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 −23 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −238 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1100 >> 3. What is the result's value R in decimal?

1101 1100>> 3
R =
3.Propositional logic & proofsMedium
a → bb ∨ ¬a¬b ∨ a¬b → ¬a

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

b → a
  1. 1a → b
  2. 2b ∨ ¬a
  3. 3¬b ∨ a
  4. 4¬b → ¬a
4.Propositional logic & proofsHard
∃x ∀y ¬R(x, y)∀x ∀y ¬R(x, y)∃x ∃y ¬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∃x ∃y ¬R(x, y)
  4. 4∀x ∃y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​CI1​C′I2​0I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = C, I₂ = C′, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = CI₂ = C′I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​FFQ4​Q4​FFQ5​Q5​CLKMOD-50

A mod-50 counter has to count through 50 different states. What is the minimum number of flip-flops n it needs?

N = 50
n =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 9) and x ≡ 7 (mod 11).

x ≡ 1 (mod 9)x ≡ 7 (mod 11)
x =
8.Number theory & modular arithmeticHard
697c = me​ mod n?m = cd​ mod nm(n, e) = (1247, 19)d

Encrypt the message m = 697 with the RSA public key (n, e) = (1247, 19). Find the ciphertext c.

n = 1247e = 19m = 697
c =
9.Algorithms & complexityMedium

About how many comparisons does binary search need in the worst case on a sorted array of 1,000,000 elements?

  1. AAbout 500,000
  2. BAbout 20
  3. CAbout 1,000
  4. DAbout 100
10.Algorithms & complexityHard

How is the travelling salesman problem classified?

  1. AIt is undecidable
  2. BThe decision version is NP-complete and the optimisation version is NP-hard
  3. CIt is in P because Dijkstra's algorithm solves it by visiting every city once
  4. DThe decision version is in P but the optimisation version is NP-complete
11.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ADouble hashing
  2. BQuadratic probing
  3. CSeparate chaining
  4. DLinear probing
12.Data structuresHard

Which property keeps every B-tree perfectly balanced?

  1. AEach node stores one key
  2. BAll leaves are at the same depth
  3. CEvery node has exactly two children
  4. DThe keys are stored only in the root
13.Graph theoryMedium

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

3 × deg 33 × deg 2
L =
14.Graph theoryHard
4712712148ABCDEF

Use Dijkstra’s algorithm to find the length d(D) of the shortest path from A to D in the weighted graph.

start: Aend: D
d(D) =
15.Databases & SQLMedium

The orders table has 5 rows, each referring to an existing customer. The customers table has 3 rows. How many rows does an INNER JOIN on customer_id return?

  1. A8
  2. B15
  3. C5
  4. D3
16.Databases & SQLHard

What is a known drawback of decomposing tables into BCNF compared with 3NF?

  1. ABCNF allows more redundancy than 3NF
  2. BSome functional dependencies may no longer be checkable within a single table
  3. CThe decomposition always loses data when joined back
  4. DBCNF tables cannot have primary keys
17.Computer networksMedium

Which TCP flag signals that a sender has finished sending data?

  1. AURG
  2. BFIN
  3. CPSH
  4. DSYN
18.Computer networksHard

In the TCP/IP model, which layer adds port numbers?

  1. AApplication layer
  2. BLink layer
  3. CTransport layer
  4. DInternet layer
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1738 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1738 hosts
p =
20.IPv4 subnettingHard
last octet = 24012806403211618040201networkhost

A host has the address 10.20.164.24/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.164.24/27
x =

3Solutions

1.Number systems

Answer: U = 233

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

23 = 0001 0111
inverted: 1110 1000
+1 → 1110 1001

Read it as unsigned

1110 1001₂ = 256 − 23 = 233

Answer

U = 233
2.Number systems

Answer: R = 27

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

Shifting right by 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1101 1100 >> 3
= 0001 1011

Value of the result

0001 1011₂ = 27

Answer

R = 27
3.Propositional logic & proofs

Answer: ¬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

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

Check with the truth table

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

Answer

¬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: N = 4

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′BC + AB′C′

Rows with F = 1

ABC = 000, 001, 011, 100
N = 4

Answer

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

Answer: n = 6

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.

Number of flip-flops

2ⁿ ≥ 50 → n = 6 (2⁶ = 64)

Answer

n = 6
7.Number theory & modular arithmetic

Answer: x = 73

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).

First congruence

x = 1 + 9k

Second congruence

1 + 9k ≡ 7 (mod 11) → 9k ≡ 6 (mod 11)
k ≡ 5·6 ≡ 8 (mod 11)

Result

x = 1 + 9·8 = 73
x = 73
check: 73 mod 9 = 1, 73 mod 11 = 7

Answer

x = 73
8.Number theory & modular arithmetic

Answer: c = 146

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.

Repeated squaring

19 = 16 + 2 + 1
6971 ≡ 697 (mod 1247)
6972 ≡ 726 (mod 1247)
6974 ≡ 842 (mod 1247)
6978 ≡ 668 (mod 1247)
69716 ≡ 1045 (mod 1247)
69719 ≡ 697·726·1045 ≡ 146 (mod 1247)

Ciphertext

c = 146

Answer

c = 146
9.Algorithms & complexity

Answer: B · About 20

Why this is the answer

220 ≈ 1,048,576, so after about 20 halvings only one element remains.

Answer

B · About 20
10.Algorithms & complexity

Answer: B · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

B · The decision version is NP-complete and the optimisation version is NP-hard
11.Data structures

Answer: C · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

C · Separate chaining
12.Data structures

Answer: B · All leaves are at the same depth

Why this is the answer

Nodes split upwards when full and the tree only gets taller when the root splits, so all leaves always stay on the same level.

Answer

B · All leaves are at the same depth
13.Graph theory

Answer: L = 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.

Handshake lemma for a tree

L + 15 = 2(L + 6) − 2
L = 5

Answer

L = 5

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

14.Graph theory

Answer: d(D) = 11

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: 4, C: 7
B (4): D: 11, E: 12
C (7)
D (11): F: 12
E (12)
F (12)

Shortest path

A → B → D
d(D) = 11

Answer

d(D) = 11
15.Databases & SQL

Answer: C · 5

Why this is the answer

Every order finds exactly one matching customer, so the join returns one row per order: 5 rows.

Answer

C · 5
16.Databases & SQL

Answer: B · Some functional dependencies may no longer be checkable within a single table

Why this is the answer

A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.

Answer

B · Some functional dependencies may no longer be checkable within a single table
17.Computer networks

Answer: B · FIN

Why this is the answer

Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.

Answer

B · FIN
18.Computer networks

Answer: C · Transport layer

Why this is the answer

TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.

Answer

C · Transport layer
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 < 1738
211 − 2 = 2046 ≥ 1738 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 31

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: ⌊24/32⌋·32 = 0
broadcast: 0 + 32 − 1 = 31
usable hosts: 1 … 30

Answer

x = 31

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