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
place values in two's complement0−12806403211608040211

The 8-bit pattern 0001 0001 stores a signed integer in two's complement. What is its value N in decimal?

0001 00018 bit
N =
2.Number systemsHard
664 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 664 items its own binary code of the same length. What is the smallest number of bits b needed?

664
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is p ↔ ¬p a tautology, a contradiction or a contingency? Click your answer.

p ↔ ¬p
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
pqrp → (q ∨ r)pq111110101100011010001000

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

p → (q ∨ r)p∴ q
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​0I1​C′I2​CI3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = C′, I₃ = C. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 0I₂ = C′I₃ = C
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 10 (decimal, Q = 1010). Which count Q (in decimal) does it show after 23 clock pulses?

Q0 = 1023n = 4
Q =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(323), the number of integers from 1 to 323 that are coprime to 323.

n = 323 = 17·19
φ(323) =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 67160 mod 78.

a = 67k = 160n = 78
67160 mod 78 =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the shortest activity first
  2. BPick the activity that finishes earliest, then repeat with the compatible ones
  3. CPick the activity that starts earliest
  4. DPick the activity with the fewest overlaps first, then repeat with the compatible ones
10.Algorithms & complexityHard

What does the recurrence T(n) = 2T(n/2) + n log n solve to?

  1. AΘ(n log n)
  2. BΘ(n log² n)
  3. CΘ(n)
  4. DΘ(n²)
11.Data structuresMedium

What is true of the adjacency matrix of an undirected graph?

  1. AIt is symmetric about the main diagonal
  2. BEach row has exactly one 1
  3. CIt is always upper triangular
  4. DIts main diagonal always contains ones
12.Data structuresHard

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

  1. AA queue (or any collection) of vertices whose in-degree is 0
  2. BA stack of all visited edges
  3. CA max-heap ordered by out-degree
  4. DAn adjacency matrix recording which vertices have already been removed
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| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
14691281315713ABCDEFG

Use Dijkstra’s algorithm to find the length d(E) of the shortest path from A to E in the weighted graph.

start: Aend: E
d(E) =
15.Databases & SQLMedium

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. ACombine the table with itself using UNION ALL and a WHERE filter
  2. BCreate a second copy of the table on disk
  3. CUse GROUP BY manager_id only
  4. DJoin the employees table with itself using two aliases
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AThe Cartesian product, pairing each row of one table with every row of the other
  2. BOnly matching rows
  3. CAll rows from both tables, matched where possible and NULL-filled otherwise
  4. DOnly unmatched rows
17.Computer networksMedium

NAT (network address translation) is mainly used to:

  1. ALet many devices with private addresses share one public IP address
  2. BEncrypt traffic between two sites
  3. CTranslate domain names into IP addresses for devices on a local network
  4. DIncrease the speed of a network cable
18.Computer networksHard

A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?

  1. AAbout 640 Mbit/s
  2. BAbout 0.5 Mbit/s
  3. CAbout 5 Mbit/s
  4. DAbout 50 Mbit/s
19.IPv4 subnettingMedium
255.255.255.24811111111111111111111111111111000network bitshost bits

A subnet uses the mask 255.255.255.248. How many usable host addresses H does it have?

255.255.255.248
H =
20.IPv4 subnettingHard
third octet = 61012806413211618140211networkhost

A host has the address 172.20.61.46/20. What is the third octet x of its network address?

172.20.61.46/20
x =

3Solutions

1.Number systems

Answer: N = 17

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.

16 + 1 = 17

Answer

N = 17
2.Number systems

Answer: b = 10 bit

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.

Powers of two

29 = 512 < 664
210 = 1024 ≥ 664

Answer

b = 10
3.Propositional logic & proofs

Answer: contradiction

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: ¬p = 0, p ↔ ¬p = 0
p = 0: ¬p = 1, p ↔ ¬p = 0

Classify

p ↔ ¬p: 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

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

Counterexample

p = 1, q = 0, r = 1

Answer

p = 1, q = 0, r = 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 = 3

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C′ + AB′C′ + ABC

Rows with F = 1

ABC = 000, 100, 111
N = 3

Answer

N = 3
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 16 pulses bring the counter back to the same state.

23 = 1·16 + 7

Count after the remaining pulses

Q = (10 − 7) mod 16 = 3
Q = 0011₂

Answer

Q = 3
7.Number theory & modular arithmetic

Answer: φ(323) = 288

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).

Prime factors

323 = 17·19

Totient

φ(323) = φ(19)·φ(17) = 18·16 = 288

Answer

φ(323) = 288
8.Number theory & modular arithmetic

Answer: 67160 mod 78 = 55

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

160 = 101000002 = 128 + 32

Repeated squaring

671 ≡ 67 (mod 78)
672 ≡ 43 (mod 78)
674 ≡ 55 (mod 78)
678 ≡ 61 (mod 78)
6716 ≡ 55 (mod 78)
6732 ≡ 61 (mod 78)
6764 ≡ 55 (mod 78)
67128 ≡ 61 (mod 78)

Multiply the needed powers

67160 ≡ 61·61 ≡ 55 (mod 78)

Answer

67160 mod 78 = 55
9.Algorithms & complexity

Answer: B · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

B · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: B · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

B · Θ(n log² n)
11.Data structures

Answer: A · It is symmetric about the main diagonal

Why this is the answer

In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.

Answer

A · It is symmetric about the main diagonal
12.Data structures

Answer: A · 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

A · A queue (or any collection) of vertices whose in-degree is 0
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) = 4, deg(C) = 2, deg(D) = 3, deg(E) = 3, deg(F) = 2, deg(G) = 3

Odd vertices

A, D, E, G → 4
neither

Answer

neither
14.Graph theory

Answer: d(E) = 18

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: 14, C: 6
C (6): E: 18, G: 21
B (14): D: 23, G: 17
G (17)
E (18): F: 19
F (19)
D (23)

Shortest path

A → C → E
d(E) = 18

Answer

d(E) = 18
15.Databases & SQL

Answer: D · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

D · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: C · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

C · All rows from both tables, matched where possible and NULL-filled otherwise
17.Computer networks

Answer: A · Let many devices with private addresses share one public IP address

Why this is the answer

NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.

Answer

A · Let many devices with private addresses share one public IP address
18.Computer networks

Answer: C · About 5 Mbit/s

Why this is the answer

Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.

Answer

C · About 5 Mbit/s
19.IPv4 subnetting

Answer: H = 6

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

255.255.255.248 = /29

Usable hosts

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 48

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 in the third octet

24 = 16

Subnet of this host

network: 172.20.48.0
broadcast: 172.20.63.255

Answer

x = 48

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