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 −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −488 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 0111 XOR 0010 1100. What is the result's value R in decimal?

0101 0111XOR0010 1100
R =
3.Propositional logic & proofsMedium
(a ∧ c) ∨ ba ∨ (c ∧ b)(a ∧ c) ∨ (a ∧ b)(a ∨ c) ∧ (a ∨ b)

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

a ∧ (c ∨ b)
  1. 1(a ∧ c) ∨ b
  2. 2a ∨ (c ∧ b)
  3. 3(a ∧ c) ∨ (a ∧ b)
  4. 4(a ∨ c) ∧ (a ∨ b)
4.Propositional logic & proofsHard
∃x (F(x) ∧ ¬G(x))∀x (F(x) → ¬G(x))∃x (¬F(x) ∧ G(x))∀x (F(x) ∧ ¬G(x))

Which statement is equivalent to ¬∀x (F(x) → G(x)), with the negation moved all the way inside? Click it.

¬∀x (F(x) → G(x))
  1. 1∃x (F(x) ∧ ¬G(x))
  2. 2∀x (F(x) → ¬G(x))
  3. 3∃x (¬F(x) ∧ G(x))
  4. 4∀x (F(x) ∧ ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011302040507060121130150140809111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 3, 10, 11, 13). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 3, 10, 11, 13)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X0X113X21415X71601201311501408X9011010

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 4, 5, 6, 15) + d(0, 1, 2, 7, 9), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(3, 4, 5, 6, 15) + d(0, 1, 2, 7, 9)
L =
7.Number theory & modular arithmeticMedium

Solve 14x ≡ 63 (mod 83). Give the solution x with 0 < x < 83.

14x ≡ 63 (mod 83)
x =
8.Number theory & modular arithmeticHard

Solve 18x ≡ 48 (mod 75). Give the smallest positive solution x.

18x ≡ 48 (mod 75)
x =
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 starts earliest
  2. BPick the activity with the fewest overlaps first, then repeat with the compatible ones
  3. CPick the shortest activity first
  4. DPick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexityHard

To show that a new problem X is NP-hard, what must you do?

  1. AReduce X to a known NP-hard problem in polynomial time
  2. BShow that X has no polynomial algorithm for small inputs
  3. CShow that X can be verified in polynomial time
  4. DReduce a known NP-hard problem to X in polynomial time
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(log n)
  2. BO(n²)
  3. CO(1)
  4. DO(n)
12.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. APrimary clustering, where long runs of occupied slots form
  2. BThe O(n) cost of reading a key's value
  3. CThe need for a hash function
  4. DThe need to store keys in the table
13.Graph theoryMedium

A connected graph has 8 vertices and 11 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 11
k =
14.Graph theoryHard
81913671891513ABCDEFG

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

|V| = 7|E| = 10
W =
15.Databases & SQLMedium

What does the "A" in ACID stand for?

  1. AAccuracy
  2. BAvailability
  3. CAggregation
  4. DAtomicity
16.Databases & SQLHard

There is a composite index on (last_name, first_name). Which condition can use it most effectively?

  1. AWHERE first_name = 'Eva'
  2. BWHERE UPPER(first_name) = 'EVA'
  3. CWHERE last_name = 'Jansen'
  4. DWHERE first_name LIKE '%a'
17.Computer networksMedium

Which IPv4 header field stops packets from circulating forever in a routing loop?

  1. AType of service
  2. BIdentification
  3. CHeader checksum
  4. DTime to live (TTL)
18.Computer networksHard

What does TCP fast retransmit do?

  1. AResends every segment twice to be safe
  2. BSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
  3. CDoubles the window after every loss
  4. DResends a segment after three duplicate ACKs, without waiting for the timeout
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1784 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1784 hosts
p =
20.IPv4 subnettingHard
third octet 42 … 46012806413201618041201commonvaries

A router has routes to these networks: 10.87.42.0/24, 10.87.43.0/24, 10.87.44.0/24, 10.87.45.0/24, 10.87.46.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.87.42.0/24 … 10.87.46.0/24
p =

3Solutions

1.Number systems

Answer: U = 208

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

48 = 0011 0000
inverted: 1100 1111
+1 → 1101 0000

Read it as unsigned

1101 0000₂ = 256 − 48 = 208

Answer

U = 208
2.Number systems

Answer: R = 123

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.

Bit by bit

XOR gives 1 where the two bits differ.

0101 0111
XOR 0010 1100
= 0111 1011

Value of the result

0111 1011₂ = 123

Answer

R = 123
3.Propositional logic & proofs

Answer: (a ∧ c) ∨ (a ∧ b)

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

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
a ∧ (c ∨ b) ≡ (a ∧ c) ∨ (a ∧ b)

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

Answer

(a ∧ c) ∨ (a ∧ b)

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

4.Propositional logic & proofs

Answer: ∃x (F(x) ∧ ¬G(x))

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 (F(x) → G(x)) ≡ ∃x (F(x) ∧ ¬G(x))

Answer

∃x (F(x) ∧ ¬G(x))

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: N = 4

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.

ABCD000111100001111010011302040507060121130150140809111110

Group the 1s

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

B′CD
AB′C
A′B′C′D′
ABC′D

Minimal SOP form

F = B′CD + AB′C + A′B′C′D′ + ABC′D
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 4

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.

BCD, A′
ABCD0001111000011110X0X113X21415X71601201311501408X9011010

Minimal cover

F = BCD + A′
L = 4

Answer

L = 4
7.Number theory & modular arithmetic

Answer: x = 46

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 14

r = 83, t = 0
r = 14, q = 5, t = 1
r = 13, q = 1, t = −5
r = 1, q = 13, t = 6
14−1 ≡ 6 (mod 83)

Solve

x ≡ 6·63 = 378 ≡ 46 (mod 83)
check: 14·46 = 644 ≡ 63 (mod 83)

Answer

x = 46
8.Number theory & modular arithmetic

Answer: x = 11

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(18, 75) = 3
6x ≡ 16 (mod 25)

Solve the reduced congruence

6−1 ≡ 21 (mod 25) → x ≡ 11 (mod 25)

Solutions modulo 75

x ∈ {11, 36, 61}
x = 11

Answer

x = 11
9.Algorithms & complexity

Answer: D · 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

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

Answer: D · Reduce a known NP-hard problem to X in polynomial time

Why this is the answer

If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.

Answer

D · Reduce a known NP-hard problem to X in polynomial time
11.Data structures

Answer: C · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

C · O(1)
12.Data structures

Answer: A · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

A · Primary clustering, where long runs of occupied slots form
13.Graph theory

Answer: k = 4

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

8 − 1 = 7

Edges to remove

k = 11 − 7 = 4

Answer

k = 4
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

GD (1): take
GE (3): take
BD (6): take
DE (7): skip, closes a cycle
AB (8): take
EF (9): take
BC (13): take

Total weight

W = 1 + 3 + 6 + 8 + 9 + 13 = 40

Answer

W = 40
15.Databases & SQL

Answer: D · Atomicity

Why this is the answer

ACID: atomicity, consistency, isolation, durability. Atomicity means a transaction is applied completely or not at all.

Answer

D · Atomicity
16.Databases & SQL

Answer: C · WHERE last_name = 'Jansen'

Why this is the answer

A composite B-tree index is ordered by last_name and then first_name, so a search on the leading column works, but first_name alone cannot use the order.

Answer

C · WHERE last_name = 'Jansen'
17.Computer networks

Answer: D · Time to live (TTL)

Why this is the answer

Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.

Answer

D · Time to live (TTL)
18.Computer networks

Answer: D · Resends a segment after three duplicate ACKs, without waiting for the timeout

Why this is the answer

Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.

Answer

D · Resends a segment after three duplicate ACKs, without waiting for the timeout
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 < 1784
211 − 2 = 2046 ≥ 1784 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: p = 21

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

42 = 00101010
46 = 00101110
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.87.40.0/21 (it also covers 3 network(s) that are not in the list).

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