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 −11 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −118 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 0110 OR 1100 0011. What is the result's value R in decimal?

1000 0110OR1100 0011
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (q ∨ r) ∧ (¬q ∧ ¬r) a tautology, a contradiction or a contingency? Click your answer.

(q ∨ r) ∧ (¬q ∧ ¬r)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((r → q) → (p ∨ ¬s)) → ¬((p ∧ r) ∨ s) true? Call this number N.

((r → q) → (p ∨ ¬s)) → ¬((p ∧ r) ∨ s)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​C′I1​CI2​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 = 0, 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 = 0I₁ = C′I₂ = CI₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 7 clock edges.

Q₃Q₂Q₁Q₀ = 11007
Q =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(104), the number of integers from 1 to 104 that are coprime to 104.

n = 104 = 23·13
φ(104) =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 976 mod 74.

a = 9k = 76n = 74
976 mod 74 =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALinear
  2. BLogarithmic
  3. CConstant
  4. DExponential
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AA decision tree with n possible outcomes must have depth at least log₂ n
  2. BEvery element must be compared at least once
  3. CBinary search is the only possible algorithm, and it needs log₂ n comparisons
  4. DComparisons take logarithmic time each
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

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n²)
  2. BO(log n) exactly
  3. CO(α(n)), the inverse Ackermann function, which is practically constant
  4. DO(n)
13.Graph theoryMedium

A connected graph has 17 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 18
k =
14.Graph theoryHard
4713121015ABCDEF

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

Second normal form (2NF) removes:

  1. AAll NULL values
  2. BTransitive dependencies between non-key columns via another non-key column
  3. CPartial dependencies of non-key columns on part of a composite key
  4. DForeign keys
16.Databases & SQLHard

Why is NATURAL JOIN considered risky in production queries?

  1. AIt joins on every column with the same name, so a new column can silently change the result
  2. BIt is not supported by any database
  3. CIt always returns a Cartesian product
  4. DIt ignores primary and foreign keys and keeps only NULL rows
17.Computer networksMedium

Which medium access method did classic shared (half-duplex) Ethernet use?

  1. ACSMA/CA
  2. BToken passing
  3. CTDMA
  4. DCSMA/CD
18.Computer networksHard

A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?

  1. AFast retransmit
  2. BSlow start
  3. CThe three-way handshake
  4. DNagle's algorithm
19.IPv4 subnettingMedium
/?11111111111111111111111110000000network bitshost bits

A department needs 91 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

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

A host has the address 10.20.251.24/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.251.24/28
x =

3Solutions

1.Number systems

Answer: U = 245

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

11 = 0000 1011
inverted: 1111 0100
+1 → 1111 0101

Read it as unsigned

1111 0101₂ = 256 − 11 = 245

Answer

U = 245
2.Number systems

Answer: R = 199

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

OR gives 1 where at least one bit is 1.

1000 0110
OR 1100 0011
= 1100 0111

Value of the result

1100 0111₂ = 199

Answer

R = 199
3.Propositional logic & proofs

Answer: contradiction

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

q = 1, r = 1: q ∨ r = 1, ¬q = 0, ¬r = 0, ¬q ∧ ¬r = 0, (q ∨ r) ∧ (¬q ∧ ¬r) = 0
q = 1, r = 0: q ∨ r = 1, ¬q = 0, ¬r = 1, ¬q ∧ ¬r = 0, (q ∨ r) ∧ (¬q ∧ ¬r) = 0
q = 0, r = 1: q ∨ r = 1, ¬q = 1, ¬r = 0, ¬q ∧ ¬r = 0, (q ∨ r) ∧ (¬q ∧ ¬r) = 0
q = 0, r = 0: q ∨ r = 0, ¬q = 1, ¬r = 1, ¬q ∧ ¬r = 1, (q ∨ r) ∧ (¬q ∧ ¬r) = 0

Classify

(q ∨ r) ∧ (¬q ∧ ¬r): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: N = 9

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

r → q; ¬s; p ∨ ¬s; (r → q) → (p ∨ ¬s); p ∧ r; (p ∧ r) ∨ s; ¬((p ∧ r) ∨ s); ((r → q) → (p ∨ ¬s)) → ¬((p ∧ r) ∨ s)

Split on p

p = 1: 2
p = 0: 7
N = 2 + 7 = 9

Answer

N = 9
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

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′BC′ + AB′C

Rows with F = 1

ABC = 010, 101
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 9

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

1100 → 0110 → 0011 → 1001 → 1100 → 0110 → 0011 → 1001

Decimal value

1001₂ = 9

Answer

Q = 9
7.Number theory & modular arithmetic

Answer: φ(104) = 48

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

104 = 23·13

Totient

φ(104) = φ(8)·φ(13) = 4·12 = 48

Answer

φ(104) = 48
8.Number theory & modular arithmetic

Answer: 976 mod 74 = 49

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.

Binary exponent

76 = 10011002 = 64 + 8 + 4

Repeated squaring

91 ≡ 9 (mod 74)
92 ≡ 7 (mod 74)
94 ≡ 49 (mod 74)
98 ≡ 33 (mod 74)
916 ≡ 53 (mod 74)
932 ≡ 71 (mod 74)
964 ≡ 9 (mod 74)

Multiply the needed powers

976 ≡ 49·33·9 ≡ 49 (mod 74)

Answer

976 mod 74 = 49
9.Algorithms & complexity

Answer: D · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

D · Exponential
10.Algorithms & complexity

Answer: A · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

A · A decision tree with n possible outcomes must have depth at least log₂ n
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: C · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

C · O(α(n)), the inverse Ackermann function, which is practically constant
13.Graph theory

Answer: k = 2

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.

Edges of a spanning tree

17 − 1 = 16

Edges to remove

k = 18 − 16 = 2

Answer

k = 2
14.Graph theory

Answer: d(F) = 6

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: 5, E: 9
D (5): F: 6
F (6)
C (7)
E (9)

Shortest path

A → B → D → F
d(F) = 6

Answer

d(F) = 6
15.Databases & SQL

Answer: C · 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

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

Answer: A · It joins on every column with the same name, so a new column can silently change the result

Why this is the answer

A natural join matches all identically named columns, for example an unrelated "name" or "created_at"; an explicit ON or USING clause is safer.

Answer

A · It joins on every column with the same name, so a new column can silently change the result
17.Computer networks

Answer: D · CSMA/CD

Why this is the answer

Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.

Answer

D · CSMA/CD
18.Computer networks

Answer: D · Nagle's algorithm

Why this is the answer

Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.

Answer

D · Nagle's algorithm
19.IPv4 subnetting

Answer: p = 25

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

26 − 2 = 62 < 91
27 − 2 = 126 ≥ 91 → h = 7

Prefix length

p = 32 − 7 = 25

Answer

p = /25
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

24 = 16

Subnet of this host

network: ⌊24/16⌋·16 = 16
broadcast: 16 + 16 − 1 = 31
usable hosts: 17 … 30

Answer

x = 31

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