Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 complement1−12816413201618141201

The 8-bit pattern 1110 1110 stores a signed integer in two's complement. What is its value N in decimal?

1110 11108 bit
N =
2.Number systemsHard
32 bits, unsigned, N = ?????????????????????????????????

What is the largest unsigned integer N that fits in 32 bits?

32 bit
N =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ((q → p) ∧ p) → q a tautology, a contradiction or a contingency? Click your answer.

((q → p) ∧ p) → q
  1. 1tautology
  2. 2contradiction
  3. 3contingency
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
ABCD000111100001111000110302141517060120131150141819011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 4, 5, 7, 8, 9, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(1, 4, 5, 7, 8, 9, 15)
N =
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. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  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.Number theory & modular arithmeticMedium

Find the inverse of 3 modulo 11: the number x with 0 < x < 11 and 3·x ≡ 1 (mod 11).

a = 3n = 11
3−1 mod 11 =
8.Number theory & modular arithmeticHard

43 is prime. Use Fermat’s little theorem to find 988444 mod 43.

p = 43k = 88444
988444 mod 43 =
9.Algorithms & complexityMedium

What does amortised analysis measure?

  1. AThe average cost per operation over a worst-case sequence of operations
  2. BThe cost of the single most expensive operation anywhere in the sequence
  3. CThe cost of an algorithm on the best input
  4. DThe expected cost over random inputs
10.Algorithms & complexityHard

For which knapsack variant does the greedy rule "take items in order of value per weight" always give an optimal answer?

  1. ABoth variants
  2. BThe 0/1 knapsack, where items must always be taken whole
  3. CNeither variant
  4. DThe fractional knapsack, where items may be split
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(1)
  2. BO(n²)
  3. CO(log n)
  4. DO(n)
12.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ATo keep the array sorted
  2. BBecause halving at one half would lose elements
  3. CTo avoid repeated resizing when pushes and pops alternate near the boundary
  4. DBecause the memory allocator can only release memory in quarters of a block
13.Graph theoryMedium

A connected graph has 7 vertices and 8 edges. How many edges k must be removed to leave a spanning tree?

|V| = 7|E| = 8
k =
14.Graph theoryHard
28121085108ABCDEF

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

Table a has 3 rows and table b has 4 rows. How many rows does SELECT * FROM a CROSS JOIN b return?

  1. A3
  2. B4
  3. C7
  4. D12
16.Databases & SQLHard

Table b contains the values 1, 2 and NULL in column y. What does SELECT * FROM a WHERE x NOT IN (SELECT y FROM b) return?

  1. ARows of a where x is not 1 or 2
  2. BNo rows at all
  3. COnly rows of a where x is NULL
  4. DEvery row of a
17.Computer networksMedium

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.2.2.3
  2. B10.0.2.3
  3. C11.1.2.3
  4. D10.1.200.5
18.Computer networksHard

What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?

  1. AThe connection stays encrypted without TLS
  2. BResponses can arrive in any order on one stream, so slow replies do not block others
  3. CSeveral requests reuse one TCP connection, avoiding repeated handshakes
  4. DThe server can push updates over UDP
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.183.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?

192.168.183.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 237112816413201618140211networkhost

A host has the address 10.20.67.237/26. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.67.237/26
x =

3Solutions

1.Number systems

Answer: N = −18

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 + 32 + 8 + 4 + 2 = −18

Answer

N = −18
2.Number systems

Answer: N = 4.295 × 10⁹

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.

Largest value

232 − 1 = 4294967295

Answer

N = 4294967295
3.Propositional logic & proofs

Answer: contingency

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

p = 1, q = 1: q → p = 1, (q → p) ∧ p = 1, ((q → p) ∧ p) → q = 1
p = 1, q = 0: q → p = 1, (q → p) ∧ p = 1, ((q → p) ∧ p) → q = 0
p = 0, q = 1: q → p = 0, (q → p) ∧ p = 0, ((q → p) ∧ p) → q = 1
p = 0, q = 0: q → p = 1, (q → p) ∧ p = 0, ((q → p) ∧ p) → q = 1

Classify

((q → p) ∧ p) → q: 1, 0, 1, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
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 = 4

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.

ABCD000111100001111000110302141517060120131150141819011010

Group the 1s

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

B′C′D
BCD
A′BC′
AB′C′

Minimal SOP form

F = B′C′D + BCD + A′BC′ + AB′C′
N = 4

Answer

N = 4

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: clock edge 4

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

First change from 1 to 0

clock edge 4

Answer

clock edge 4
7.Number theory & modular arithmetic

Answer: 3−1 mod 11 = 4

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·3 (mod 11).

r = 11, t = 0
r = 3, q = 3, t = 1
r = 2, q = 1, t = −3
r = 1, q = 2, t = 4

Inverse

1 ≡ 4·3 (mod 11)
x = 4
check: 3·4 = 12 = 1·11 + 1

Answer

3−1 mod 11 = 4
8.Number theory & modular arithmetic

Answer: 988444 mod 43 = 15

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.

Reduce the exponent

88444 mod 42 = 34
988444 ≡ 934 (mod 43)

Small power

934 mod 43 = 15

Answer

988444 mod 43 = 15
9.Algorithms & complexity

Answer: A · The average cost per operation over a worst-case sequence of operations

Why this is the answer

Amortised analysis spreads the cost of rare expensive operations over many cheap ones, giving a guaranteed average for any sequence.

Answer

A · The average cost per operation over a worst-case sequence of operations
10.Algorithms & complexity

Answer: D · The fractional knapsack, where items may be split

Why this is the answer

When items can be split, the best ratio first is optimal; in 0/1 knapsack a high-ratio item can block a better combination, so dynamic programming is needed.

Answer

D · The fractional knapsack, where items may be split
11.Data structures

Answer: A · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

A · O(1)
12.Data structures

Answer: C · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

C · To avoid repeated resizing when pushes and pops alternate near the boundary
13.Graph theory

Answer: k = 2

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

7 − 1 = 6

Edges to remove

k = 8 − 6 = 2

Answer

k = 2
14.Graph theory

Answer: d(F) = 19

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: 2, C: 8
B (2): D: 14, E: 10
C (8)
E (10): F: 20
D (14): F: 19
F (19)

Shortest path

A → B → D → F
d(F) = 19

Answer

d(F) = 19
15.Databases & SQL

Answer: D · 12

Why this is the answer

A cross join gives the Cartesian product, 3 × 4 = 12 rows.

Answer

D · 12
16.Databases & SQL

Answer: B · No rows at all

Why this is the answer

The comparison with NULL is UNKNOWN, so the whole condition is never TRUE; NOT EXISTS or filtering out NULLs avoids this trap.

Answer

B · No rows at all
17.Computer networks

Answer: D · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

D · 10.1.200.5
18.Computer networks

Answer: C · Several requests reuse one TCP connection, avoiding repeated handshakes

Why this is the answer

Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.

Answer

C · Several requests reuse one TCP connection, avoiding repeated handshakes
19.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

27 = 128

Subnet number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 128
20.IPv4 subnetting

Answer: x = 255

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: ⌊237/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254

Answer

x = 255

ExamProblems · examproblems.com/courses/computer-science-fundamentals