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 complement1−12816413211608141201

The 8-bit pattern 1111 0110 stores a signed integer in two's complement. What is its value N in decimal?

1111 01108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 1111 >> 2. What is the result's value R in decimal?

1011 1111>> 2
R =
3.Propositional logic & proofsMedium
pqr(q ∧ p) ∨ q((¬p ∧ ¬r) ∨ r) → ((q ∧ p) ∨ q)¬(((¬p ∧ ¬r) ∨ r) → ((q ∧ p) ∨ q))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is ¬(((¬p ∧ ¬r) ∨ r) → ((q ∧ p) ∨ q)) true? Call this number N.

¬(((¬p ∧ ¬r) ∨ r) → ((q ∧ p) ∨ q))
N =
4.Propositional logic & proofsHard
∀x (A(x) → ¬B(x))∃x (A(x) ∧ ¬B(x))∃x (¬A(x) ∧ B(x))∀x (A(x) ∧ ¬B(x))

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

¬∀x (A(x) → B(x))
  1. 1∀x (A(x) → ¬B(x))
  2. 2∃x (A(x) ∧ ¬B(x))
  3. 3∃x (¬A(x) ∧ B(x))
  4. 4∀x (A(x) ∧ ¬B(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011031204151716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 1, 2, 5, 6, 7). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 1, 2, 5, 6, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKT123456

A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

Q₀ = 0T
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(27), the number of integers from 1 to 27 that are coprime to 27.

n = 27 = 33
φ(27) =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (85, 17)d = ?

An RSA key uses the primes p = 5 and q = 17 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).

p = 5q = 17e = 17
d =
9.Algorithms & complexityMedium

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

  1. ALinear
  2. BLogarithmic
  3. CExponential
  4. DConstant
10.Algorithms & complexityHard

How is the travelling salesman problem classified?

  1. AThe decision version is NP-complete and the optimisation version is NP-hard
  2. BIt is undecidable
  3. CThe decision version is in P but the optimisation version is NP-complete
  4. DIt is in P because Dijkstra's algorithm solves it by visiting every city once
11.Data structuresMedium

How does extract-min work in a binary min-heap?

  1. ASearch all leaves for the new minimum
  2. BRemove the root and shift all elements one place left
  3. CMove the last element to the root and sift it down
  4. DRemove the last element of the array
12.Data structuresHard

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

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

A connected graph has 9 vertices and 12 edges. How many edges k must be removed to leave a spanning tree?

|V| = 9|E| = 12
k =
14.Graph theoryHard
53156814161918ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 3; skipped edges do not count.

k = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge GB
  8. 8edge GD
  9. 9edge GE
15.Databases & SQLMedium

Why can a table have only one clustered index?

  1. AClustered indexes can only be on the primary key
  2. BThe SQL standard forbids more than one index on a table
  3. CThe rows can be stored physically in only one order
  4. DClustered indexes use too much memory
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index on the primary key only
  2. BAn index that is rebuilt every night
  3. CAn index that covers every column of every table
  4. DAn index that contains all columns the query needs, so the table itself need not be read
17.Computer networksMedium

What is the main difference between the HTTP methods GET and POST?

  1. AGET deletes a resource, POST reads it from the server's database
  2. BGET uses UDP, POST uses TCP
  3. CGET is encrypted, POST is not
  4. DGET retrieves a resource, POST submits data in the request body
18.Computer networksHard

What does TCP fast retransmit do?

  1. AResends every segment twice to be safe
  2. BSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
  3. CResends a segment after three duplicate ACKs, without waiting for the timeout
  4. DDoubles the window after every loss
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1849 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1849 hosts
p =
20.IPv4 subnettingHard
last octet = 182112806413211608141201networkhost

A host has the address 10.20.155.182/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.155.182/29
x =

3Solutions

1.Number systems

Answer: N = −10

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 + 32 + 16 + 4 + 2 = −10

Answer

N = −10
2.Number systems

Answer: R = 47

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

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1011 1111 >> 2
= 0010 1111

Value of the result

0010 1111₂ = 47

Answer

R = 47
3.Propositional logic & proofs

Answer: N = 3

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.

Truth table row by row

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

Count the rows with value 1

¬(((¬p ∧ ¬r) ∨ r) → ((q ∧ p) ∨ q)): 0, 0, 1, 0, 0, 0, 1, 1
N = 3

Answer

N = 3
4.Propositional logic & proofs

Answer: ∃x (A(x) ∧ ¬B(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) → B(x)) ≡ ∃x (A(x) ∧ ¬B(x))

Answer

∃x (A(x) ∧ ¬B(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: N = 2

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.

ABC00011110011011031204151716

Group the 0s

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

(A + B′ + C′)
(A′ + B + C)

Minimal POS form

F = (A + B′ + C′)(A′ + B + C)
N = 2

Answer

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

Answer: N = 4

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.

Q after every edge

A T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

1: T = 1 → Q = 1
2: T = 1 → Q = 0
3: T = 0 → Q = 0
4: T = 1 → Q = 1
5: T = 0 → Q = 1
6: T = 1 → Q = 0

Count the changes

0 → 1 → 0 → 0 → 1 → 1 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: φ(27) = 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

27 = 33

Totient

φ(27) = 27 − 9 = 18

Answer

φ(27) = 18
8.Number theory & modular arithmetic

Answer: d = 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.

Modulus and totient

n = 5·17 = 85
φ(n) = 4·16 = 64

Extended Euclidean algorithm

r = 64, t = 0
r = 17, q = 3, t = 1
r = 13, q = 1, t = −3
r = 4, q = 3, t = 4
r = 1, q = 4, t = −15

Private exponent

d = 49
check: 17·49 = 833 = 13·64 + 1

Answer

d = 49
9.Algorithms & complexity

Answer: C · Exponential

Why this is the answer

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

Answer

C · Exponential
10.Algorithms & complexity

Answer: A · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

A · The decision version is NP-complete and the optimisation version is NP-hard
11.Data structures

Answer: C · Move the last element to the root and sift it down

Why this is the answer

The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).

Answer

C · Move the last element to the root and sift it down
12.Data structures

Answer: B · 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

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

Answer: k = 4

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

9 − 1 = 8

Edges to remove

k = 12 − 8 = 4

Answer

k = 4
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.

Kruskal’s algorithm

1. AC (3)
2. AB (5)
3. CE (6)
4. DE (8)
5. DF (14)
6. GB (16)

edge number 3

CE

Answer

edge CE
15.Databases & SQL

Answer: C · The rows can be stored physically in only one order

Why this is the answer

A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.

Answer

C · The rows can be stored physically in only one order
16.Databases & SQL

Answer: D · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

D · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: D · GET retrieves a resource, POST submits data in the request body

Why this is the answer

GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.

Answer

D · GET retrieves a resource, POST submits data in the request body
18.Computer networks

Answer: C · Resends a segment after three duplicate ACKs, without waiting for the timeout

Why this is the answer

Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.

Answer

C · Resends a segment after three duplicate ACKs, without waiting for the timeout
19.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1849
211 − 2 = 2046 ≥ 1849 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 183

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

23 = 8

Subnet of this host

network: ⌊182/8⌋·8 = 176
broadcast: 176 + 8 − 1 = 183
usable hosts: 177 … 182

Answer

x = 183

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