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 −52 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −528 bit
U =
2.Number systemsHard
32 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

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

¬((p ∧ q) → p)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x ¬A(x)∃x ¬A(x)∀x A(x)∃x A(x)

Which statement is equivalent to ¬∃x A(x), with the negation moved all the way inside? Click it.

¬∃x A(x)
  1. 1∀x ¬A(x)
  2. 2∃x ¬A(x)
  3. 3∀x A(x)
  4. 4∃x A(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 1, 2, 5, 13, 14)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKUP

A 4-bit binary up counter starts at 13 (decimal, Q = 1101). Which count Q (in decimal) does it show after 37 clock pulses?

Q0 = 1337n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 13 modulo 64: the number x with 0 < x < 64 and 13·x ≡ 1 (mod 64).

a = 13n = 64
13−1 mod 64 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (1591, 23)d = ?

An RSA key uses the primes p = 37 and q = 43 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 37q = 43e = 23
d =
9.Algorithms & complexityMedium

How many levels does the recursion of merge sort have for an array of n elements?

  1. AAbout n / 2
  2. BAbout n
  3. CAbout log₂ n
  4. DAbout √n
10.Algorithms & complexityHard

Which definition of the class P is correct?

  1. AProblems that need exponential time on every computer
  2. BProblems whose solutions can be checked but never found in polynomial time
  3. CProblems that can be solved in parallel
  4. DDecision problems solvable by a deterministic algorithm in polynomial time
11.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the postorder traversal?

  1. A1 2 4 5 3
  2. B4 5 2 3 1
  3. C5 4 3 2 1
  4. D4 2 5 1 3
12.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. AAnswers only for numbers
  2. BPossible false negatives but no false positives
  3. CPossible false positives but no false negatives
  4. DExact answers always
13.Graph theoryMedium

A connected graph has 14 vertices and 14 edges. How many edges k must be removed to leave a spanning tree?

|V| = 14|E| = 14
k =
14.Graph theoryHard
10142013178ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 2.

start: Dk = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

Which ACID property guarantees that committed changes survive a system crash?

  1. AConsistency
  2. BDurability
  3. CIsolation
  4. DAtomicity
16.Databases & SQLHard

Two-phase locking (2PL) ensures:

  1. AConflict-serializable schedules
  2. BThat no locks are needed
  3. CThat deadlocks can never happen
  4. DThat transactions never wait
17.Computer networksMedium

Using even parity, which parity bit is added to the data bits 1011001?

  1. A1
  2. B0
  3. C2
  4. D4
18.Computer networksHard

A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?

  1. ADNS needs ICMP to resolve names
  2. BPath MTU discovery fails because "fragmentation needed" messages are blocked
  3. CTLS certificates are sent over ICMP
  4. DTCP cannot open connections without ICMP
19.IPv4 subnettingMedium
/16 → /2411111111111111111111111100000000network bitshost bits

The network 172.24.0.0/16 is divided into equal subnets with prefix /24. How many subnets N are there?

172.24.0.0/16/24
N =
20.IPv4 subnettingHard
third octet = 184112806413211618040201networkhost

A host has the address 172.20.184.221/22. What is the third octet x of its network address?

172.20.184.221/22
x =

3Solutions

1.Number systems

Answer: U = 204

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

52 = 0011 0100
inverted: 1100 1011
+1 → 1100 1100

Read it as unsigned

1100 1100₂ = 256 − 52 = 204

Answer

U = 204
2.Number systems

Answer: M = 1024 KiB

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.

Bits and bytes

512·512·32 = 8388608 bit = 1048576 B

Convert to KiB

1048576/1024 = 1024 KiB

Answer

M = 1024 KiB
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

p = 1, q = 1: p ∧ q = 1, (p ∧ q) → p = 1, ¬((p ∧ q) → p) = 0
p = 1, q = 0: p ∧ q = 0, (p ∧ q) → p = 1, ¬((p ∧ q) → p) = 0
p = 0, q = 1: p ∧ q = 0, (p ∧ q) → p = 1, ¬((p ∧ q) → p) = 0
p = 0, q = 0: p ∧ q = 0, (p ∧ q) → p = 1, ¬((p ∧ q) → p) = 0

Classify

¬((p ∧ q) → p): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ∀x ¬A(x)

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.

Move the negation inwards

¬∃x A(x) ≡ ∀x ¬A(x)

Answer

∀x ¬A(x)

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: AB = 11, CD = 10

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m14: m15, m12, m10, m6 → 0
ABCD0001111000011110111111

The isolated 1

m14 = ABCD′ → AB = 11, CD = 10

Answer

AB = 11, CD = 10
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 2

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 16 pulses bring the counter back to the same state.

37 = 2·16 + 5

Count after the remaining pulses

Q = (13 + 5) mod 16 = 2
Q = 0010₂

Answer

Q = 2
7.Number theory & modular arithmetic

Answer: 13−1 mod 64 = 5

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·13 (mod 64).

r = 64, t = 0
r = 13, q = 4, t = 1
r = 12, q = 1, t = −4
r = 1, q = 12, t = 5

Inverse

1 ≡ 5·13 (mod 64)
x = 5
check: 13·5 = 65 = 1·64 + 1

Answer

13−1 mod 64 = 5
8.Number theory & modular arithmetic

Answer: d = 263

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.

Modulus and totient

n = 37·43 = 1591
φ(n) = 36·42 = 1512

Extended Euclidean algorithm

r = 1512, t = 0
r = 23, q = 65, t = 1
r = 17, q = 1, t = −65
r = 6, q = 2, t = 66
r = 5, q = 1, t = −197
r = 1, q = 5, t = 263

Private exponent

d = 263
check: 23·263 = 6049 = 4·1512 + 1

Answer

d = 263
9.Algorithms & complexity

Answer: C · About log₂ n

Why this is the answer

Halving from n down to 1 takes about log₂ n levels, and each level does O(n) merging work, giving O(n log n).

Answer

C · About log₂ n
10.Algorithms & complexity

Answer: D · Decision problems solvable by a deterministic algorithm in polynomial time

Why this is the answer

P contains decision problems with a deterministic polynomial-time algorithm, such as checking whether a graph is connected.

Answer

D · Decision problems solvable by a deterministic algorithm in polynomial time
11.Data structures

Answer: B · 4 5 2 3 1

Why this is the answer

Postorder processes 4 and 5 before 2, then 3, and finally the root 1.

Answer

B · 4 5 2 3 1
12.Data structures

Answer: C · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

C · Possible false positives but no false negatives
13.Graph theory

Answer: k = 1

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

14 − 1 = 13

Edges to remove

k = 14 − 13 = 1

Answer

k = 1
14.Graph theory

Answer: edge CE

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.

Prim’s algorithm

1. DE (7)
2. CE (1)
3. BE (8)
4. AB (10)

edge number 2

CE

Answer

edge CE
15.Databases & SQL

Answer: B · Durability

Why this is the answer

Durability means that once a transaction commits, its effects are kept, usually by writing a log to stable storage.

Answer

B · Durability
16.Databases & SQL

Answer: A · Conflict-serializable schedules

Why this is the answer

2PL guarantees serializability, but deadlocks remain possible and must be detected or prevented.

Answer

A · Conflict-serializable schedules
17.Computer networks

Answer: B · 0

Why this is the answer

1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.

Answer

B · 0
18.Computer networks

Answer: B · Path MTU discovery fails because "fragmentation needed" messages are blocked

Why this is the answer

PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.

Answer

B · Path MTU discovery fails because "fragmentation needed" messages are blocked
19.IPv4 subnetting

Answer: N = 256

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

24 − 16 = 8

Number of subnets

N = 28 = 256

Answer

N = 256

Each /24 subnet has 28 − 2 = 254 usable hosts.

20.IPv4 subnetting

Answer: x = 184

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

22 = 4

Subnet of this host

network: 172.20.184.0
broadcast: 172.20.187.255

Answer

x = 184

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