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 −66 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −668 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1110 << 2. What is the result's value R in decimal?

1000 1110<< 2
R =
3.Propositional logic & proofsMedium
(p ∨ r) → q¬q → ¬(p ∨ r)¬(p ∨ r) → ¬qq ∧ ¬(p ∨ r)

You want to prove q → (p ∨ r) by contrapositive. Which implication do you prove instead? Click it.

q → (p ∨ r)
  1. 1(p ∨ r) → q
  2. 2¬q → ¬(p ∨ r)
  3. 3¬(p ∨ r) → ¬q
  4. 4q ∧ ¬(p ∨ r)
4.Propositional logic & proofsHard
∀x (P(x) ∧ ∀y ¬R(x, y))∃x (P(x) ∧ ∀y ¬R(x, y))∃x (¬P(x) ∧ ∃y R(x, y))∃x (P(x) → ∀y ¬R(x, y))

Which statement is equivalent to ¬∀x (P(x) → ∃y R(x, y)), with the negation moved all the way inside? Click it.

¬∀x (P(x) → ∃y R(x, y))
  1. 1∀x (P(x) ∧ ∀y ¬R(x, y))
  2. 2∃x (P(x) ∧ ∀y ¬R(x, y))
  3. 3∃x (¬P(x) ∧ ∃y R(x, y))
  4. 4∃x (P(x) → ∀y ¬R(x, y))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110302140507161120130150141809011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 4, 6, 8, 10, 12). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(1, 4, 6, 8, 10, 12)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK123456

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Solve 9x ≡ 5 (mod 61). Give the solution x with 0 < x < 61.

9x ≡ 5 (mod 61)
x =
8.Number theory & modular arithmeticHard

Solve 40x ≡ 12 (mod 44). Give the smallest positive solution x.

40x ≡ 12 (mod 44)
x =
9.Algorithms & complexityMedium

A function runs one loop of n iterations and then, separately, another loop of n iterations. What is its time complexity?

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

Which bound is correct for log(n!)?

  1. AΘ(n)
  2. BΘ(n²)
  3. CΘ(log n)
  4. DΘ(n log n)
11.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 3
  2. BIndex 0
  3. CIndex 4
  4. DIndex 6
12.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AAn adjacency matrix recording which vertices have already been removed
  2. BA stack of all visited edges
  3. CA max-heap ordered by out-degree
  4. DA queue (or any collection) of vertices whose in-degree is 0
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| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
111131782076315ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 2; skipped edges do not count.

k = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge EF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GD
  10. 10edge GE
15.Databases & SQLMedium

Which ACID property guarantees that committed changes survive a system crash?

  1. AIsolation
  2. BAtomicity
  3. CConsistency
  4. DDurability
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. BIndexes never work with dates
  3. CYEAR is not valid SQL
  4. DApplying a function to the column prevents a simple index range lookup
17.Computer networksMedium

Which IPv4 header field stops packets from circulating forever in a routing loop?

  1. AType of service
  2. BTime to live (TTL)
  3. CHeader checksum
  4. DIdentification
18.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A24 kbit/s
  2. B12 kbit/s
  3. C3 kbit/s
  4. D6 kbit/s
19.IPv4 subnettingMedium
255.255.255.12811111111111111111111111110000000network bitshost bits

A subnet uses the mask 255.255.255.128. How many usable host addresses H does it have?

255.255.255.128
H =
20.IPv4 subnettingHard

The network 192.168.154.0/24 is split with VLSM for these LANs: A: 18 hosts, B: 90 hosts, C: 14 hosts, D: 41 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 C?

A: 18B: 90C: 14D: 41
x =

3Solutions

1.Number systems

Answer: U = 190

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

66 = 0100 0010
inverted: 1011 1101
+1 → 1011 1110

Read it as unsigned

1011 1110₂ = 256 − 66 = 190

Answer

U = 190
2.Number systems

Answer: R = 56

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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1000 1110 << 2
= 0011 1000

Value of the result

0011 1000₂ = 56

Answer

R = 56
3.Propositional logic & proofs

Answer: ¬(p ∨ 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).

Parts of the implication

a = q, b = p ∨ r

Contrapositive

¬b → ¬a = ¬(p ∨ r) → ¬q

Answer

¬(p ∨ r) → ¬q

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: ∃x (P(x) ∧ ∀y ¬R(x, y))

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) → ∃y R(x, y)) ≡ ∃x (P(x) ∧ ∀y ¬R(x, y))

Answer

∃x (P(x) ∧ ∀y ¬R(x, y))

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 = 13

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.

ABCD000111100001111000110302140507161120130150141809011110

Group the 1s

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

BC′D′
A′BD′
AB′D′
A′B′C′D

Minimal SOP form

F = BC′D′ + A′BD′ + AB′D′ + A′B′C′D
L = 13

Answer

L = 13

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: N = 2

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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 0 → Q = 0
2: J = 0, K = 1 → Q = 0
3: J = 0, K = 0 → Q = 0
4: J = 0, K = 1 → Q = 0
5: J = 1, K = 0 → Q = 1
6: J = 1, K = 1 → Q = 0

Count the changes

0 → 0 → 0 → 0 → 0 → 1 → 0
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: x = 48

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 9

r = 61, t = 0
r = 9, q = 6, t = 1
r = 7, q = 1, t = −6
r = 2, q = 3, t = 7
r = 1, q = 2, t = −27
9−1 ≡ 34 (mod 61)

Solve

x ≡ 34·5 = 170 ≡ 48 (mod 61)
check: 9·48 = 432 ≡ 5 (mod 61)

Answer

x = 48
8.Number theory & modular arithmetic

Answer: x = 8

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(40, 44) = 4
10x ≡ 3 (mod 11)

Solve the reduced congruence

10−1 ≡ 10 (mod 11) → x ≡ 8 (mod 11)

Solutions modulo 44

x ∈ {8, 19, 30, 41}
x = 8

Answer

x = 8
9.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

n + n = 2n, and constant factors are dropped, so the function is O(n).

Answer

B · O(n)
10.Algorithms & complexity

Answer: D · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

D · Θ(n log n)
11.Data structures

Answer: C · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

C · Index 4
12.Data structures

Answer: D · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

D · A queue (or any collection) of vertices whose in-degree is 0
13.Graph theory

Answer: Euler 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) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: edge GD

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

1. AC (1)
2. GD (3)
3. GC (6)
4. GB (7)
5. DE (8)
6. EF (20)

edge number 2

GD

Answer

edge GD
15.Databases & SQL

Answer: D · Durability

Why this is the answer

Durability means that once a transaction commits, its effects are kept, usually by writing a log to stable storage.

Answer

D · Durability
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: B · Time to live (TTL)

Why this is the answer

Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.

Answer

B · Time to live (TTL)
18.Computer networks

Answer: B · 12 kbit/s

Why this is the answer

2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.

Answer

B · 12 kbit/s
19.IPv4 subnetting

Answer: H = 126

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

255.255.255.128 = /25

Usable hosts

27 − 2 = 126

Answer

H = 126
20.IPv4 subnetting

Answer: x = 224

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

B: 90 hosts → 27 = 128 (/25)
D: 41 hosts → 26 = 64 (/26)
A: 18 hosts → 25 = 32 (/27)
C: 14 hosts → 24 = 16 (/28)

Allocation, largest first

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

Answer

x = 224

LAN C: 192.168.154.224/28.

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