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 complement?−128?64?32?16?8?4?2?1

Write −92 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −928 bit
U =
2.Number systemsHard
4866 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4866 items its own binary code of the same length. What is the smallest number of bits b needed?

4866
b = bit
3.Propositional logic & proofsMedium
¬a → b¬¬a → ¬b¬b → ¬¬ab ∧ ¬¬a

You want to prove b → ¬a by contrapositive. Which implication do you prove instead? Click it.

b → ¬a
  1. 1¬a → b
  2. 2¬¬a → ¬b
  3. 3¬b → ¬¬a
  4. 4b ∧ ¬¬a
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ¬(q → r) ∨ ((q ∨ s) → ¬p) true? Call this number N.

¬(q → r) ∨ ((q ∨ s) → ¬p)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010111312040517161120130151141809011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 3, 6, 7, 8, 12, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 3, 6, 7, 8, 12, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKUP

A 3-bit binary up counter starts at 5 (decimal, Q = 101). Which count Q (in decimal) does it show after 18 clock pulses?

Q0 = 518n = 3
Q =
7.Number theory & modular arithmeticMedium

Solve 39x ≡ 45 (mod 59). Give the solution x with 0 < x < 59.

39x ≡ 45 (mod 59)
x =
8.Number theory & modular arithmeticHard

Solve 38x ≡ 48 (mod 52). Give the smallest positive solution x.

38x ≡ 48 (mod 52)
x =
9.Algorithms & complexityMedium

Which statement about a recursive function without a reachable base case is correct?

  1. AIt keeps calling itself until the call stack overflows
  2. BThe compiler always turns it into a loop that ends safely
  3. CIt returns 0 automatically
  4. DIt runs in O(1) time
10.Algorithms & complexityHard

Why can Dijkstra's algorithm give wrong results when some edge weights are negative?

  1. AA vertex is finalised when extracted, but a later negative edge could still give it a shorter path
  2. BIt only works on undirected graphs
  3. CIt goes into an infinite loop on every negative edge
  4. DNegative numbers cannot be stored in a priority queue
11.Data structuresMedium

How many nodes does a perfect binary tree of height h have, with height counted in edges?

  1. Ah2
  2. B2h + 1
  3. C2h+1 − 1
  4. D2h
12.Data structuresHard

How long does it take to search for an arbitrary key in a binary min-heap of n elements?

  1. AO(log log n)
  2. BO(n)
  3. CO(log n)
  4. DO(1)
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
18126178119ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 3.

start: Ak = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

A LEFT JOIN returns:

  1. ARows that match in neither table
  2. BAll rows from the left table and matching rows from the right, with NULLs where there is no match
  3. COnly matching rows from both tables
  4. DAll rows from the right table only
16.Databases & SQLHard

What is a known drawback of decomposing tables into BCNF compared with 3NF?

  1. ASome functional dependencies may no longer be checkable within a single table
  2. BThe decomposition always loses data when joined back
  3. CBCNF tables cannot have primary keys
  4. DBCNF allows more redundancy than 3NF
17.Computer networksMedium

What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?

  1. A/24
  2. B/8
  3. C/16
  4. D/32
18.Computer networksHard

A DNS cache poisoning attack aims to:

  1. AMake a resolver store a false IP address for a domain
  2. BSteal Wi-Fi passwords
  3. CDelete DNS servers
  4. DOverload a server with traffic so that it stops answering queries
19.IPv4 subnettingMedium
255.255.255.25211111111111111111111111111111100network bitshost bits

A subnet uses the mask 255.255.255.252. How many usable host addresses H does it have?

255.255.255.252
H =
20.IPv4 subnettingHard
third octet = 145112806403211608040211networkhost

A host has the address 172.20.145.147/21. What is the third octet x of its network address?

172.20.145.147/21
x =

3Solutions

1.Number systems

Answer: U = 164

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.

Invert and add one

92 = 0101 1100
inverted: 1010 0011
+1 → 1010 0100

Read it as unsigned

1010 0100₂ = 256 − 92 = 164

Answer

U = 164
2.Number systems

Answer: b = 13 bit

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.

Powers of two

212 = 4096 < 4866
213 = 8192 ≥ 4866

Answer

b = 13
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).

Parts of the implication

a = b, b = ¬a

Contrapositive

¬b → ¬a = ¬¬a → ¬b

Answer

¬¬a → ¬b

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: N = 12

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

q → r; ¬(q → r); q ∨ s; ¬p; (q ∨ s) → ¬p; ¬(q → r) ∨ ((q ∨ s) → ¬p)

Split on p

p = 1: 4
p = 0: 8
N = 4 + 8 = 12

Answer

N = 12
5.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.

ABCD000111100001111010111312040517161120130151141809011010

Group the 0s

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

(A′ + D′)
(A + B′ + C)
(A′ + B + C′)

Minimal POS form

F = (A′ + D′)(A + B′ + C)(A′ + B + C′)
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 7

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 8 pulses bring the counter back to the same state.

18 = 2·8 + 2

Count after the remaining pulses

Q = (5 + 2) mod 8 = 7
Q = 111₂

Answer

Q = 7
7.Number theory & modular arithmetic

Answer: x = 42

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).

Inverse of 39

r = 59, t = 0
r = 39, q = 1, t = 1
r = 20, q = 1, t = −1
r = 19, q = 1, t = 2
r = 1, q = 19, t = −3
39−1 ≡ 56 (mod 59)

Solve

x ≡ 56·45 = 2520 ≡ 42 (mod 59)
check: 39·42 = 1638 ≡ 45 (mod 59)

Answer

x = 42
8.Number theory & modular arithmetic

Answer: x = 4

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).

Divide by the gcd

gcd(38, 52) = 2
19x ≡ 24 (mod 26)

Solve the reduced congruence

19−1 ≡ 11 (mod 26) → x ≡ 4 (mod 26)

Solutions modulo 52

x ∈ {4, 30}
x = 4

Answer

x = 4
9.Algorithms & complexity

Answer: A · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

A · It keeps calling itself until the call stack overflows
10.Algorithms & complexity

Answer: A · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path

Why this is the answer

The greedy step relies on non-negative weights so that extracted distances are final; Bellman–Ford is used when weights may be negative.

Answer

A · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path
11.Data structures

Answer: C · 2h+1 − 1

Why this is the answer

Summing 1 + 2 + 4 + ... + 2h gives 2h+1 − 1; for example, height 2 gives 7 nodes.

Answer

C · 2h+1 − 1
12.Data structures

Answer: B · O(n)

Why this is the answer

A heap gives no rule for which subtree contains a key, so in the worst case all nodes must be checked; only the minimum is fast to find.

Answer

B · O(n)
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: edge CE

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.

Prim’s algorithm

1. AC (12)
2. BC (6)
3. CE (8)
4. DE (11)

edge number 3

CE

Answer

edge CE
15.Databases & SQL

Answer: B · All rows from the left table and matching rows from the right, with NULLs where there is no match

Why this is the answer

A left outer join keeps every left row; unmatched right columns are filled with NULL.

Answer

B · All rows from the left table and matching rows from the right, with NULLs where there is no match
16.Databases & SQL

Answer: A · Some functional dependencies may no longer be checkable within a single table

Why this is the answer

A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.

Answer

A · Some functional dependencies may no longer be checkable within a single table
17.Computer networks

Answer: A · /24

Why this is the answer

Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.

Answer

A · /24
18.Computer networks

Answer: A · Make a resolver store a false IP address for a domain

Why this is the answer

Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.

Answer

A · Make a resolver store a false IP address for a domain
19.IPv4 subnetting

Answer: H = 2

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

255.255.255.252 = /30

Usable hosts

22 − 2 = 2

Answer

H = 2
20.IPv4 subnetting

Answer: x = 144

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 in the third octet

23 = 8

Subnet of this host

network: 172.20.144.0
broadcast: 172.20.151.255

Answer

x = 144

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