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 −85 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −858 bit
U =
2.Number systemsHard
32-bit two's complement, N = ??sign???????????????????????????????

What is the most negative integer N that can be stored in 32-bit two's complement?

32 bit
N =
3.Propositional logic & proofsMedium
(c → a) ∧ (b → a)c → (b → a)(c → b) → ac ∧ (b → a)

Which formula is logically equivalent to (c ∧ b) → a? Click it.

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

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

¬∀x ∃y R(x, y)
  1. 1∃x ∀y ¬R(x, y)
  2. 2∀x ∃y ¬R(x, y)
  3. 3∃y ∀x ¬R(x, y)
  4. 4∀x ∀y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111

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(5, 7, 11, 13)
  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
CLKD123456

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

Q₀ = 0D
N =
7.Number theory & modular arithmeticMedium

Solve 9x ≡ 13 (mod 47). Give the solution x with 0 < x < 47.

9x ≡ 13 (mod 47)
x =
8.Number theory & modular arithmeticHard

Solve 6x ≡ 2 (mod 28). Give the smallest positive solution x.

6x ≡ 2 (mod 28)
x =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. AThe base only matters for small n
  2. BLogarithms with different bases differ only by a constant factor
  3. CLogarithms are always rounded to integers
  4. DAll logarithms in computer science have base 10, so it is implied
10.Algorithms & complexityHard

A problem is NP-complete if it is:

  1. ASolvable in polynomial time
  2. BOutside NP but harder than all NP problems, such as the halting problem
  3. CIn NP and every NP problem reduces to it in polynomial time
  4. DImpossible to solve
11.Data structuresMedium

The keys 50, 30, 70, 20, 40 are inserted in this order into an empty binary search tree. What is the left child of 30?

  1. A70
  2. B40
  3. C50
  4. D20
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA stack only
  2. BA single sorted array
  3. CA binary search tree only
  4. DA hash map combined with a doubly linked list
13.Graph theoryMedium

A connected graph has 13 vertices and 17 edges. How many edges k must be removed to leave a spanning tree?

|V| = 13|E| = 17
k =
14.Graph theoryHard
912138151611141ABCDEF

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 3; skipped edges do not count.

k = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF
  9. 9edge BE
15.Databases & SQLMedium

Third normal form (3NF) removes:

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

What does a FULL OUTER JOIN return?

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

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.2.2.3
  2. B11.1.2.3
  3. C10.0.2.3
  4. D10.1.200.5
18.Computer networksHard

A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?

  1. ANagle's algorithm
  2. BThe three-way handshake
  3. CFast retransmit
  4. DSlow start
19.IPv4 subnettingMedium
255.255.252.011111111111111111111110000000000network bitshost bits

A subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?

255.255.252.0
H =
20.IPv4 subnettingHard
last octet = 194112816403201608041201networkhost

A host has the address 10.20.235.194/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.235.194/27
x =

3Solutions

1.Number systems

Answer: U = 171

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

85 = 0101 0101
inverted: 1010 1010
+1 → 1010 1011

Read it as unsigned

1010 1011₂ = 256 − 85 = 171

Answer

U = 171
2.Number systems

Answer: N = −2.147 × 10⁹

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

−231 = −2147483648

Answer

N = −2147483648
3.Propositional logic & proofs

Answer: c → (b → a)

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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(c ∧ b) → a ≡ c → (b → a)

Check with the truth table

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

Answer

c → (b → a)

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

4.Propositional logic & proofs

Answer: ∃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 ∃y R(x, y) ≡ ∃x ∀y ¬R(x, y)

Answer

∃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 = 10, CD = 11

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.

m11: m10, m9, m15, m3 → 0
ABCD00011110000111101111

The isolated 1

m11 = AB′CD → AB = 10, CD = 11

Answer

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

Answer: N = 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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 1 → Q = 1
3: D = 0 → Q = 0
4: D = 0 → Q = 0
5: D = 0 → Q = 0
6: D = 1 → Q = 1

Count the changes

0 → 1 → 1 → 0 → 0 → 0 → 1
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 38

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).

Inverse of 9

r = 47, t = 0
r = 9, q = 5, t = 1
r = 2, q = 4, t = −5
r = 1, q = 2, t = 21
9−1 ≡ 21 (mod 47)

Solve

x ≡ 21·13 = 273 ≡ 38 (mod 47)
check: 9·38 = 342 ≡ 13 (mod 47)

Answer

x = 38
8.Number theory & modular arithmetic

Answer: x = 5

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(6, 28) = 2
3x ≡ 1 (mod 14)

Solve the reduced congruence

3−1 ≡ 5 (mod 14) → x ≡ 5 (mod 14)

Solutions modulo 28

x ∈ {5, 19}
x = 5

Answer

x = 5
9.Algorithms & complexity

Answer: B · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

B · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: C · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

C · In NP and every NP problem reduces to it in polynomial time
11.Data structures

Answer: D · 20

Why this is the answer

20 is smaller than 50 and smaller than 30, so it becomes the left child of 30; 40 becomes its right child.

Answer

D · 20
12.Data structures

Answer: D · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

D · A hash map combined with a doubly linked list
13.Graph theory

Answer: k = 5

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.

Edges of a spanning tree

13 − 1 = 12

Edges to remove

k = 17 − 12 = 5

Answer

k = 5
14.Graph theory

Answer: edge AB

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

1. BE (1)
2. BD (8)
3. AB (9)
4. DF (11)
5. AC (12)

edge number 3

AB

Answer

edge AB
15.Databases & SQL

Answer: A · 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

A · Transitive dependencies between non-key attributes
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: D · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

D · 10.1.200.5
18.Computer networks

Answer: A · Nagle's algorithm

Why this is the answer

Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.

Answer

A · Nagle's algorithm
19.IPv4 subnetting

Answer: H = 1022

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

255.255.252.0 = /22

Usable hosts

210 − 2 = 1022

Answer

H = 1022
20.IPv4 subnetting

Answer: x = 223

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: ⌊194/32⌋·32 = 192
broadcast: 192 + 32 − 1 = 223
usable hosts: 193 … 222

Answer

x = 223

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