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 −58 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −588 bit
U =
2.Number systemsHard
32 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (c ∨ a) → (c ∧ a) a tautology, a contradiction or a contingency? Click your answer.

(c ∨ a) → (c ∧ a)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
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
ABCD000111100001111000011312140517160120131150141809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(2, 3, 4, 6, 7, 8, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(2, 3, 4, 6, 7, 8, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​CLKMOD-13

A mod-13 counter has to count through 13 different states. What is the minimum number of flip-flops n it needs?

N = 13
n =
7.Number theory & modular arithmeticMedium

Find the inverse of 77 modulo 80: the number x with 0 < x < 80 and 77·x ≡ 1 (mod 80).

a = 77n = 80
77−1 mod 80 =
8.Number theory & modular arithmeticHard
172c = me​ mod n?m = cd​ mod nm(n, e) = (205, 7)d

Encrypt the message m = 172 with the RSA public key (n, e) = (205, 7). Find the ciphertext c.

n = 205e = 7m = 172
c =
9.Algorithms & complexityMedium

Why do many library sorting routines switch to insertion sort for very small subarrays?

  1. ARecursive algorithms cannot handle arrays below 10 elements
  2. BInsertion sort is O(n log n) in the worst case
  3. CInsertion sort has little overhead and is fast on small or nearly sorted inputs
  4. DInsertion sort uses less memory than any other sorting algorithm on large inputs
10.Algorithms & complexityHard

How is the travelling salesman problem classified?

  1. AThe decision version is in P but the optimisation version is NP-complete
  2. BIt is undecidable
  3. CIt is in P because Dijkstra's algorithm solves it by visiting every city once
  4. DThe decision version is NP-complete and the optimisation version is NP-hard
11.Data structuresMedium

What is the time complexity of searching for a key in a balanced binary search tree?

  1. AO(log n)
  2. BO(1)
  3. CO(n log n)
  4. DO(n)
12.Data structuresHard

For a hash table to work correctly, which rule must hash and equality follow?

  1. AEqual hash values imply equal keys
  2. BDifferent keys must have different hash values
  3. CHash values must be prime numbers
  4. DEqual keys must have equal hash values
13.Graph theoryMedium

A connected graph has 19 vertices and 28 edges. How many edges k must be removed to leave a spanning tree?

|V| = 19|E| = 28
k =
14.Graph theoryHard
1491211214964375ABCDEFG

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

Table a has 3 rows and table b has 4 rows. How many rows does SELECT * FROM a CROSS JOIN b return?

  1. A4
  2. B3
  3. C12
  4. D7
16.Databases & SQLHard

A reporting database copies customer names into the orders table to avoid joins. What is the main cost of this denormalisation?

  1. ARedundant copies must be kept consistent whenever customer data changes
  2. BThe database cannot use indexes any more
  3. CPrimary keys can no longer be used
  4. DQueries on the orders table become slower
17.Computer networksMedium

A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?

  1. A10.1.0.0/16, because it is the longest matching prefix
  2. B10.0.0.0/8, because it covers more addresses than the /16
  3. CBoth routes, splitting the traffic equally
  4. D10.0.0.0/8, because it was added first
18.Computer networksHard

In a TCP connection, what does the receive window (rwnd) control?

  1. AThe number of open ports
  2. BThe size of each IP packet header
  3. CHow much unacknowledged data the sender may send, based on the receiver's buffer
  4. DThe retransmission time-out, calculated from the measured round-trip time of each segment
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.91.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 5?

192.168.91.0/24/28subnet 5 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 43012806413201618041211networkhost

A host has the address 10.20.177.43/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.177.43/28
x =

3Solutions

1.Number systems

Answer: U = 198

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

58 = 0011 1010
inverted: 1100 0101
+1 → 1100 0110

Read it as unsigned

1100 0110₂ = 256 − 58 = 198

Answer

U = 198
2.Number systems

Answer: M = 1024 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

512·512·32 = 8388608 bit = 1048576 B

Convert to KiB

1048576/1024 = 1024 KiB

Answer

M = 1024 KiB
3.Propositional logic & proofs

Answer: contingency

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).

Truth table row by row

a = 1, c = 1: c ∨ a = 1, c ∧ a = 1, (c ∨ a) → (c ∧ a) = 1
a = 1, c = 0: c ∨ a = 1, c ∧ a = 0, (c ∨ a) → (c ∧ a) = 0
a = 0, c = 1: c ∨ a = 1, c ∧ a = 0, (c ∨ a) → (c ∧ a) = 0
a = 0, c = 0: c ∨ a = 0, c ∧ a = 0, (c ∨ a) → (c ∧ a) = 1

Classify

(c ∨ a) → (c ∧ a): 1, 0, 0, 1
1 in 2 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
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: 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.

ABCD000111100001111000011312140517160120131150141809011010

Group the 1s

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

BCD
A′C
A′BD′
AB′C′D′

Minimal SOP form

F = BCD + A′C + A′BD′ + AB′C′D′
L = 12

Answer

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

Answer: n = 4

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.

Number of flip-flops

2ⁿ ≥ 13 → n = 4 (2⁴ = 16)

Answer

n = 4
7.Number theory & modular arithmetic

Answer: 77−1 mod 80 = 53

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 80).

r = 80, t = 0
r = 77, q = 1, t = 1
r = 3, q = 25, t = −1
r = 2, q = 1, t = 26
r = 1, q = 2, t = −27

Inverse

1 ≡ −27·77 (mod 80)
x = 53
check: 77·53 = 4081 = 51·80 + 1

Answer

77−1 mod 80 = 53
8.Number theory & modular arithmetic

Answer: c = 43

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.

Repeated squaring

7 = 4 + 2 + 1
1721 ≡ 172 (mod 205)
1722 ≡ 64 (mod 205)
1724 ≡ 201 (mod 205)
1727 ≡ 172·64·201 ≡ 43 (mod 205)

Ciphertext

c = 43

Answer

c = 43
9.Algorithms & complexity

Answer: C · Insertion sort has little overhead and is fast on small or nearly sorted inputs

Why this is the answer

For small n the simple loop of insertion sort beats the recursion overhead of quicksort or merge sort, even though it is O(n²).

Answer

C · Insertion sort has little overhead and is fast on small or nearly sorted inputs
10.Algorithms & complexity

Answer: D · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

D · The decision version is NP-complete and the optimisation version is NP-hard
11.Data structures

Answer: A · O(log n)

Why this is the answer

A balanced tree has height about log₂ n, so a search follows one path from root to leaf.

Answer

A · O(log n)
12.Data structures

Answer: D · Equal keys must have equal hash values

Why this is the answer

If two equal keys hashed differently they would be stored in different buckets and lookups would fail; unequal keys may still share a hash.

Answer

D · Equal keys must have equal hash values
13.Graph theory

Answer: k = 10

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.

Edges of a spanning tree

19 − 1 = 18

Edges to remove

k = 28 − 18 = 10

Answer

k = 10
14.Graph theory

Answer: d(F) = 17

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: 14, C: 9
C (9): E: 11, G: 12
E (11): D: 25, F: 17
G (12): D: 19
B (14)
F (17)
D (19)

Shortest path

A → C → E → F
d(F) = 17

Answer

d(F) = 17
15.Databases & SQL

Answer: C · 12

Why this is the answer

A cross join gives the Cartesian product, 3 × 4 = 12 rows.

Answer

C · 12
16.Databases & SQL

Answer: A · Redundant copies must be kept consistent whenever customer data changes

Why this is the answer

Denormalisation speeds up reads but brings back redundancy, so updates must change every copy or anomalies appear.

Answer

A · Redundant copies must be kept consistent whenever customer data changes
17.Computer networks

Answer: A · 10.1.0.0/16, because it is the longest matching prefix

Why this is the answer

IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.

Answer

A · 10.1.0.0/16, because it is the longest matching prefix
18.Computer networks

Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer

Why this is the answer

The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.

Answer

C · How much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnetting

Answer: x = 64

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

24 = 16

Subnet number 5

network: (5 − 1)·16 = 64
broadcast: 64 + 16 − 1 = 79

Answer

x = 64
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

24 = 16

Subnet of this host

network: ⌊43/16⌋·16 = 32
broadcast: 32 + 16 − 1 = 47
usable hosts: 33 … 46

Answer

x = 47

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