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 −59 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −598 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0001 1011 AND 1000 1001. What is the result's value R in decimal?

0001 1011AND1000 1001
R =
3.Propositional logic & proofsMedium
pqrr → ¬qp ↔ r(r → ¬q) ∧ (p ↔ r)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (r → ¬q) ∧ (p ↔ r) true? Call this number N.

(r → ¬q) ∧ (p ↔ r)
N =
4.Propositional logic & proofsHard
∃x (F(x) ∧ ∀y ¬R(x, y))∀x (F(x) ∧ ∀y ¬R(x, y))∃x (¬F(x) ∧ ∃y R(x, y))∃x (F(x) → ∀y ¬R(x, y))

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

¬∀x (F(x) → ∃y R(x, y))
  1. 1∃x (F(x) ∧ ∀y ¬R(x, y))
  2. 2∀x (F(x) ∧ ∀y ¬R(x, y))
  3. 3∃x (¬F(x) ∧ ∃y R(x, y))
  4. 4∃x (F(x) → ∀y ¬R(x, y))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010110302041507061121130151140819111110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 5, 9, 10, 11, 12, 13, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 5, 9, 10, 11, 12, 13, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111010X1130204051706012013X15014X819011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 3, 7, 9, 10) + d(1, 8, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 3, 7, 9, 10) + d(1, 8, 15)
N =
7.Number theory & modular arithmeticMedium

Solve 3x ≡ 8 (mod 11). Give the solution x with 0 < x < 11.

3x ≡ 8 (mod 11)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (713, 7)d = ?

An RSA key uses the primes p = 23 and q = 31 and the public exponent e = 7. Find the private exponent d (0 < d < φ(n)).

p = 23q = 31e = 7
d =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALinear
  2. BExponential
  3. CConstant
  4. DLogarithmic
10.Algorithms & complexityHard

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

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

What balance condition does an AVL tree maintain at every node?

  1. AEvery node is coloured red or black
  2. BAll leaves are at exactly the same depth
  3. CThe heights of the left and right subtrees differ by at most 1
  4. DBoth subtrees of every node contain exactly the same number of nodes
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA single sorted array
  2. BA binary search tree only
  3. CA hash map combined with a doubly linked list
  4. DA stack only
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 10
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
97912142103ABCDEF

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

A view in SQL is:

  1. AAn index type
  2. BA copy of a table on disk
  3. CA screen in the database client for browsing tables
  4. DA stored query that behaves like a virtual table
16.Databases & SQLHard

After customers LEFT JOIN orders, why use COUNT(o.id) rather than COUNT(*) to count orders per customer?

  1. ACOUNT(*) is not allowed after a join
  2. BCOUNT(*) counts the NULL-filled row as 1 for customers without orders
  3. CCOUNT(*) ignores rows with NULL values
  4. DCOUNT(o.id) counts only distinct customers
17.Computer networksMedium

What uniquely identifies one end of a TCP connection on the internet?

  1. AA MAC address alone
  2. BAn IP address together with a port number
  3. CA domain name alone
  4. DA subnet mask and gateway
18.Computer networksHard

A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?

  1. AFast retransmit
  2. BSlow start
  3. CNagle's algorithm
  4. DThe three-way handshake
19.IPv4 subnettingMedium
255.255.255.24011111111111111111111111111110000network bitshost bits

A subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?

255.255.255.240
H =
20.IPv4 subnettingHard
last octet = 120012816413211618040201networkhost

A host has the address 10.20.143.120/27. What is the fourth octet x of its network address?

10.20.143.120/27
x =

3Solutions

1.Number systems

Answer: U = 197

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

59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101

Read it as unsigned

1100 0101₂ = 256 − 59 = 197

Answer

U = 197
2.Number systems

Answer: R = 9

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

AND gives 1 only where both bits are 1.

0001 1011
AND 1000 1001
= 0000 1001

Value of the result

0000 1001₂ = 9

Answer

R = 9
3.Propositional logic & proofs

Answer: N = 3

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: ¬q = 0, r → ¬q = 0, p ↔ r = 1, (r → ¬q) ∧ (p ↔ r) = 0
p = 1, q = 1, r = 0: ¬q = 0, r → ¬q = 1, p ↔ r = 0, (r → ¬q) ∧ (p ↔ r) = 0
p = 1, q = 0, r = 1: ¬q = 1, r → ¬q = 1, p ↔ r = 1, (r → ¬q) ∧ (p ↔ r) = 1
p = 1, q = 0, r = 0: ¬q = 1, r → ¬q = 1, p ↔ r = 0, (r → ¬q) ∧ (p ↔ r) = 0
p = 0, q = 1, r = 1: ¬q = 0, r → ¬q = 0, p ↔ r = 0, (r → ¬q) ∧ (p ↔ r) = 0
p = 0, q = 1, r = 0: ¬q = 0, r → ¬q = 1, p ↔ r = 1, (r → ¬q) ∧ (p ↔ r) = 1
p = 0, q = 0, r = 1: ¬q = 1, r → ¬q = 1, p ↔ r = 0, (r → ¬q) ∧ (p ↔ r) = 0
p = 0, q = 0, r = 0: ¬q = 1, r → ¬q = 1, p ↔ r = 1, (r → ¬q) ∧ (p ↔ r) = 1

Count the rows with value 1

(r → ¬q) ∧ (p ↔ r): 0, 0, 1, 0, 0, 1, 0, 1
N = 3

Answer

N = 3
4.Propositional logic & proofs

Answer: ∃x (F(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 (F(x) → ∃y R(x, y)) ≡ ∃x (F(x) ∧ ∀y ¬R(x, y))

Answer

∃x (F(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: 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.

ABCD000111100001111010110302041507061121130151140819111110

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′)
(A + B′ + D)
(A′ + B + C + D)

Minimal POS form

F = (B′ + C′ + D′)(A + C′)(A + B′ + D)(A′ + B + C + D)
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

B′C′, BCD, A′CD, A′B′D, AB′D′
ABCD000111100001111010X1130204051706012013X15014X819011110

Minimal cover

F = B′C′ + A′CD + AB′D′
N = 3

Answer

N = 3
7.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).

Inverse of 3

r = 11, t = 0
r = 3, q = 3, t = 1
r = 2, q = 1, t = −3
r = 1, q = 2, t = 4
3−1 ≡ 4 (mod 11)

Solve

x ≡ 4·8 = 32 ≡ 10 (mod 11)
check: 3·10 = 30 ≡ 8 (mod 11)

Answer

x = 10
8.Number theory & modular arithmetic

Answer: d = 283

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.

Modulus and totient

n = 23·31 = 713
φ(n) = 22·30 = 660

Extended Euclidean algorithm

r = 660, t = 0
r = 7, q = 94, t = 1
r = 2, q = 3, t = −94
r = 1, q = 2, t = 283

Private exponent

d = 283
check: 7·283 = 1981 = 3·660 + 1

Answer

d = 283
9.Algorithms & complexity

Answer: B · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

B · Exponential
10.Algorithms & complexity

Answer: C · 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

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

Answer: C · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

C · The heights of the left and right subtrees differ by at most 1
12.Data structures

Answer: C · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

C · A hash map combined with a doubly linked list
13.Graph theory

Answer: neither

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

Odd vertices

A, B, C, G → 4
neither

Answer

neither
14.Graph theory

Answer: d(F) = 24

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: 9, C: 7
C (7): E: 21
B (9): D: 21, E: 12
E (12): D: 14
D (14): F: 24
F (24)

Shortest path

A → B → E → D → F
d(F) = 24

Answer

d(F) = 24
15.Databases & SQL

Answer: D · A stored query that behaves like a virtual table

Why this is the answer

A view stores a SELECT statement; querying it runs the underlying query (materialised views store the result).

Answer

D · A stored query that behaves like a virtual table
16.Databases & SQL

Answer: B · COUNT(*) counts the NULL-filled row as 1 for customers without orders

Why this is the answer

The left join gives one row with NULL order columns for such customers; COUNT(o.id) skips the NULL and correctly reports 0.

Answer

B · COUNT(*) counts the NULL-filled row as 1 for customers without orders
17.Computer networks

Answer: B · An IP address together with a port number

Why this is the answer

A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.

Answer

B · An IP address together with a port number
18.Computer networks

Answer: C · Nagle's algorithm

Why this is the answer

Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.

Answer

C · Nagle's algorithm
19.IPv4 subnetting

Answer: H = 14

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

255.255.255.240 = /28

Usable hosts

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 96

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

25 = 32

Subnet of this host

network: ⌊120/32⌋·32 = 96
broadcast: 96 + 32 − 1 = 127
usable hosts: 97 … 126

Answer

x = 96

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