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 −116 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1168 bit
U =
2.Number systemsHard
10 bits, unsigned, N = ???????????

What is the largest unsigned integer N that fits in 10 bits?

10 bit
N =
3.Propositional logic & proofsMedium
(a ∧ b) ∨ c(a ∧ b) ∨ (a ∧ c)a ∨ (b ∧ c)(a ∨ b) ∧ (a ∨ c)

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

a ∧ (b ∨ c)
  1. 1(a ∧ b) ∨ c
  2. 2(a ∧ b) ∨ (a ∧ c)
  3. 3a ∨ (b ∧ c)
  4. 4(a ∨ b) ∧ (a ∨ c)
4.Propositional logic & proofsHard
pqrp → rq → rp → q111110101100011010001000

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

p → rq → r∴ p → 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
MUX0I0​1I1​C′I2​C′I3​ABS1​S0​F

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

I₀ = 0I₁ = 1I₂ = C′I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111010110302X41507X6012113015114X8X9011X10

For F(A, B, C, D) = Σm(0, 1, 5, 13, 14) + d(4, 6, 8, 9, 10), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 5, 13, 14) + d(4, 6, 8, 9, 10)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 48 modulo 59: the number x with 0 < x < 59 and 48·x ≡ 1 (mod 59).

a = 48n = 59
48−1 mod 59 =
8.Number theory & modular arithmeticHard
51c = me​ mod n?m = cd​ mod nm(n, e) = (247, 17)d

Encrypt the message m = 51 with the RSA public key (n, e) = (247, 17). Find the ciphertext c.

n = 247e = 17m = 51
c =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: for j = 1 to i: do O(1) work?

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

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

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

In a binary heap stored in an array starting at index 0, where is the parent of the node at index i?

  1. AAt index i − 1
  2. BAt index ⌊(i − 1) / 2⌋
  3. CAt index 2i + 1
  4. DAt index ⌊i / 2⌋ + 1
12.Data structuresHard

Which statement about heapsort is correct?

  1. AIt runs in O(n) time on any input
  2. BIt sorts in place in O(n log n) worst-case time but is not stable
  3. CIt is stable and needs O(n) extra memory
  4. DIt needs a binary search tree as extra storage, so it uses O(n) extra memory
13.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
4711592647ABCDEFG

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

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

Which query returns the number of orders per customer?

  1. ASELECT customer_id FROM orders WHERE COUNT(*) > 0
  2. BSELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
  3. CSELECT customer_id, COUNT(*) FROM orders ORDER BY customer_id
  4. DSELECT DISTINCT COUNT(*) FROM orders
16.Databases & SQLHard

An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?

  1. AStrict two-phase locking with row locks
  2. BOptimistic concurrency control
  3. CDatabase sharding
  4. DWrite-ahead logging
17.Computer networksMedium

Which protocol does the "ping" command use?

  1. ATCP
  2. BUDP
  3. CHTTP
  4. DICMP
18.Computer networksHard

What is the maximum transmission unit (MTU) of standard Ethernet payload?

  1. A9000 bytes in every network
  2. B1500 bytes
  3. C65,535 bytes
  4. D64 bytes
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/248 subnets
H =
20.IPv4 subnettingHard
third octet 43 … 50012806413201618041211commonvaries

A router has routes to these networks: 10.13.43.0/24, 10.13.44.0/24, 10.13.45.0/24, 10.13.46.0/24, 10.13.47.0/24, 10.13.48.0/24, 10.13.49.0/24, 10.13.50.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.13.43.0/24 … 10.13.50.0/24
p =

3Solutions

1.Number systems

Answer: U = 140

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

116 = 0111 0100
inverted: 1000 1011
+1 → 1000 1100

Read it as unsigned

1000 1100₂ = 256 − 116 = 140

Answer

U = 140
2.Number systems

Answer: N = 1023

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.

Largest value

210 − 1 = 1023

Answer

N = 1023
3.Propositional logic & proofs

Answer: (a ∧ b) ∨ (a ∧ c)

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 ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c)

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

Answer

(a ∧ b) ∨ (a ∧ c)

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

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: 0 0 | 1
p = 1, q = 0, r = 1: 1 1 | 0
p = 1, q = 0, r = 0: 0 1 | 0
p = 0, q = 1, r = 1: 1 1 | 1
p = 0, q = 1, r = 0: 1 0 | 1
p = 0, q = 0, r = 1: 1 1 | 1
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 = 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 + AB′C′ + ABC′

Rows with F = 1

ABC = 010, 011, 100, 110
N = 4

Answer

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

Answer: N = 1

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.

C′D, B′C′, BCD′, A′C′, ACD′
ABCD000111100001111010110302X41507X6012113015114X8X9011X10

Essential prime implicants

C′D
N = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: 48−1 mod 59 = 16

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·48 (mod 59).

r = 59, t = 0
r = 48, q = 1, t = 1
r = 11, q = 4, t = −1
r = 4, q = 2, t = 5
r = 3, q = 1, t = −11
r = 1, q = 3, t = 16

Inverse

1 ≡ 16·48 (mod 59)
x = 16
check: 48·16 = 768 = 13·59 + 1

Answer

48−1 mod 59 = 16
8.Number theory & modular arithmetic

Answer: c = 155

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
511 ≡ 51 (mod 247)
512 ≡ 131 (mod 247)
514 ≡ 118 (mod 247)
518 ≡ 92 (mod 247)
5116 ≡ 66 (mod 247)
5117 ≡ 51·66 ≡ 155 (mod 247)

Ciphertext

c = 155

Answer

c = 155
9.Algorithms & complexity

Answer: D · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

D · O(n²)
10.Algorithms & complexity

Answer: A · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

A · O(n log n)
11.Data structures

Answer: B · At index ⌊(i − 1) / 2⌋

Why this is the answer

Children of i are at 2i + 1 and 2i + 2, so the parent of i is ⌊(i − 1) / 2⌋; with 1-based indexing it is ⌊i / 2⌋.

Answer

B · At index ⌊(i − 1) / 2⌋
12.Data structures

Answer: B · It sorts in place in O(n log n) worst-case time but is not stable

Why this is the answer

Heapsort builds a max-heap and extracts the maximum n times within the same array; swaps over long distances can reorder equal keys.

Answer

B · It sorts in place in O(n log n) worst-case time but is not stable
13.Graph theory

Answer: Euler circuit

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.

Degrees

deg(A) = 2, deg(B) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2, deg(F) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(E) = 13

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): C: 5, G: 6
C (5)
G (6): D: 10, E: 13
D (10): F: 19
E (13)
F (19)

Shortest path

A → B → G → E
d(E) = 13

Answer

d(E) = 13
15.Databases & SQL

Answer: B · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id

Why this is the answer

GROUP BY customer_id makes one group per customer, and COUNT(*) counts the rows in each group.

Answer

B · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
16.Databases & SQL

Answer: B · Optimistic concurrency control

Why this is the answer

Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.

Answer

B · Optimistic concurrency control
17.Computer networks

Answer: D · ICMP

Why this is the answer

Ping sends ICMP echo requests and measures the time until echo replies return.

Answer

D · ICMP
18.Computer networks

Answer: B · 1500 bytes

Why this is the answer

Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.

Answer

B · 1500 bytes
19.IPv4 subnetting

Answer: H = 30

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).

Bits to borrow

23 = 8 ≥ 8 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: p = 19

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

43 = 00101011
50 = 00110010
common leading bits: 3

Prefix length

p = 16 + 3 = 19

Answer

p = /19

Summary route: 10.13.32.0/19 (it also covers 24 network(s) that are not in the list).

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