10, 11, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100001131204050706X12X13015114X819111010

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 9, 11, 14) + d(8, 12, 13), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(2, 3, 9, 11, 14) + d(8, 12, 13)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 83 modulo 93: the number x with 0 < x < 93 and 83·x ≡ 1 (mod 93).

a = 83n = 93
83−1 mod 93 =
8.Number theory & modular arithmeticHard
499c = me​ mod n?m = cd​ mod nm(n, e) = (611, 17)d

Encrypt the message m = 499 with the RSA public key (n, e) = (611, 17). Find the ciphertext c.

n = 611e = 17m = 499
c =
9.Algorithms & complexityMedium

About how many comparisons does binary search need in the worst case on a sorted array of 1,000,000 elements?

  1. AAbout 100
  2. BAbout 20
  3. CAbout 500,000
  4. DAbout 1,000
10.Algorithms & complexityHard

What is the time complexity of building a binary heap from n unsorted elements with the bottom-up method?

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

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe oldest keys are deleted to make room
  2. BOnly the colliding keys are moved to a new overflow list at the end of the table
  3. CThe existing keys are sorted in place
  4. DA larger table is allocated and every key is inserted again
12.Data structuresHard

Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?

  1. AO(1)
  2. BO(log n)
  3. CO(n)
  4. DO(α(n))
13.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
14.Graph theoryHard
431132713615ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 6; A itself is round 1.

start: Around 6
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?

  1. AThe number of students who passed and failed
  2. BAn error because CASE is not allowed in SELECT
  3. COnly the students who passed
  4. DEach name with the label 'pass' or 'fail' depending on the grade
16.Databases & SQLHard

When is decomposing R into R1 and R2 guaranteed to be lossless?

  1. AWhen the common attributes of R1 and R2 form a superkey of R1 or of R2
  2. BWhen R1 and R2 have no attributes in common
  3. CWhen R1 and R2 have the same number of columns
  4. DWhen every attribute of R appears in both R1 and R2
17.Computer networksMedium

Why is UDP often preferred for live video calls?

  1. AUDP guarantees every packet arrives, so the picture never freezes
  2. BUDP is required for all video
  3. CLow delay matters more than retransmitting every lost packet
  4. DUDP encrypts data automatically
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 when the domain name is longer than ten characters
  3. COnly for reverse lookups of IP addresses
  4. DFor responses too large for UDP and for zone transfers
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2430 subnets
H =
20.IPv4 subnettingHard
last octet = 69012816403201608140211networkhost

A host has the address 10.20.73.69/30. What is the fourth octet x of its network address?

10.20.73.69/30
x =

3Solutions

1.Number systems

Answer: U = 184

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

72 = 0100 1000
inverted: 1011 0111
+1 → 1011 1000

Read it as unsigned

1011 1000₂ = 256 − 72 = 184

Answer

U = 184
2.Number systems

Answer: b = 13 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

212 = 4096 < 4635
213 = 8192 ≥ 4635

Answer

b = 13
3.Propositional logic & proofs

Answer: (q ∧ p) ∧ ¬¬r

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).

Parts of the implication

a = q ∧ p, b = ¬r

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = (q ∧ p) ∧ ¬¬r

Answer

(q ∧ p) ∧ ¬¬r

If the assumption leads to a contradiction, the implication must be true.

4.Propositional logic & proofs

Answer: ∀x ¬P(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 P(x) ≡ ∀x ¬P(x)

Answer

∀x ¬P(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.

ABCD000111100001111000110312141517060120131150140819111110

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
B′CD′
BCD
ACD
A′BC′

Minimal SOP form

F = B′C′D + B′CD′ + BCD + ACD + A′BC′
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: 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.

B′CD, AC′, AB′D, ABD′, A′B′C
ABCD00011110000111100001131204050706X12X13015114X819111010

Minimal cover

F = AB′D + ABD′ + A′B′C
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 83−1 mod 93 = 65

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·83 (mod 93).

r = 93, t = 0
r = 83, q = 1, t = 1
r = 10, q = 8, t = −1
r = 3, q = 3, t = 9
r = 1, q = 3, t = −28

Inverse

1 ≡ −28·83 (mod 93)
x = 65
check: 83·65 = 5395 = 58·93 + 1

Answer

83−1 mod 93 = 65
8.Number theory & modular arithmetic

Answer: c = 44

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

17 = 16 + 1
4991 ≡ 499 (mod 611)
4992 ≡ 324 (mod 611)
4994 ≡ 495 (mod 611)
4998 ≡ 14 (mod 611)
49916 ≡ 196 (mod 611)
49917 ≡ 499·196 ≡ 44 (mod 611)

Ciphertext

c = 44

Answer

c = 44
9.Algorithms & complexity

Answer: B · About 20

Why this is the answer

220 ≈ 1,048,576, so after about 20 halvings only one element remains.

Answer

B · About 20
10.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

Summing sift-down costs over the levels gives a convergent series, so heapify runs in linear time.

Answer

C · O(n)
11.Data structures

Answer: D · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

D · A larger table is allocated and every key is inserted again
12.Data structures

Answer: C · O(n)

Why this is the answer

Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.

Answer

C · O(n)
13.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) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 3, deg(F) = 1

Odd vertices

B, D, E, F → 4
neither

Answer

neither
14.Graph theory

Answer: G

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

1. A (0): B: 4, C: 3
2. C (3): E: 16, G: 16
3. B (4): D: 5
4. D (5): F: 7, G: 11
5. F (7): E: 14
6. G (11)
7. E (14)

round 6

G

Answer

G
15.Databases & SQL

Answer: D · Each name with the label 'pass' or 'fail' depending on the grade

Why this is the answer

A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.

Answer

D · Each name with the label 'pass' or 'fail' depending on the grade
16.Databases & SQL

Answer: A · When the common attributes of R1 and R2 form a superkey of R1 or of R2

Why this is the answer

If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.

Answer

A · When the common attributes of R1 and R2 form a superkey of R1 or of R2
17.Computer networks

Answer: C · Low delay matters more than retransmitting every lost packet

Why this is the answer

Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.

Answer

C · Low delay matters more than retransmitting every lost packet
18.Computer networks

Answer: D · 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

D · For responses too large for UDP and for zone transfers
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).

Bits to borrow

25 = 32 ≥ 30 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 68

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

22 = 4

Subnet of this host

network: ⌊69/4⌋·4 = 68
broadcast: 68 + 4 − 1 = 71
usable hosts: 69 … 70

Answer

x = 68

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