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 −93 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −938 bit
U =
2.Number systemsHard
16 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬a → ¬c¬a ∧ c¬a ∨ ca ∨ ¬c

Which formula is logically equivalent to a → c? Click it.

a → c
  1. 1¬a → ¬c
  2. 2¬a ∧ c
  3. 3¬a ∨ c
  4. 4a ∨ ¬c
4.Propositional logic & proofsHard
∃x (P(x) → ∀y ¬R(x, y))∃x (P(x) ∧ ∀y ¬R(x, y))∃x (¬P(x) ∧ ∃y R(x, y))∀x (P(x) ∧ ∀y ¬R(x, y))

Which statement is equivalent to ¬∀x (P(x) → ∃y R(x, y)), with the negation moved all the way inside? Click it.

¬∀x (P(x) → ∃y R(x, y))
  1. 1∃x (P(x) → ∀y ¬R(x, y))
  2. 2∃x (P(x) ∧ ∀y ¬R(x, y))
  3. 3∃x (¬P(x) ∧ ∃y R(x, y))
  4. 4∀x (P(x) ∧ ∀y ¬R(x, y))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 4, 9, 13, 14)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X0X113X204050706X12113015X140819X11110

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 9, 10, 13) + d(0, 1, 2, 11, 12, 14), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(3, 9, 10, 13) + d(0, 1, 2, 11, 12, 14)
L =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(338), the number of integers from 1 to 338 that are coprime to 338.

n = 338 = 2·132
φ(338) =
8.Number theory & modular arithmeticHard

Solve 8x ≡ 20 (mod 42). Give the smallest positive solution x.

8x ≡ 20 (mod 42)
x =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. AConstant
  2. BLinear
  3. CLogarithmic
  4. DExponential
10.Algorithms & complexityHard

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

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

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i + 1) mod n
  2. Bi + 1
  3. C2i + 1
  4. D(i − 1) mod n
12.Data structuresHard

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

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

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
14.Graph theoryHard
10102121512141542ABCDEFG

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

Which query returns the number of different cities in the customers table?

  1. ASELECT COUNT(DISTINCT city) FROM customers
  2. BSELECT DISTINCT COUNT(city) FROM customers
  3. CSELECT SUM(DISTINCT city) FROM customers
  4. DSELECT COUNT(city) FROM customers GROUP BY city
16.Databases & SQLHard

For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?

  1. ACross join
  2. BBitmap join on the primary key
  3. CNested loop join with index lookups
  4. DHash join
17.Computer networksMedium

Which of these is a private IPv4 address range?

  1. A192.168.0.0/16
  2. B8.8.8.0/24
  3. C100.0.0.0/8
  4. D200.1.1.0/24
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIPv6 only works on local networks, so addresses never need to be shared globally
  2. BIPv6 does not use addresses
  3. CIPv6 encrypts all addresses
  4. DIts 128-bit address space is large enough to give every device a global address
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.59.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?

192.168.59.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 120012816413211618040201networkhost

A host has the address 10.20.23.120/30. What is the fourth octet x of its network address?

10.20.23.120/30
x =

3Solutions

1.Number systems

Answer: U = 163

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

93 = 0101 1101
inverted: 1010 0010
+1 → 1010 0011

Read it as unsigned

1010 0011₂ = 256 − 93 = 163

Answer

U = 163
2.Number systems

Answer: M = 512 KiB

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.

Bits and bytes

512·512·16 = 4194304 bit = 524288 B

Convert to KiB

524288/1024 = 512 KiB

Answer

M = 512 KiB
3.Propositional logic & proofs

Answer: ¬a ∨ c

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.
a → c ≡ ¬a ∨ c

Check with the truth table

a = 1, c = 1: 1 | 1 0 1 1
a = 1, c = 0: 0 | 1 0 0 1
a = 0, c = 1: 1 | 0 1 1 0
a = 0, c = 0: 1 | 1 0 1 1

Answer

¬a ∨ c

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

4.Propositional logic & proofs

Answer: ∃x (P(x) ∧ ∀y ¬R(x, y))

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.

Move the negation inwards

¬∀x (P(x) → ∃y R(x, y)) ≡ ∃x (P(x) ∧ ∀y ¬R(x, y))

Answer

∃x (P(x) ∧ ∀y ¬R(x, y))

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: AB = 11, CD = 10

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m14: m15, m12, m10, m6 → 0
ABCD000111100001111011111

The isolated 1

m14 = ABCD′ → AB = 11, CD = 10

Answer

AB = 11, CD = 10
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 5

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′D, B′C, AC′D, ACD′, A′B′, ABC′
ABCD0001111000011110X0X113X204050706X12113015X140819X11110

Minimal cover

F = B′C + AC′D
L = 5

Answer

L = 5
7.Number theory & modular arithmetic

Answer: φ(338) = 156

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

338 = 2·132

Totient

φ(338) = φ(2)·φ(169) = 1·156 = 156

Answer

φ(338) = 156
8.Number theory & modular arithmetic

Answer: x = 13

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).

Divide by the gcd

gcd(8, 42) = 2
4x ≡ 10 (mod 21)

Solve the reduced congruence

4−1 ≡ 16 (mod 21) → x ≡ 13 (mod 21)

Solutions modulo 42

x ∈ {13, 34}
x = 13

Answer

x = 13
9.Algorithms & complexity

Answer: D · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

D · Exponential
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 · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

A · (i + 1) mod n
12.Data structures

Answer: A · 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

A · To make it very unlikely that an attacker can pick many keys that all collide
13.Graph theory

Answer: H = 60

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.

Count

H = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
14.Graph theory

Answer: d(F) = 30

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: 10, C: 10
B (10): D: 22, G: 25
C (10): E: 25, G: 14
G (14): E: 16
E (16): F: 30
D (22)
F (30)

Shortest path

A → C → G → E → F
d(F) = 30

Answer

d(F) = 30
15.Databases & SQL

Answer: A · SELECT COUNT(DISTINCT city) FROM customers

Why this is the answer

COUNT(DISTINCT city) counts unique non-NULL cities; DISTINCT outside COUNT only removes duplicate result rows.

Answer

A · SELECT COUNT(DISTINCT city) FROM customers
16.Databases & SQL

Answer: D · Hash join

Why this is the answer

A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.

Answer

D · Hash join
17.Computer networks

Answer: A · 192.168.0.0/16

Why this is the answer

Private ranges are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16; they are not routed on the public internet.

Answer

A · 192.168.0.0/16
18.Computer networks

Answer: D · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

D · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: x = 255

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

27 = 128

Subnet number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 255
20.IPv4 subnetting

Answer: x = 120

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

22 = 4

Subnet of this host

network: ⌊120/4⌋·4 = 120
broadcast: 120 + 4 − 1 = 123
usable hosts: 121 … 122

Answer

x = 120

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