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
place values in two's complement1−12816403201618040211

The 8-bit pattern 1100 1001 stores a signed integer in two's complement. What is its value N in decimal?

1100 10018 bit
N =
2.Number systemsHard
16 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
a ∧ ¬b¬a ∨ ¬ba ↔ ¬b¬a ↔ ¬b

Which formula is logically equivalent to ¬(a ↔ b)? Click it.

¬(a ↔ b)
  1. 1a ∧ ¬b
  2. 2¬a ∨ ¬b
  3. 3a ↔ ¬b
  4. 4¬a ↔ ¬b
4.Propositional logic & proofsHard
prr → ppr11100100

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

r → pp∴ r
  1. 1p = 1, r = 1
  2. 2p = 1, r = 0
  3. 3p = 0, r = 1
  4. 4p = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110302040507061120130150141809111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 8, 11, 12). How many product terms N does it have?

F(A, B, C, D) = Σm(1, 8, 11, 12)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011X302140517160121130150141819X11110

For F(A, B, C, D) = Σm(0, 1, 4, 6, 7, 8, 9, 10, 13) + d(3, 11), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 4, 6, 7, 8, 9, 10, 13) + d(3, 11)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 27 modulo 29: the number x with 0 < x < 29 and 27·x ≡ 1 (mod 29).

a = 27n = 29
27−1 mod 29 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 61128 mod 70.

a = 61k = 128n = 70
61128 mod 70 =
9.Algorithms & complexityMedium

What is the amortised time complexity of appending to a dynamic array that doubles its capacity when full?

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

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, because insertion sort is not a comparison sort
  2. BYes, the lower bound is therefore wrong
  3. CYes, but only for arrays shorter than 100
  4. DNo, the lower bound applies to the worst case over all inputs, not to every input
11.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. AAscending sorted order
  2. BLevel order
  3. CInsertion order
  4. DDescending order always
12.Data structuresHard

A segment tree is typically used to:

  1. AAnswer range queries (such as range sums) and updates in O(log n)
  2. BSort strings alphabetically by storing each prefix in a separate tree node
  3. CImplement a stack
  4. DStore a graph's edges
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
ABCDEF

How many walks of length 3 go from E to C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

In relational algebra, the selection operator (σ) corresponds to which SQL clause?

  1. ASELECT list
  2. BWHERE
  3. CORDER BY
  4. DGROUP BY
16.Databases & SQLHard

What is a phantom read?

  1. AA repeated query returns new rows inserted by another committed transaction
  2. BA query that returns no rows
  3. CReading data that was never written
  4. DReading uncommitted changes
17.Computer networksMedium

How does a stateful firewall differ from a simple packet filter?

  1. AIt blocks all incoming traffic without exceptions, including replies to outgoing requests
  2. BIt encrypts every packet it forwards
  3. CIt tracks connections and allows replies that belong to established sessions
  4. DIt only checks the MAC address of each frame
18.Computer networksHard

After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?

  1. ATo make the cable carry signals further
  2. BTo lower the chance of colliding again when many stations compete
  3. CTo synchronise all stations to one clock
  4. DTo give the station with the highest MAC address priority on the shared cable
19.IPv4 subnettingMedium
255.255.255.24011111111111111111111111111110000network bitshost bits

A subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?

255.255.255.240
H =
20.IPv4 subnettingHard
last octet = 85012816403211608140211networkhost

A host has the address 10.20.88.85/25. What is the fourth octet x of the last usable host address in its subnet?

10.20.88.85/25
x =

3Solutions

1.Number systems

Answer: N = −55

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 8 + 1 = −55

Answer

N = −55
2.Number systems

Answer: M = 1536 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

1024·768·16 = 12582912 bit = 1572864 B

Convert to KiB

1572864/1024 = 1536 KiB

Answer

M = 1536 KiB
3.Propositional logic & proofs

Answer: a ↔ ¬b

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

¬(a ↔ b) is 1 exactly when a and b differ, which is a ↔ ¬b.
¬(a ↔ b) ≡ a ↔ ¬b

Check with the truth table

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

Answer

a ↔ ¬b

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

4.Propositional logic & proofs

Answer: p = 1, r = 0

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

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

Counterexample

p = 1, r = 0

Answer

p = 1, r = 0

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 = 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.

ABCD000111100001111000110302040507061120130150141809111010

Group the 1s

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

AC′D′
A′B′C′D
AB′CD

Minimal SOP form

F = AC′D′ + A′B′C′D + AB′CD
N = 3

Answer

N = 3
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′D, B′C′, A′C′D′, A′CD, AC′D, AB′, A′BD′, A′BC
ABCD00011110000111101011X302140517160121130150141819X11110

Essential prime implicants

AC′D, AB′
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: 27−1 mod 29 = 14

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·27 (mod 29).

r = 29, t = 0
r = 27, q = 1, t = 1
r = 2, q = 13, t = −1
r = 1, q = 2, t = 14

Inverse

1 ≡ 14·27 (mod 29)
x = 14
check: 27·14 = 378 = 13·29 + 1

Answer

27−1 mod 29 = 14
8.Number theory & modular arithmetic

Answer: 61128 mod 70 = 11

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.

Binary exponent

128 = 100000002 = 128

Repeated squaring

611 ≡ 61 (mod 70)
612 ≡ 11 (mod 70)
614 ≡ 51 (mod 70)
618 ≡ 11 (mod 70)
6116 ≡ 51 (mod 70)
6132 ≡ 11 (mod 70)
6164 ≡ 51 (mod 70)
61128 ≡ 11 (mod 70)

Multiply the needed powers

61128 ≡ 11 ≡ 11 (mod 70)

Answer

61128 mod 70 = 11
9.Algorithms & complexity

Answer: A · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

A · O(1)
10.Algorithms & complexity

Answer: D · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

D · No, the lower bound applies to the worst case over all inputs, not to every input
11.Data structures

Answer: A · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

A · Ascending sorted order
12.Data structures

Answer: A · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

A · Answer range queries (such as range sums) and updates in O(log 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: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0 0
B: 1 0 1 1 1 0
C: 1 1 0 0 0 0
D: 0 1 0 0 1 1
E: 0 1 0 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)EC = 1·1 + 1·1 + 1·0 + 2·0 + 3·0 + 1·0 = 2

Answer

N = 2
15.Databases & SQL

Answer: B · WHERE

Why this is the answer

σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.

Answer

B · WHERE
16.Databases & SQL

Answer: A · A repeated query returns new rows inserted by another committed transaction

Why this is the answer

Phantoms occur when another transaction inserts rows that match a range condition; serializable isolation prevents them.

Answer

A · A repeated query returns new rows inserted by another committed transaction
17.Computer networks

Answer: C · It tracks connections and allows replies that belong to established sessions

Why this is the answer

A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.

Answer

C · It tracks connections and allows replies that belong to established sessions
18.Computer networks

Answer: B · To lower the chance of colliding again when many stations compete

Why this is the answer

Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.

Answer

B · To lower the chance of colliding again when many stations compete
19.IPv4 subnetting

Answer: H = 14

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

255.255.255.240 = /28

Usable hosts

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 126

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

27 = 128

Subnet of this host

network: ⌊85/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126

Answer

x = 126

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