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 −63 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −638 bit
U =
2.Number systemsHard
16 bit1 px800 px600 pxM = ? KiB

An uncompressed image is 800 × 600 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

800 × 600 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬r → ¬qq → rr ∧ ¬q¬q → ¬r

You want to prove r → q by contrapositive. Which implication do you prove instead? Click it.

r → q
  1. 1¬r → ¬q
  2. 2q → r
  3. 3r ∧ ¬q
  4. 4¬q → ¬r
4.Propositional logic & proofsHard

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

¬(((r ∨ s) ∧ s) ∧ (q ↔ ¬(p → r)))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000011302140517161120131151140809111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(3, 4, 6, 7, 10, 11, 12, 14, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(3, 4, 6, 7, 10, 11, 12, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110000113X2X4051716112X13X150141819011010

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 6, 7, 8, 9, 12) + d(2, 4, 13, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(3, 6, 7, 8, 9, 12) + d(2, 4, 13, 15)
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(1625), the number of integers from 1 to 1625 that are coprime to 1625.

n = 1625 = 53·13
φ(1625) =
8.Number theory & modular arithmeticHard

23 is prime. Use Fermat’s little theorem to find 3068021 mod 23.

p = 23k = 68021
3068021 mod 23 =
9.Algorithms & complexityMedium

Which algorithm finds the shortest paths from one source in a graph with non-negative edge weights?

  1. AKruskal's algorithm
  2. BDijkstra's algorithm
  3. CDepth-first search
  4. DBellman–Ford is required in all cases
10.Algorithms & complexityHard

To show that a new problem X is NP-hard, what must you do?

  1. AShow that X has no polynomial algorithm for small inputs
  2. BShow that X can be verified in polynomial time
  3. CReduce X to a known NP-hard problem in polynomial time
  4. DReduce a known NP-hard problem to X in polynomial time
11.Data structuresMedium

A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?

  1. AAdjacency matrix
  2. BA binary heap of edges
  3. CAdjacency list
  4. DA single sorted array of vertices
12.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. AAnswers only for numbers
  2. BPossible false negatives but no false positives
  3. CExact answers always
  4. DPossible false positives but no false negatives
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
159165174137121810ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 11
W =
15.Databases & SQLMedium

What does SELECT SUM(amount) FROM orders return when the WHERE clause matches no rows?

  1. AAn empty string
  2. BAn error
  3. C0
  4. DNULL
16.Databases & SQLHard

Which isolation level prevents dirty reads but still allows non-repeatable reads?

  1. ARead committed
  2. BRepeatable read
  3. CSerializable
  4. DRead uncommitted
17.Computer networksMedium

What is the IPv6 loopback address?

  1. Afe80::1
  2. B::1
  3. Cff02::1
  4. D::
18.Computer networksHard

The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?

  1. ADistance-vector; split horizon and poison reverse
  2. BStatic routing; adding default routes
  3. CPath-vector; longer AS paths
  4. DLink-state; flooding link updates
19.IPv4 subnettingMedium
255.255.248.011111111111111111111100000000000network bitshost bits

A subnet uses the mask 255.255.248.0. How many usable host addresses H does it have?

255.255.248.0
H =
20.IPv4 subnettingHard
third octet = 2012806403201608041201networkhost

A host has the address 172.20.2.111/17. What is the third octet x of its subnet's broadcast address?

172.20.2.111/17
x =

3Solutions

1.Number systems

Answer: U = 193

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

63 = 0011 1111
inverted: 1100 0000
+1 → 1100 0001

Read it as unsigned

1100 0001₂ = 256 − 63 = 193

Answer

U = 193
2.Number systems

Answer: M = 937.5 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

800·600·16 = 7680000 bit = 960000 B

Convert to KiB

960000/1024 = 937.5 KiB

Answer

M = 937.5 KiB
3.Propositional logic & proofs

Answer: ¬q → ¬r

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 = r, b = q

Contrapositive

¬b → ¬a = ¬q → ¬r

Answer

¬q → ¬r

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

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

Split on p

p = 1: 6
p = 0: 6
N = 6 + 6 = 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.

ABCD000111100001111000011302140517161120131151140809111110

Group the 1s

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

CD
BD′
AC

Minimal SOP form

F = CD + BD′ + AC
N = 3

Answer

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

Answer: N = 2

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

BC′D′, BCD, A′C, AC′, A′BD′
ABCD0001111000011110000113X2X4051716112X13X150141819011010

Minimal cover

F = A′C + AC′
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: φ(1625) = 1200

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).

Prime factors

1625 = 53·13

Totient

φ(1625) = φ(125)·φ(13) = 100·12 = 1200

Answer

φ(1625) = 1200
8.Number theory & modular arithmetic

Answer: 3068021 mod 23 = 11

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

68021 mod 22 = 19
3068021 ≡ 3019 (mod 23)

Small power

3019 mod 23 = 11

Answer

3068021 mod 23 = 11
9.Algorithms & complexity

Answer: B · Dijkstra's algorithm

Why this is the answer

Dijkstra's algorithm works with non-negative weights; Bellman–Ford handles negative weights; Kruskal builds minimum spanning trees.

Answer

B · Dijkstra's algorithm
10.Algorithms & complexity

Answer: D · Reduce a known NP-hard problem to X in polynomial time

Why this is the answer

If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.

Answer

D · Reduce a known NP-hard problem to X in polynomial time
11.Data structures

Answer: C · Adjacency list

Why this is the answer

A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.

Answer

C · Adjacency list
12.Data structures

Answer: D · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

D · Possible false positives but no false negatives
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) = 3, deg(C) = 3, deg(D) = 4, deg(E) = 2, deg(F) = 1, deg(G) = 3

Odd vertices

B, C, F, G → 4
neither

Answer

neither
14.Graph theory

Answer: W = 50

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.

Kruskal’s algorithm

DE (4): take
BD (5): take
GB (7): take
AC (9): take
GE (10): skip, closes a cycle
GC (12): take
DF (13): take

Total weight

W = 4 + 5 + 7 + 9 + 12 + 13 = 50

Answer

W = 50
15.Databases & SQL

Answer: D · NULL

Why this is the answer

SUM, AVG, MIN and MAX return NULL on an empty set, while COUNT returns 0; COALESCE(SUM(amount), 0) gives 0 instead.

Answer

D · NULL
16.Databases & SQL

Answer: A · Read committed

Why this is the answer

Read committed hides uncommitted changes; repeatable read also prevents non-repeatable reads; serializable prevents phantoms too.

Answer

A · Read committed
17.Computer networks

Answer: B · ::1

Why this is the answer

::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.

Answer

B · ::1
18.Computer networks

Answer: A · Distance-vector; split horizon and poison reverse

Why this is the answer

After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.

Answer

A · Distance-vector; split horizon and poison reverse
19.IPv4 subnetting

Answer: H = 2046

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

255.255.248.0 = /21

Usable hosts

211 − 2 = 2046

Answer

H = 2046
20.IPv4 subnetting

Answer: x = 127

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

27 = 128

Subnet of this host

network: 172.20.0.0
broadcast: 172.20.127.255

Answer

x = 127

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