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 −63 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −638 bit
U =
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
r ↔ ¬q¬r ↔ ¬qr ∧ ¬q¬r ∨ ¬q

Which formula is logically equivalent to ¬(r ↔ q)? Click it.

¬(r ↔ q)
  1. 1r ↔ ¬q
  2. 2¬r ↔ ¬q
  3. 3r ∧ ¬q
  4. 4¬r ∨ ¬q
4.Propositional logic & proofsHard
pqp → qqp11100100

The argument with premises p → q; q and conclusion p is invalid. Click the row of the truth table that is a counterexample.

p → qq∴ p
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010001130214150706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(3, 4, 5). How many product terms N does it have?

F(A, B, C) = Σm(3, 4, 5)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-12

A synchronous mod-12 counter counts 0, 1, …, 11 and then returns to 0. It starts at 9. Which count Q (in decimal) does it show after 30 clock pulses?

Q0 = 930N = 12
Q =
7.Number theory & modular arithmeticMedium

Solve 68x ≡ 30 (mod 71). Give the solution x with 0 < x < 71.

68x ≡ 30 (mod 71)
x =
8.Number theory & modular arithmeticHard
75c = me​ mod n?m = cd​ mod nm(n, e) = (1081, 19)d

Encrypt the message m = 75 with the RSA public key (n, e) = (1081, 19). Find the ciphertext c.

n = 1081e = 19m = 75
c =
9.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(n)
  2. BΩ(n log n)
  3. CΩ(n²)
  4. DΩ(log n)
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 fractional knapsack, where items may be split
  3. CNeither variant
  4. DThe 0/1 knapsack, where items must always be taken whole
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(n²)
  2. BO(n)
  3. CO(log n)
  4. DO(1)
12.Data structuresHard

What guarantee do the red-black tree rules give about the height of a tree with n keys?

  1. AThe height is exactly log2 n
  2. BThe height is at most n / 2
  3. CThe height is at most log2 n + 1
  4. DThe height is at most 2 log2(n + 1)
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
132414710121558ABCDEFG

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

Why can a table have only one clustered index?

  1. AClustered indexes use too much memory
  2. BThe SQL standard forbids more than one index on a table
  3. CThe rows can be stored physically in only one order
  4. DClustered indexes can only be on the primary key
16.Databases & SQLHard

What does EXPLAIN (or EXPLAIN ANALYZE in PostgreSQL) show for a query?

  1. AThe execution plan chosen by the optimiser, such as scans and join methods
  2. BThe list of users who ran the query
  3. CA rewrite of the query in relational algebra notation only
  4. DThe meaning of each SQL keyword in the query
17.Computer networksMedium

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

  1. AHeader checksum
  2. BIdentification
  3. CTime to live (TTL)
  4. DType of service
18.Computer networksHard

Which routing protocol is a link-state protocol?

  1. AARP
  2. BBGP
  3. COSPF
  4. DRIP
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 785 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

785 hosts
p =
20.IPv4 subnettingHard
last octet = 154112806403211618041201networkhost

A host has the address 10.20.122.154/26. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.122.154/26
x =

3Solutions

1.Number systems

Answer: U = 193

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

63 = 0011 1111
inverted: 1100 0000
+1 → 1100 0001

Read it as unsigned

1100 0001₂ = 256 − 63 = 193

Answer

U = 193
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: 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 1 exactly when a and b differ, which is a ↔ ¬b.
¬(r ↔ q) ≡ r ↔ ¬q

Check with the truth table

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

Answer

r ↔ ¬q

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

4.Propositional logic & proofs

Answer: p = 0, q = 1

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.

Premises and conclusion row by row

p = 1, q = 1: 1 1 | 1
p = 1, q = 0: 0 0 | 1
p = 0, q = 1: 1 1 | 0
p = 0, q = 0: 1 0 | 0

Counterexample

p = 0, q = 1

Answer

p = 0, q = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: N = 2

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.

ABC00011110010001130214150706

Group the 1s

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

AB′
A′BC

Minimal SOP form

F = AB′ + A′BC
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 3

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.

Full cycles

Every 12 pulses bring the counter back to the same state.

30 = 2·12 + 6

Count after the remaining pulses

Q = (9 + 6) mod 12 = 3
Q = 0011₂

Answer

Q = 3
7.Number theory & modular arithmetic

Answer: x = 61

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 68

r = 71, t = 0
r = 68, q = 1, t = 1
r = 3, q = 22, t = −1
r = 2, q = 1, t = 23
r = 1, q = 2, t = −24
68−1 ≡ 47 (mod 71)

Solve

x ≡ 47·30 = 1410 ≡ 61 (mod 71)
check: 68·61 = 4148 ≡ 30 (mod 71)

Answer

x = 61
8.Number theory & modular arithmetic

Answer: c = 202

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

19 = 16 + 2 + 1
751 ≡ 75 (mod 1081)
752 ≡ 220 (mod 1081)
754 ≡ 836 (mod 1081)
758 ≡ 570 (mod 1081)
7516 ≡ 600 (mod 1081)
7519 ≡ 75·220·600 ≡ 202 (mod 1081)

Ciphertext

c = 202

Answer

c = 202
9.Algorithms & complexity

Answer: B · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

B · Ω(n log n)
10.Algorithms & complexity

Answer: B · 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

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

Answer: D · 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

D · O(1)
12.Data structures

Answer: D · The height is at most 2 log2(n + 1)

Why this is the answer

Every path has the same number of black nodes and red nodes cannot be adjacent, so the longest path is at most twice the shortest, giving height O(log n).

Answer

D · The height is at most 2 log2(n + 1)
13.Graph theory

Answer: Euler path, no 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) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 2

Odd vertices

B, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

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: 13, C: 2
C (2): B: 6, E: 9
B (6): D: 20, G: 11
E (9): D: 19, F: 24
G (11)
D (19)
F (24)

Shortest path

A → C → E → F
d(F) = 24

Answer

d(F) = 24
15.Databases & SQL

Answer: C · The rows can be stored physically in only one order

Why this is the answer

A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.

Answer

C · The rows can be stored physically in only one order
16.Databases & SQL

Answer: A · The execution plan chosen by the optimiser, such as scans and join methods

Why this is the answer

EXPLAIN displays the plan (index or sequential scans, join algorithms, estimated costs); EXPLAIN ANALYZE also runs the query and reports actual times.

Answer

A · The execution plan chosen by the optimiser, such as scans and join methods
17.Computer networks

Answer: C · 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

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

Answer: C · OSPF

Why this is the answer

OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.

Answer

C · OSPF
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 785
210 − 2 = 1022 ≥ 785 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 191

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: ⌊154/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190

Answer

x = 191

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