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
place values in two's complement1−12816413211608141211

The 8-bit pattern 1111 0111 stores a signed integer in two's complement. What is its value N in decimal?

1111 01118 bit
N =
2.Number systemsHard
24 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(q ∨ r) → pp ∧ ¬(q ∨ r)¬p → ¬(q ∨ r)¬(q ∨ r) → ¬p

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

p → (q ∨ r)
  1. 1(q ∨ r) → p
  2. 2p ∧ ¬(q ∨ r)
  3. 3¬p → ¬(q ∨ r)
  4. 4¬(q ∨ r) → ¬p
4.Propositional logic & proofsHard
∀x ¬F(x)∃x ¬F(x)∀x F(x)∃x F(x)

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

¬∀x F(x)
  1. 1∀x ¬F(x)
  2. 2∃x ¬F(x)
  3. 3∀x F(x)
  4. 4∃x F(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

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(4, 6, 9, 12, 14)
  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
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLK

A 3-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 8 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 3tpd = 8 ns
fmax = MHz
7.Number theory & modular arithmeticMedium

Solve 23x ≡ 10 (mod 53). Give the solution x with 0 < x < 53.

23x ≡ 10 (mod 53)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 2743 mod 85.

a = 27k = 43n = 85
2743 mod 85 =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(n) solve to?

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

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. AGeneral SAT on formulas in conjunctive normal form
  2. B2-SAT, where each clause has at most two literals
  3. CCircuit satisfiability
  4. D3-SAT, where each clause has exactly three literals
11.Data structuresMedium

In a binary min-heap stored in an array (index starting at 0), the children of the node at index i are at:

  1. A2i + 1 and 2i + 2
  2. B2i and 2i + 1
  3. Ci + 1 and i + 2
  4. Di/2 and i/2 + 1
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 stack that keeps elements waiting for their answer
  2. BA binary search tree rebuilt for each element
  3. CA queue that holds every pair of elements
  4. DA hash table of all elements
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
7186101617121203ABCDEFG

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

|V| = 7|E| = 10
W =
15.Databases & SQLMedium

In relational algebra, the selection operator (σ) corresponds to which SQL clause?

  1. AGROUP BY
  2. BORDER BY
  3. CSELECT list
  4. DWHERE
16.Databases & SQLHard

R(A, B, C, D) has the functional dependencies A → B and C → D. What is the candidate key?

  1. A{A}
  2. B{A, C}
  3. C{C}
  4. D{B, D}
17.Computer networksMedium

Which medium access method did classic shared (half-duplex) Ethernet use?

  1. ACSMA/CD
  2. BToken passing
  3. CCSMA/CA
  4. DTDMA
18.Computer networksHard

A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?

  1. ATurn off the PC's DNS client
  2. BChange the PC's subnet mask to /32
  3. CGive the PC a different MAC address so the router can find it from outside
  4. DConfigure port forwarding on the router to the PC's private address
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.208.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?

192.168.208.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet 234 … 237112816413201618041201commonvaries

A router has routes to these networks: 10.137.234.0/24, 10.137.235.0/24, 10.137.236.0/24, 10.137.237.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.137.234.0/24 … 10.137.237.0/24
p =

3Solutions

1.Number systems

Answer: N = −9

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 + 64 + 32 + 16 + 4 + 2 + 1 = −9

Answer

N = −9
2.Number systems

Answer: M = 900 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

640·480·24 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB
3.Propositional logic & proofs

Answer: ¬(q ∨ r) → ¬p

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 = p, b = q ∨ r

Contrapositive

¬b → ¬a = ¬(q ∨ r) → ¬p

Answer

¬(q ∨ r) → ¬p

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

4.Propositional logic & proofs

Answer: ∃x ¬F(x)

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 F(x) ≡ ∃x ¬F(x)

Answer

∃x ¬F(x)

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: AB = 10, 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.

m9: m8, m11, m13, m1 → 0
ABCD000111100001111011111

The isolated 1

m9 = AB′C′D → AB = 10, CD = 01

Answer

AB = 10, CD = 01
6.Digital logic: gates, K-maps & flip-flops

Answer: fmax = 41.67 MHz

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.

Worst-case settling time

T = 3·8 ns = 24 ns

Maximum frequency

fmax = 1/24 ns = 41.67 MHz

Answer

fmax = 41.67 MHz
7.Number theory & modular arithmetic

Answer: x = 35

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 23

r = 53, t = 0
r = 23, q = 2, t = 1
r = 7, q = 3, t = −2
r = 2, q = 3, t = 7
r = 1, q = 2, t = −23
23−1 ≡ 30 (mod 53)

Solve

x ≡ 30·10 = 300 ≡ 35 (mod 53)
check: 23·35 = 805 ≡ 10 (mod 53)

Answer

x = 35
8.Number theory & modular arithmetic

Answer: 2743 mod 85 = 3

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.

Binary exponent

43 = 1010112 = 32 + 8 + 2 + 1

Repeated squaring

271 ≡ 27 (mod 85)
272 ≡ 49 (mod 85)
274 ≡ 21 (mod 85)
278 ≡ 16 (mod 85)
2716 ≡ 1 (mod 85)
2732 ≡ 1 (mod 85)

Multiply the needed powers

2743 ≡ 27·49·16·1 ≡ 3 (mod 85)

Answer

2743 mod 85 = 3
9.Algorithms & complexity

Answer: A · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

A · O(n²)
10.Algorithms & complexity

Answer: B · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

B · 2-SAT, where each clause has at most two literals
11.Data structures

Answer: A · 2i + 1 and 2i + 2

Why this is the answer

For 0-based arrays, the left child is at 2i + 1, the right child at 2i + 2, and the parent at ⌊(i − 1)/2⌋.

Answer

A · 2i + 1 and 2i + 2
12.Data structures

Answer: A · 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

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

Answer: Euler path, no circuit

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) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 2, deg(E) = 4, deg(F) = 2

Odd vertices

A, B → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: W = 49

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

GC (1): take
GE (3): take
CE (6): skip, closes a cycle
AB (7): take
DE (10): take
GB (12): take
DF (16): take

Total weight

W = 1 + 3 + 7 + 10 + 12 + 16 = 49

Answer

W = 49
15.Databases & SQL

Answer: D · WHERE

Why this is the answer

σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.

Answer

D · WHERE
16.Databases & SQL

Answer: B · {A, C}

Why this is the answer

A and C are not determined by anything, so every key must contain both, and {A, C}+ = {A, B, C, D}.

Answer

B · {A, C}
17.Computer networks

Answer: A · CSMA/CD

Why this is the answer

Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.

Answer

A · CSMA/CD
18.Computer networks

Answer: D · Configure port forwarding on the router to the PC's private address

Why this is the answer

NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.

Answer

D · Configure port forwarding on the router to the PC's private address
19.IPv4 subnetting

Answer: x = 255

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

27 = 128

Subnet number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 255
20.IPv4 subnetting

Answer: p = 21

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

234 = 11101010
237 = 11101101
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.137.232.0/21 (it also covers 4 network(s) that are not in the list).

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