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 −73 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −738 bit
U =
2.Number systemsHard
16 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
q → rr ∨ ¬q¬r → ¬q¬r ∨ q

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

r → q
  1. 1q → r
  2. 2r ∨ ¬q
  3. 3¬r → ¬q
  4. 4¬r ∨ q
4.Propositional logic & proofsHard
∃x (F(x) ∧ ¬G(x))∀x (F(x) ∧ ¬G(x))∀x (F(x) → ¬G(x))∃x (F(x) → ¬G(x))

Which statement is equivalent to ¬∀x (F(x) → G(x)), with the negation moved all the way inside? Click it.

¬∀x (F(x) → G(x))
  1. 1∃x (F(x) ∧ ¬G(x))
  2. 2∀x (F(x) ∧ ¬G(x))
  3. 3∀x (F(x) → ¬G(x))
  4. 4∃x (F(x) → ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(6, 7, 11, 12, 13)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​TQ4​Q4​TQ5​Q5​CLK

A 6-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 15 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 6tpd = 15 ns
fmax = MHz
7.Number theory & modular arithmeticMedium

Solve 44x ≡ 35 (mod 53). Give the solution x with 0 < x < 53.

44x ≡ 35 (mod 53)
x =
8.Number theory & modular arithmeticHard
423c = me​ mod n?m = cd​ mod nm(n, e) = (703, 17)d

Encrypt the message m = 423 with the RSA public key (n, e) = (703, 17). Find the ciphertext c.

n = 703e = 17m = 423
c =
9.Algorithms & complexityMedium

What is the worst-case time complexity of quicksort?

  1. AO(n²)
  2. BO(n log n)
  3. CO(n)
  4. DO(log n)
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AEvery element must be compared at least once
  2. BA decision tree with n possible outcomes must have depth at least log₂ n
  3. CComparisons take logarithmic time each
  4. DBinary search is the only possible algorithm, and it needs log₂ n comparisons
11.Data structuresMedium

How long does a search for a word of length L take in a trie with a fixed alphabet?

  1. AO(L), independent of how many words are stored
  2. BO(n), where n is the number of words
  3. CO(log n), where n is the number of words
  4. DO(L × n)
12.Data structuresHard

A Fenwick tree (binary indexed tree) over an array of n numbers supports which operations in O(log n)?

  1. AInserting elements at any position
  2. BSorting the array and binary search
  3. CPrefix-sum queries and single-element updates
  4. DFinding the median in O(1)
13.Graph theoryMedium

A connected graph has 17 vertices and 21 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 21
k =
14.Graph theoryHard
310114411113ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

First normal form (1NF) requires that:

  1. AEvery table has exactly one column
  2. BEvery non-key column depends on the whole key and on nothing but the key
  3. CEach column contains atomic values and there are no repeating groups
  4. DThere are no foreign keys
16.Databases & SQLHard

What does SELECT name, dept, AVG(salary) OVER (PARTITION BY dept) FROM emp return?

  1. AOne row per department with its average salary
  2. BOnly employees who earn more than their department average
  3. CAn error because name is not in a GROUP BY
  4. DEvery employee row together with the average salary of their department
17.Computer networksMedium

What does ARP do?

  1. ARoutes packets on the internet
  2. BFinds the MAC address that belongs to an IP address on the local network
  3. CAssigns IP addresses automatically to hosts when they join the local network
  4. DResolves domain names
18.Computer networksHard

DNS queries normally use UDP. When does DNS typically use TCP instead?

  1. AFor responses too large for UDP and for zone transfers
  2. BOnly for reverse lookups of IP addresses
  3. COnly when the domain name is longer than ten characters
  4. DFor every query sent from a web browser
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1702 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1702 hosts
p =
20.IPv4 subnettingHard

The network 192.168.104.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 25 hosts, C: 87 hosts, D: 59 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?

A: 10B: 25C: 87D: 59
x =

3Solutions

1.Number systems

Answer: U = 183

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

73 = 0100 1001
inverted: 1011 0110
+1 → 1011 0111

Read it as unsigned

1011 0111₂ = 256 − 73 = 183

Answer

U = 183
2.Number systems

Answer: M = 1536 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

1024·768·16 = 12582912 bit = 1572864 B

Convert to KiB

1572864/1024 = 1536 KiB

Answer

M = 1536 KiB
3.Propositional logic & proofs

Answer: ¬r ∨ 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

a → b is 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
r → q ≡ ¬r ∨ q

Check with the truth table

q = 1, r = 1: 1 | 1 1 1 1
q = 1, r = 0: 1 | 0 0 0 1
q = 0, r = 1: 0 | 1 1 1 0
q = 0, r = 0: 1 | 1 1 1 1

Answer

¬r ∨ q

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

4.Propositional logic & proofs

Answer: ∃x (F(x) ∧ ¬G(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 (F(x) → G(x)) ≡ ∃x (F(x) ∧ ¬G(x))

Answer

∃x (F(x) ∧ ¬G(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: AB = 10, CD = 11

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m11: m10, m9, m15, m3 → 0
ABCD000111100001111011111

The isolated 1

m11 = AB′CD → AB = 10, CD = 11

Answer

AB = 10, CD = 11
6.Digital logic: gates, K-maps & flip-flops

Answer: fmax = 11.11 MHz

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.

Worst-case settling time

T = 6·15 ns = 90 ns

Maximum frequency

fmax = 1/90 ns = 11.11 MHz

Answer

fmax = 11.11 MHz
7.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).

Inverse of 44

r = 53, t = 0
r = 44, q = 1, t = 1
r = 9, q = 4, t = −1
r = 8, q = 1, t = 5
r = 1, q = 8, t = −6
44−1 ≡ 47 (mod 53)

Solve

x ≡ 47·35 = 1645 ≡ 2 (mod 53)
check: 44·2 = 88 ≡ 35 (mod 53)

Answer

x = 2
8.Number theory & modular arithmetic

Answer: c = 118

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
4231 ≡ 423 (mod 703)
4232 ≡ 367 (mod 703)
4234 ≡ 416 (mod 703)
4238 ≡ 118 (mod 703)
42316 ≡ 567 (mod 703)
42317 ≡ 423·567 ≡ 118 (mod 703)

Ciphertext

c = 118

Answer

c = 118
9.Algorithms & complexity

Answer: A · O(n²)

Why this is the answer

With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).

Answer

A · O(n²)
10.Algorithms & complexity

Answer: B · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

B · A decision tree with n possible outcomes must have depth at least log₂ n
11.Data structures

Answer: A · O(L), independent of how many words are stored

Why this is the answer

Each character selects one child, so a lookup takes L steps no matter how many words the trie contains.

Answer

A · O(L), independent of how many words are stored
12.Data structures

Answer: C · Prefix-sum queries and single-element updates

Why this is the answer

A Fenwick tree stores partial sums over ranges determined by the low bits of each index, so both prefix sums and point updates take O(log n).

Answer

C · Prefix-sum queries and single-element updates
13.Graph theory

Answer: k = 5

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 = 21 − 16 = 5

Answer

k = 5
14.Graph theory

Answer: d(F) = 12

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 3, C: 10
B (3): D: 7, G: 14
D (7): E: 11
C (10): G: 11
E (11): F: 12
G (11)
F (12)

Shortest path

A → B → D → E → F
d(F) = 12

Answer

d(F) = 12
15.Databases & SQL

Answer: C · Each column contains atomic values and there are no repeating groups

Why this is the answer

1NF requires atomic values per cell and a unique row identification, without repeating groups.

Answer

C · Each column contains atomic values and there are no repeating groups
16.Databases & SQL

Answer: D · Every employee row together with the average salary of their department

Why this is the answer

Unlike GROUP BY, a window function keeps every input row and adds a value computed over its partition, here the department.

Answer

D · Every employee row together with the average salary of their department
17.Computer networks

Answer: B · Finds the MAC address that belongs to an IP address on the local network

Why this is the answer

ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.

Answer

B · Finds the MAC address that belongs to an IP address on the local network
18.Computer networks

Answer: A · 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

A · For responses too large for UDP and for zone transfers
19.IPv4 subnetting

Answer: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1702
211 − 2 = 2046 ≥ 1702 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 128

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

C: 87 hosts → 27 = 128 (/25)
D: 59 hosts → 26 = 64 (/26)
B: 25 hosts → 25 = 32 (/27)
A: 10 hosts → 24 = 16 (/28)

Allocation, largest first

C: .0 – .127
D: .128 – .191
B: .192 – .223
A: .224 – .239

Answer

x = 128

LAN D: 192.168.104.128/26.

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