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 −3 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −38 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1100 XOR 0010 1111. What is the result's value R in decimal?

1101 1100XOR0010 1111
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is r ∨ ¬r a tautology, a contradiction or a contingency? Click your answer.

r ∨ ¬r
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
prp ∨ rp¬r11100100

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

p ∨ rp∴ ¬r
  1. 1p = 1, r = 1
  2. 2p = 1, r = 0
  3. 3p = 0, r = 1
  4. 4p = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​CI1​C′I2​CI3​ABS1​S0​F

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

I₀ = C′I₁ = CI₂ = C′I₃ = C
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-10

A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 9. Which count Q (in decimal) does it show after 19 clock pulses?

Q0 = 919N = 10
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 9 modulo 17: the number x with 0 < x < 17 and 9·x ≡ 1 (mod 17).

a = 9n = 17
9−1 mod 17 =
8.Number theory & modular arithmeticHard
111c = me​ mod n?m = cd​ mod nm(n, e) = (203, 19)d

Encrypt the message m = 111 with the RSA public key (n, e) = (203, 19). Find the ciphertext c.

n = 203e = 19m = 111
c =
9.Algorithms & complexityMedium

A greedy algorithm:

  1. AAlways gives the optimal solution
  2. BMakes the locally best choice at each step
  3. CUses only recursion
  4. DTries every possible solution
10.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is NP-hard
  2. BIt is undecidable
  3. CIt is always solvable in O(V + E)
  4. DIt can be solved by Dijkstra's algorithm with negated weights
11.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AThat keys are stored in insertion order
  2. BConstant-time search in every case
  3. CO(log n) height through rebalancing
  4. DThat no node has two children
12.Data structuresHard

Which statement about heapsort is correct?

  1. AIt needs a binary search tree as extra storage, so it uses O(n) extra memory
  2. BIt runs in O(n) time on any input
  3. CIt is stable and needs O(n) extra memory
  4. DIt sorts in place in O(n log n) worst-case time but is not stable
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDE

How many walks of length 2 go from D to E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Second normal form (2NF) removes:

  1. AAll NULL values
  2. BForeign keys
  3. CTransitive dependencies between non-key columns via another non-key column
  4. DPartial dependencies of non-key columns on part of a composite key
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. ANone, it is in BCNF
  2. BSecond normal form, because student_name depends on student_id only
  3. CThird normal form only, because grade depends on student_name
  4. DFirst normal form, because the grade column does not hold atomic values
17.Computer networksMedium

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

  1. A0.1 ms
  2. B1 ms
  3. C100 ms
  4. D10 ms
18.Computer networksHard

DNS queries normally use UDP. When does DNS typically use TCP instead?

  1. AOnly for reverse lookups of IP addresses
  2. BFor responses too large for UDP and for zone transfers
  3. COnly when the domain name is longer than ten characters
  4. DFor every query sent from a web browser
19.IPv4 subnettingMedium
/16 → /2111111111111111111111100000000000network bitshost bits

The network 172.30.0.0/16 is divided into equal subnets with prefix /21. How many subnets N are there?

172.30.0.0/16/21
N =
20.IPv4 subnettingHard
third octet = 243112816413211608041211networkhost

A host has the address 172.20.243.115/23. What is the third octet x of its subnet's broadcast address?

172.20.243.115/23
x =

3Solutions

1.Number systems

Answer: U = 253

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

3 = 0000 0011
inverted: 1111 1100
+1 → 1111 1101

Read it as unsigned

1111 1101₂ = 256 − 3 = 253

Answer

U = 253
2.Number systems

Answer: R = 243

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.

1101 1100
XOR 0010 1111
= 1111 0011

Value of the result

1111 0011₂ = 243

Answer

R = 243
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

r = 1: ¬r = 0, r ∨ ¬r = 1
r = 0: ¬r = 1, r ∨ ¬r = 1

Classify

r ∨ ¬r: 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

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

Counterexample

p = 1, r = 1

Answer

p = 1, 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′C′ + A′BC + AB′C′ + ABC

Rows with F = 1

ABC = 000, 011, 100, 111
N = 4

Answer

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

Answer: Q = 8

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.

Full cycles

Every 10 pulses bring the counter back to the same state.

19 = 1·10 + 9

Count after the remaining pulses

Q = (9 + 9) mod 10 = 8
Q = 1000₂

Answer

Q = 8
7.Number theory & modular arithmetic

Answer: 9−1 mod 17 = 2

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·9 (mod 17).

r = 17, t = 0
r = 9, q = 1, t = 1
r = 8, q = 1, t = −1
r = 1, q = 8, t = 2

Inverse

1 ≡ 2·9 (mod 17)
x = 2
check: 9·2 = 18 = 1·17 + 1

Answer

9−1 mod 17 = 2
8.Number theory & modular arithmetic

Answer: c = 181

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
1111 ≡ 111 (mod 203)
1112 ≡ 141 (mod 203)
1114 ≡ 190 (mod 203)
1118 ≡ 169 (mod 203)
11116 ≡ 141 (mod 203)
11119 ≡ 111·141·141 ≡ 181 (mod 203)

Ciphertext

c = 181

Answer

c = 181
9.Algorithms & complexity

Answer: B · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

B · Makes the locally best choice at each step
10.Algorithms & complexity

Answer: A · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

A · It is NP-hard
11.Data structures

Answer: C · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

C · O(log n) height through rebalancing
12.Data structures

Answer: D · 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

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

Answer: neither

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) = 1, deg(B) = 3, deg(C) = 1, deg(D) = 3, deg(E) = 4, deg(F) = 1, deg(G) = 3

Odd vertices

A, B, C, D, F, G → 6
neither

Answer

neither
14.Graph theory

Answer: N = 1

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 0 1 1
C: 1 0 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

Common neighbours

(M²)DE = 0·0 + 1·1 + 0·1 + 0·1 + 1·0 = 1

Answer

N = 1
15.Databases & SQL

Answer: D · Partial dependencies of non-key columns on part of a composite key

Why this is the answer

2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.

Answer

D · Partial dependencies of non-key columns on part of a composite key
16.Databases & SQL

Answer: B · 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

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

Answer: D · 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

D · 10 ms
18.Computer networks

Answer: B · For responses too large for UDP and for zone transfers

Why this is the answer

If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.

Answer

B · For responses too large for UDP and for zone transfers
19.IPv4 subnetting

Answer: N = 32

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).

Borrowed bits

21 − 16 = 5

Number of subnets

N = 25 = 32

Answer

N = 32

Each /21 subnet has 211 − 2 = 2046 usable hosts.

20.IPv4 subnetting

Answer: x = 243

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: 172.20.242.0
broadcast: 172.20.243.255

Answer

x = 243

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