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
place values in two's complement1−12816403211608140201

The 8-bit pattern 1101 0100 stores a signed integer in two's complement. What is its value N in decimal?

1101 01008 bit
N =
2.Number systemsHard
1066 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 1066 items its own binary code of the same length. What is the smallest number of bits b needed?

1066
b = bit
3.Propositional logic & proofsMedium
(p ∧ r) ∨ qp ∨ (r ∧ q)(p ∧ r) ∨ (p ∧ q)(p ∨ r) ∧ (p ∨ q)

Which formula is logically equivalent to p ∧ (r ∨ q)? Click it.

p ∧ (r ∨ q)
  1. 1(p ∧ r) ∨ q
  2. 2p ∨ (r ∧ q)
  3. 3(p ∧ r) ∨ (p ∧ q)
  4. 4(p ∨ r) ∧ (p ∨ q)
4.Propositional logic & proofsHard
abca → cb → ca → b111110101100011010001000

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

a → cb → c∴ a → b
  1. 1a = 1, b = 1, c = 1
  2. 2a = 1, b = 1, c = 0
  3. 3a = 1, b = 0, c = 1
  4. 4a = 1, b = 0, c = 0
  5. 5a = 0, b = 1, c = 1
  6. 6a = 0, b = 1, c = 0
  7. 7a = 0, b = 0, c = 1
  8. 8a = 0, b = 0, c = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000011302040507160121130150140819111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(3, 6, 9, 11, 13). How many product terms N does it have?

F(A, B, C, D) = Σm(3, 6, 9, 11, 13)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 10 clock edges.

Q₃Q₂Q₁Q₀ = 000010
Q =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(19), the number of integers from 1 to 19 that are coprime to 19.

n = 19 = 19
φ(19) =
8.Number theory & modular arithmeticHard
209c = me​ mod n?m = cd​ mod nm(n, e) = (299, 17)d

Encrypt the message m = 209 with the RSA public key (n, e) = (299, 17). Find the ciphertext c.

n = 299e = 17m = 209
c =
9.Algorithms & complexityMedium

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

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

What did the Cook–Levin theorem establish?

  1. AThe halting problem is undecidable
  2. BP is not equal to NP
  3. CEvery NP problem can be solved in exponential time
  4. DThe Boolean satisfiability problem (SAT) is NP-complete
11.Data structuresMedium

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA binary search tree
  2. BA stack
  3. CA hash table
  4. DA queue
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA singly linked list of the window's elements
  2. BA min-heap rebuilt for every window
  3. CA sorted array updated by shifting
  4. DA deque holding indices in decreasing order of value
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
14.Graph theoryHard
97313171611ABCDE

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

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. ACombine the table with itself using UNION ALL and a WHERE filter
  2. BUse GROUP BY manager_id only
  3. CJoin the employees table with itself using two aliases
  4. DCreate a second copy of the table on disk
16.Databases & SQLHard

Table enrolment(student_id, course_id, student_name, grade) has the key (student_id, course_id). Which normal form does it violate?

  1. AFirst normal form, because the grade column does not hold atomic values
  2. BNone, it is in BCNF
  3. CThird normal form only, because grade depends on student_name
  4. DSecond normal form, because student_name depends on student_id only
17.Computer networksMedium

How does TLS typically use asymmetric and symmetric cryptography?

  1. AOnly asymmetric cryptography is used throughout
  2. BSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
  3. COnly symmetric cryptography is used, with a fixed shared password
  4. DAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
18.Computer networksHard

Head-of-line blocking in HTTP/2 over TCP occurs because:

  1. AA lost TCP segment delays all multiplexed streams until it is retransmitted
  2. BServers cannot handle parallel requests
  3. CEach request needs its own new TCP connection, so later requests wait in a queue
  4. DHTTP/2 uses UDP
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
third octet 166 … 168112806413201608141201commonvaries

A router has routes to these networks: 10.23.166.0/24, 10.23.167.0/24, 10.23.168.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.23.166.0/24 … 10.23.168.0/24
p =

3Solutions

1.Number systems

Answer: N = −44

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 16 + 4 = −44

Answer

N = −44
2.Number systems

Answer: b = 11 bit

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.

Powers of two

210 = 1024 < 1066
211 = 2048 ≥ 1066

Answer

b = 11
3.Propositional logic & proofs

Answer: (p ∧ r) ∨ (p ∧ q)

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).
p ∧ (r ∨ q) ≡ (p ∧ r) ∨ (p ∧ q)

Check with the truth table

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

Answer

(p ∧ r) ∨ (p ∧ q)

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

4.Propositional logic & proofs

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

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

Counterexample

a = 1, b = 0, c = 1

Answer

a = 1, b = 0, c = 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

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.

ABCD000111100001111000011302040507160121130150140819111010

Group the 1s

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

B′CD
AC′D
A′BCD′

Minimal SOP form

F = B′CD + AC′D + A′BCD′
N = 3

Answer

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

Answer: Q = 12

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 → 1000 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: φ(19) = 18

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

19 = 19

Totient

φ(19) = 19 − 1 = 18

Answer

φ(19) = 18
8.Number theory & modular arithmetic

Answer: c = 248

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

17 = 16 + 1
2091 ≡ 209 (mod 299)
2092 ≡ 27 (mod 299)
2094 ≡ 131 (mod 299)
2098 ≡ 118 (mod 299)
20916 ≡ 170 (mod 299)
20917 ≡ 209·170 ≡ 248 (mod 299)

Ciphertext

c = 248

Answer

c = 248
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: D · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

D · The Boolean satisfiability problem (SAT) is NP-complete
11.Data structures

Answer: B · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

B · A stack
12.Data structures

Answer: D · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

D · A deque holding indices in decreasing order of value
13.Graph theory

Answer: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
14.Graph theory

Answer: W = 34

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

BC (3): take
AC (7): take
AB (9): skip, closes a cycle
BE (11): take
BD (13): take

Total weight

W = 3 + 7 + 11 + 13 = 34

Answer

W = 34
15.Databases & SQL

Answer: C · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

C · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: D · Second normal form, because student_name depends on student_id only

Why this is the answer

student_name depends on only part of the key, which is a partial dependency; moving it to a students table fixes the 2NF violation.

Answer

D · Second normal form, because student_name depends on student_id only
17.Computer networks

Answer: D · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data

Why this is the answer

Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.

Answer

D · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
18.Computer networks

Answer: A · A lost TCP segment delays all multiplexed streams until it is retransmitted

Why this is the answer

TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.

Answer

A · A lost TCP segment delays all multiplexed streams until it is retransmitted
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: p = 20

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

166 = 10100110
168 = 10101000
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.23.160.0/20 (it also covers 13 network(s) that are not in the list).

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