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 −24 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −248 bit
U =
2.Number systemsHard
4057 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4057 items its own binary code of the same length. What is the smallest number of bits b needed?

4057
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

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

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

How many of the 16 rows of the truth table make (¬q ∨ s) ∨ ((q → ¬r) ∨ p) true? Call this number N.

(¬q ∨ s) ∨ ((q → ¬r) ∨ p)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001131214050706

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(0, 2, 3, 4). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(0, 2, 3, 4)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011130204X50716X120131150141809111010

For F(A, B, C, D) = Σm(0, 1, 3, 6, 8, 11, 15) + d(5, 12), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 3, 6, 8, 11, 15) + d(5, 12)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 86 modulo 93: the number x with 0 < x < 93 and 86·x ≡ 1 (mod 93).

a = 86n = 93
86−1 mod 93 =
8.Number theory & modular arithmeticHard

79 is prime. Use Fermat’s little theorem to find 291184 mod 79.

p = 79k = 91184
291184 mod 79 =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(1) solve to?

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

What is the difference between an undecidable problem and an intractable one?

  1. AUndecidable problems are in NP, while intractable problems are in P
  2. BIntractable problems cannot be solved by any algorithm, while undecidable ones only need exponential time
  3. CAn undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
  4. DBoth mean the problem has no algorithm at all
11.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe existing keys are sorted in place
  2. BThe oldest keys are deleted to make room
  3. COnly the colliding keys are moved to a new overflow list at the end of the table
  4. DA larger table is allocated and every key is inserted again
12.Data structuresHard

For a hash table to work correctly, which rule must hash and equality follow?

  1. AEqual hash values imply equal keys
  2. BEqual keys must have equal hash values
  3. CHash values must be prime numbers
  4. DDifferent keys must have different hash values
13.Graph theoryMedium

A tree has 2 vertices of degree 3, 2 vertices of degree 4, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 32 × deg 4
L =
14.Graph theoryHard
1991661410181215ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 5; skipped edges do not count.

k = 5
  1. 1edge AC
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge GB
  9. 9edge GD
15.Databases & SQLMedium

A column holds the values 10, 20 and NULL. What does AVG(col) return?

  1. A15
  2. B10
  3. CNULL
  4. D30
16.Databases & SQLHard

R(A, B, C, D) has the functional dependencies A → B and C → D. What is the candidate key?

  1. A{B, D}
  2. B{A}
  3. C{A, C}
  4. D{C}
17.Computer networksMedium

What is the role of a default gateway on a host?

  1. AIt is the router used for traffic to other networks
  2. BIt assigns MAC addresses
  3. CIt is the DNS server
  4. DIt is the switch port the host is connected to
18.Computer networksHard

An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?

  1. AThe switch shuts down all of its ports
  2. BAll DNS queries fail across the internet
  3. CThe gateway's IP address is permanently changed to the attacker's address on every router
  4. DVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 696 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

696 hosts
p =
20.IPv4 subnettingHard

The network 192.168.198.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 41 hosts, C: 62 hosts, D: 19 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN B?

A: 13B: 41C: 62D: 19
x =

3Solutions

1.Number systems

Answer: U = 232

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

24 = 0001 1000
inverted: 1110 0111
+1 → 1110 1000

Read it as unsigned

1110 1000₂ = 256 − 24 = 232

Answer

U = 232
2.Number systems

Answer: b = 12 bit

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.

Powers of two

211 = 2048 < 4057
212 = 4096 ≥ 4057

Answer

b = 12
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

p = 1, r = 1: p → r = 1, ¬r = 0, p ∧ ¬r = 0, (p → r) ∧ (p ∧ ¬r) = 0
p = 1, r = 0: p → r = 0, ¬r = 1, p ∧ ¬r = 1, (p → r) ∧ (p ∧ ¬r) = 0
p = 0, r = 1: p → r = 1, ¬r = 0, p ∧ ¬r = 0, (p → r) ∧ (p ∧ ¬r) = 0
p = 0, r = 0: p → r = 1, ¬r = 1, p ∧ ¬r = 0, (p → r) ∧ (p ∧ ¬r) = 0

Classify

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

Answer

contradiction
4.Propositional logic & proofs

Answer: N = 15

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.

Columns

¬q; ¬q ∨ s; ¬r; q → ¬r; (q → ¬r) ∨ p; (¬q ∨ s) ∨ ((q → ¬r) ∨ p)

Split on p

p = 1: 8
p = 0: 7
N = 8 + 7 = 15

Answer

N = 15
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.

ABC00011110011001131214050706

Group the 0s

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

(B + C′)
(A′ + B′)

Minimal POS form

F = (B + C′)(A′ + B′)
N = 2

Answer

N = 2
6.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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′C′D′, B′CD, A′C′D, AC′D′, ACD, A′B′D, A′B′C′, A′BCD′
ABCD00011110000111101011130204X50716X120131150141809111010

Essential prime implicants

ACD, A′BCD′
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: 86−1 mod 93 = 53

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·86 (mod 93).

r = 93, t = 0
r = 86, q = 1, t = 1
r = 7, q = 12, t = −1
r = 2, q = 3, t = 13
r = 1, q = 2, t = −40

Inverse

1 ≡ −40·86 (mod 93)
x = 53
check: 86·53 = 4558 = 49·93 + 1

Answer

86−1 mod 93 = 53
8.Number theory & modular arithmetic

Answer: 291184 mod 79 = 4

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.

Reduce the exponent

91184 mod 78 = 2
291184 ≡ 22 (mod 79)

Small power

22 mod 79 = 4

Answer

291184 mod 79 = 4
9.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

C · O(n)
10.Algorithms & complexity

Answer: C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow

Why this is the answer

The halting problem has no algorithm at all; an NP-hard problem such as TSP can be solved by brute force, just not in known polynomial time.

Answer

C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
11.Data structures

Answer: D · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

D · A larger table is allocated and every key is inserted again
12.Data structures

Answer: B · Equal keys must have equal hash values

Why this is the answer

If two equal keys hashed differently they would be stored in different buckets and lookups would fail; unequal keys may still share a hash.

Answer

B · Equal keys must have equal hash values
13.Graph theory

Answer: L = 8

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.

Handshake lemma for a tree

L + 14 = 2(L + 4) − 2
L = 8

Answer

L = 8

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: edge DE

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

1. CE (6)
2. BC (9)
3. DF (10)
4. GB (12)
5. DE (14)
6. AC (19)

edge number 5

DE

Answer

edge DE
15.Databases & SQL

Answer: A · 15

Why this is the answer

AVG ignores NULL, so it averages 10 and 20 to get 15, not 30 / 3 = 10.

Answer

A · 15
16.Databases & SQL

Answer: C · {A, C}

Why this is the answer

A and C are not determined by anything, so every key must contain both, and {A, C}+ = {A, B, C, D}.

Answer

C · {A, C}
17.Computer networks

Answer: A · It is the router used for traffic to other networks

Why this is the answer

The host sends packets for non-local destinations to the default gateway, usually its local router.

Answer

A · It is the router used for traffic to other networks
18.Computer networks

Answer: D · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack

Why this is the answer

ARP spoofing poisons hosts' ARP caches with the attacker's MAC for the gateway IP. The attacker can then read, change or drop traffic; dynamic ARP inspection helps prevent it.

Answer

D · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
19.IPv4 subnetting

Answer: p = 22

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).

Host bits needed

29 − 2 = 510 < 696
210 − 2 = 1022 ≥ 696 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 64

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

C: 62 hosts → 26 = 64 (/26)
B: 41 hosts → 26 = 64 (/26)
D: 19 hosts → 25 = 32 (/27)
A: 13 hosts → 24 = 16 (/28)

Allocation, largest first

C: .0 – .63
B: .64 – .127
D: .128 – .159
A: .160 – .175

Answer

x = 64

LAN B: 192.168.198.64/26.

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