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 complement?−128?64?32?16?8?4?2?1

Write −54 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −548 bit
U =
2.Number systemsHard
8 bit1 px1280 px720 pxM = ? KiB

An uncompressed image is 1280 × 720 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1280 × 720 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(a → c) ∧ (b → c)a → (b → c)(a → b) → ca ∧ (b → c)

Which formula is logically equivalent to (a ∧ b) → c? Click it.

(a ∧ b) → c
  1. 1(a → c) ∧ (b → c)
  2. 2a → (b → c)
  3. 3(a → b) → c
  4. 4a ∧ (b → c)
4.Propositional logic & proofsHard
∀x A(x)∃x A(x)∀x ¬A(x)∃x ¬A(x)

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

¬∃x A(x)
  1. 1∀x A(x)
  2. 2∃x A(x)
  3. 3∀x ¬A(x)
  4. 4∃x ¬A(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​C′I1​C′I2​CI3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = C′, I₂ = C′, I₃ = C. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = C′I₂ = C′I₃ = C
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKMOD-5

A synchronous mod-5 counter counts 0, 1, …, 4 and then returns to 0. It starts at 1. Which count Q (in decimal) does it show after 13 clock pulses?

Q0 = 113N = 5
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 3) and x ≡ 10 (mod 11).

x ≡ 2 (mod 3)x ≡ 10 (mod 11)
x =
8.Number theory & modular arithmeticHard

61 is prime. Use Fermat’s little theorem to find 343972 mod 61.

p = 61k = 43972
343972 mod 61 =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms are always rounded to integers
  2. BAll logarithms in computer science have base 10, so it is implied
  3. CLogarithms with different bases differ only by a constant factor
  4. DThe base only matters for small n
10.Algorithms & complexityHard

How is the travelling salesman problem classified?

  1. AIt is in P because Dijkstra's algorithm solves it by visiting every city once
  2. BThe decision version is in P but the optimisation version is NP-complete
  3. CThe decision version is NP-complete and the optimisation version is NP-hard
  4. DIt is undecidable
11.Data structuresMedium

What is true of the adjacency matrix of an undirected graph?

  1. AIt is always upper triangular
  2. BEach row has exactly one 1
  3. CIts main diagonal always contains ones
  4. DIt is symmetric about the main diagonal
12.Data structuresHard

Deleting a node with two children from a binary search tree is usually done by:

  1. ARebuilding the subtree below it from scratch and reinserting every key it contained
  2. BReplacing it with its in-order successor (or predecessor) and deleting that node
  3. CRemoving the whole subtree
  4. DSwapping it with the root
13.Graph theoryMedium

A connected graph has 9 vertices and 12 edges. How many edges k must be removed to leave a spanning tree?

|V| = 9|E| = 12
k =
14.Graph theoryHard
34141010151337119ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

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

The orders table has 5 rows, each referring to an existing customer. The customers table has 3 rows. How many rows does an INNER JOIN on customer_id return?

  1. A15
  2. B5
  3. C8
  4. D3
16.Databases & SQLHard

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

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

How many usable host addresses does a /26 IPv4 subnet have?

  1. A64
  2. B126
  3. C62
  4. D30
18.Computer networksHard

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt serves cached content from servers close to the users
  2. BIt replaces TCP with a faster transport protocol
  3. CIt removes the need for DNS lookups
  4. DIt increases the bandwidth of the users' own home connections
19.IPv4 subnettingMedium
255.255.255.22411111111111111111111111111100000network bitshost bits

A subnet uses the mask 255.255.255.224. How many usable host addresses H does it have?

255.255.255.224
H =
20.IPv4 subnettingHard
third octet = 152112806403211618040201networkhost

A host has the address 172.20.152.105/22. What is the third octet x of its network address?

172.20.152.105/22
x =

3Solutions

1.Number systems

Answer: U = 202

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.

Invert and add one

54 = 0011 0110
inverted: 1100 1001
+1 → 1100 1010

Read it as unsigned

1100 1010₂ = 256 − 54 = 202

Answer

U = 202
2.Number systems

Answer: M = 900 KiB

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.

Bits and bytes

1280·720·8 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB
3.Propositional logic & proofs

Answer: a → (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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(a ∧ b) → c ≡ a → (b → c)

Check with the truth table

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

Answer

a → (b → c)

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

4.Propositional logic & proofs

Answer: ∀x ¬A(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 A(x) ≡ ∀x ¬A(x)

Answer

∀x ¬A(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

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′BC′ + AB′C′ + ABC

Rows with F = 1

ABC = 000, 001, 010, 100, 111
N = 5

Answer

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

Answer: Q = 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.

Full cycles

Every 5 pulses bring the counter back to the same state.

13 = 2·5 + 3

Count after the remaining pulses

Q = (1 + 3) mod 5 = 4
Q = 100₂

Answer

Q = 4
7.Number theory & modular arithmetic

Answer: x = 32

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).

First congruence

x = 2 + 3k

Second congruence

2 + 3k ≡ 10 (mod 11) → 3k ≡ 8 (mod 11)
k ≡ 4·8 ≡ 10 (mod 11)

Result

x = 2 + 3·10 = 32
x = 32
check: 32 mod 3 = 2, 32 mod 11 = 10

Answer

x = 32
8.Number theory & modular arithmetic

Answer: 343972 mod 61 = 9

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.

Reduce the exponent

43972 mod 60 = 52
343972 ≡ 352 (mod 61)

Small power

352 mod 61 = 9

Answer

343972 mod 61 = 9
9.Algorithms & complexity

Answer: C · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

C · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: C · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

C · The decision version is NP-complete and the optimisation version is NP-hard
11.Data structures

Answer: D · It is symmetric about the main diagonal

Why this is the answer

In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.

Answer

D · It is symmetric about the main diagonal
12.Data structures

Answer: B · Replacing it with its in-order successor (or predecessor) and deleting that node

Why this is the answer

The in-order successor has at most one child, so it can be removed easily after copying its key into the node being deleted.

Answer

B · Replacing it with its in-order successor (or predecessor) and deleting that node
13.Graph theory

Answer: k = 4

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

9 − 1 = 8

Edges to remove

k = 12 − 8 = 4

Answer

k = 4
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: 3, C: 4
2. B (3): D: 13, G: 6
3. C (4)
4. G (6): E: 15
5. D (13): F: 28
6. E (15)
7. F (28)

round 3

C

Answer

C
15.Databases & SQL

Answer: B · 5

Why this is the answer

Every order finds exactly one matching customer, so the join returns one row per order: 5 rows.

Answer

B · 5
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: C · 62

Why this is the answer

A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.

Answer

C · 62
18.Computer networks

Answer: A · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

A · It serves cached content from servers close to the users
19.IPv4 subnetting

Answer: H = 30

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

255.255.255.224 = /27

Usable hosts

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 152

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: 172.20.152.0
broadcast: 172.20.155.255

Answer

x = 152

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