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 −61 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −618 bit
U =
2.Number systemsHard
862 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 862 items its own binary code of the same length. What is the smallest number of bits b needed?

862
b = bit
3.Propositional logic & proofsMedium
(q → p) → rq → (p → r)(q → r) ∧ (p → r)q ∧ (p → r)

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

(q ∧ p) → r
  1. 1(q → p) → r
  2. 2q → (p → r)
  3. 3(q → r) ∧ (p → r)
  4. 4q ∧ (p → r)
4.Propositional logic & proofsHard
∃x ¬F(x)∀x F(x)∃x F(x)∀x ¬F(x)

Which statement is equivalent to ¬∀x F(x), with the negation moved all the way inside? Click it.

¬∀x F(x)
  1. 1∃x ¬F(x)
  2. 2∀x F(x)
  3. 3∃x F(x)
  4. 4∀x ¬F(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 3, 5, 6, 7, 11, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110101103020405171611201311511418X9011X10

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 6, 7, 8, 12, 14, 15) + d(9, 10), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 6, 7, 8, 12, 14, 15) + d(9, 10)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 26 modulo 29: the number x with 0 < x < 29 and 26·x ≡ 1 (mod 29).

a = 26n = 29
26−1 mod 29 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 7137 mod 34.

a = 7k = 137n = 34
7137 mod 34 =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?

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

Which bound is correct for log(n!)?

  1. AΘ(n²)
  2. BΘ(n)
  3. CΘ(n log n)
  4. DΘ(log n)
11.Data structuresMedium

How is a new element inserted into a binary heap?

  1. APlace it at the end of the array and sift it up while it violates the heap property
  2. BRebuild the whole heap from scratch
  3. CPlace it at the root and sift it down
  4. DInsert it at its sorted position by shifting all of the larger elements one place right
12.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. AEmptying the slot could break the probe sequence for other keys
  2. BIt makes the table sorted
  3. CTombstones speed up hashing
  4. DDeleted keys must be kept forever for security, so slots can never be reused
13.Graph theoryMedium
ABCDEFGEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 7|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDEF

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

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

A table has rows where col is NULL. Which rows does WHERE col = NULL return?

  1. AAn error in every database
  2. BNo rows, because the comparison is unknown
  3. COnly the rows where col is NULL, as with IS NULL
  4. DAll rows of the table
16.Databases & SQLHard

A table has a million rows and a Boolean column is_active that is TRUE for 95% of them. Why might the optimiser ignore an index on is_active for WHERE is_active = TRUE?

  1. ATRUE is treated as NULL by the index
  2. BThe optimiser never uses indexes in WHERE clauses
  3. CMost rows match, so a full table scan is cheaper than many index lookups
  4. DIndexes cannot be created on Boolean columns
17.Computer networksMedium

What is the main purpose of a VLAN?

  1. ATo increase the speed of every switch port
  2. BTo give each device a public IP address
  3. CTo split one physical switched network into separate broadcast domains
  4. DTo encrypt all traffic between switches so other departments cannot read it
18.Computer networksHard

TCP congestion control's "slow start" phase:

  1. AHalves the congestion window every round trip until a packet loss is detected
  2. BKeeps the window fixed at one segment
  3. CIs used only by UDP
  4. DIncreases the congestion window exponentially until a threshold or loss
19.IPv4 subnettingMedium
255.255.255.19211111111111111111111111111000000network bitshost bits

A subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?

255.255.255.192
H =
20.IPv4 subnettingHard
third octet 66 … 67012816403201608041201commonvaries

A router has routes to these networks: 10.4.66.0/24, 10.4.67.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.4.66.0/24 … 10.4.67.0/24
p =

3Solutions

1.Number systems

Answer: U = 195

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

61 = 0011 1101
inverted: 1100 0010
+1 → 1100 0011

Read it as unsigned

1100 0011₂ = 256 − 61 = 195

Answer

U = 195
2.Number systems

Answer: b = 10 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

29 = 512 < 862
210 = 1024 ≥ 862

Answer

b = 10
3.Propositional logic & proofs

Answer: q → (p → r)

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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(q ∧ p) → r ≡ q → (p → r)

Check with the truth table

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

Answer

q → (p → r)

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

4.Propositional logic & proofs

Answer: ∃x ¬F(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 F(x) ≡ ∃x ¬F(x)

Answer

∃x ¬F(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: AB = 00, CD = 00

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m0: m1, m2, m4, m8 → 0
ABCD00011110000111101111111

The isolated 1

m0 = A′B′C′D′ → AB = 00, CD = 00

Answer

AB = 00, CD = 00
6.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.

Prime implicants

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

B′C′, BC, AD′
ABCD0001111000011110101103020405171611201311511418X9011X10

Minimal cover

F = B′C′ + BC + AD′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 26−1 mod 29 = 19

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·26 (mod 29).

r = 29, t = 0
r = 26, q = 1, t = 1
r = 3, q = 8, t = −1
r = 2, q = 1, t = 9
r = 1, q = 2, t = −10

Inverse

1 ≡ −10·26 (mod 29)
x = 19
check: 26·19 = 494 = 17·29 + 1

Answer

26−1 mod 29 = 19
8.Number theory & modular arithmetic

Answer: 7137 mod 34 = 27

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

137 = 100010012 = 128 + 8 + 1

Repeated squaring

71 ≡ 7 (mod 34)
72 ≡ 15 (mod 34)
74 ≡ 21 (mod 34)
78 ≡ 33 (mod 34)
716 ≡ 1 (mod 34)
732 ≡ 1 (mod 34)
764 ≡ 1 (mod 34)
7128 ≡ 1 (mod 34)

Multiply the needed powers

7137 ≡ 7·33·1 ≡ 27 (mod 34)

Answer

7137 mod 34 = 27
9.Algorithms & complexity

Answer: D · O(n log n)

Why this is the answer

The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).

Answer

D · O(n log n)
10.Algorithms & complexity

Answer: C · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

C · Θ(n log n)
11.Data structures

Answer: A · Place it at the end of the array and sift it up while it violates the heap property

Why this is the answer

Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.

Answer

A · Place it at the end of the array and sift it up while it violates the heap property
12.Data structures

Answer: A · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

A · Emptying the slot could break the probe sequence for other keys
13.Graph theory

Answer: neither

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.

Degrees

deg(A) = 1, deg(B) = 2, deg(C) = 2, deg(D) = 3, deg(E) = 3, deg(F) = 2, deg(G) = 1

Odd vertices

A, D, E, G → 4
neither

Answer

neither
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 0 1 0 0
C: 1 0 0 0 1 0
D: 0 1 0 0 1 1
E: 0 0 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)AF = 2·0 + 0·0 + 0·0 + 1·1 + 1·1 + 0·0 = 2

Answer

N = 2
15.Databases & SQL

Answer: B · No rows, because the comparison is unknown

Why this is the answer

Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.

Answer

B · No rows, because the comparison is unknown
16.Databases & SQL

Answer: C · Most rows match, so a full table scan is cheaper than many index lookups

Why this is the answer

An index helps when it narrows the search to few rows; when most rows qualify, random index lookups cost more than reading the table sequentially.

Answer

C · Most rows match, so a full table scan is cheaper than many index lookups
17.Computer networks

Answer: C · To split one physical switched network into separate broadcast domains

Why this is the answer

VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.

Answer

C · To split one physical switched network into separate broadcast domains
18.Computer networks

Answer: D · Increases the congestion window exponentially until a threshold or loss

Why this is the answer

Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.

Answer

D · Increases the congestion window exponentially until a threshold or loss
19.IPv4 subnetting

Answer: H = 62

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

255.255.255.192 = /26

Usable hosts

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: p = 23

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

66 = 01000010
67 = 01000011
common leading bits: 7

Prefix length

p = 16 + 7 = 23

Answer

p = /23

Summary route: 10.4.66.0/23

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