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 complement1−12816403211608040201

The 8-bit pattern 1101 0000 stores a signed integer in two's complement. What is its value N in decimal?

1101 00008 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1100 0110 << 3. What is the result's value R in decimal?

1100 0110<< 3
R =
3.Propositional logic & proofsMedium
p → q¬q → ¬p¬p → ¬q¬p ∧ ¬q

Which formula is logically equivalent to q → p? Click it.

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

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

¬∃x (A(x) ∧ B(x))
  1. 1∀x (A(x) → ¬B(x))
  2. 2∀x (A(x) ∧ ¬B(x))
  3. 3∃x (A(x) → ¬B(x))
  4. 4∃x (¬A(x) ∧ ¬B(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001030204151716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 5, 6, 7). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 5, 6, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110100103X20415X706X120130150141809111110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 5, 8, 10, 11) + d(2, 7, 12), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 5, 8, 10, 11) + d(2, 7, 12)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 7 modulo 25: the number x with 0 < x < 25 and 7·x ≡ 1 (mod 25).

a = 7n = 25
7−1 mod 25 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 34108 mod 41.

a = 34k = 108n = 41
34108 mod 41 =
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. BLogarithms with different bases differ only by a constant factor
  3. CAll logarithms in computer science have base 10, so it is implied
  4. DThe base only matters for small n
10.Algorithms & complexityHard

Why can Dijkstra's algorithm give wrong results when some edge weights are negative?

  1. ANegative numbers cannot be stored in a priority queue
  2. BA vertex is finalised when extracted, but a later negative edge could still give it a shorter path
  3. CIt only works on undirected graphs
  4. DIt goes into an infinite loop on every negative edge
11.Data structuresMedium

A circular buffer (ring buffer) is commonly used to implement:

  1. AA binary search tree
  2. BA hash function
  3. CA graph traversal
  4. DA fixed-size queue
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AThe root must be red
  2. BEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  3. CEvery node must be red
  4. DEvery path from a node to its descendant leaves has the same number of black nodes
13.Graph theoryMedium

A tree has 2 vertices of degree 3, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 3
L =
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

What does SELECT COALESCE(phone, 'unknown') FROM customers return for a customer whose phone is NULL?

  1. AAn empty string
  2. BAn error
  3. C'unknown'
  4. DNULL
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. BIndexes never work with dates
  3. CAn index on order_date is only used for ORDER BY clauses, never for WHERE
  4. DApplying a function to the column prevents a simple index range lookup
17.Computer networksMedium

How does a stateful firewall differ from a simple packet filter?

  1. AIt blocks all incoming traffic without exceptions, including replies to outgoing requests
  2. BIt only checks the MAC address of each frame
  3. CIt encrypts every packet it forwards
  4. DIt tracks connections and allows replies that belong to established sessions
18.Computer networksHard

What does BGP mainly do on the internet?

  1. AAssigns IP addresses to home users through their internet providers
  2. BEncrypts data between routers
  3. CResolves names to addresses
  4. DExchanges routing information between autonomous systems
19.IPv4 subnettingMedium
/20 → /2311111111111111111111111000000000network bitshost bits

The network 10.146.224.0/20 is divided into equal subnets with prefix /23. How many subnets N are there?

10.146.224.0/20/23
N =
20.IPv4 subnettingHard
third octet = 158112806403211618141201networkhost

A host has the address 172.20.158.183/19. What is the third octet x of its network address?

172.20.158.183/19
x =

3Solutions

1.Number systems

Answer: N = −48

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 = −48

Answer

N = −48
2.Number systems

Answer: R = 48

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

Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1100 0110 << 3
= 0011 0000

Value of the result

0011 0000₂ = 48

Answer

R = 48
3.Propositional logic & proofs

Answer: ¬p → ¬q

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

An implication is equivalent to its contrapositive ¬b → ¬a, not to its converse b → a or its inverse ¬a → ¬b.
q → p ≡ ¬p → ¬q

Check with the truth table

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

Answer

¬p → ¬q

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

4.Propositional logic & proofs

Answer: ∀x (A(x) → ¬B(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) ∧ B(x)) ≡ ∀x (A(x) → ¬B(x))

Answer

∀x (A(x) → ¬B(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 = 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.

ABC00011110011001030204151716

Group the 0s

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

(A + C′)
(A + B′)
(A′ + B + C)

Minimal POS form

F = (A + C′)(A + B′)(A′ + B + C)
N = 3

Answer

N = 3
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′D′, AC′D′, A′BD, AB′C
ABCD0001111000011110100103X20415X706X120130150141809111110

Minimal cover

F = B′D′ + A′BD + AB′C
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 7−1 mod 25 = 18

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·7 (mod 25).

r = 25, t = 0
r = 7, q = 3, t = 1
r = 4, q = 1, t = −3
r = 3, q = 1, t = 4
r = 1, q = 3, t = −7

Inverse

1 ≡ −7·7 (mod 25)
x = 18
check: 7·18 = 126 = 5·25 + 1

Answer

7−1 mod 25 = 18
8.Number theory & modular arithmetic

Answer: 34108 mod 41 = 4

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.

Binary exponent

108 = 11011002 = 64 + 32 + 8 + 4

Repeated squaring

341 ≡ 34 (mod 41)
342 ≡ 8 (mod 41)
344 ≡ 23 (mod 41)
348 ≡ 37 (mod 41)
3416 ≡ 16 (mod 41)
3432 ≡ 10 (mod 41)
3464 ≡ 18 (mod 41)

Multiply the needed powers

34108 ≡ 23·37·10·18 ≡ 4 (mod 41)

Answer

34108 mod 41 = 4
9.Algorithms & complexity

Answer: B · 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

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

Answer: B · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path

Why this is the answer

The greedy step relies on non-negative weights so that extracted distances are final; Bellman–Ford is used when weights may be negative.

Answer

B · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path
11.Data structures

Answer: D · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

D · A fixed-size queue
12.Data structures

Answer: D · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

D · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: L = 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.

Handshake lemma for a tree

L + 6 = 2(L + 2) − 2
L = 4

Answer

L = 4

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

14.Graph theory

Answer: N = 4

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 0
E: 0 1 1 0 0

One more step

(M³)CC = 1·1 + 2·1 + 3·0 + 1·0 + 1·1 = 4

Answer

N = 4
15.Databases & SQL

Answer: C · 'unknown'

Why this is the answer

COALESCE checks its arguments from left to right and returns the first that is not NULL, here 'unknown'.

Answer

C · 'unknown'
16.Databases & SQL

Answer: D · 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

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

Answer: D · It tracks connections and allows replies that belong to established sessions

Why this is the answer

A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.

Answer

D · It tracks connections and allows replies that belong to established sessions
18.Computer networks

Answer: D · Exchanges routing information between autonomous systems

Why this is the answer

BGP is the inter-domain routing protocol; autonomous systems advertise reachable prefixes and select paths based on policies.

Answer

D · Exchanges routing information between autonomous systems
19.IPv4 subnetting

Answer: N = 8

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).

Borrowed bits

23 − 20 = 3

Number of subnets

N = 23 = 8

Answer

N = 8

Each /23 subnet has 29 − 2 = 510 usable hosts.

20.IPv4 subnetting

Answer: x = 128

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

25 = 32

Subnet of this host

network: 172.20.128.0
broadcast: 172.20.159.255

Answer

x = 128

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