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 complement?−128?64?32?16?8?4?2?1

Write −82 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −828 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1010 1011 << 3. What is the result's value R in decimal?

1010 1011<< 3
R =
3.Propositional logic & proofsMedium
abcb ↔ ca ∧ (b ↔ c)((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c))111?110?101?100?011?010?001?000?

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

((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c))
N =
4.Propositional logic & proofsHard
∀x (F(x) → ¬G(x))∃x (F(x) ∧ ¬G(x))∃x (¬F(x) ∧ G(x))∃x (F(x) → ¬G(x))

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

¬∀x (F(x) → G(x))
  1. 1∀x (F(x) → ¬G(x))
  2. 2∃x (F(x) ∧ ¬G(x))
  3. 3∃x (¬F(x) ∧ G(x))
  4. 4∃x (F(x) → ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B′C + BC + A′C
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKSHIFT

A 4-bit shift register Q₃Q₂Q₁Q₀ holds 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀, Q₀ is lost) and the serial input enters Q₃. The serial input is 0, 1, 0, 0 at the 4 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.

Q₃Q₂Q₁Q₀ = 11000, 1, 0, 0
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 8 modulo 15: the number x with 0 < x < 15 and 8·x ≡ 1 (mod 15).

a = 8n = 15
8−1 mod 15 =
8.Number theory & modular arithmeticHard

Solve 77x ≡ 22 (mod 88). Give the smallest positive solution x.

77x ≡ 22 (mod 88)
x =
9.Algorithms & complexityMedium

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. AYes, because Big O is an upper bound and n² grows no faster than n³
  2. BYes, but only if n is smaller than 100
  3. CNo, because Big O must always state the exact growth rate of the function
  4. DNo, because n³ is smaller than n² for large n
10.Algorithms & complexityHard

Which definition of the class P is correct?

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

A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?

  1. AA binary heap of edges
  2. BA single sorted array of vertices
  3. CAdjacency list
  4. DAdjacency matrix
12.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CBecause halving at one half would lose elements
  4. DTo keep the array sorted
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| = 11
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
57188102119ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 8
W =
15.Databases & SQLMedium

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

  1. ADurability
  2. BConsistency
  3. CAtomicity
  4. DIsolation
16.Databases & SQLHard

Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?

  1. A1, 1, 3
  2. B1, 1, 2
  3. C2, 2, 3
  4. D1, 2, 3
17.Computer networksMedium

How many usable host addresses does a /26 IPv4 subnet have?

  1. A30
  2. B126
  3. C62
  4. D64
18.Computer networksHard

An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?

  1. A1 collision domain, 8 broadcast domains
  2. B8 collision domains, 8 broadcast domains
  3. C8 collision domains, 1 broadcast domain
  4. D1 collision domain, 1 broadcast domain
19.IPv4 subnettingMedium
255.255.255.24011111111111111111111111111110000network bitshost bits

A subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?

255.255.255.240
H =
20.IPv4 subnettingHard
last octet = 99012816413201608041211networkhost

A host has the address 10.20.98.99/25. What is the fourth octet x of the last usable host address in its subnet?

10.20.98.99/25
x =

3Solutions

1.Number systems

Answer: U = 174

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.

Invert and add one

82 = 0101 0010
inverted: 1010 1101
+1 → 1010 1110

Read it as unsigned

1010 1110₂ = 256 − 82 = 174

Answer

U = 174
2.Number systems

Answer: R = 88

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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1010 1011 << 3
= 0101 1000

Value of the result

0101 1000₂ = 88

Answer

R = 88
3.Propositional logic & proofs

Answer: N = 5

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

a = 1, b = 1, c = 1: a ∧ b = 1, ¬c = 0, (a ∧ b) ∨ ¬c = 1, b ↔ c = 1, a ∧ (b ↔ c) = 1, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 1
a = 1, b = 1, c = 0: a ∧ b = 1, ¬c = 1, (a ∧ b) ∨ ¬c = 1, b ↔ c = 0, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 1
a = 1, b = 0, c = 1: a ∧ b = 0, ¬c = 0, (a ∧ b) ∨ ¬c = 0, b ↔ c = 0, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 0
a = 1, b = 0, c = 0: a ∧ b = 0, ¬c = 1, (a ∧ b) ∨ ¬c = 1, b ↔ c = 1, a ∧ (b ↔ c) = 1, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 1
a = 0, b = 1, c = 1: a ∧ b = 0, ¬c = 0, (a ∧ b) ∨ ¬c = 0, b ↔ c = 1, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 0
a = 0, b = 1, c = 0: a ∧ b = 0, ¬c = 1, (a ∧ b) ∨ ¬c = 1, b ↔ c = 0, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 1
a = 0, b = 0, c = 1: a ∧ b = 0, ¬c = 0, (a ∧ b) ∨ ¬c = 0, b ↔ c = 0, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 0
a = 0, b = 0, c = 0: a ∧ b = 0, ¬c = 1, (a ∧ b) ∨ ¬c = 1, b ↔ c = 1, a ∧ (b ↔ c) = 0, ((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)) = 1

Count the rows with value 1

((a ∧ b) ∨ ¬c) ∨ (a ∧ (b ↔ c)): 1, 1, 0, 1, 0, 1, 0, 1
N = 5

Answer

N = 5
4.Propositional logic & proofs

Answer: ∃x (F(x) ∧ ¬G(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 (F(x) → G(x)) ≡ ∃x (F(x) ∧ ¬G(x))

Answer

∃x (F(x) ∧ ¬G(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: L = 4

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

A′B′C → m(1)
BC → m(3, 7)
A′C → m(1, 3)
ABC00011110010011130204051706

Group and read off

F = BC + A′C

Count the literals

L = 4

Answer

L = 4
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.

State after every edge

1100 → 0110 → 1011 → 0101 → 0010

Decimal value

0010₂ = 2

Answer

Q = 2
7.Number theory & modular arithmetic

Answer: 8−1 mod 15 = 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·8 (mod 15).

r = 15, t = 0
r = 8, q = 1, t = 1
r = 7, q = 1, t = −1
r = 1, q = 7, t = 2

Inverse

1 ≡ 2·8 (mod 15)
x = 2
check: 8·2 = 16 = 1·15 + 1

Answer

8−1 mod 15 = 2
8.Number theory & modular arithmetic

Answer: x = 6

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).

Divide by the gcd

gcd(77, 88) = 11
7x ≡ 2 (mod 8)

Solve the reduced congruence

7−1 ≡ 7 (mod 8) → x ≡ 6 (mod 8)

Solutions modulo 88

x ∈ {6, 14, 22, 30, 38, 46, 54, 62, 70, 78, 86}
x = 6

Answer

x = 6
9.Algorithms & complexity

Answer: A · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

A · Yes, because Big O is an upper bound and n² grows no faster than n³
10.Algorithms & complexity

Answer: A · 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

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

Answer: C · Adjacency list

Why this is the answer

A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.

Answer

C · Adjacency list
12.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
13.Graph theory

Answer: Euler path, no 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) = 3, deg(F) = 2, deg(G) = 3

Odd vertices

E, G → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: W = 32

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

DF (2): take
AB (5): take
AC (7): take
CE (8): take
BE (9): skip, closes a cycle
DE (10): take

Total weight

W = 2 + 5 + 7 + 8 + 10 = 32

Answer

W = 32
15.Databases & SQL

Answer: A · 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

A · Durability
16.Databases & SQL

Answer: A · 1, 1, 3

Why this is the answer

RANK gives tied rows the same rank and skips the following numbers (1, 1, 3); DENSE_RANK would give 1, 1, 2.

Answer

A · 1, 1, 3
17.Computer networks

Answer: C · 62

Why this is the answer

A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.

Answer

C · 62
18.Computer networks

Answer: C · 8 collision domains, 1 broadcast domain

Why this is the answer

Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.

Answer

C · 8 collision domains, 1 broadcast domain
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).

Prefix length

255.255.255.240 = /28

Usable hosts

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 126

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

27 = 128

Subnet of this host

network: ⌊99/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126

Answer

x = 126

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