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 −41 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −418 bit
U =
2.Number systemsHard
20-bit two's complement, N = ??sign???????????????????

What is the most negative integer N that can be stored in 20-bit two's complement?

20 bit
N =
3.Propositional logic & proofsMedium
¬r ∨ p¬r → ¬pr ∨ ¬p¬r ∧ p

Which formula is logically equivalent to r → p? Click it.

r → p
  1. 1¬r ∨ p
  2. 2¬r → ¬p
  3. 3r ∨ ¬p
  4. 4¬r ∧ p
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make (p ↔ (p ∧ ¬s)) ↔ ¬(¬(¬r ∧ q) → ¬q) true? Call this number N.

(p ↔ (p ∧ ¬s)) ↔ ¬(¬(¬r ∧ q) → ¬q)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011130214050706

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(1, 3, 4). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(1, 3, 4)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 4 clock edges.

Q₃Q₂Q₁Q₀ = 10004
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 8 modulo 71: the number x with 0 < x < 71 and 8·x ≡ 1 (mod 71).

a = 8n = 71
8−1 mod 71 =
8.Number theory & modular arithmeticHard
mc = me​ mod n138m = cd​ mod n?(n, e) = (473, 11)d

An RSA key uses p = 11, q = 43 and e = 11. Decrypt the ciphertext c = 138: find the message m.

p = 11q = 43e = 11c = 138
m =
9.Algorithms & complexityMedium

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

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

What is the time complexity of building a binary heap from n unsorted elements with the bottom-up method?

  1. AO(n²)
  2. BO(n)
  3. CO(n log n)
  4. DO(log n)
11.Data structuresMedium

An application needs fast lookups and also needs to list keys in sorted order and answer range queries. Which structure fits best?

  1. AA hash table
  2. BAn unsorted linked list
  3. CA stack
  4. DA balanced binary search tree
12.Data structuresHard

Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?

  1. AThose bits may also belong to other elements, which would cause false negatives
  2. BBloom filters store the full elements, and these cannot be located again once hashed
  3. CBits in a Bloom filter cannot be changed after creation
  4. DDeleting would increase the false positive rate to 100%
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
141751516812ABCDEFG

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

Third normal form (3NF) removes:

  1. ARedundant indexes on columns that are not part of a key
  2. BAtomic values
  3. CComposite keys
  4. DTransitive dependencies between non-key attributes
16.Databases & SQLHard

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

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

How many usable host addresses does a /24 IPv4 network have?

  1. A256
  2. B254
  3. C252
  4. D255
18.Computer networksHard

TCP congestion control's "slow start" phase:

  1. AHalves the congestion window every round trip until a packet loss is detected
  2. BIncreases the congestion window exponentially until a threshold or loss
  3. CIs used only by UDP
  4. DKeeps the window fixed at one segment
19.IPv4 subnettingMedium
255.255.255.12811111111111111111111111110000000network bitshost bits

A subnet uses the mask 255.255.255.128. How many usable host addresses H does it have?

255.255.255.128
H =
20.IPv4 subnettingHard
third octet 14 … 17012806403201618141201commonvaries

A router has routes to these networks: 10.14.14.0/24, 10.14.15.0/24, 10.14.16.0/24, 10.14.17.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.14.14.0/24 … 10.14.17.0/24
p =

3Solutions

1.Number systems

Answer: U = 215

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

41 = 0010 1001
inverted: 1101 0110
+1 → 1101 0111

Read it as unsigned

1101 0111₂ = 256 − 41 = 215

Answer

U = 215
2.Number systems

Answer: N = −524288

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.

Most negative value

−219 = −524288

Answer

N = −524288
3.Propositional logic & proofs

Answer: ¬r ∨ p

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 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
r → p ≡ ¬r ∨ p

Check with the truth table

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

Answer

¬r ∨ p

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

4.Propositional logic & proofs

Answer: N = 6

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.

Columns

¬s; p ∧ ¬s; p ↔ (p ∧ ¬s); ¬r; ¬r ∧ q; ¬(¬r ∧ q); ¬q; ¬(¬r ∧ q) → ¬q; ¬(¬(¬r ∧ q) → ¬q); (p ↔ (p ∧ ¬s)) ↔ ¬(¬(¬r ∧ q) → ¬q)

Split on p

p = 1: 4
p = 0: 2
N = 4 + 2 = 6

Answer

N = 6
5.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.

ABC00011110010011130214050706

Group the 0s

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

(B′ + C)
(A + C)
(A′ + C′)

Minimal POS form

F = (B′ + C)(A + C)(A′ + C′)
N = 3

Answer

N = 3

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: Q = 8

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.

State after every edge

1000 → 0100 → 0010 → 0001 → 1000

Decimal value

1000₂ = 8

Answer

Q = 8
7.Number theory & modular arithmetic

Answer: 8−1 mod 71 = 9

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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·8 (mod 71).

r = 71, t = 0
r = 8, q = 8, t = 1
r = 7, q = 1, t = −8
r = 1, q = 7, t = 9

Inverse

1 ≡ 9·8 (mod 71)
x = 9
check: 8·9 = 72 = 1·71 + 1

Answer

8−1 mod 71 = 9
8.Number theory & modular arithmetic

Answer: m = 468

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.

Private exponent

n = 473, φ(n) = 420
d = 11−1 mod 420 = 191

Decrypt by repeated squaring

191 = 128 + 32 + 16 + 8 + 4 + 2 + 1
1381 ≡ 138 (mod 473)
1382 ≡ 124 (mod 473)
1384 ≡ 240 (mod 473)
1388 ≡ 367 (mod 473)
13816 ≡ 357 (mod 473)
13832 ≡ 212 (mod 473)
13864 ≡ 9 (mod 473)
138128 ≡ 81 (mod 473)
138191 ≡ 138·124·240·367·357·212·81 ≡ 468 (mod 473)

Message

m = 468
check: 46811 mod 473 = 138

Answer

m = 468
9.Algorithms & complexity

Answer: A · 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

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

Answer: B · O(n)

Why this is the answer

Summing sift-down costs over the levels gives a convergent series, so heapify runs in linear time.

Answer

B · O(n)
11.Data structures

Answer: D · A balanced binary search tree

Why this is the answer

A balanced BST gives O(log n) lookups and keeps keys ordered, so in-order traversal and range queries are easy; a hash table scatters keys.

Answer

D · A balanced binary search tree
12.Data structures

Answer: A · Those bits may also belong to other elements, which would cause false negatives

Why this is the answer

Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.

Answer

A · Those bits may also belong to other elements, which would cause false negatives
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) = 1, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 3

Odd vertices

A, E → 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) = 14

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: 1, C: 4
B (1): C: 2, D: 8
C (2): G: 8
D (8): E: 13, F: 23
G (8)
E (13): F: 14
F (14)

Shortest path

A → B → D → E → F
d(F) = 14

Answer

d(F) = 14
15.Databases & SQL

Answer: D · Transitive dependencies between non-key attributes

Why this is the answer

In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.

Answer

D · Transitive dependencies between non-key attributes
16.Databases & SQL

Answer: A · 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

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

Answer: B · 254

Why this is the answer

A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.

Answer

B · 254
18.Computer networks

Answer: B · Increases the congestion window exponentially until a threshold or loss

Why this is the answer

Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.

Answer

B · Increases the congestion window exponentially until a threshold or loss
19.IPv4 subnetting

Answer: H = 126

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

255.255.255.128 = /25

Usable hosts

27 − 2 = 126

Answer

H = 126
20.IPv4 subnetting

Answer: p = 19

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

14 = 00001110
17 = 00010001
common leading bits: 3

Prefix length

p = 16 + 3 = 19

Answer

p = /19

Summary route: 10.14.0.0/19 (it also covers 28 network(s) that are not in the list).

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