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 −49 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −498 bit
U =
2.Number systemsHard
3123 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3123 items its own binary code of the same length. What is the smallest number of bits b needed?

3123
b = bit
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
ABCD000111100001111010110302040507161120131150140809011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 6, 10, 12, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 6, 10, 12, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKDOWN

A 3-bit binary down counter starts at 6 (decimal, Q = 110). Which count Q (in decimal) does it show after 19 clock pulses?

Q0 = 619n = 3
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 19 modulo 37: the number x with 0 < x < 37 and 19·x ≡ 1 (mod 37).

a = 19n = 37
19−1 mod 37 =
8.Number theory & modular arithmeticHard
523c = me​ mod n?m = cd​ mod nm(n, e) = (731, 5)d

Encrypt the message m = 523 with the RSA public key (n, e) = (731, 5). Find the ciphertext c.

n = 731e = 5m = 523
c =
9.Algorithms & complexityMedium

A function runs one loop of n iterations and then, separately, another loop of n iterations. What is its time complexity?

  1. AO(log n)
  2. BO(2ⁿ)
  3. CO(n²)
  4. DO(n)
10.Algorithms & complexityHard

Which two things must be shown to prove that a decision problem X is NP-complete?

  1. AX is in P, and X reduces to SAT
  2. BX is in NP, and a known NP-complete problem reduces to X in polynomial time
  3. CX reduces to a problem in P, and X is in NP
  4. DX has no polynomial algorithm, and X is decidable
11.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe oldest keys are deleted to make room
  2. BA larger table is allocated and every key is inserted again
  3. CThe existing keys are sorted in place
  4. DOnly the colliding keys are moved to a new overflow list at the end of the table
12.Data structuresHard

How does a radix tree (compressed trie) improve on a plain trie?

  1. AIt merges chains of single-child nodes into one edge labelled with a string
  2. BIt sorts the children of each node by frequency
  3. CIt stores each word in a hash table at the root instead of in separate nodes
  4. DIt keeps only the first letter of each word
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 4 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 34 × deg 53 × deg 2
L =
14.Graph theoryHard
ABCDEF

How many walks of length 3 go from C to F? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

What does SELECT COALESCE(phone, 'unknown') FROM customers return for a customer whose phone is NULL?

  1. AAn error
  2. B'unknown'
  3. CAn empty string
  4. DNULL
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. BFirst normal form, because the grade column does not hold atomic values
  3. CSecond normal form, because student_name depends on student_id only
  4. DThird normal form only, because grade depends on student_name
17.Computer networksMedium

HTTP is stateless. How do websites usually keep a user logged in across requests?

  1. ABy storing the password in the URL
  2. BBy giving each user a fixed IP address
  3. CWith cookies holding a session identifier or token
  4. DBy keeping the TCP connection open for days
18.Computer networksHard

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt removes the need for DNS lookups
  2. BIt increases the bandwidth of the users' own home connections
  3. CIt serves cached content from servers close to the users
  4. DIt replaces TCP with a faster transport protocol
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/246 subnets
H =
20.IPv4 subnettingHard
third octet = 236112816413201618140201networkhost

A host has the address 172.20.236.176/22. What is the third octet x of its network address?

172.20.236.176/22
x =

3Solutions

1.Number systems

Answer: U = 207

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

49 = 0011 0001
inverted: 1100 1110
+1 → 1100 1111

Read it as unsigned

1100 1111₂ = 256 − 49 = 207

Answer

U = 207
2.Number systems

Answer: b = 12 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

211 = 2048 < 3123
212 = 4096 ≥ 3123

Answer

b = 12
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 = 1, ¬q = 0, p ∧ ¬q = 0, (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: N = 7

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.

ABCD000111100001111010110302040507161120131150140809011110

Group the 0s

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

(B + C′ + D′)
(B′ + C + D′)
(A + C′ + D′)
(A + B + C′)
(A + B′ + C)
(A′ + B + C)
(A′ + B′ + C′ + D)

Minimal POS form

F = (B + C′ + D′)(B′ + C + D′)(A + C′ + D′)(A + B + C′)(A + B′ + C)(A′ + B + C)(A′ + B′ + C′ + D)
N = 7

Answer

N = 7

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: Q = 3

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.

19 = 2·8 + 3

Count after the remaining pulses

Q = (6 − 3) mod 8 = 3
Q = 011₂

Answer

Q = 3
7.Number theory & modular arithmetic

Answer: 19−1 mod 37 = 2

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·19 (mod 37).

r = 37, t = 0
r = 19, q = 1, t = 1
r = 18, q = 1, t = −1
r = 1, q = 18, t = 2

Inverse

1 ≡ 2·19 (mod 37)
x = 2
check: 19·2 = 38 = 1·37 + 1

Answer

19−1 mod 37 = 2
8.Number theory & modular arithmetic

Answer: c = 166

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

5 = 4 + 1
5231 ≡ 523 (mod 731)
5232 ≡ 135 (mod 731)
5234 ≡ 681 (mod 731)
5235 ≡ 523·681 ≡ 166 (mod 731)

Ciphertext

c = 166

Answer

c = 166
9.Algorithms & complexity

Answer: D · O(n)

Why this is the answer

n + n = 2n, and constant factors are dropped, so the function is O(n).

Answer

D · O(n)
10.Algorithms & complexity

Answer: B · X is in NP, and a known NP-complete problem reduces to X in polynomial time

Why this is the answer

Membership in NP shows X is not harder than NP; the reduction from a known NP-complete problem shows X is at least as hard as all of NP.

Answer

B · X is in NP, and a known NP-complete problem reduces to X in polynomial time
11.Data structures

Answer: B · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

B · A larger table is allocated and every key is inserted again
12.Data structures

Answer: A · It merges chains of single-child nodes into one edge labelled with a string

Why this is the answer

Collapsing paths with no branching reduces the number of nodes and the memory used while keeping prefix search.

Answer

A · It merges chains of single-child nodes into one edge labelled with a string
13.Graph theory

Answer: L = 18

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.

Handshake lemma for a tree

L + 38 = 2(L + 11) − 2
L = 18

Answer

L = 18

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: N = 1

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0 0
B: 1 0 0 0 1 0
C: 1 0 0 0 1 0
D: 0 0 0 0 1 1
E: 0 1 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)CF = 0·0 + 2·0 + 2·0 + 1·1 + 0·1 + 1·0 = 1

Answer

N = 1
15.Databases & SQL

Answer: B · 'unknown'

Why this is the answer

COALESCE checks its arguments from left to right and returns the first that is not NULL, here 'unknown'.

Answer

B · 'unknown'
16.Databases & SQL

Answer: C · 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

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

Answer: C · With cookies holding a session identifier or token

Why this is the answer

The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.

Answer

C · With cookies holding a session identifier or token
18.Computer networks

Answer: C · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

C · It serves cached content from servers close to the users
19.IPv4 subnetting

Answer: H = 30

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

23 = 8 ≥ 6 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 236

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

22 = 4

Subnet of this host

network: 172.20.236.0
broadcast: 172.20.239.255

Answer

x = 236

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