Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 −103 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1038 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0010 0100 >> 2. What is the result's value R in decimal?

0010 0100>> 2
R =
3.Propositional logic & proofsMedium
¬a → ¬ba ∨ ¬b¬a ∨ b¬a ∧ b

Which formula is logically equivalent to a → b? Click it.

a → b
  1. 1¬a → ¬b
  2. 2a ∨ ¬b
  3. 3¬a ∨ b
  4. 4¬a ∧ b
4.Propositional logic & proofsHard
∀x ∀y (R(x, y) ∧ ¬R(y, x))∃x ∃y (R(x, y) → ¬R(y, x))∃x ∃y (¬R(x, y) ∧ ¬R(y, x))∃x ∃y (R(x, y) ∧ ¬R(y, x))

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

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∀x ∀y (R(x, y) ∧ ¬R(y, x))
  2. 2∃x ∃y (R(x, y) → ¬R(y, x))
  3. 3∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
  4. 4∃x ∃y (R(x, y) ∧ ¬R(y, x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000111302041507060121130151141819011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 3, 5, 8, 9, 13, 14). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(1, 3, 5, 8, 9, 13, 14)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKMOD-6

A synchronous mod-6 counter counts 0, 1, …, 5 and then returns to 0. It starts at 1. Which count Q (in decimal) does it show after 7 clock pulses?

Q0 = 17N = 6
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 5) and x ≡ 5 (mod 6).

x ≡ 2 (mod 5)x ≡ 5 (mod 6)
x =
8.Number theory & modular arithmeticHard

53 is prime. Use Fermat’s little theorem to find 1567432 mod 53.

p = 53k = 67432
1567432 mod 53 =
9.Algorithms & complexityMedium

What does amortised analysis measure?

  1. AThe average cost per operation over a worst-case sequence of operations
  2. BThe cost of an algorithm on the best input
  3. CThe expected cost over random inputs
  4. DThe cost of the single most expensive operation anywhere in the sequence
10.Algorithms & complexityHard

The halting problem shows that:

  1. ASome programs run faster on faster computers, so halting depends on hardware
  2. BLoops should be avoided
  3. CEvery program eventually halts
  4. DNo algorithm can decide for every program and input whether it halts
11.Data structuresMedium

In a binary min-heap stored in an array (index starting at 0), the children of the node at index i are at:

  1. A2i + 1 and 2i + 2
  2. B2i and 2i + 1
  3. Ci/2 and i/2 + 1
  4. Di + 1 and i + 2
12.Data structuresHard

Which statement about heapsort is correct?

  1. AIt needs a binary search tree as extra storage, so it uses O(n) extra memory
  2. BIt runs in O(n) time on any input
  3. CIt is stable and needs O(n) extra memory
  4. DIt sorts in place in O(n log n) worst-case time but is not stable
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
11971594315ABCDEF

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

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. AUse GROUP BY manager_id only
  2. BCreate a second copy of the table on disk
  3. CCombine the table with itself using UNION ALL and a WHERE filter
  4. DJoin the employees table with itself using two aliases
16.Databases & SQLHard

Which window function assigns 1, 2, 3, ... to rows within each partition without ties?

  1. ADENSE_RANK()
  2. BRANK()
  3. CROW_NUMBER()
  4. DCOUNT()
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ALink bandwidth
  2. BMeasured delay
  3. CHop count
  4. DBusiness policy
18.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A6 kbit/s
  2. B12 kbit/s
  3. C24 kbit/s
  4. D3 kbit/s
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1331 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1331 hosts
p =
20.IPv4 subnettingHard
last octet = 46012806413201618141201networkhost

A host has the address 10.20.225.46/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.225.46/29
x =

3Solutions

1.Number systems

Answer: U = 153

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

103 = 0110 0111
inverted: 1001 1000
+1 → 1001 1001

Read it as unsigned

1001 1001₂ = 256 − 103 = 153

Answer

U = 153
2.Number systems

Answer: R = 9

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.

Bit by bit

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0010 0100 >> 2
= 0000 1001

Value of the result

0000 1001₂ = 9

Answer

R = 9
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 | 1 1 1 0
a = 1, b = 0: 0 | 1 1 0 0
a = 0, b = 1: 1 | 0 0 1 1
a = 0, b = 0: 1 | 1 1 1 0

Answer

¬a ∨ b

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

4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, 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 ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, 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: L = 12

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.

ABCD000111100001111000111302041507060121130151141819011010

Group the 1s

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

C′D
A′B′D
AB′C′
ABCD′

Minimal SOP form

F = C′D + A′B′D + AB′C′ + ABCD′
L = 12

Answer

L = 12
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 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.

Full cycles

Every 6 pulses bring the counter back to the same state.

7 = 1·6 + 1

Count after the remaining pulses

Q = (1 + 1) mod 6 = 2
Q = 010₂

Answer

Q = 2
7.Number theory & modular arithmetic

Answer: x = 17

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 = 2 + 5k

Second congruence

2 + 5k ≡ 5 (mod 6) → 5k ≡ 3 (mod 6)
k ≡ 5·3 ≡ 3 (mod 6)

Result

x = 2 + 5·3 = 17
x = 17
check: 17 mod 5 = 2, 17 mod 6 = 5

Answer

x = 17
8.Number theory & modular arithmetic

Answer: 1567432 mod 53 = 15

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.

Reduce the exponent

67432 mod 52 = 40
1567432 ≡ 1540 (mod 53)

Small power

1540 mod 53 = 15

Answer

1567432 mod 53 = 15
9.Algorithms & complexity

Answer: A · The average cost per operation over a worst-case sequence of operations

Why this is the answer

Amortised analysis spreads the cost of rare expensive operations over many cheap ones, giving a guaranteed average for any sequence.

Answer

A · The average cost per operation over a worst-case sequence of operations
10.Algorithms & complexity

Answer: D · No algorithm can decide for every program and input whether it halts

Why this is the answer

Turing showed the halting problem is undecidable: assuming a decider exists leads to a contradiction with a program that does the opposite.

Answer

D · No algorithm can decide for every program and input whether it halts
11.Data structures

Answer: A · 2i + 1 and 2i + 2

Why this is the answer

For 0-based arrays, the left child is at 2i + 1, the right child at 2i + 2, and the parent at ⌊(i − 1)/2⌋.

Answer

A · 2i + 1 and 2i + 2
12.Data structures

Answer: D · It sorts in place in O(n log n) worst-case time but is not stable

Why this is the answer

Heapsort builds a max-heap and extracts the maximum n times within the same array; swaps over long distances can reorder equal keys.

Answer

D · It sorts in place in O(n log n) worst-case time but is not stable
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) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 1, deg(E) = 3, deg(F) = 1

Odd vertices

C, D, E, F → 4
neither

Answer

neither
14.Graph theory

Answer: d(F) = 21

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: 11, C: 9
C (9): E: 18
B (11): D: 26
E (18): F: 21
F (21): D: 25
D (25)

Shortest path

A → C → E → F
d(F) = 21

Answer

d(F) = 21
15.Databases & SQL

Answer: D · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

D · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: C · ROW_NUMBER()

Why this is the answer

ROW_NUMBER gives unique sequential numbers; RANK leaves gaps after ties; DENSE_RANK does not.

Answer

C · ROW_NUMBER()
17.Computer networks

Answer: C · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

C · Hop count
18.Computer networks

Answer: B · 12 kbit/s

Why this is the answer

2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.

Answer

B · 12 kbit/s
19.IPv4 subnetting

Answer: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1331
211 − 2 = 2046 ≥ 1331 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 47

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

23 = 8

Subnet of this host

network: ⌊46/8⌋·8 = 40
broadcast: 40 + 8 − 1 = 47
usable hosts: 41 … 46

Answer

x = 47

ExamProblems · examproblems.com/courses/computer-science-fundamentals