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 complement0−12806403201608040201

Set the bits so they store −46 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −468 bit
U =
2.Number systemsHard
8 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px8 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. 1¬a ∧ b
  2. 2¬a → ¬b
  3. 3a ∨ ¬b
  4. 4¬a ∨ b
4.Propositional logic & proofsHard
qrr → q¬r¬q11100100

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

r → q¬r∴ ¬q
  1. 1q = 1, r = 1
  2. 2q = 1, r = 0
  3. 3q = 0, r = 1
  4. 4q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111111

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(1, 4, 6, 8, 10, 11, 12, 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
CLKT12345678

A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

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

Find the inverse of 28 modulo 53: the number x with 0 < x < 53 and 28·x ≡ 1 (mod 53).

a = 28n = 53
28−1 mod 53 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (253, 23)d = ?

An RSA key uses the primes p = 11 and q = 23 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 11q = 23e = 23
d =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the shortest activity first
  2. BPick the activity with the fewest overlaps first, then repeat with the compatible ones
  3. CPick the activity that starts earliest
  4. DPick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexityHard

Which bound is correct for log(n!)?

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

A circular buffer (ring buffer) is commonly used to implement:

  1. AA binary search tree
  2. BA fixed-size queue
  3. CA graph traversal
  4. DA hash function
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. ALooking up the object may fail because it sits in the bucket for its old hash
  2. BThe map throws an error at the moment the field changes
  3. CNothing changes, because a key's hash is computed only once when the program starts
  4. DThe map automatically moves the key to its new bucket
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
4119731175122ABCDEFG

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

Which statement about HAVING is correct?

  1. AIt can use aggregate functions to filter groups
  2. BIt must come before WHERE in the SELECT statement
  3. CIt sorts the results
  4. DIt cannot be used with GROUP BY
16.Databases & SQLHard

Table enrolment(student_id, course_id, student_name, grade) has the key (student_id, course_id). Which normal form does it violate?

  1. ANone, it is in BCNF
  2. BSecond normal form, because student_name depends on student_id only
  3. CFirst normal form, because the grade column does not hold atomic values
  4. DThird normal form only, because grade depends on student_name
17.Computer networksMedium

What is a cyclic redundancy check (CRC) used for in Ethernet frames?

  1. ACorrecting every bit error automatically
  2. BEncrypting the frame
  3. CDetecting transmission errors
  4. DChoosing the best route
18.Computer networksHard

An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?

  1. AThe switch shuts down all of its ports
  2. BVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
  3. CAll DNS queries fail across the internet
  4. DThe gateway's IP address is permanently changed to the attacker's address on every router
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2430 subnets
H =
20.IPv4 subnettingHard
last octet = 21012806403211608140211networkhost

A host has the address 10.20.10.21/30. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.10.21/30
x =

3Solutions

1.Number systems

Answer: U = 210

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

46 = 0010 1110
inverted: 1101 0001
+1 → 1101 0010

Read it as unsigned

1101 0010₂ = 256 − 46 = 210

Answer

U = 210
2.Number systems

Answer: M = 768 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

1024·768·8 = 6291456 bit = 786432 B

Convert to KiB

786432/1024 = 768 KiB

Answer

M = 768 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 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
a → b ≡ ¬a ∨ b

Check with the truth table

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

Answer

¬a ∨ b

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

4.Propositional logic & proofs

Answer: q = 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

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

Counterexample

q = 1, r = 0

Answer

q = 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: AB = 00, CD = 01

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.

m1: m0, m3, m5, m9 → 0
ABCD000111100001111011111111

The isolated 1

m1 = A′B′C′D → AB = 00, CD = 01

Answer

AB = 00, CD = 01
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

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

Count the changes

0 → 1 → 0 → 0 → 0 → 0 → 0 → 0 → 0
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: 28−1 mod 53 = 36

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·28 (mod 53).

r = 53, t = 0
r = 28, q = 1, t = 1
r = 25, q = 1, t = −1
r = 3, q = 8, t = 2
r = 1, q = 3, t = −17

Inverse

1 ≡ −17·28 (mod 53)
x = 36
check: 28·36 = 1008 = 19·53 + 1

Answer

28−1 mod 53 = 36
8.Number theory & modular arithmetic

Answer: d = 67

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.

Modulus and totient

n = 11·23 = 253
φ(n) = 10·22 = 220

Extended Euclidean algorithm

r = 220, t = 0
r = 23, q = 9, t = 1
r = 13, q = 1, t = −9
r = 10, q = 1, t = 10
r = 3, q = 3, t = −19
r = 1, q = 3, t = 67

Private exponent

d = 67
check: 23·67 = 1541 = 7·220 + 1

Answer

d = 67
9.Algorithms & complexity

Answer: D · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

D · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: A · 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

A · Θ(n log n)
11.Data structures

Answer: B · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

B · A fixed-size queue
12.Data structures

Answer: A · 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

A · Looking up the object may fail because it sits in the bucket for its old hash
13.Graph theory

Answer: Euler path, no circuit

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

Odd vertices

B, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: d(F) = 18

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: 4
B (4): C: 15, D: 13, G: 9
G (9): E: 11
E (11): F: 18
D (13)
C (15)
F (18)

Shortest path

A → B → G → E → F
d(F) = 18

Answer

d(F) = 18
15.Databases & SQL

Answer: A · It can use aggregate functions to filter groups

Why this is the answer

HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.

Answer

A · It can use aggregate functions to filter groups
16.Databases & SQL

Answer: B · Second normal form, because student_name depends on student_id only

Why this is the answer

student_name depends on only part of the key, which is a partial dependency; moving it to a students table fixes the 2NF violation.

Answer

B · Second normal form, because student_name depends on student_id only
17.Computer networks

Answer: C · Detecting transmission errors

Why this is the answer

The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.

Answer

C · Detecting transmission errors
18.Computer networks

Answer: B · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack

Why this is the answer

ARP spoofing poisons hosts' ARP caches with the attacker's MAC for the gateway IP. The attacker can then read, change or drop traffic; dynamic ARP inspection helps prevent it.

Answer

B · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
19.IPv4 subnetting

Answer: H = 6

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).

Bits to borrow

25 = 32 ≥ 30 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 23

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: ⌊21/4⌋·4 = 20
broadcast: 20 + 4 − 1 = 23
usable hosts: 21 … 22

Answer

x = 23

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