Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 complement1−12816413211608141211

The 8-bit pattern 1111 0111 stores a signed integer in two's complement. What is its value N in decimal?

1111 01118 bit
N =
2.Number systemsHard
32 bit1 px1920 px1080 pxM = ? KiB

An uncompressed image is 1920 × 1080 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1920 × 1080 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
pqr¬(q ↔ r)¬(q ↔ r) ∨ p((r ↔ q) ∧ p) ∨ (¬(q ↔ r) ∨ p)111?110?101?100?011?010?001?000?

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

((r ↔ q) ∧ p) ∨ (¬(q ↔ r) ∨ p)
N =
4.Propositional logic & proofsHard
¬pp ∨ qp ∧ qr ∨ q

The premises r ∨ p; ¬r ∨ q are given. Which conclusion follows from them? Click it.

r ∨ p¬r ∨ q
  1. 1¬p
  2. 2p ∨ q
  3. 3p ∧ q
  4. 4r ∨ q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = AB + AB′C + B′C′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = AB + AB′C + B′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110100113X214X507161121130151141809011010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 3, 4, 6, 8, 12, 13, 14) + d(2, 5), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 3, 4, 6, 8, 12, 13, 14) + d(2, 5)
N =
7.Number theory & modular arithmeticMedium

Solve 29x ≡ 23 (mod 83). Give the solution x with 0 < x < 83.

29x ≡ 23 (mod 83)
x =
8.Number theory & modular arithmeticHard
96c = me​ mod n?m = cd​ mod nm(n, e) = (187, 17)d

Encrypt the message m = 96 with the RSA public key (n, e) = (187, 17). Find the ciphertext c.

n = 187e = 17m = 96
c =
9.Algorithms & complexityMedium

You sort an array of n elements once and then answer k membership queries with binary search. What is the total cost?

  1. AO(k n log n)
  2. BO(nk)
  3. CO(n + k)
  4. DO(n log n + k log n)
10.Algorithms & complexityHard

Which statement about selecting the k-th smallest element of an unsorted array is correct?

  1. AQuickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
  2. BQuickselect runs in O(log n) worst case, because it recurses on only one side of the pivot
  3. CSelection always needs Ω(n log n) because it requires sorting
  4. DSelection is only possible in O(n²)
11.Data structuresMedium

The keys 50, 30, 70, 20, 40 are inserted in this order into an empty binary search tree. What is the left child of 30?

  1. A50
  2. B20
  3. C70
  4. D40
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA deque holding indices in decreasing order of value
  2. BA singly linked list of the window's elements
  3. CA min-heap rebuilt for every window
  4. DA sorted array updated by shifting
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| = 12
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
51010751110913ABCDEFG

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

What does the functional dependency A → B mean?

  1. AA is a foreign key referencing B
  2. BEach value of B is associated with exactly one value of A
  3. CEach value of A is associated with exactly one value of B
  4. DA and B are both primary keys
16.Databases & SQLHard

Table b contains the values 1, 2 and NULL in column y. What does SELECT * FROM a WHERE x NOT IN (SELECT y FROM b) return?

  1. ANo rows at all
  2. BOnly rows of a where x is NULL
  3. CEvery row of a
  4. DRows of a where x is not 1 or 2
17.Computer networksMedium

What is the purpose of the TCP three-way handshake?

  1. ATo encrypt the data
  2. BTo close a connection quickly
  3. CTo find the IP address of a host before data can be sent to it
  4. DTo establish a connection and synchronise sequence numbers
18.Computer networksHard

Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?

  1. AOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
  2. BBy routing A's frames through C
  3. CBy increasing the TTL of the frames
  4. DBy making A and C use collision detection so they stop sending when frames overlap
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 16 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2416 subnets
H =
20.IPv4 subnettingHard
last octet = 195112816403201608041211networkhost

A host has the address 10.20.167.195/26. What is the fourth octet x of its network address?

10.20.167.195/26
x =

3Solutions

1.Number systems

Answer: N = −9

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 + 1 = −9

Answer

N = −9
2.Number systems

Answer: M = 8100 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

1920·1080·32 = 66355200 bit = 8294400 B

Convert to KiB

8294400/1024 = 8100 KiB

Answer

M = 8100 KiB
3.Propositional logic & proofs

Answer: N = 6

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

Count the rows with value 1

((r ↔ q) ∧ p) ∨ (¬(q ↔ r) ∨ p): 1, 1, 1, 1, 0, 1, 1, 0
N = 6

Answer

N = 6
4.Propositional logic & proofs

Answer: p ∨ q

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.

Rows with all premises true

p = 1, q = 1, r = 1: ¬p = 0, p ∨ q = 1, p ∧ q = 1, r ∨ q = 1
p = 1, q = 1, r = 0: ¬p = 0, p ∨ q = 1, p ∧ q = 1, r ∨ q = 1
p = 1, q = 0, r = 0: ¬p = 0, p ∨ q = 1, p ∧ q = 0, r ∨ q = 0
p = 0, q = 1, r = 1: ¬p = 1, p ∨ q = 1, p ∧ q = 0, r ∨ q = 1

Conclusion

p ∨ q (resolution)

Answer

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

Answer: L = 3

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

AB → m(6, 7)
AB′C → m(5)
B′C′ → m(0, 4)
ABC00011110011001030214151716

Group and read off

F = B′C′ + A

Count the literals

L = 3

Answer

L = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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′, BD′, BC′, A′D′, A′B′C
ABCD0001111000011110100113X214X507161121130151141809011010

Minimal cover

F = C′D′ + BD′ + BC′ + A′B′C
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 38

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).

Inverse of 29

r = 83, t = 0
r = 29, q = 2, t = 1
r = 25, q = 1, t = −2
r = 4, q = 6, t = 3
r = 1, q = 4, t = −20
29−1 ≡ 63 (mod 83)

Solve

x ≡ 63·23 = 1449 ≡ 38 (mod 83)
check: 29·38 = 1102 ≡ 23 (mod 83)

Answer

x = 38
8.Number theory & modular arithmetic

Answer: c = 79

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
961 ≡ 96 (mod 187)
962 ≡ 53 (mod 187)
964 ≡ 4 (mod 187)
968 ≡ 16 (mod 187)
9616 ≡ 69 (mod 187)
9617 ≡ 96·69 ≡ 79 (mod 187)

Ciphertext

c = 79

Answer

c = 79
9.Algorithms & complexity

Answer: D · O(n log n + k log n)

Why this is the answer

Sorting costs O(n log n) and each query O(log n); this beats k linear scans (O(nk)) when k is large.

Answer

D · O(n log n + k log n)
10.Algorithms & complexity

Answer: A · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case

Why this is the answer

Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.

Answer

A · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
11.Data structures

Answer: B · 20

Why this is the answer

20 is smaller than 50 and smaller than 30, so it becomes the left child of 30; 40 becomes its right child.

Answer

B · 20
12.Data structures

Answer: A · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

A · A deque holding indices in decreasing order of value
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) = 4, deg(C) = 4, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 4

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(F) = 27

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): C: 5
C (5): B: 15, E: 15, G: 15
B (15)
E (15): D: 22
G (15)
D (22): F: 27
F (27)

Shortest path

A → C → E → D → F
d(F) = 27

Answer

d(F) = 27
15.Databases & SQL

Answer: C · Each value of A is associated with exactly one value of B

Why this is the answer

A → B means any two rows with the same A value must have the same B value; A determines B.

Answer

C · Each value of A is associated with exactly one value of B
16.Databases & SQL

Answer: A · No rows at all

Why this is the answer

The comparison with NULL is UNKNOWN, so the whole condition is never TRUE; NOT EXISTS or filtering out NULLs avoids this trap.

Answer

A · No rows at all
17.Computer networks

Answer: D · To establish a connection and synchronise sequence numbers

Why this is the answer

The handshake sets up a connection and agrees initial sequence numbers on both sides before data is sent.

Answer

D · To establish a connection and synchronise sequence numbers
18.Computer networks

Answer: A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait

Why this is the answer

A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.

Answer

A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
19.IPv4 subnetting

Answer: H = 14

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

24 = 16 ≥ 16 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 192

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

26 = 64

Subnet of this host

network: ⌊195/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254

Answer

x = 192

ExamProblems · examproblems.com/courses/computer-science-fundamentals