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 −94 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −948 bit
U =
2.Number systemsHard
1421 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 1421 items its own binary code of the same length. What is the smallest number of bits b needed?

1421
b = bit
3.Propositional logic & proofsMedium
¬(a ∨ b) → ¬c¬c → ¬(a ∨ b)c ∧ ¬(a ∨ b)(a ∨ b) → c

You want to prove c → (a ∨ b) by contrapositive. Which implication do you prove instead? Click it.

c → (a ∨ b)
  1. 1¬(a ∨ b) → ¬c
  2. 2¬c → ¬(a ∨ b)
  3. 3c ∧ ¬(a ∨ b)
  4. 4(a ∨ b) → c
4.Propositional logic & proofsHard
aba → b¬a¬b11100100

The argument with premises a → b; ¬a and conclusion ¬b is invalid. Click the row of the truth table that is a counterexample.

a → b¬a∴ ¬b
  1. 1a = 1, b = 1
  2. 2a = 1, b = 0
  3. 3a = 0, b = 1
  4. 4a = 0, b = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​C′I1​CI2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = C′, I₂ = C, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = C′I₂ = CI₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110100103021405X706X121130151141809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 4, 8, 10, 13, 14) + d(7, 12), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 4, 8, 10, 13, 14) + d(7, 12)
L =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 6 (mod 10) and x ≡ 7 (mod 13).

x ≡ 6 (mod 10)x ≡ 7 (mod 13)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n96m = cd​ mod n?(n, e) = (713, 23)d

An RSA key uses p = 23, q = 31 and e = 23. Decrypt the ciphertext c = 96: find the message m.

p = 23q = 31e = 23c = 96
m =
9.Algorithms & complexityMedium

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

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

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

  1. AO(n √n)
  2. BO(n)
  3. CO(n log n)
  4. DO(n²)
11.Data structuresMedium

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo prevent the list from growing beyond a fixed maximum number of nodes
  2. BTo simplify code by removing special cases at the head or tail
  3. CTo store the number of elements
  4. DTo make the list sorted
12.Data structuresHard

An object is stored as a key in a hash map, and then a field used by its hash function is changed. What is the likely result?

  1. AThe map automatically moves the key to its new bucket
  2. BLooking up the object may fail because it sits in the bucket for its old hash
  3. CNothing changes, because a key's hash is computed only once when the program starts
  4. DThe map throws an error at the moment the field changes
13.Graph theoryMedium

A connected graph has 18 vertices and 25 edges. How many edges k must be removed to leave a spanning tree?

|V| = 18|E| = 25
k =
14.Graph theoryHard
137815512811ABCDEF

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

What does INTERSECT return when used between two SELECT queries?

  1. AAll rows from both results
  2. BThe Cartesian product of the results
  3. CRows in the first result but not the second
  4. DRows that appear in both results
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index that covers every column of every table
  2. BAn index that contains all columns the query needs, so the table itself need not be read
  3. CAn index on the primary key only
  4. DAn index that is rebuilt every night
17.Computer networksMedium

Using even parity, which parity bit is added to the data bits 1011001?

  1. A1
  2. B4
  3. C2
  4. D0
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIPv6 does not use addresses
  2. BIts 128-bit address space is large enough to give every device a global address
  3. CIPv6 only works on local networks, so addresses never need to be shared globally
  4. DIPv6 encrypts all addresses
19.IPv4 subnettingMedium
255.255.248.011111111111111111111100000000000network bitshost bits

A subnet uses the mask 255.255.248.0. How many usable host addresses H does it have?

255.255.248.0
H =
20.IPv4 subnettingHard
third octet 196 … 200112816403201608140201commonvaries

A router has routes to these networks: 10.85.196.0/24, 10.85.197.0/24, 10.85.198.0/24, 10.85.199.0/24, 10.85.200.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.85.196.0/24 … 10.85.200.0/24
p =

3Solutions

1.Number systems

Answer: U = 162

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

94 = 0101 1110
inverted: 1010 0001
+1 → 1010 0010

Read it as unsigned

1010 0010₂ = 256 − 94 = 162

Answer

U = 162
2.Number systems

Answer: b = 11 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

210 = 1024 < 1421
211 = 2048 ≥ 1421

Answer

b = 11
3.Propositional logic & proofs

Answer: ¬(a ∨ b) → ¬c

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).

Parts of the implication

a = c, b = a ∨ b

Contrapositive

¬b → ¬a = ¬(a ∨ b) → ¬c

Answer

¬(a ∨ b) → ¬c

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: a = 0, b = 1

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.

Premises and conclusion row by row

a = 1, b = 1: 1 0 | 0
a = 1, b = 0: 0 0 | 1
a = 0, b = 1: 1 1 | 0
a = 0, b = 0: 1 1 | 1

Counterexample

a = 0, b = 1

Answer

a = 0, b = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: N = 5

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′BC′ + AB′C + ABC′

Rows with F = 1

ABC = 000, 001, 010, 101, 110
N = 5

Answer

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

Answer: L = 7

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.

C′D′, AD′, ABC′
ABCD0001111000011110100103021405X706X121130151141809011110

Minimal cover

F = C′D′ + AD′ + ABC′
L = 7

Answer

L = 7
7.Number theory & modular arithmetic

Answer: x = 46

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).

First congruence

x = 6 + 10k

Second congruence

6 + 10k ≡ 7 (mod 13) → 10k ≡ 1 (mod 13)
k ≡ 4·1 ≡ 4 (mod 13)

Result

x = 6 + 10·4 = 46
x = 46
check: 46 mod 10 = 6, 46 mod 13 = 7

Answer

x = 46
8.Number theory & modular arithmetic

Answer: m = 487

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 = 713, φ(n) = 660
d = 23−1 mod 660 = 287

Decrypt by repeated squaring

287 = 256 + 16 + 8 + 4 + 2 + 1
961 ≡ 96 (mod 713)
962 ≡ 660 (mod 713)
964 ≡ 670 (mod 713)
968 ≡ 423 (mod 713)
9616 ≡ 679 (mod 713)
9632 ≡ 443 (mod 713)
9664 ≡ 174 (mod 713)
96128 ≡ 330 (mod 713)
96256 ≡ 524 (mod 713)
96287 ≡ 96·660·670·423·679·524 ≡ 487 (mod 713)

Message

m = 487
check: 48723 mod 713 = 96

Answer

m = 487
9.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

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

Answer

B · O(n)
10.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

C · O(n log n)
11.Data structures

Answer: B · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

B · To simplify code by removing special cases at the head or tail
12.Data structures

Answer: B · Looking up the object may fail because it sits in the bucket for its old hash

Why this is the answer

The key's location was chosen from its old hash; with a new hash, lookups search the wrong bucket, so keys should be immutable.

Answer

B · Looking up the object may fail because it sits in the bucket for its old hash
13.Graph theory

Answer: k = 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.

Edges of a spanning tree

18 − 1 = 17

Edges to remove

k = 25 − 17 = 8

Answer

k = 8
14.Graph theory

Answer: d(F) = 20

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: 3
B (1): D: 9, E: 12
C (3)
D (9): F: 21
E (12): F: 20
F (20)

Shortest path

A → B → E → F
d(F) = 20

Answer

d(F) = 20
15.Databases & SQL

Answer: D · Rows that appear in both results

Why this is the answer

INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.

Answer

D · Rows that appear in both results
16.Databases & SQL

Answer: B · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

B · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: D · 0

Why this is the answer

1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.

Answer

D · 0
18.Computer networks

Answer: B · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

B · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: H = 2046

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

255.255.248.0 = /21

Usable hosts

211 − 2 = 2046

Answer

H = 2046
20.IPv4 subnetting

Answer: p = 20

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

196 = 11000100
200 = 11001000
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.85.192.0/20 (it also covers 11 network(s) that are not in the list).

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