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 complement1−12816403211618140201

The 8-bit pattern 1101 1100 stores a signed integer in two's complement. What is its value N in decimal?

1101 11008 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0001 1010 XOR 1000 0110. What is the result's value R in decimal?

0001 1010XOR1000 0110
R =
3.Propositional logic & proofsMedium
c ∧ b(c → b) ∨ (b → c)(c → b) ∧ (¬b → ¬c)(c → b) ∧ (b → c)

Which formula is logically equivalent to c ↔ b? Click it.

c ↔ b
  1. 1c ∧ b
  2. 2(c → b) ∨ (b → c)
  3. 3(c → b) ∧ (¬b → ¬c)
  4. 4(c → b) ∧ (b → c)
4.Propositional logic & proofsHard
¬c ∧ b¬cbc

The premises c → b; ¬b are given. Which conclusion follows from them? Click it.

c → b¬b
  1. 1¬c ∧ b
  2. 2¬c
  3. 3b
  4. 4c
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011130204150716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 3, 5, 6). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 1, 3, 5, 6)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100011130204X5X716012X13015X141809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(1, 3, 6, 8, 10) + d(5, 7, 13, 14), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(1, 3, 6, 8, 10) + d(5, 7, 13, 14)
L =
7.Number theory & modular arithmeticMedium

Solve 63x ≡ 38 (mod 73). Give the solution x with 0 < x < 73.

63x ≡ 38 (mod 73)
x =
8.Number theory & modular arithmeticHard

Solve 62x ≡ 50 (mod 74). Give the smallest positive solution x.

62x ≡ 50 (mod 74)
x =
9.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(log n)
  2. BΩ(n²)
  3. CΩ(n log n)
  4. DΩ(n)
10.Algorithms & complexityHard

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AMany keys may collide into the same bucket
  2. BArrays have O(n) access
  3. CHash functions are slow
  4. DKeys must be sorted first
11.Data structuresMedium

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA hash table
  2. BA stack
  3. CA binary search tree
  4. DA queue
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA hash map combined with a doubly linked list
  2. BA stack only
  3. CA binary search tree only
  4. DA single sorted array
13.Graph theoryMedium

A connected graph has 17 vertices and 22 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 22
k =
14.Graph theoryHard
15913612613ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

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

A table stores the customer's address on every order row. Changing the address in one row but not in the others is an example of:

  1. AA phantom read
  2. BAn insertion anomaly
  3. CAn update anomaly
  4. DA deletion anomaly
16.Databases & SQLHard

Why might a query with WHERE YEAR(order_date) = 2025 not use an index on order_date?

  1. AIndexes never work with dates
  2. BApplying a function to the column prevents a simple index range lookup
  3. CAn index on order_date is only used for ORDER BY clauses, never for WHERE
  4. DYEAR is not valid SQL
17.Computer networksMedium

How many usable host addresses does a /24 IPv4 network have?

  1. A255
  2. B254
  3. C256
  4. D252
18.Computer networksHard

A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?

  1. ATLS certificates are sent over ICMP
  2. BPath MTU discovery fails because "fragmentation needed" messages are blocked
  3. CTCP cannot open connections without ICMP
  4. DDNS needs ICMP to resolve names
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.228.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 4?

192.168.228.0/24/26subnet 4 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet = 14012806403201618141201networkhost

A host has the address 172.20.14.240/22. What is the third octet x of its subnet's broadcast address?

172.20.14.240/22
x =

3Solutions

1.Number systems

Answer: N = −36

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.

−128 + 64 + 16 + 8 + 4 = −36

Answer

N = −36
2.Number systems

Answer: R = 156

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

XOR gives 1 where the two bits differ.

0001 1010
XOR 1000 0110
= 1001 1100

Value of the result

1001 1100₂ = 156

Answer

R = 156
3.Propositional logic & proofs

Answer: (c → b) ∧ (b → c)

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 means the implication holds in both directions: (a → b) ∧ (b → a).
c ↔ b ≡ (c → b) ∧ (b → c)

Check with the truth table

b = 1, c = 1: 1 | 1 1 1 1
b = 1, c = 0: 0 | 0 1 1 0
b = 0, c = 1: 0 | 0 1 0 0
b = 0, c = 0: 1 | 0 1 1 1

Answer

(c → b) ∧ (b → c)

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

4.Propositional logic & proofs

Answer: ¬c

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.

Rows with all premises true

b = 0, c = 0: ¬c ∧ b = 0, ¬c = 1, b = 0, c = 0

Conclusion

¬c (modus tollens)

Answer

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

Answer: L = 9

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.

ABC00011110011011130204150716

Group the 1s

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

B′C
A′C
A′B′
ABC′

Minimal SOP form

F = B′C + A′C + A′B′ + ABC′
L = 9

Answer

L = 9
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 8

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.

BCD′, A′D, ACD′, AB′D′, A′BC
ABCD00011110000111100011130204X5X716012X13015X141809011110

Minimal cover

F = BCD′ + A′D + AB′D′
L = 8

Answer

L = 8

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

7.Number theory & modular arithmetic

Answer: x = 40

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 63

r = 73, t = 0
r = 63, q = 1, t = 1
r = 10, q = 6, t = −1
r = 3, q = 3, t = 7
r = 1, q = 3, t = −22
63−1 ≡ 51 (mod 73)

Solve

x ≡ 51·38 = 1938 ≡ 40 (mod 73)
check: 63·40 = 2520 ≡ 38 (mod 73)

Answer

x = 40
8.Number theory & modular arithmetic

Answer: x = 2

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(62, 74) = 2
31x ≡ 25 (mod 37)

Solve the reduced congruence

31−1 ≡ 6 (mod 37) → x ≡ 2 (mod 37)

Solutions modulo 74

x ∈ {2, 39}
x = 2

Answer

x = 2
9.Algorithms & complexity

Answer: C · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

C · Ω(n log n)
10.Algorithms & complexity

Answer: A · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

A · Many keys may collide into the same bucket
11.Data structures

Answer: B · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

B · A stack
12.Data structures

Answer: A · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

A · A hash map combined with a doubly linked list
13.Graph theory

Answer: k = 6

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

17 − 1 = 16

Edges to remove

k = 22 − 16 = 6

Answer

k = 6
14.Graph theory

Answer: C

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: 15
2. B (15): C: 24, D: 28, E: 18
3. E (18): F: 19
4. F (19): D: 25
5. C (24)
6. D (25)

round 5

C

Answer

C
15.Databases & SQL

Answer: C · An update anomaly

Why this is the answer

Redundant storage means one change must be made in many rows; missing some leaves the data inconsistent, which normalisation prevents.

Answer

C · An update anomaly
16.Databases & SQL

Answer: B · Applying a function to the column prevents a simple index range lookup

Why this is the answer

The expression is not "sargable"; using order_date >= '2025-01-01' AND order_date < '2026-01-01' allows an index range scan.

Answer

B · Applying a function to the column prevents a simple index range lookup
17.Computer networks

Answer: B · 254

Why this is the answer

A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.

Answer

B · 254
18.Computer networks

Answer: B · Path MTU discovery fails because "fragmentation needed" messages are blocked

Why this is the answer

PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.

Answer

B · Path MTU discovery fails because "fragmentation needed" messages are blocked
19.IPv4 subnetting

Answer: x = 192

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

26 = 64

Subnet number 4

network: (4 − 1)·64 = 192
broadcast: 192 + 64 − 1 = 255

Answer

x = 192
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 in the third octet

22 = 4

Subnet of this host

network: 172.20.12.0
broadcast: 172.20.15.255

Answer

x = 15

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