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 −48 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −488 bit
U =
2.Number systemsHard
16 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(p ∧ r) → q(p → q) ∧ (r → q)p ∨ (r → q)(p → q) ∨ (r → q)

Which formula is logically equivalent to (p ∨ r) → q? Click it.

(p ∨ r) → q
  1. 1(p ∧ r) → q
  2. 2(p → q) ∧ (r → q)
  3. 3p ∨ (r → q)
  4. 4(p → q) ∨ (r → q)
4.Propositional logic & proofsHard
∃x ¬P(x)∃x P(x)∀x P(x)∀x ¬P(x)

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

¬∃x P(x)
  1. 1∃x ¬P(x)
  2. 2∃x P(x)
  3. 3∀x P(x)
  4. 4∀x ¬P(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011031214151706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 2, 4, 5, 7). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 1, 2, 4, 5, 7)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD1234567

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

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

Find the inverse of 13 modulo 54: the number x with 0 < x < 54 and 13·x ≡ 1 (mod 54).

a = 13n = 54
13−1 mod 54 =
8.Number theory & modular arithmeticHard

Solve 35x ≡ 15 (mod 40). Give the smallest positive solution x.

35x ≡ 15 (mod 40)
x =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: for j = 1 to i: do O(1) work?

  1. AO(n³)
  2. BO(n log n)
  3. CO(n)
  4. DO(n²)
10.Algorithms & complexityHard

What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?

  1. AO(n) time and O(n) space
  2. BO(log n) time and O(n) space
  3. CO(n) time and O(1) space
  4. DO(2ⁿ) time and O(1) space
11.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ALinear probing
  2. BDouble hashing
  3. CSeparate chaining
  4. DQuadratic probing
12.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA stack of all visited edges
  2. BA queue (or any collection) of vertices whose in-degree is 0
  3. CA max-heap ordered by out-degree
  4. DAn adjacency matrix recording which vertices have already been removed
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
ABCDE

How many walks of length 2 go from B to E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

What does ROLLBACK TO SAVEPOINT s1 do?

  1. ADeletes the savepoint and all tables changed since then
  2. BUndoes the changes made after savepoint s1, keeping the transaction open
  3. CEnds the transaction and commits everything
  4. DUndoes the whole transaction from the beginning
16.Databases & SQLHard

x is NULL and y = 1. What is the value of the condition (x > 5 OR y = 1)?

  1. AUNKNOWN
  2. BFALSE
  3. CTRUE
  4. DNULL, so the row is always rejected
17.Computer networksMedium

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.1.200.5
  2. B10.0.2.3
  3. C11.1.2.3
  4. D10.2.2.3
18.Computer networksHard

Which of these addresses is a public (not RFC 1918 private) IPv4 address?

  1. A172.32.0.1
  2. B172.31.0.1
  3. C192.168.255.1
  4. D10.255.0.1
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2420 subnets
H =
20.IPv4 subnettingHard
last octet = 204112816403201618140201networkhost

A host has the address 10.20.165.204/26. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.165.204/26
x =

3Solutions

1.Number systems

Answer: U = 208

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

48 = 0011 0000
inverted: 1100 1111
+1 → 1101 0000

Read it as unsigned

1101 0000₂ = 256 − 48 = 208

Answer

U = 208
2.Number systems

Answer: M = 512 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

512·512·16 = 4194304 bit = 524288 B

Convert to KiB

524288/1024 = 512 KiB

Answer

M = 512 KiB
3.Propositional logic & proofs

Answer: (p → q) ∧ (r → q)

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).

The law

(a ∨ b) → c needs c whenever either a or b holds: (a → c) ∧ (b → c).
(p ∨ r) → q ≡ (p → q) ∧ (r → q)

Check with the truth table

p = 1, q = 1, r = 1: 1 | 1 1 1 1
p = 1, q = 1, r = 0: 1 | 1 1 1 1
p = 1, q = 0, r = 1: 0 | 0 0 1 0
p = 1, q = 0, r = 0: 0 | 1 0 1 1
p = 0, q = 1, r = 1: 1 | 1 1 1 1
p = 0, q = 1, r = 0: 1 | 1 1 1 1
p = 0, q = 0, r = 1: 0 | 1 0 0 1
p = 0, q = 0, r = 0: 1 | 1 1 1 1

Answer

(p → q) ∧ (r → q)

The columns after the bar are the options in the order shown.

4.Propositional logic & proofs

Answer: ∀x ¬P(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 P(x) ≡ ∀x ¬P(x)

Answer

∀x ¬P(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 = 5

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.

ABC00011110011011031214151706

Group the 1s

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

B′
A′C′
AC

Minimal SOP form

F = B′ + A′C′ + AC
L = 5

Answer

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

Answer: N = 3

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 D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 0 → Q = 0
2: D = 0 → Q = 0
3: D = 0 → Q = 0
4: D = 1 → Q = 1
5: D = 0 → Q = 0
6: D = 1 → Q = 1
7: D = 1 → Q = 1

Count the changes

0 → 0 → 0 → 0 → 1 → 0 → 1 → 1
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 13−1 mod 54 = 25

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 54).

r = 54, t = 0
r = 13, q = 4, t = 1
r = 2, q = 6, t = −4
r = 1, q = 2, t = 25

Inverse

1 ≡ 25·13 (mod 54)
x = 25
check: 13·25 = 325 = 6·54 + 1

Answer

13−1 mod 54 = 25
8.Number theory & modular arithmetic

Answer: x = 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).

Divide by the gcd

gcd(35, 40) = 5
7x ≡ 3 (mod 8)

Solve the reduced congruence

7−1 ≡ 7 (mod 8) → x ≡ 5 (mod 8)

Solutions modulo 40

x ∈ {5, 13, 21, 29, 37}
x = 5

Answer

x = 5
9.Algorithms & complexity

Answer: D · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

D · O(n²)
10.Algorithms & complexity

Answer: C · O(n) time and O(1) space

Why this is the answer

Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.

Answer

C · O(n) time and O(1) space
11.Data structures

Answer: C · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

C · Separate chaining
12.Data structures

Answer: B · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

B · A queue (or any collection) of vertices whose in-degree is 0
13.Graph theory

Answer: H = 2520

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.

Count

H = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: N = 1

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 0 1 1
C: 1 0 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

Common neighbours

(M²)BE = 1·0 + 0·1 + 0·1 + 1·1 + 1·0 = 1

Answer

N = 1
15.Databases & SQL

Answer: B · Undoes the changes made after savepoint s1, keeping the transaction open

Why this is the answer

A savepoint marks a point inside a transaction; rolling back to it cancels later work while earlier changes remain pending until COMMIT.

Answer

B · Undoes the changes made after savepoint s1, keeping the transaction open
16.Databases & SQL

Answer: C · TRUE

Why this is the answer

x > 5 is UNKNOWN, but OR with TRUE is TRUE regardless, so the row is kept; UNKNOWN AND FALSE would likewise be FALSE.

Answer

C · TRUE
17.Computer networks

Answer: A · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

A · 10.1.200.5
18.Computer networks

Answer: A · 172.32.0.1

Why this is the answer

172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.

Answer

A · 172.32.0.1
19.IPv4 subnetting

Answer: H = 6

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

25 = 32 ≥ 20 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 255

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: ⌊204/64⌋·64 = 192
broadcast: 192 + 64 − 1 = 255
usable hosts: 193 … 254

Answer

x = 255

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