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 complement?−128?64?32?16?8?4?2?1

Write −15 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −158 bit
U =
2.Number systemsHard
32 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬q ∧ p¬q ∨ p¬q ∨ ¬p¬q ∧ ¬p

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

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

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

¬∃x (A(x) → B(x))
  1. 1∃x (A(x) ∧ ¬B(x))
  2. 2∀x (A(x) ∧ ¬B(x))
  3. 3∀x (¬A(x) → ¬B(x))
  4. 4∀x (A(x) → ¬B(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302140517061121131150141809111010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 4, 7, 8, 11, 12, 13, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 4, 7, 8, 11, 12, 13, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110001113021405X716X12113115X140819011010

Find the minimal SOP form of F(A, B, C, D) = Σm(1, 3, 4, 6, 9, 13, 15) + d(7, 12, 14), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(1, 3, 4, 6, 9, 13, 15) + d(7, 12, 14)
L =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 9) and x ≡ 0 (mod 11).

x ≡ 2 (mod 9)x ≡ 0 (mod 11)
x =
8.Number theory & modular arithmeticHard
148c = me​ mod n?m = cd​ mod nm(n, e) = (323, 13)d

Encrypt the message m = 148 with the RSA public key (n, e) = (323, 13). Find the ciphertext c.

n = 323e = 13m = 148
c =
9.Algorithms & complexityMedium

What is the time complexity of multiplying two n × n matrices with the standard triple loop?

  1. AO(n³)
  2. BO(n)
  3. CO(n² log 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(1) space
  2. BO(log n) time and O(n) space
  3. CO(2ⁿ) time and O(1) space
  4. DO(n) time and O(n) space
11.Data structuresMedium

An application needs fast lookups and also needs to list keys in sorted order and answer range queries. Which structure fits best?

  1. AA hash table
  2. BA stack
  3. CA balanced binary search tree
  4. DAn unsorted linked list
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. CThe map throws an error at the moment the field changes
  4. DNothing changes, because a key's hash is computed only once when the program starts
13.Graph theoryMedium

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

|V| = 18|E| = 18
k =
14.Graph theoryHard
1310216151212142ABCDEFG

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

In relational algebra, the projection operator (π) corresponds to:

  1. AChoosing which rows satisfy a given condition
  2. BSorting the result
  3. CCombining two tables on a key
  4. DChoosing which columns appear in the result
16.Databases & SQLHard

Which isolation level prevents dirty reads but still allows non-repeatable reads?

  1. ARepeatable read
  2. BRead uncommitted
  3. CRead committed
  4. DSerializable
17.Computer networksMedium

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

  1. AAAAA
  2. BMX
  3. CA
  4. DCNAME
18.Computer networksHard

Why does a browser warn when a site presents a self-signed certificate?

  1. AThe certificate is always expired
  2. BSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
  3. CThe site must be using HTTP instead of HTTPS
  4. DNo trusted certificate authority vouches that the key belongs to that site
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1942 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1942 hosts
p =
20.IPv4 subnettingHard
last octet = 163112806413201608041211networkhost

A host has the address 10.20.143.163/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.143.163/29
x =

3Solutions

1.Number systems

Answer: U = 241

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.

Invert and add one

15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001

Read it as unsigned

1111 0001₂ = 256 − 15 = 241

Answer

U = 241
2.Number systems

Answer: M = 1024 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

512·512·32 = 8388608 bit = 1048576 B

Convert to KiB

1048576/1024 = 1024 KiB

Answer

M = 1024 KiB
3.Propositional logic & proofs

Answer: ¬q ∨ ¬p

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.
¬(q ∧ p) ≡ ¬q ∨ ¬p

Check with the truth table

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

Answer

¬q ∨ ¬p

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

4.Propositional logic & proofs

Answer: ∀x (A(x) ∧ ¬B(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 (A(x) → B(x)) ≡ ∀x (A(x) ∧ ¬B(x))

Answer

∀x (A(x) ∧ ¬B(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 = 4

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.

ABCD000111100001111010010302140517061121131150141809111010

Group the 0s

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

(C′ + D)
(B + C + D′)
(A + C + D′)
(A + B + D′)

Minimal POS form

F = (C′ + D)(B + C + D′)(A + C + D′)(A + B + D′)
N = 4

Answer

N = 4

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: L = 10

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.

BD′, BC, B′C′D, A′CD, AC′D, AB, A′B′D
ABCD0001111000011110001113021405X716X12113115X140819011010

Minimal cover

F = BD′ + B′C′D + A′CD + AB
L = 10

Answer

L = 10

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

7.Number theory & modular arithmetic

Answer: x = 11

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 = 2 + 9k

Second congruence

2 + 9k ≡ 0 (mod 11) → 9k ≡ 9 (mod 11)
k ≡ 5·9 ≡ 1 (mod 11)

Result

x = 2 + 9·1 = 11
x = 11
check: 11 mod 9 = 2, 11 mod 11 = 0

Answer

x = 11
8.Number theory & modular arithmetic

Answer: c = 48

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

13 = 8 + 4 + 1
1481 ≡ 148 (mod 323)
1482 ≡ 263 (mod 323)
1484 ≡ 47 (mod 323)
1488 ≡ 271 (mod 323)
14813 ≡ 148·47·271 ≡ 48 (mod 323)

Ciphertext

c = 48

Answer

c = 48
9.Algorithms & complexity

Answer: A · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

A · O(n³)
10.Algorithms & complexity

Answer: A · 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

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

Answer: C · A balanced binary search tree

Why this is the answer

A balanced BST gives O(log n) lookups and keeps keys ordered, so in-order traversal and range queries are easy; a hash table scatters keys.

Answer

C · A balanced binary search tree
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 = 1

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 = 18 − 17 = 1

Answer

k = 1
14.Graph theory

Answer: d(F) = 28

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: 13, C: 10
C (10): B: 12, G: 22
B (12): D: 13
D (13): E: 19, F: 28
E (19): G: 21
G (21)
F (28)

Shortest path

A → C → B → D → F
d(F) = 28

Answer

d(F) = 28
15.Databases & SQL

Answer: D · Choosing which columns appear in the result

Why this is the answer

Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).

Answer

D · Choosing which columns appear in the result
16.Databases & SQL

Answer: C · Read committed

Why this is the answer

Read committed hides uncommitted changes; repeatable read also prevents non-repeatable reads; serializable prevents phantoms too.

Answer

C · Read committed
17.Computer networks

Answer: A · 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

A · AAAA
18.Computer networks

Answer: D · No trusted certificate authority vouches that the key belongs to that site

Why this is the answer

Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.

Answer

D · No trusted certificate authority vouches that the key belongs to that site
19.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1942
211 − 2 = 2046 ≥ 1942 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 167

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: ⌊163/8⌋·8 = 160
broadcast: 160 + 8 − 1 = 167
usable hosts: 161 … 166

Answer

x = 167

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