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 −26 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −268 bit
U =
2.Number systemsHard
24 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(a → b) ∧ (c → b)a ∧ (c → b)(a → c) → ba → (c → b)

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

(a ∧ c) → b
  1. 1(a → b) ∧ (c → b)
  2. 2a ∧ (c → b)
  3. 3(a → c) → b
  4. 4a → (c → b)
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
ABCD000111100001111010111302141507061121131150140819011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 3, 4, 5, 9, 10, 12, 13, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 3, 4, 5, 9, 10, 12, 13, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKT1234567

A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0T
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
7.Number theory & modular arithmeticMedium

Solve 30x ≡ 4 (mod 53). Give the solution x with 0 < x < 53.

30x ≡ 4 (mod 53)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 5191 mod 14.

a = 5k = 191n = 14
5191 mod 14 =
9.Algorithms & complexityMedium

Breadth-first search on an unweighted graph finds:

  1. AShortest paths in graphs that contain negative edge weights
  2. BA topological order in all graphs
  3. CThe minimum spanning tree
  4. DShortest paths in number of edges from the start node
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. CAlways choosing the first element
  4. DSorting the input before running quicksort
11.Data structuresMedium

An application needs fast lookups and also needs to list keys in sorted order and answer range queries. Which structure fits best?

  1. AAn unsorted linked list
  2. BA hash table
  3. CA stack
  4. DA balanced binary search tree
12.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(α(n)), the inverse Ackermann function, which is practically constant
  2. BO(n²)
  3. CO(n)
  4. DO(log n) exactly
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
142018101116ABCDE

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

A foreign key is declared with ON DELETE CASCADE. What happens when a parent row is deleted?

  1. AThe child rows that reference it are deleted automatically
  2. BThe foreign key in the child rows is set to NULL automatically
  3. CThe parent row is copied to an archive table
  4. DThe delete is always rejected
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. BAn index on order_date is only used for ORDER BY clauses, never for WHERE
  3. CIndexes never work with dates
  4. DApplying a function to the column prevents a simple index range lookup
17.Computer networksMedium

TLS provides which security properties for HTTPS?

  1. AOnly faster loading
  2. BProtection against all malware and phishing on the visited site
  3. CAnonymity of the user
  4. DConfidentiality, integrity and server authentication
18.Computer networksHard

After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?

  1. ATo give the station with the highest MAC address priority on the shared cable
  2. BTo synchronise all stations to one clock
  3. CTo make the cable carry signals further
  4. DTo lower the chance of colliding again when many stations compete
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/244 subnets
H =
20.IPv4 subnettingHard
third octet 129 … 134112806403201608040211commonvaries

A router has routes to these networks: 10.63.129.0/24, 10.63.130.0/24, 10.63.131.0/24, 10.63.132.0/24, 10.63.133.0/24, 10.63.134.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.63.129.0/24 … 10.63.134.0/24
p =

3Solutions

1.Number systems

Answer: U = 230

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

26 = 0001 1010
inverted: 1110 0101
+1 → 1110 0110

Read it as unsigned

1110 0110₂ = 256 − 26 = 230

Answer

U = 230
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

640·480·24 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB
3.Propositional logic & proofs

Answer: a → (c → b)

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 ∧ c) → b ≡ a → (c → b)

Check with the truth table

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

Answer

a → (c → b)

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

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.

ABCD000111100001111010111302141507061121131150140819011110

Group the 0s

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

(B′ + C′ + D)
(A + C′ + D)
(A + B′ + C′)
(A′ + B + C + D)
(A′ + B + C′ + D′)

Minimal POS form

F = (B′ + C′ + D)(A + C′ + D)(A + B′ + C′)(A′ + B + C + D)(A′ + B + C′ + D′)
N = 5

Answer

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

Answer: clock edge 6

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.

Q after every edge

A T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

1: T = 0 → Q = 0
2: T = 0 → Q = 0
3: T = 0 → Q = 0
4: T = 1 → Q = 1
5: T = 0 → Q = 1
6: T = 1 → Q = 0
7: T = 0 → Q = 0

First change from 1 to 0

clock edge 6

Answer

clock edge 6
7.Number theory & modular arithmetic

Answer: x = 39

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 30

r = 53, t = 0
r = 30, q = 1, t = 1
r = 23, q = 1, t = −1
r = 7, q = 3, t = 2
r = 2, q = 3, t = −7
r = 1, q = 2, t = 23
30−1 ≡ 23 (mod 53)

Solve

x ≡ 23·4 = 92 ≡ 39 (mod 53)
check: 30·39 = 1170 ≡ 4 (mod 53)

Answer

x = 39
8.Number theory & modular arithmetic

Answer: 5191 mod 14 = 3

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

191 = 101111112 = 128 + 32 + 16 + 8 + 4 + 2 + 1

Repeated squaring

51 ≡ 5 (mod 14)
52 ≡ 11 (mod 14)
54 ≡ 9 (mod 14)
58 ≡ 11 (mod 14)
516 ≡ 9 (mod 14)
532 ≡ 11 (mod 14)
564 ≡ 9 (mod 14)
5128 ≡ 11 (mod 14)

Multiply the needed powers

5191 ≡ 5·11·9·11·9·11·11 ≡ 3 (mod 14)

Answer

5191 mod 14 = 3
9.Algorithms & complexity

Answer: D · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

D · Shortest paths in number of edges from the start node
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: D · A balanced binary search tree

Why this is the answer

A balanced BST gives O(log n) lookups and keeps keys ordered, so in-order traversal and range queries are easy; a hash table scatters keys.

Answer

D · A balanced binary search tree
12.Data structures

Answer: A · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

A · O(α(n)), the inverse Ackermann function, which is practically constant
13.Graph theory

Answer: H = 2520

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.

Count

H = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: W = 32

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.

Kruskal’s algorithm

DE (1): take
BE (6): take
BD (10): skip, closes a cycle
CE (11): take
AB (14): take

Total weight

W = 1 + 6 + 11 + 14 = 32

Answer

W = 32
15.Databases & SQL

Answer: A · The child rows that reference it are deleted automatically

Why this is the answer

CASCADE propagates the delete to referencing rows; other options are RESTRICT/NO ACTION (reject) and SET NULL.

Answer

A · The child rows that reference it are deleted automatically
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 · Confidentiality, integrity and server authentication

Why this is the answer

TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).

Answer

D · Confidentiality, integrity and server authentication
18.Computer networks

Answer: D · To lower the chance of colliding again when many stations compete

Why this is the answer

Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.

Answer

D · To lower the chance of colliding again when many stations compete
19.IPv4 subnetting

Answer: H = 62

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

22 = 4 ≥ 4 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: p = 21

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

129 = 10000001
134 = 10000110
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.63.128.0/21 (it also covers 2 network(s) that are not in the list).

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