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 complement1−12806403211618041201

The 8-bit pattern 1001 1010 stores a signed integer in two's complement. What is its value N in decimal?

1001 10108 bit
N =
2.Number systemsHard
2700 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 2700 items its own binary code of the same length. What is the smallest number of bits b needed?

2700
b = bit
3.Propositional logic & proofsMedium
abcc ∧ b((b ↔ a) ∧ c) ∧ (c ∧ b)¬(((b ↔ a) ∧ c) ∧ (c ∧ b))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) true? Call this number N.

¬(((b ↔ a) ∧ c) ∧ (c ∧ b))
N =
4.Propositional logic & proofsHard
r¬q ∧ r¬qq

The premises q → p; p → r; ¬r are given. Which conclusion follows from them? Click it.

q → pp → r¬r
  1. 1r
  2. 2¬q ∧ r
  3. 3¬q
  4. 4q
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​C′I1​0I2​C′I3​ABS1​S0​F

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

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

A 3-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 10 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 3tpd = 10 ns
fmax = MHz
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(56), the number of integers from 1 to 56 that are coprime to 56.

n = 56 = 23·7
φ(56) =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (155, 23)d = ?

An RSA key uses the primes p = 5 and q = 31 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 5q = 31e = 23
d =
9.Algorithms & complexityMedium

What does the recurrence T(n) = 2T(n − 1) + O(1) solve to?

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

Which two things must be shown to prove that a decision problem X is NP-complete?

  1. AX has no polynomial algorithm, and X is decidable
  2. BX is in P, and X reduces to SAT
  3. CX reduces to a problem in P, and X is in NP
  4. DX is in NP, and a known NP-complete problem reduces to X in polynomial time
11.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA singly linked list
  2. BA doubly linked list
  3. CA sorted array
  4. DA dynamic array
12.Data structuresHard

What is the running time of breadth-first search on a graph with V vertices stored as an adjacency matrix?

  1. AO(V + E)
  2. BO(V²)
  3. CO(E log V)
  4. DO(V)
13.Graph theoryMedium

A tree has 3 vertices of degree 3, 2 vertices of degree 4, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 32 × deg 4
L =
14.Graph theoryHard
1620176392124ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 9
W =
15.Databases & SQLMedium

A view in SQL is:

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

The query FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.status = 'paid' unexpectedly drops customers with no orders. Why?

  1. AThe join should use USING instead of ON
  2. Bstatus is a reserved word
  3. CThe WHERE condition on o rejects the NULL-filled rows, turning it into an inner join
  4. DLEFT JOIN never keeps unmatched rows
17.Computer networksMedium

Which DNS record type maps a host name to an IPv6 address?

  1. AAAAA
  2. BCNAME
  3. CMX
  4. DA
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 50 Mbit/s
  2. BAbout 640 Mbit/s
  3. CAbout 5 Mbit/s
  4. DAbout 0.5 Mbit/s
19.IPv4 subnettingMedium
/?11111111111111111111111111100000network bitshost bits

A department needs 22 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

22 hosts
p =
20.IPv4 subnettingHard
last octet = 217112816403211618040211networkhost

A host has the address 10.20.224.217/27. What is the fourth octet x of the last usable host address in its subnet?

10.20.224.217/27
x =

3Solutions

1.Number systems

Answer: N = −102

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 + 16 + 8 + 2 = −102

Answer

N = −102
2.Number systems

Answer: b = 12 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

211 = 2048 < 2700
212 = 4096 ≥ 2700

Answer

b = 12
3.Propositional logic & proofs

Answer: N = 7

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

a = 1, b = 1, c = 1: b ↔ a = 1, (b ↔ a) ∧ c = 1, c ∧ b = 1, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 1, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 0
a = 1, b = 1, c = 0: b ↔ a = 1, (b ↔ a) ∧ c = 0, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 1, b = 0, c = 1: b ↔ a = 0, (b ↔ a) ∧ c = 0, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 1, b = 0, c = 0: b ↔ a = 0, (b ↔ a) ∧ c = 0, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 0, b = 1, c = 1: b ↔ a = 0, (b ↔ a) ∧ c = 0, c ∧ b = 1, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 0, b = 1, c = 0: b ↔ a = 0, (b ↔ a) ∧ c = 0, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 0, b = 0, c = 1: b ↔ a = 1, (b ↔ a) ∧ c = 1, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1
a = 0, b = 0, c = 0: b ↔ a = 1, (b ↔ a) ∧ c = 0, c ∧ b = 0, ((b ↔ a) ∧ c) ∧ (c ∧ b) = 0, ¬(((b ↔ a) ∧ c) ∧ (c ∧ b)) = 1

Count the rows with value 1

¬(((b ↔ a) ∧ c) ∧ (c ∧ b)): 0, 1, 1, 1, 1, 1, 1, 1
N = 7

Answer

N = 7
4.Propositional logic & proofs

Answer: ¬q

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.

Rows with all premises true

p = 0, q = 0, r = 0: r = 0, ¬q ∧ r = 0, ¬q = 1, q = 0

Conclusion

¬q (modus tollens twice)

Answer

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

Answer: N = 2

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′BC′ + ABC′

Rows with F = 1

ABC = 010, 110
N = 2

Answer

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

Answer: fmax = 33.33 MHz

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.

Worst-case settling time

T = 3·10 ns = 30 ns

Maximum frequency

fmax = 1/30 ns = 33.33 MHz

Answer

fmax = 33.33 MHz
7.Number theory & modular arithmetic

Answer: φ(56) = 24

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

56 = 23·7

Totient

φ(56) = φ(8)·φ(7) = 4·6 = 24

Answer

φ(56) = 24
8.Number theory & modular arithmetic

Answer: d = 47

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 = 5·31 = 155
φ(n) = 4·30 = 120

Extended Euclidean algorithm

r = 120, t = 0
r = 23, q = 5, t = 1
r = 5, q = 4, t = −5
r = 3, q = 1, t = 21
r = 2, q = 1, t = −26
r = 1, q = 2, t = 47

Private exponent

d = 47
check: 23·47 = 1081 = 9·120 + 1

Answer

d = 47
9.Algorithms & complexity

Answer: A · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

A · O(2ⁿ)
10.Algorithms & complexity

Answer: D · X is in NP, and a known NP-complete problem reduces to X in polynomial time

Why this is the answer

Membership in NP shows X is not harder than NP; the reduction from a known NP-complete problem shows X is at least as hard as all of NP.

Answer

D · X is in NP, and a known NP-complete problem reduces to X in polynomial time
11.Data structures

Answer: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
12.Data structures

Answer: B · O(V²)

Why this is the answer

Each of the V vertices scans a row of V entries, giving O(V²); with adjacency lists BFS is O(V + E).

Answer

B · O(V²)
13.Graph theory

Answer: L = 9

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.

Handshake lemma for a tree

L + 17 = 2(L + 5) − 2
L = 9

Answer

L = 9

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: W = 40

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

GC (2): take
EF (3): take
GE (4): take
DF (6): take
GB (9): take
GD (12): skip, closes a cycle
AB (16): take

Total weight

W = 2 + 3 + 4 + 6 + 9 + 16 = 40

Answer

W = 40
15.Databases & SQL

Answer: C · 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

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

Answer: C · The WHERE condition on o rejects the NULL-filled rows, turning it into an inner join

Why this is the answer

Unmatched rows have o.status NULL, so the WHERE test is not TRUE; moving the condition into the ON clause keeps every customer.

Answer

C · The WHERE condition on o rejects the NULL-filled rows, turning it into an inner join
17.Computer networks

Answer: A · AAAA

Why this is the answer

An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.

Answer

A · AAAA
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: p = 27

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

24 − 2 = 14 < 22
25 − 2 = 30 ≥ 22 → h = 5

Prefix length

p = 32 − 5 = 27

Answer

p = /27
20.IPv4 subnetting

Answer: x = 222

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: ⌊217/32⌋·32 = 192
broadcast: 192 + 32 − 1 = 223
usable hosts: 193 … 222

Answer

x = 222

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