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

N = −88 bit
U =
2.Number systemsHard
4237 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4237 items its own binary code of the same length. What is the smallest number of bits b needed?

4237
b = bit
3.Propositional logic & proofsMedium
¬r ∧ ¬(q ∨ p)¬r → ¬(q ∨ p)r ∧ ¬(q ∨ p)¬r ∧ (q ∨ p)

You want to prove r → (q ∨ p) by contradiction. Which statement do you assume at the start? Click it.

r → (q ∨ p)
  1. 1¬r ∧ ¬(q ∨ p)
  2. 2¬r → ¬(q ∨ p)
  3. 3r ∧ ¬(q ∨ p)
  4. 4¬r ∧ (q ∨ p)
4.Propositional logic & proofsHard

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

(r ∨ p) ∧ ((r → s) ∨ ¬q)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010001031214151706

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(2, 4, 5, 7). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(2, 4, 5, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKT1234567

A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

Q₀ = 0T
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 10) and x ≡ 9 (mod 11).

x ≡ 1 (mod 10)x ≡ 9 (mod 11)
x =
8.Number theory & modular arithmeticHard

How many solutions N does 30x ≡ 16 (mod 32) have in {0, 1, …, 31}?

30x ≡ 16 (mod 32)
N =
9.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(n²)
  2. BΩ(n log n)
  3. CΩ(log n)
  4. DΩ(n)
10.Algorithms & complexityHard

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n²)
  2. BΘ(n³)
  3. CΘ(n log n)
  4. DΘ(n² log n)
11.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the preorder traversal?

  1. A1 2 4 5 3
  2. B4 2 5 1 3
  3. C4 5 2 3 1
  4. D1 2 3 4 5
12.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(n) time and O(n) extra space
  2. BO(1) time and O(1) extra space
  3. CO(n) time and O(1) extra space
  4. DO(n log n) time and O(1) extra space
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.

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

Why does SELECT price * 2 AS double_price FROM t WHERE double_price > 10 fail in most databases?

  1. AArithmetic is not allowed in the SELECT list
  2. BWHERE is evaluated before the SELECT list, so the alias does not exist yet
  3. CAliases may not contain an underscore
  4. DWHERE can only compare text columns
16.Databases & SQLHard

For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?

  1. AHash join
  2. BBitmap join on the primary key
  3. CCross join
  4. DNested loop join with index lookups
17.Computer networksMedium

What uniquely identifies one end of a TCP connection on the internet?

  1. AAn IP address together with a port number
  2. BA domain name alone
  3. CA subnet mask and gateway
  4. DA MAC address alone
18.Computer networksHard

A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?

  1. AA missing or wrong default gateway
  2. BThe TCP window is too small
  3. CA duplicate MAC address on the internet
  4. DA faulty DNS server setting
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.159.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?

192.168.159.0/24/27subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 31012806403211618141211networkhost

A host has the address 10.20.255.31/30. What is the fourth octet x of its network address?

10.20.255.31/30
x =

3Solutions

1.Number systems

Answer: U = 248

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

8 = 0000 1000
inverted: 1111 0111
+1 → 1111 1000

Read it as unsigned

1111 1000₂ = 256 − 8 = 248

Answer

U = 248
2.Number systems

Answer: b = 13 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

212 = 4096 < 4237
213 = 8192 ≥ 4237

Answer

b = 13
3.Propositional logic & proofs

Answer: r ∧ ¬(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).

Parts of the implication

a = r, b = q ∨ p

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = r ∧ ¬(q ∨ p)

Answer

r ∧ ¬(q ∨ p)

If the assumption leads to a contradiction, the implication must be true.

4.Propositional logic & proofs

Answer: N = 10

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

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

Split on p

p = 1: 7
p = 0: 3
N = 7 + 3 = 10

Answer

N = 10
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.

ABC00011110010001031214151706

Group the 0s

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

(A + C′)
(A + B)
(A′ + B′ + C)

Minimal POS form

F = (A + C′)(A + B)(A′ + B′ + C)
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Q after every edge

A T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

1: T = 0 → Q = 0
2: T = 1 → Q = 1
3: T = 1 → Q = 0
4: T = 0 → Q = 0
5: T = 1 → Q = 1
6: T = 1 → Q = 0
7: T = 1 → Q = 1

Count the changes

0 → 0 → 1 → 0 → 0 → 1 → 0 → 1
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: x = 31

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 = 1 + 10k

Second congruence

1 + 10k ≡ 9 (mod 11) → 10k ≡ 8 (mod 11)
k ≡ 10·8 ≡ 3 (mod 11)

Result

x = 1 + 10·3 = 31
x = 31
check: 31 mod 10 = 1, 31 mod 11 = 9

Answer

x = 31
8.Number theory & modular arithmetic

Answer: N = 2

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).

Divide by the gcd

gcd(30, 32) = 2
15x ≡ 8 (mod 16)

Solve the reduced congruence

15−1 ≡ 15 (mod 16) → x ≡ 8 (mod 16)

Solutions modulo 32

x ∈ {8, 24}
N = 2

Answer

N = 2
9.Algorithms & complexity

Answer: B · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

B · Ω(n log n)
10.Algorithms & complexity

Answer: B · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

B · Θ(n³)
11.Data structures

Answer: A · 1 2 4 5 3

Why this is the answer

Preorder visits 1, then its left subtree (2, 4, 5), then 3; the other options are inorder, postorder and level-order.

Answer

A · 1 2 4 5 3
12.Data structures

Answer: C · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

C · O(n) time and O(1) extra space
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) = 3, deg(C) = 3, deg(D) = 1, deg(E) = 2

Odd vertices

A, B, C, D → 4
neither

Answer

neither
14.Graph theory

Answer: N = 4

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
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)EE = 2·0 + 2·1 + 1·1 + 1·1 + 3·0 = 4

Answer

N = 4
15.Databases & SQL

Answer: B · WHERE is evaluated before the SELECT list, so the alias does not exist yet

Why this is the answer

Logically FROM and WHERE run before SELECT, so repeat the expression in WHERE or use a subquery; ORDER BY can use the alias because it runs after SELECT.

Answer

B · WHERE is evaluated before the SELECT list, so the alias does not exist yet
16.Databases & SQL

Answer: A · Hash join

Why this is the answer

A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.

Answer

A · Hash join
17.Computer networks

Answer: A · An IP address together with a port number

Why this is the answer

A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.

Answer

A · An IP address together with a port number
18.Computer networks

Answer: A · A missing or wrong default gateway

Why this is the answer

Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.

Answer

A · A missing or wrong default gateway
19.IPv4 subnetting

Answer: x = 95

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

25 = 32

Subnet number 3

network: (3 − 1)·32 = 64
broadcast: 64 + 32 − 1 = 95

Answer

x = 95
20.IPv4 subnetting

Answer: x = 28

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

22 = 4

Subnet of this host

network: ⌊31/4⌋·4 = 28
broadcast: 28 + 4 − 1 = 31
usable hosts: 29 … 30

Answer

x = 28

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