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 −105 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1058 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1100 1111 << 3. What is the result's value R in decimal?

1100 1111<< 3
R =
3.Propositional logic & proofsMedium
¬p ∨ r¬p ∨ ¬r¬p ∧ r¬p ∧ ¬r

Which formula is logically equivalent to ¬(p ∧ r)? Click it.

¬(p ∧ r)
  1. 1¬p ∨ r
  2. 2¬p ∨ ¬r
  3. 3¬p ∧ r
  4. 4¬p ∧ ¬r
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
ABCD000111100001111000110312141507161121130151140819011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 2, 4, 5, 6, 9, 10, 12, 13, 14). How many product terms N does it have?

F(A, B, C, D) = Σm(1, 2, 4, 5, 6, 9, 10, 12, 13, 14)
N =
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 9 (decimal, Q = 1001). Which count Q (in decimal) does it show after 38 clock pulses?

Q0 = 938n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 19 modulo 23: the number x with 0 < x < 23 and 19·x ≡ 1 (mod 23).

a = 19n = 23
19−1 mod 23 =
8.Number theory & modular arithmeticHard
640c = me​ mod n?m = cd​ mod nm(n, e) = (1073, 5)d

Encrypt the message m = 640 with the RSA public key (n, e) = (1073, 5). Find the ciphertext c.

n = 1073e = 5m = 640
c =
9.Algorithms & complexityMedium

Which sorting algorithm is stable?

  1. ASelection sort (typical version)
  2. BMerge sort
  3. CQuicksort (typical in-place version)
  4. DHeap sort
10.Algorithms & complexityHard

Which definition of the class P is correct?

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

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(n²)
  2. BO(log n)
  3. CO(1)
  4. DO(n)
12.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(1) time and O(1) extra space
  2. BO(n) time and O(1) extra space
  3. CO(n) time and O(n) extra space
  4. DO(n log n) time and O(1) extra space
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
149111013413ABCDEF

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

A table stores the customer's address on every order row. Changing the address in one row but not in the others is an example of:

  1. AAn update anomaly
  2. BAn insertion anomaly
  3. CA phantom read
  4. DA deletion anomaly
16.Databases & SQLHard

With multiversion concurrency control (MVCC), how can readers avoid blocking writers?

  1. AReaders read uncommitted data directly
  2. BReaders always wait until all writers have committed
  3. CWriters are not allowed while any reader is active
  4. DReaders see a consistent snapshot of committed row versions while writers create new versions
17.Computer networksMedium

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

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

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt replaces TCP with a faster transport protocol
  2. BIt removes the need for DNS lookups
  3. CIt serves cached content from servers close to the users
  4. DIt increases the bandwidth of the users' own home connections
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.88.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 9?

192.168.88.0/24/29subnet 9 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 51012806413211608041211networkhost

A host has the address 10.20.60.51/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.60.51/29
x =

3Solutions

1.Number systems

Answer: U = 151

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

105 = 0110 1001
inverted: 1001 0110
+1 → 1001 0111

Read it as unsigned

1001 0111₂ = 256 − 105 = 151

Answer

U = 151
2.Number systems

Answer: R = 120

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.

1100 1111 << 3
= 0111 1000

Value of the result

0111 1000₂ = 120

Answer

R = 120
3.Propositional logic & proofs

Answer: ¬p ∨ ¬r

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

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(p ∧ r) ≡ ¬p ∨ ¬r

Check with the truth table

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

Answer

¬p ∨ ¬r

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

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: N = 3

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.

ABCD000111100001111000110312141507161121130151140819011110

Group the 1s

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

C′D
CD′
BD′

Minimal SOP form

F = C′D + CD′ + BD′
N = 3

Answer

N = 3

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: Q = 15

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.

38 = 2·16 + 6

Count after the remaining pulses

Q = (9 + 6) mod 16 = 15
Q = 1111₂

Answer

Q = 15
7.Number theory & modular arithmetic

Answer: 19−1 mod 23 = 17

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·19 (mod 23).

r = 23, t = 0
r = 19, q = 1, t = 1
r = 4, q = 4, t = −1
r = 3, q = 1, t = 5
r = 1, q = 3, t = −6

Inverse

1 ≡ −6·19 (mod 23)
x = 17
check: 19·17 = 323 = 14·23 + 1

Answer

19−1 mod 23 = 17
8.Number theory & modular arithmetic

Answer: c = 989

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

5 = 4 + 1
6401 ≡ 640 (mod 1073)
6402 ≡ 787 (mod 1073)
6404 ≡ 248 (mod 1073)
6405 ≡ 640·248 ≡ 989 (mod 1073)

Ciphertext

c = 989

Answer

c = 989
9.Algorithms & complexity

Answer: B · Merge sort

Why this is the answer

Merge sort is stable when merging takes from the left half first on ties. In-place quicksort, heap sort and selection sort are usually not stable.

Answer

B · Merge sort
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 · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

C · O(1)
12.Data structures

Answer: B · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

B · O(n) time and O(1) extra space
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: d(F) = 16

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): B: 1, C: 4
B (1): D: 12
C (4): E: 14
D (12): F: 16
E (14)
F (16)

Shortest path

A → B → D → F
d(F) = 16

Answer

d(F) = 16
15.Databases & SQL

Answer: A · An update anomaly

Why this is the answer

Redundant storage means one change must be made in many rows; missing some leaves the data inconsistent, which normalisation prevents.

Answer

A · An update anomaly
16.Databases & SQL

Answer: D · Readers see a consistent snapshot of committed row versions while writers create new versions

Why this is the answer

MVCC keeps old row versions so a reader can see the data as of its snapshot, while a writer creates a new version instead of overwriting in place.

Answer

D · Readers see a consistent snapshot of committed row versions while writers create new versions
17.Computer networks

Answer: B · 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

B · 10.1.200.5
18.Computer networks

Answer: C · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

C · It serves cached content from servers close to the users
19.IPv4 subnetting

Answer: x = 64

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 number 9

network: (9 − 1)·8 = 64
broadcast: 64 + 8 − 1 = 71

Answer

x = 64
20.IPv4 subnetting

Answer: x = 55

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: ⌊51/8⌋·8 = 48
broadcast: 48 + 8 − 1 = 55
usable hosts: 49 … 54

Answer

x = 55

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