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 −100 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1008 bit
U =
2.Number systemsHard
966 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 966 items its own binary code of the same length. What is the smallest number of bits b needed?

966
b = bit
3.Propositional logic & proofsMedium
¬q ∨ r¬q ∧ r¬q ∨ ¬r¬q ∧ ¬r

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

¬(q ∧ r)
  1. 1¬q ∨ r
  2. 2¬q ∧ r
  3. 3¬q ∨ ¬r
  4. 4¬q ∧ ¬r
4.Propositional logic & proofsHard
pqrq → (p ∨ r)qp111110101100011010001000

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

q → (p ∨ r)q∴ p
  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
ABCD000111100001111010010302141507061120131150141819011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 4, 5, 8, 9, 10, 12, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 4, 5, 8, 9, 10, 12, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKT1234567

A positive-edge-triggered T 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 = 0T
  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. 7clock edge 7
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(361), the number of integers from 1 to 361 that are coprime to 361.

n = 361 = 192
φ(361) =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (91, 13)d = ?

An RSA key uses the primes p = 7 and q = 13 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).

p = 7q = 13e = 13
d =
9.Algorithms & complexityMedium

What does amortised analysis measure?

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

What did the Cook–Levin theorem establish?

  1. AThe Boolean satisfiability problem (SAT) is NP-complete
  2. BEvery NP problem can be solved in exponential time
  3. CThe halting problem is undecidable
  4. DP is not equal to NP
11.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n), because later elements must shift
  2. BO(1), because the index is known
  3. CO(n²), because each element moves n times
  4. DO(log n), because of binary search
12.Data structuresHard

Why might a server choose its hash function at random from a universal family?

  1. ATo guarantee that no two keys can ever collide, whatever keys an attacker sends
  2. BTo make the hash table sorted
  3. CTo remove the need to resize the table
  4. DTo make it very unlikely that an attacker can pick many keys that all collide
13.Graph theoryMedium

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

3 × deg 32 × deg 44 × deg 2
L =
14.Graph theoryHard
1314131239131210ABCDEFG

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

Students can take many courses and courses have many students. How is this modelled in tables?

  1. AWith a foreign key column course_id in the students table pointing to courses
  2. BBy merging students and courses into one table
  3. CWith a list of course ids stored in one column
  4. DWith a junction table holding foreign keys to both students and courses
16.Databases & SQLHard

With multiversion concurrency control (MVCC), how can readers avoid blocking writers?

  1. AReaders always wait until all writers have committed
  2. BReaders see a consistent snapshot of committed row versions while writers create new versions
  3. CReaders read uncommitted data directly
  4. DWriters are not allowed while any reader is active
17.Computer networksMedium

What characterises a distributed denial-of-service (DDoS) attack?

  1. AMany compromised machines flood a target with traffic
  2. BOne attacker guesses a user's password
  3. CA virus deletes files on a single PC
  4. DAn attacker reads unencrypted Wi-Fi traffic from a nearby car
18.Computer networksHard

The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?

  1. ALink-state; flooding link updates
  2. BPath-vector; longer AS paths
  3. CStatic routing; adding default routes
  4. DDistance-vector; split horizon and poison reverse
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1773 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1773 hosts
p =
20.IPv4 subnettingHard

The network 192.168.33.0/24 is split with VLSM for these LANs: A: 18 hosts, B: 24 hosts, C: 35 hosts, D: 12 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?

A: 18B: 24C: 35D: 12E: 3
x =

3Solutions

1.Number systems

Answer: U = 156

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

100 = 0110 0100
inverted: 1001 1011
+1 → 1001 1100

Read it as unsigned

1001 1100₂ = 256 − 100 = 156

Answer

U = 156
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 < 966
210 = 1024 ≥ 966

Answer

b = 10
3.Propositional logic & proofs

Answer: ¬q ∨ ¬r

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

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(q ∧ r) ≡ ¬q ∨ ¬r

Check with the truth table

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

Answer

¬q ∨ ¬r

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

4.Propositional logic & proofs

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

Counterexample

p = 0, q = 1, r = 1

Answer

p = 0, q = 1, 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: L = 15

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.

ABCD000111100001111010010302141507061120131150141819011110

Group the 1s

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

C′D′
AB′D′
A′BC′
AB′C′
ABCD

Minimal SOP form

F = C′D′ + AB′D′ + A′BC′ + AB′C′ + ABCD
L = 15

Answer

L = 15
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 7

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 T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

1: T = 0 → Q = 0
2: T = 0 → Q = 0
3: T = 0 → Q = 0
4: T = 0 → Q = 0
5: T = 1 → Q = 1
6: T = 0 → Q = 1
7: T = 1 → Q = 0

First change from 1 to 0

clock edge 7

Answer

clock edge 7
7.Number theory & modular arithmetic

Answer: φ(361) = 342

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

361 = 192

Totient

φ(361) = 361 − 19 = 342

Answer

φ(361) = 342
8.Number theory & modular arithmetic

Answer: d = 61

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 = 7·13 = 91
φ(n) = 6·12 = 72

Extended Euclidean algorithm

r = 72, t = 0
r = 13, q = 5, t = 1
r = 7, q = 1, t = −5
r = 6, q = 1, t = 6
r = 1, q = 6, t = −11

Private exponent

d = 61
check: 13·61 = 793 = 11·72 + 1

Answer

d = 61
9.Algorithms & complexity

Answer: B · 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

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

Answer: A · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

A · The Boolean satisfiability problem (SAT) is NP-complete
11.Data structures

Answer: A · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

A · O(n), because later elements must shift
12.Data structures

Answer: D · To make it very unlikely that an attacker can pick many keys that all collide

Why this is the answer

With universal hashing the expected number of collisions is small for any set of keys, since the attacker does not know which function was chosen.

Answer

D · To make it very unlikely that an attacker can pick many keys that all collide
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 + 25 = 2(L + 9) − 2
L = 9

Answer

L = 9

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

14.Graph theory

Answer: d(F) = 20

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: 14
B (13): D: 26, G: 26
C (14): E: 26, G: 15
G (15): D: 17, E: 25
D (17): F: 20
F (20)
E (25)

Shortest path

A → C → G → D → F
d(F) = 20

Answer

d(F) = 20
15.Databases & SQL

Answer: D · With a junction table holding foreign keys to both students and courses

Why this is the answer

A many-to-many relationship needs an associative (junction) table, for example enrolment(student_id, course_id), often with a composite primary key.

Answer

D · With a junction table holding foreign keys to both students and courses
16.Databases & SQL

Answer: B · Readers see a consistent snapshot of committed row versions while writers create new versions

Why this is the answer

MVCC keeps old row versions so a reader can see the data as of its snapshot, while a writer creates a new version instead of overwriting in place.

Answer

B · Readers see a consistent snapshot of committed row versions while writers create new versions
17.Computer networks

Answer: A · Many compromised machines flood a target with traffic

Why this is the answer

In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.

Answer

A · Many compromised machines flood a target with traffic
18.Computer networks

Answer: D · Distance-vector; split horizon and poison reverse

Why this is the answer

After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.

Answer

D · Distance-vector; split horizon and poison reverse
19.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1773
211 − 2 = 2046 ≥ 1773 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 128

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

C: 35 hosts → 26 = 64 (/26)
B: 24 hosts → 25 = 32 (/27)
A: 18 hosts → 25 = 32 (/27)
D: 12 hosts → 24 = 16 (/28)
E: 3 hosts → 23 = 8 (/29)

Allocation, largest first

C: .0 – .63
B: .64 – .95
A: .96 – .127
D: .128 – .143
E: .144 – .151

Answer

x = 128

LAN D: 192.168.33.128/28.

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