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 −107 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1078 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1000 << 1. What is the result's value R in decimal?

1101 1000<< 1
R =
3.Propositional logic & proofsMedium
pqrr ∧ ¬q¬p ∨ (r ∧ ¬q)(¬p ∧ q) ↔ (¬p ∨ (r ∧ ¬q))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (¬p ∧ q) ↔ (¬p ∨ (r ∧ ¬q)) true? Call this number N.

(¬p ∧ q) ↔ (¬p ∨ (r ∧ ¬q))
N =
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ¬((b ↔ c) ↔ (b → c)) → ((d ↔ a) → (a ∨ d)) true? Call this number N.

¬((b ↔ c) ↔ (b → c)) → ((d ↔ a) → (a ∨ d))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(5, 6, 8, 12, 14, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​CLKMOD-9

A mod-9 counter has to count through 9 different states. What is the minimum number of flip-flops n it needs?

N = 9
n =
7.Number theory & modular arithmeticMedium

Solve 8x ≡ 5 (mod 11). Give the solution x with 0 < x < 11.

8x ≡ 5 (mod 11)
x =
8.Number theory & modular arithmeticHard

67 is prime. Use Fermat’s little theorem to find 572581 mod 67.

p = 67k = 72581
572581 mod 67 =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?

  1. AO(log n)
  2. BO(n²)
  3. CO(n log n)
  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 has no polynomial algorithm, and X is decidable
  4. DX reduces to a problem in P, and X is in NP
11.Data structuresMedium

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

  1. AAdjacency list
  2. BA single sorted array of vertices
  3. CAdjacency matrix
  4. DA binary heap of edges
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(1)
  4. DO(log n)
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
21310412515ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
15.Databases & SQLMedium

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

  1. ANULL
  2. BAn empty string
  3. CAn error
  4. D0
16.Databases & SQLHard

When is a schedule of concurrent transactions called serializable?

  1. AWhen its effect is equivalent to some serial execution of the same transactions
  2. BWhen every transaction is given a serial number
  3. CWhen the transactions run on one CPU only
  4. DWhen no transaction ever reads data
17.Computer networksMedium

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.1.200.5
  2. B10.2.2.3
  3. C11.1.2.3
  4. D10.0.2.3
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIts 128-bit address space is large enough to give every device a global address
  2. BIPv6 does not use addresses
  3. CIPv6 encrypts all addresses
  4. DIPv6 only works on local networks, so addresses never need to be shared globally
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2420 subnets
H =
20.IPv4 subnettingHard
third octet 223 … 225112816403211618141211commonvaries

A router has routes to these networks: 10.1.223.0/24, 10.1.224.0/24, 10.1.225.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.1.223.0/24 … 10.1.225.0/24
p =

3Solutions

1.Number systems

Answer: U = 149

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

107 = 0110 1011
inverted: 1001 0100
+1 → 1001 0101

Read it as unsigned

1001 0101₂ = 256 − 107 = 149

Answer

U = 149
2.Number systems

Answer: R = 176

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

1101 1000 << 1
= 1011 0000

Value of the result

1011 0000₂ = 176

Answer

R = 176
3.Propositional logic & proofs

Answer: N = 5

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.

Truth table row by row

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

Count the rows with value 1

(¬p ∧ q) ↔ (¬p ∨ (r ∧ ¬q)): 1, 1, 0, 1, 1, 1, 0, 0
N = 5

Answer

N = 5
4.Propositional logic & proofs

Answer: N = 15

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

b ↔ c; b → c; (b ↔ c) ↔ (b → c); ¬((b ↔ c) ↔ (b → c)); d ↔ a; a ∨ d; (d ↔ a) → (a ∨ d); ¬((b ↔ c) ↔ (b → c)) → ((d ↔ a) → (a ∨ d))

Split on a

a = 1: 8
a = 0: 7
N = 8 + 7 = 15

Answer

N = 15
5.Digital logic: gates, K-maps & flip-flops

Answer: AB = 01, CD = 01

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m5: m4, m7, m1, m13 → 0
ABCD0001111000011110111111

The isolated 1

m5 = A′BC′D → AB = 01, CD = 01

Answer

AB = 01, CD = 01
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 ≥ 9 → n = 4 (2⁴ = 16)

Answer

n = 4
7.Number theory & modular arithmetic

Answer: x = 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).

Inverse of 8

r = 11, t = 0
r = 8, q = 1, t = 1
r = 3, q = 2, t = −1
r = 2, q = 1, t = 3
r = 1, q = 2, t = −4
8−1 ≡ 7 (mod 11)

Solve

x ≡ 7·5 = 35 ≡ 2 (mod 11)
check: 8·2 = 16 ≡ 5 (mod 11)

Answer

x = 2
8.Number theory & modular arithmetic

Answer: 572581 mod 67 = 58

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

72581 mod 66 = 47
572581 ≡ 547 (mod 67)

Small power

547 mod 67 = 58

Answer

572581 mod 67 = 58
9.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).

Answer

C · O(n log 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: A · 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

A · Adjacency list
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 = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: B

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

1. A (0): B: 2, C: 1
2. C (1): E: 5
3. B (2): D: 12
4. E (5)
5. D (12): F: 17
6. F (17)

round 3

B

Answer

B
15.Databases & SQL

Answer: A · 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

A · NULL
16.Databases & SQL

Answer: A · When its effect is equivalent to some serial execution of the same transactions

Why this is the answer

Serializability is the main correctness criterion for concurrency: the interleaved result must equal the result of some one-at-a-time order.

Answer

A · When its effect is equivalent to some serial execution of the same transactions
17.Computer networks

Answer: A · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

A · 10.1.200.5
18.Computer networks

Answer: A · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

A · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: H = 6

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

25 = 32 ≥ 20 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: p = 18

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

223 = 11011111
225 = 11100001
common leading bits: 2

Prefix length

p = 16 + 2 = 18

Answer

p = /18

Summary route: 10.1.192.0/18 (it also covers 61 network(s) that are not in the list).

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