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 −94 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −948 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 0100 AND 0110 1011. What is the result's value R in decimal?

0111 0100AND0110 1011
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ((c ∨ b) ∧ ¬c) → b a tautology, a contradiction or a contingency? Click your answer.

((c ∨ b) ∧ ¬c) → b
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
pqrr → pq → pr → q111110101100011010001000

The argument with premises r → p; q → p and conclusion r → q is invalid. Click the row of the truth table that is a counterexample.

r → pq → p∴ r → q
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​C′I1​0I2​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 = 0, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = C′I₂ = 0I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​DQ4​Q4​DQ5​Q5​DQ6​Q6​DQ7​Q7​CLKJOHNSON

A Johnson (twisted-ring) counter is made of 8 D flip-flops, with the inverted last output fed back to the first input, and starts at all zeros. How many different states N does it go through?

n = 8
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 15 modulo 41: the number x with 0 < x < 41 and 15·x ≡ 1 (mod 41).

a = 15n = 41
15−1 mod 41 =
8.Number theory & modular arithmeticHard

83 is prime. Use Fermat’s little theorem to find 429087 mod 83.

p = 83k = 29087
429087 mod 83 =
9.Algorithms & complexityMedium

What is the time complexity of multiplying two n × n matrices with the standard triple loop?

  1. AO(n)
  2. BO(n³)
  3. CO(n²)
  4. DO(n² log n)
10.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(2n²)
  2. BO(n³)
  3. CO(n² 2ⁿ)
  4. DO(n!)
11.Data structuresMedium

How does extract-min work in a binary min-heap?

  1. AMove the last element to the root and sift it down
  2. BRemove the last element of the array
  3. CRemove the root and shift all elements one place left
  4. DSearch all leaves for the new minimum
12.Data structuresHard

Why do databases often use B+ trees rather than plain B-trees for indexes?

  1. AB+ trees use binary internal nodes, which reduces the number of disk reads
  2. BB+ trees do not need to be balanced
  3. CAll records are in linked leaves, which makes range scans efficient
  4. DB+ trees store data only in the root
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 4 vertices of degree 4, 4 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 34 × deg 44 × deg 5
L =
14.Graph theoryHard
2136715848ABCDEF

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

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. A5
  2. B15
  3. C8
  4. D3
16.Databases & SQLHard

Table b contains the values 1, 2 and NULL in column y. What does SELECT * FROM a WHERE x NOT IN (SELECT y FROM b) return?

  1. AOnly rows of a where x is NULL
  2. BRows of a where x is not 1 or 2
  3. CNo rows at all
  4. DEvery row of a
17.Computer networksMedium

A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?

  1. A10 ms
  2. B1 ms
  3. C0.1 ms
  4. D100 ms
18.Computer networksHard

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

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

A department needs 1449 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1449 hosts
p =
20.IPv4 subnettingHard
third octet = 3012806403201608041211networkhost

A host has the address 172.20.3.224/23. What is the third octet x of its subnet's broadcast address?

172.20.3.224/23
x =

3Solutions

1.Number systems

Answer: U = 162

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

94 = 0101 1110
inverted: 1010 0001
+1 → 1010 0010

Read it as unsigned

1010 0010₂ = 256 − 94 = 162

Answer

U = 162
2.Number systems

Answer: R = 96

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

AND gives 1 only where both bits are 1.

0111 0100
AND 0110 1011
= 0110 0000

Value of the result

0110 0000₂ = 96

Answer

R = 96
3.Propositional logic & proofs

Answer: tautology

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).

Truth table row by row

b = 1, c = 1: c ∨ b = 1, ¬c = 0, (c ∨ b) ∧ ¬c = 0, ((c ∨ b) ∧ ¬c) → b = 1
b = 1, c = 0: c ∨ b = 1, ¬c = 1, (c ∨ b) ∧ ¬c = 1, ((c ∨ b) ∧ ¬c) → b = 1
b = 0, c = 1: c ∨ b = 1, ¬c = 0, (c ∨ b) ∧ ¬c = 0, ((c ∨ b) ∧ ¬c) → b = 1
b = 0, c = 0: c ∨ b = 0, ¬c = 1, (c ∨ b) ∧ ¬c = 0, ((c ∨ b) ∧ ¬c) → b = 1

Classify

((c ∨ b) ∧ ¬c) → b: 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: p = 1, q = 0, r = 1

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.

Premises and conclusion row by row

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

Counterexample

p = 1, q = 0, r = 1

Answer

p = 1, q = 0, r = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: N = 3

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′

Rows with F = 1

ABC = 000, 001, 010
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 16

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.

Sequence for 8 flip-flops

00000000 → 10000000 → 11000000 → 11100000 → 11110000 → 11111000 → 11111100 → 11111110 → 11111111 → 01111111 → 00111111 → 00011111 → 00001111 → 00000111 → 00000011 → 00000001

Number of states

N = 16

Answer

N = 16
7.Number theory & modular arithmetic

Answer: 15−1 mod 41 = 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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·15 (mod 41).

r = 41, t = 0
r = 15, q = 2, t = 1
r = 11, q = 1, t = −2
r = 4, q = 2, t = 3
r = 3, q = 1, t = −8
r = 1, q = 3, t = 11

Inverse

1 ≡ 11·15 (mod 41)
x = 11
check: 15·11 = 165 = 4·41 + 1

Answer

15−1 mod 41 = 11
8.Number theory & modular arithmetic

Answer: 429087 mod 83 = 70

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.

Reduce the exponent

29087 mod 82 = 59
429087 ≡ 459 (mod 83)

Small power

459 mod 83 = 70

Answer

429087 mod 83 = 70
9.Algorithms & complexity

Answer: B · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

B · O(n³)
10.Algorithms & complexity

Answer: C · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

C · O(n² 2ⁿ)
11.Data structures

Answer: A · Move the last element to the root and sift it down

Why this is the answer

The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).

Answer

A · Move the last element to the root and sift it down
12.Data structures

Answer: C · All records are in linked leaves, which makes range scans efficient

Why this is the answer

In a B+ tree internal nodes only guide the search and the leaves are linked in key order, so a range query finds the first key and then walks along the leaves.

Answer

C · All records are in linked leaves, which makes range scans efficient
13.Graph theory

Answer: L = 26

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 + 48 = 2(L + 12) − 2
L = 26

Answer

L = 26

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

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: 2
B (2): C: 15, D: 8, E: 10
D (8): F: 16
E (10): F: 14
F (14)
C (15)

Shortest path

A → B → E → F
d(F) = 14

Answer

d(F) = 14
15.Databases & SQL

Answer: A · 5

Why this is the answer

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

Answer

A · 5
16.Databases & SQL

Answer: C · No rows at all

Why this is the answer

The comparison with NULL is UNKNOWN, so the whole condition is never TRUE; NOT EXISTS or filtering out NULLs avoids this trap.

Answer

C · No rows at all
17.Computer networks

Answer: A · 10 ms

Why this is the answer

2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.

Answer

A · 10 ms
18.Computer networks

Answer: D · Transport layer

Why this is the answer

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

Answer

D · 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 < 1449
211 − 2 = 2046 ≥ 1449 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 3

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 in the third octet

21 = 2

Subnet of this host

network: 172.20.2.0
broadcast: 172.20.3.255

Answer

x = 3

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