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
place values in two's complement1−12806403201608041201

The 8-bit pattern 1000 0010 stores a signed integer in two's complement. What is its value N in decimal?

1000 00108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 0000 >> 1. What is the result's value R in decimal?

1011 0000>> 1
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (p ↔ q) ∧ (p ↔ ¬q) a tautology, a contradiction or a contingency? Click your answer.

(p ↔ q) ∧ (p ↔ ¬q)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x ∀y ¬R(x, y)∀x ∃y ¬R(x, y)∃x ∃y ¬R(x, y)∃x ∀y ¬R(x, y)

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

¬∃x ∀y R(x, y)
  1. 1∀x ∀y ¬R(x, y)
  2. 2∀x ∃y ¬R(x, y)
  3. 3∃x ∃y ¬R(x, y)
  4. 4∃x ∀y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′C′ + A′BC′ + BC′ + A′BC to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′C′ + A′BC′ + BC′ + A′BC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD123456

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

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

Find the inverse of 14 modulo 23: the number x with 0 < x < 23 and 14·x ≡ 1 (mod 23).

a = 14n = 23
14−1 mod 23 =
8.Number theory & modular arithmeticHard

53 is prime. Use Fermat’s little theorem to find 2343210 mod 53.

p = 53k = 43210
2343210 mod 53 =
9.Algorithms & complexityMedium

Quicksort always uses the first element as pivot. On which input does it show its worst-case behaviour?

  1. AAn array in random order
  2. BAn array of length 2
  3. CAn array whose elements are all different
  4. DAn array that is already sorted
10.Algorithms & complexityHard

Which algorithm finds a minimum spanning tree by repeatedly adding the cheapest edge that does not form a cycle?

  1. ADijkstra's algorithm
  2. BPrim's algorithm starting from one node
  3. CKruskal's algorithm
  4. DFloyd–Warshall
11.Data structuresMedium

How long does a search for a word of length L take in a trie with a fixed alphabet?

  1. AO(L × n)
  2. BO(n), where n is the number of words
  3. CO(log n), where n is the number of words
  4. DO(L), independent of how many words are stored
12.Data structuresHard

For each element of an array you want the next greater element to its right, in O(n) total time. Which structure helps?

  1. AA hash table of all elements
  2. BA binary search tree rebuilt for each element
  3. CA stack that keeps elements waiting for their answer
  4. DA queue that holds every pair of elements
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
1571913131114ABCDEF

Use Dijkstra’s algorithm to find the length d(D) of the shortest path from A to D in the weighted graph.

start: Aend: D
d(D) =
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. A7
  2. B4
  3. C3
  4. D12
16.Databases & SQLHard

There is a composite index on (last_name, first_name). Which condition can use it most effectively?

  1. AWHERE last_name = 'Jansen'
  2. BWHERE first_name = 'Eva'
  3. CWHERE UPPER(first_name) = 'EVA'
  4. DWHERE first_name LIKE '%a'
17.Computer networksMedium

Which IPv4 header field stops packets from circulating forever in a routing loop?

  1. ATime to live (TTL)
  2. BHeader checksum
  3. CIdentification
  4. DType of service
18.Computer networksHard

Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?

  1. AA longer cable between the hosts
  2. BA router or layer 3 switch to route between the VLANs
  3. CA hub connected to both VLANs
  4. DThe same MAC address on both hosts
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 10.43.240.0/24 is divided into equal subnets with prefix /28. How many subnets N are there?

10.43.240.0/24/28
N =
20.IPv4 subnettingHard
last octet = 194112816403201608041201networkhost

A host has the address 10.20.123.194/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.123.194/29
x =

3Solutions

1.Number systems

Answer: N = −126

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 + 2 = −126

Answer

N = −126
2.Number systems

Answer: R = 88

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 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1011 0000 >> 1
= 0101 1000

Value of the result

0101 1000₂ = 88

Answer

R = 88
3.Propositional logic & proofs

Answer: contradiction

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

p = 1, q = 1: p ↔ q = 1, ¬q = 0, p ↔ ¬q = 0, (p ↔ q) ∧ (p ↔ ¬q) = 0
p = 1, q = 0: p ↔ q = 0, ¬q = 1, p ↔ ¬q = 1, (p ↔ q) ∧ (p ↔ ¬q) = 0
p = 0, q = 1: p ↔ q = 0, ¬q = 0, p ↔ ¬q = 1, (p ↔ q) ∧ (p ↔ ¬q) = 0
p = 0, q = 0: p ↔ q = 1, ¬q = 1, p ↔ ¬q = 0, (p ↔ q) ∧ (p ↔ ¬q) = 0

Classify

(p ↔ q) ∧ (p ↔ ¬q): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ∀x ∃y ¬R(x, y)

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) ≡ ∀x ∃y ¬R(x, y)

Answer

∀x ∃y ¬R(x, y)

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 = 6

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′C′ → m(0, 2)
A′BC′ → m(2)
BC′ → m(2, 6)
A′BC → m(3)
ABC00011110011001131204050716

Group and read off

F = BC′ + A′C′ + A′B

Count the literals

L = 6

Answer

L = 6
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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 0 → Q = 0
3: D = 1 → Q = 1
4: D = 0 → Q = 0
5: D = 0 → Q = 0
6: D = 0 → Q = 0

Count the changes

0 → 1 → 0 → 1 → 0 → 0 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: 14−1 mod 23 = 5

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·14 (mod 23).

r = 23, t = 0
r = 14, q = 1, t = 1
r = 9, q = 1, t = −1
r = 5, q = 1, t = 2
r = 4, q = 1, t = −3
r = 1, q = 4, t = 5

Inverse

1 ≡ 5·14 (mod 23)
x = 5
check: 14·5 = 70 = 3·23 + 1

Answer

14−1 mod 23 = 5
8.Number theory & modular arithmetic

Answer: 2343210 mod 53 = 52

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

43210 mod 52 = 50
2343210 ≡ 2350 (mod 53)

Small power

2350 mod 53 = 52

Answer

2343210 mod 53 = 52
9.Algorithms & complexity

Answer: D · An array that is already sorted

Why this is the answer

On sorted input the pivot is always the minimum, so one side is empty and the other has n − 1 elements, giving O(n²).

Answer

D · An array that is already sorted
10.Algorithms & complexity

Answer: C · Kruskal's algorithm

Why this is the answer

Kruskal sorts edges by weight and adds them if they connect different components; Prim grows one tree from a start vertex.

Answer

C · Kruskal's algorithm
11.Data structures

Answer: D · O(L), independent of how many words are stored

Why this is the answer

Each character selects one child, so a lookup takes L steps no matter how many words the trie contains.

Answer

D · O(L), independent of how many words are stored
12.Data structures

Answer: C · A stack that keeps elements waiting for their answer

Why this is the answer

Scanning left to right, pop every stacked element smaller than the current one (the current element is their answer), then push it; each element moves at most twice.

Answer

C · A stack that keeps elements waiting for their answer
13.Graph theory

Answer: H = 2520

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 = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: d(D) = 23

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): C: 15
C (15): B: 22, E: 24
B (22): D: 23
D (23): F: 36
E (24): F: 35
F (35)

Shortest path

A → C → B → D
d(D) = 23

Answer

d(D) = 23
15.Databases & SQL

Answer: D · 12

Why this is the answer

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

Answer

D · 12
16.Databases & SQL

Answer: A · WHERE last_name = 'Jansen'

Why this is the answer

A composite B-tree index is ordered by last_name and then first_name, so a search on the leading column works, but first_name alone cannot use the order.

Answer

A · WHERE last_name = 'Jansen'
17.Computer networks

Answer: A · Time to live (TTL)

Why this is the answer

Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.

Answer

A · Time to live (TTL)
18.Computer networks

Answer: B · A router or layer 3 switch to route between the VLANs

Why this is the answer

VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.

Answer

B · A router or layer 3 switch to route between the VLANs
19.IPv4 subnetting

Answer: N = 16

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).

Borrowed bits

28 − 24 = 4

Number of subnets

N = 24 = 16

Answer

N = 16

Each /28 subnet has 24 − 2 = 14 usable hosts.

20.IPv4 subnetting

Answer: x = 198

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: ⌊194/8⌋·8 = 192
broadcast: 192 + 8 − 1 = 199
usable hosts: 193 … 198

Answer

x = 198

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