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 −25 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −258 bit
U =
2.Number systemsHard
8 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ¬(c ∨ ¬c) a tautology, a contradiction or a contingency? Click your answer.

¬(c ∨ ¬c)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∃x (¬P(x) ∧ ¬Q(x))∃x (P(x) → ¬Q(x))∀x (P(x) → ¬Q(x))∀x (¬P(x) ∧ ¬Q(x))

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

¬∃x (P(x) ∧ Q(x))
  1. 1∃x (¬P(x) ∧ ¬Q(x))
  2. 2∃x (P(x) → ¬Q(x))
  3. 3∀x (P(x) → ¬Q(x))
  4. 4∀x (¬P(x) ∧ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001130214150706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 3, 4, 5). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 3, 4, 5)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110100113X2141507X61121131150141819X11110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 3, 4, 5, 8, 9, 10, 12, 13, 15) + d(2, 6, 11), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 3, 4, 5, 8, 9, 10, 12, 13, 15) + d(2, 6, 11)
L =
7.Number theory & modular arithmeticMedium

Find the inverse of 11 modulo 83: the number x with 0 < x < 83 and 11·x ≡ 1 (mod 83).

a = 11n = 83
11−1 mod 83 =
8.Number theory & modular arithmeticHard

Solve 9x ≡ 12 (mod 39). Give the smallest positive solution x.

9x ≡ 12 (mod 39)
x =
9.Algorithms & complexityMedium

Which sorting algorithm is stable?

  1. ASelection sort (typical version)
  2. BHeap sort
  3. CMerge sort
  4. DQuicksort (typical in-place version)
10.Algorithms & complexityHard

Which technique gives an expected O(n log n) running time for quicksort on any input?

  1. AUsing recursion only on the left side
  2. BChoosing the pivot uniformly at random
  3. CSorting the input before running quicksort
  4. DAlways choosing the first element
11.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ASeparate chaining
  2. BDouble hashing
  3. CQuadratic probing
  4. DLinear probing
12.Data structuresHard

Why do databases often use B+ trees rather than plain B-trees for indexes?

  1. AB+ trees use binary internal nodes, which reduces the number of disk reads
  2. BB+ trees store data only in the root
  3. CAll records are in linked leaves, which makes range scans efficient
  4. DB+ trees do not need to be balanced
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1214136151014105ABCDEFG

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
  7. 7G
15.Databases & SQLMedium

Why can a table have only one clustered index?

  1. AClustered indexes use too much memory
  2. BThe SQL standard forbids more than one index on a table
  3. CClustered indexes can only be on the primary key
  4. DThe rows can be stored physically in only one order
16.Databases & SQLHard

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

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

What is a cyclic redundancy check (CRC) used for in Ethernet frames?

  1. AEncrypting the frame
  2. BCorrecting every bit error automatically
  3. CChoosing the best route
  4. DDetecting transmission errors
18.Computer networksHard

What does the Diffie-Hellman key exchange achieve on its own, and what is its weakness?

  1. AIt authenticates both users with passwords, but needs a central server
  2. BIt encrypts bulk data directly over a public channel, but it is far too slow for large files and video
  3. CIt signs certificates for websites, but only works with short keys
  4. DIt agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/245 subnets
H =
20.IPv4 subnettingHard
last octet = 202112816403201618041201networkhost

A host has the address 10.20.138.202/26. What is the fourth octet x of the last usable host address in its subnet?

10.20.138.202/26
x =

3Solutions

1.Number systems

Answer: U = 231

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

25 = 0001 1001
inverted: 1110 0110
+1 → 1110 0111

Read it as unsigned

1110 0111₂ = 256 − 25 = 231

Answer

U = 231
2.Number systems

Answer: M = 768 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

1024·768·8 = 6291456 bit = 786432 B

Convert to KiB

786432/1024 = 768 KiB

Answer

M = 768 KiB
3.Propositional logic & proofs

Answer: contradiction

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).

Truth table row by row

c = 1: ¬c = 0, c ∨ ¬c = 1, ¬(c ∨ ¬c) = 0
c = 0: ¬c = 1, c ∨ ¬c = 1, ¬(c ∨ ¬c) = 0

Classify

¬(c ∨ ¬c): 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ∀x (P(x) → ¬Q(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) ∧ Q(x)) ≡ ∀x (P(x) → ¬Q(x))

Answer

∀x (P(x) → ¬Q(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: L = 7

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.

ABC00011110011001130214150706

Group the 1s

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

B′C′
AB′
A′BC

Minimal SOP form

F = B′C′ + AB′ + A′BC
L = 7

Answer

L = 7
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.

C′D′, B′D′, B′C, BC′, A′D′, AD, AC′, AB′
ABCD0001111000011110100113X2141507X61121131150141819X11110

Minimal cover

F = C′D′ + B′C + BC′ + AD
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: 11−1 mod 83 = 68

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·11 (mod 83).

r = 83, t = 0
r = 11, q = 7, t = 1
r = 6, q = 1, t = −7
r = 5, q = 1, t = 8
r = 1, q = 5, t = −15

Inverse

1 ≡ −15·11 (mod 83)
x = 68
check: 11·68 = 748 = 9·83 + 1

Answer

11−1 mod 83 = 68
8.Number theory & modular arithmetic

Answer: x = 10

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(9, 39) = 3
3x ≡ 4 (mod 13)

Solve the reduced congruence

3−1 ≡ 9 (mod 13) → x ≡ 10 (mod 13)

Solutions modulo 39

x ∈ {10, 23, 36}
x = 10

Answer

x = 10
9.Algorithms & complexity

Answer: C · Merge sort

Why this is the answer

Merge sort is stable when merging takes from the left half first on ties. In-place quicksort, heap sort and selection sort are usually not stable.

Answer

C · Merge sort
10.Algorithms & complexity

Answer: B · Choosing the pivot uniformly at random

Why this is the answer

With a random pivot, the expected running time is O(n log n) for every input; bad cases still exist but are very unlikely.

Answer

B · Choosing the pivot uniformly at random
11.Data structures

Answer: A · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

A · Separate chaining
12.Data structures

Answer: C · All records are in linked leaves, which makes range scans efficient

Why this is the answer

In a B+ tree internal nodes only guide the search and the leaves are linked in key order, so a range query finds the first key and then walks along the leaves.

Answer

C · All records are in linked leaves, which makes range scans efficient
13.Graph theory

Answer: Euler path, no circuit

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) = 4, deg(C) = 3, deg(D) = 1, deg(E) = 2

Odd vertices

C, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

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: 12, C: 14
2. B (12): D: 18
3. C (14): E: 29
4. D (18): E: 28, F: 32, G: 23
5. G (23)
6. E (28)
7. F (32)

round 5

G

Answer

G
15.Databases & SQL

Answer: D · The rows can be stored physically in only one order

Why this is the answer

A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.

Answer

D · The rows can be stored physically in only one order
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 · Detecting transmission errors

Why this is the answer

The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.

Answer

D · Detecting transmission errors
18.Computer networks

Answer: D · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle

Why this is the answer

Diffie-Hellman lets two parties derive the same secret that eavesdroppers cannot compute. An attacker in the middle can run separate exchanges with each side unless the values are signed or otherwise authenticated.

Answer

D · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
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).

Bits to borrow

23 = 8 ≥ 5 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 254

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 of this host

network: ⌊202/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254

Answer

x = 254

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