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 −115 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1158 bit
U =
2.Number systemsHard
24 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (a → c) ∧ (a ∧ ¬c) a tautology, a contradiction or a contingency? Click your answer.

(a → c) ∧ (a ∧ ¬c)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
pr ∧ pr¬p

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

r ∨ p¬r
  1. 1p
  2. 2r ∧ p
  3. 3r
  4. 4¬p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011031204050716

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

F(A, B, C) = Σm(1, 2, 6)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK123456

A positive-edge-triggered JK 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 = 0JK
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 4 modulo 19: the number x with 0 < x < 19 and 4·x ≡ 1 (mod 19).

a = 4n = 19
4−1 mod 19 =
8.Number theory & modular arithmeticHard
mc = me​ mod n251m = cd​ mod n?(n, e) = (703, 13)d

An RSA key uses p = 19, q = 37 and e = 13. Decrypt the ciphertext c = 251: find the message m.

p = 19q = 37e = 13c = 251
m =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the activity with the fewest overlaps first, then repeat with the compatible ones
  2. BPick the activity that finishes earliest, then repeat with the compatible ones
  3. CPick the shortest activity first
  4. DPick the activity that starts earliest
10.Algorithms & complexityHard

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AArrays have O(n) access
  2. BHash functions are slow
  3. CKeys must be sorted first
  4. DMany keys may collide into the same bucket
11.Data structuresMedium

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(V + E)
  2. BO(E²)
  3. CO(V × E)
  4. DO(V²)
12.Data structuresHard

How long does it take to search for an arbitrary key in a binary min-heap of n elements?

  1. AO(1)
  2. BO(log n)
  3. CO(n)
  4. DO(log log n)
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
14.Graph theoryHard
912513117151ABCDEF

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

Why can a table have only one clustered index?

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

What is the result of SELECT COUNT(col) FROM t if all values in col are NULL and the table has 10 rows?

  1. A10
  2. BNULL
  3. CAn error
  4. D0
17.Computer networksMedium

Which DNS record type maps a host name to an IPv6 address?

  1. ACNAME
  2. BA
  3. CMX
  4. DAAAA
18.Computer networksHard

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

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

The network 192.168.252.0/24 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?

192.168.252.0/24/29subnet 9 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 82012816403211608041201networkhost

A host has the address 10.20.45.82/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.45.82/28
x =

3Solutions

1.Number systems

Answer: U = 141

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

115 = 0111 0011
inverted: 1000 1100
+1 → 1000 1101

Read it as unsigned

1000 1101₂ = 256 − 115 = 141

Answer

U = 141
2.Number systems

Answer: M = 900 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

640·480·24 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB
3.Propositional logic & proofs

Answer: contradiction

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).

Truth table row by row

a = 1, c = 1: a → c = 1, ¬c = 0, a ∧ ¬c = 0, (a → c) ∧ (a ∧ ¬c) = 0
a = 1, c = 0: a → c = 0, ¬c = 1, a ∧ ¬c = 1, (a → c) ∧ (a ∧ ¬c) = 0
a = 0, c = 1: a → c = 1, ¬c = 0, a ∧ ¬c = 0, (a → c) ∧ (a ∧ ¬c) = 0
a = 0, c = 0: a → c = 1, ¬c = 1, a ∧ ¬c = 0, (a → c) ∧ (a ∧ ¬c) = 0

Classify

(a → c) ∧ (a ∧ ¬c): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: p

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

Conclusion

p (disjunctive syllogism)

Answer

p
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.

ABC00011110010011031204050716

Group the 1s

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

BC′
A′B′C

Minimal SOP form

F = BC′ + A′B′C
N = 2

Answer

N = 2
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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 1 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 0 → Q = 1
4: J = 1, K = 1 → Q = 0
5: J = 1, K = 0 → Q = 1
6: J = 1, K = 0 → Q = 1

Count the changes

0 → 1 → 1 → 1 → 0 → 1 → 1
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 4−1 mod 19 = 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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·4 (mod 19).

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

Inverse

1 ≡ 5·4 (mod 19)
x = 5
check: 4·5 = 20 = 1·19 + 1

Answer

4−1 mod 19 = 5
8.Number theory & modular arithmetic

Answer: m = 177

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.

Private exponent

n = 703, φ(n) = 648
d = 13−1 mod 648 = 349

Decrypt by repeated squaring

349 = 256 + 64 + 16 + 8 + 4 + 1
2511 ≡ 251 (mod 703)
2512 ≡ 434 (mod 703)
2514 ≡ 655 (mod 703)
2518 ≡ 195 (mod 703)
25116 ≡ 63 (mod 703)
25132 ≡ 454 (mod 703)
25164 ≡ 137 (mod 703)
251128 ≡ 491 (mod 703)
251256 ≡ 655 (mod 703)
251349 ≡ 251·655·195·63·137·655 ≡ 177 (mod 703)

Message

m = 177
check: 17713 mod 703 = 251

Answer

m = 177
9.Algorithms & complexity

Answer: B · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

B · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: D · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

D · Many keys may collide into the same bucket
11.Data structures

Answer: A · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

A · O(V + E)
12.Data structures

Answer: C · O(n)

Why this is the answer

A heap gives no rule for which subtree contains a key, so in the worst case all nodes must be checked; only the minimum is fast to find.

Answer

C · O(n)
13.Graph theory

Answer: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
14.Graph theory

Answer: d(F) = 25

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: 9, C: 12
B (9): E: 10
E (10): D: 21, F: 25
C (12)
D (21)
F (25)

Shortest path

A → B → E → F
d(F) = 25

Answer

d(F) = 25
15.Databases & SQL

Answer: B · 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

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

Answer: D · 0

Why this is the answer

COUNT(col) counts non-NULL values, so it returns 0; COUNT(*) would return 10.

Answer

D · 0
17.Computer networks

Answer: D · AAAA

Why this is the answer

An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.

Answer

D · AAAA
18.Computer networks

Answer: D · 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

D · 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 = 95

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

24 = 16

Subnet of this host

network: ⌊82/16⌋·16 = 80
broadcast: 80 + 16 − 1 = 95
usable hosts: 81 … 94

Answer

x = 95

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