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 complement1−12806413201608140201

The 8-bit pattern 1010 0100 stores a signed integer in two's complement. What is its value N in decimal?

1010 01008 bit
N =
2.Number systemsHard
16 bit1 px256 px256 pxM = ? KiB

An uncompressed image is 256 × 256 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

256 × 256 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬b ∨ ¬c¬b ∧ c¬b ∧ ¬cb ∨ ¬c

Which formula is logically equivalent to ¬(b ∨ c)? Click it.

¬(b ∨ c)
  1. 1¬b ∨ ¬c
  2. 2¬b ∧ c
  3. 3¬b ∧ ¬c
  4. 4b ∨ ¬c
4.Propositional logic & proofsHard
∃x (P(x) → ¬Q(x))∀x (P(x) ∧ ¬Q(x))∃x (¬P(x) ∧ Q(x))∃x (P(x) ∧ ¬Q(x))

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

¬∀x (P(x) → Q(x))
  1. 1∃x (P(x) → ¬Q(x))
  2. 2∀x (P(x) ∧ ¬Q(x))
  3. 3∃x (¬P(x) ∧ Q(x))
  4. 4∃x (P(x) ∧ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

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(3, 4, 12, 13, 14, 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
ABCD0001111000011110101113X204151706012113015014X8X9011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 3, 5, 7, 10, 13) + d(2, 8, 9), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 3, 5, 7, 10, 13) + d(2, 8, 9)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 77 modulo 79: the number x with 0 < x < 79 and 77·x ≡ 1 (mod 79).

a = 77n = 79
77−1 mod 79 =
8.Number theory & modular arithmeticHard
mc = me​ mod n60m = cd​ mod n?(n, e) = (161, 19)d

An RSA key uses p = 7, q = 23 and e = 19. Decrypt the ciphertext c = 60: find the message m.

p = 7q = 23e = 19c = 60
m =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Topological sorting is possible only for:

  1. AGraphs with negative weights
  2. BComplete graphs
  3. CDirected acyclic graphs
  4. DUndirected graphs
11.Data structuresMedium

The keys 50, 30, 70, 20, 40 are inserted in this order into an empty binary search tree. What is the left child of 30?

  1. A50
  2. B70
  3. C40
  4. D20
12.Data structuresHard

A Fenwick tree (binary indexed tree) over an array of n numbers supports which operations in O(log n)?

  1. ASorting the array and binary search
  2. BInserting elements at any position
  3. CFinding the median in O(1)
  4. DPrefix-sum queries and single-element updates
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
ABCDEF

How many walks of length 2 go from D to B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

What does the "A" in ACID stand for?

  1. AAvailability
  2. BAtomicity
  3. CAccuracy
  4. DAggregation
16.Databases & SQLHard

Table employee(emp_id, dept_id, dept_name) has the key emp_id, and dept_id → dept_name. Which statement is correct?

  1. AIt is in BCNF and therefore in 3NF
  2. BIt is not in 2NF
  3. CIt is in 2NF but not in 3NF
  4. DIt is not in 1NF
17.Computer networksMedium

What is the main purpose of a VLAN?

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

Why was the TCP window scale option introduced?

  1. AWithout it, TCP could not detect lost segments on long-distance links
  2. BThe 32-bit window field was too large and wasted header space in every TCP segment sent
  3. CUDP needed a way to control its sending rate on congested paths
  4. DThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
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 = 54012806413211608141201networkhost

A host has the address 10.20.205.54/27. What is the fourth octet x of the last usable host address in its subnet?

10.20.205.54/27
x =

3Solutions

1.Number systems

Answer: N = −92

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 + 32 + 4 = −92

Answer

N = −92
2.Number systems

Answer: M = 128 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

256·256·16 = 1048576 bit = 131072 B

Convert to KiB

131072/1024 = 128 KiB

Answer

M = 128 KiB
3.Propositional logic & proofs

Answer: ¬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).

The law

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(b ∨ c) ≡ ¬b ∧ ¬c

Check with the truth table

b = 1, c = 1: 0 | 0 0 0 1
b = 1, c = 0: 0 | 1 0 0 1
b = 0, c = 1: 0 | 1 1 0 0
b = 0, c = 0: 1 | 1 0 1 1

Answer

¬b ∧ ¬c

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

4.Propositional logic & proofs

Answer: ∃x (P(x) ∧ ¬Q(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 (P(x) → Q(x)) ≡ ∃x (P(x) ∧ ¬Q(x))

Answer

∃x (P(x) ∧ ¬Q(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 = 11

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.

m3: m2, m1, m7, m11 → 0
ABCD0001111000011110111111

The isolated 1

m3 = A′B′CD → AB = 00, CD = 11

Answer

AB = 00, CD = 11
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.

C′D, B′D′, B′C′, A′D, A′B′
ABCD0001111000011110101113X204151706012113015014X8X9011110

Minimal cover

F = C′D + B′D′ + A′D
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 77−1 mod 79 = 39

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·77 (mod 79).

r = 79, t = 0
r = 77, q = 1, t = 1
r = 2, q = 38, t = −1
r = 1, q = 2, t = 39

Inverse

1 ≡ 39·77 (mod 79)
x = 39
check: 77·39 = 3003 = 38·79 + 1

Answer

77−1 mod 79 = 39
8.Number theory & modular arithmetic

Answer: m = 88

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 = 161, φ(n) = 132
d = 19−1 mod 132 = 7

Decrypt by repeated squaring

7 = 4 + 2 + 1
601 ≡ 60 (mod 161)
602 ≡ 58 (mod 161)
604 ≡ 144 (mod 161)
607 ≡ 60·58·144 ≡ 88 (mod 161)

Message

m = 88
check: 8819 mod 161 = 60

Answer

m = 88
9.Algorithms & complexity

Answer: D · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

D · O(mn)
10.Algorithms & complexity

Answer: C · Directed acyclic graphs

Why this is the answer

A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.

Answer

C · Directed acyclic graphs
11.Data structures

Answer: D · 20

Why this is the answer

20 is smaller than 50 and smaller than 30, so it becomes the left child of 30; 40 becomes its right child.

Answer

D · 20
12.Data structures

Answer: D · Prefix-sum queries and single-element updates

Why this is the answer

A Fenwick tree stores partial sums over ranges determined by the low bits of each index, so both prefix sums and point updates take O(log n).

Answer

D · Prefix-sum queries and single-element updates
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: N = 1

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

Common neighbours

(M²)DB = 0·0 + 1·0 + 0·1 + 0·1 + 1·1 + 0·0 = 1

Answer

N = 1
15.Databases & SQL

Answer: B · Atomicity

Why this is the answer

ACID: atomicity, consistency, isolation, durability. Atomicity means a transaction is applied completely or not at all.

Answer

B · Atomicity
16.Databases & SQL

Answer: C · It is in 2NF but not in 3NF

Why this is the answer

The key has one column, so 2NF holds, but emp_id → dept_id → dept_name is a transitive dependency, which violates 3NF.

Answer

C · It is in 2NF but not in 3NF
17.Computer networks

Answer: D · 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

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

Answer: D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths

Why this is the answer

The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.

Answer

D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
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 = 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).

Block size

25 = 32

Subnet of this host

network: ⌊54/32⌋·32 = 32
broadcast: 32 + 32 − 1 = 63
usable hosts: 33 … 62

Answer

x = 62

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