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 −114 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1148 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1010 XOR 0010 1100. What is the result's value R in decimal?

1000 1010XOR0010 1100
R =
3.Propositional logic & proofsMedium
(a → c) ∧ (¬c → ¬a)(a → c) ∨ (c → a)(a → c) ∧ (c → a)a ∧ c

Which formula is logically equivalent to a ↔ c? Click it.

a ↔ c
  1. 1(a → c) ∧ (¬c → ¬a)
  2. 2(a → c) ∨ (c → a)
  3. 3(a → c) ∧ (c → a)
  4. 4a ∧ c
4.Propositional logic & proofsHard
cb¬c ∧ b¬c

The premises c → b; ¬b are given. Which conclusion follows from them? Click it.

c → b¬b
  1. 1c
  2. 2b
  3. 3¬c ∧ b
  4. 4¬c
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011031204050716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 2, 6). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 1, 2, 6)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 9 clock edges.

Q₃Q₂Q₁Q₀ = 00009
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 7 (mod 9) and x ≡ 2 (mod 11).

x ≡ 7 (mod 9)x ≡ 2 (mod 11)
x =
8.Number theory & modular arithmeticHard

How many solutions N does 16x ≡ 4 (mod 38) have in {0, 1, …, 37}?

16x ≡ 4 (mod 38)
N =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms are always rounded to integers
  2. BLogarithms with different bases differ only by a constant factor
  3. CAll logarithms in computer science have base 10, so it is implied
  4. DThe base only matters for small n
10.Algorithms & complexityHard

What would follow if someone found a polynomial-time algorithm for one NP-complete problem?

  1. AAll exponential-time algorithms would immediately run in linear time
  2. BOnly that one problem would move into P
  3. CEvery problem in NP could be solved in polynomial time, so P = NP
  4. DThe halting problem would become decidable
11.Data structuresMedium

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

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

What guarantee do the red-black tree rules give about the height of a tree with n keys?

  1. AThe height is at most n / 2
  2. BThe height is at most log2 n + 1
  3. CThe height is at most 2 log2(n + 1)
  4. DThe height is exactly log2 n
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
14415653984ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

Which ACID property means concurrent transactions do not see each other's unfinished changes?

  1. AIsolation
  2. BDurability
  3. CAtomicity
  4. DAvailability
16.Databases & SQLHard

What does the following return: SELECT name FROM students WHERE grade > (SELECT AVG(grade) FROM students)?

  1. AStudents with the highest grade only
  2. BThe average grade of all students
  3. CAn error because subqueries are not allowed in WHERE
  4. DStudents with a grade above the average grade
17.Computer networksMedium

Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?

  1. A192.168.0.0/25
  2. B192.168.1.0/23
  3. C192.168.0.0/23
  4. D192.168.0.0/22
18.Computer networksHard

An attacker on a LAN sends forged ARP replies claiming the gateway's IP address. What is the likely result?

  1. AAll DNS queries fail across the internet
  2. BThe gateway's IP address is permanently changed to the attacker's address on every router
  3. CVictims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
  4. DThe switch shuts down all of its ports
19.IPv4 subnettingMedium
255.255.255.19211111111111111111111111111000000network bitshost bits

A subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?

255.255.255.192
H =
20.IPv4 subnettingHard
third octet 96 … 102012816413201608040201commonvaries

A router has routes to these networks: 10.158.96.0/24, 10.158.97.0/24, 10.158.98.0/24, 10.158.99.0/24, 10.158.100.0/24, 10.158.101.0/24, 10.158.102.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.158.96.0/24 … 10.158.102.0/24
p =

3Solutions

1.Number systems

Answer: U = 142

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

114 = 0111 0010
inverted: 1000 1101
+1 → 1000 1110

Read it as unsigned

1000 1110₂ = 256 − 114 = 142

Answer

U = 142
2.Number systems

Answer: R = 166

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

XOR gives 1 where the two bits differ.

1000 1010
XOR 0010 1100
= 1010 0110

Value of the result

1010 0110₂ = 166

Answer

R = 166
3.Propositional logic & proofs

Answer: (a → c) ∧ (c → a)

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).

The law

a ↔ b means the implication holds in both directions: (a → b) ∧ (b → a).
a ↔ c ≡ (a → c) ∧ (c → a)

Check with the truth table

a = 1, c = 1: 1 | 1 1 1 1
a = 1, c = 0: 0 | 0 1 0 0
a = 0, c = 1: 0 | 1 1 0 0
a = 0, c = 0: 1 | 1 1 1 0

Answer

(a → c) ∧ (c → a)

The columns after the bar are the options in the order shown.

4.Propositional logic & proofs

Answer: ¬c

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.

Rows with all premises true

b = 0, c = 0: c = 0, b = 0, ¬c ∧ b = 0, ¬c = 1

Conclusion

¬c (modus tollens)

Answer

¬c
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 4

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.

ABC00011110011011031204050716

Group the 1s

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

BC′
A′B′

Minimal SOP form

F = BC′ + A′B′
L = 4

Answer

L = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 8

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 → 1000

Decimal value

1000₂ = 8

Answer

Q = 8
7.Number theory & modular arithmetic

Answer: x = 79

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).

First congruence

x = 7 + 9k

Second congruence

7 + 9k ≡ 2 (mod 11) → 9k ≡ 6 (mod 11)
k ≡ 5·6 ≡ 8 (mod 11)

Result

x = 7 + 9·8 = 79
x = 79
check: 79 mod 9 = 7, 79 mod 11 = 2

Answer

x = 79
8.Number theory & modular arithmetic

Answer: N = 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).

Divide by the gcd

gcd(16, 38) = 2
8x ≡ 2 (mod 19)

Solve the reduced congruence

8−1 ≡ 12 (mod 19) → x ≡ 5 (mod 19)

Solutions modulo 38

x ∈ {5, 24}
N = 2

Answer

N = 2
9.Algorithms & complexity

Answer: B · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

B · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: C · Every problem in NP could be solved in polynomial time, so P = NP

Why this is the answer

Since all of NP reduces to any NP-complete problem in polynomial time, a fast algorithm for one gives a fast algorithm for all.

Answer

C · Every problem in NP could be solved in polynomial time, so P = NP
11.Data structures

Answer: A · 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

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

Answer: C · The height is at most 2 log2(n + 1)

Why this is the answer

Every path has the same number of black nodes and red nodes cannot be adjacent, so the longest path is at most twice the shortest, giving height O(log n).

Answer

C · The height is at most 2 log2(n + 1)
13.Graph theory

Answer: Euler 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) = 2, deg(B) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2, deg(F) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(F) = 17

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 14, C: 4
C (4): E: 9
E (9): D: 12, F: 17, B: 13
D (12)
B (13)
F (17)

Shortest path

A → C → E → F
d(F) = 17

Answer

d(F) = 17
15.Databases & SQL

Answer: A · Isolation

Why this is the answer

Isolation keeps concurrent transactions apart; isolation levels decide how strictly this is enforced.

Answer

A · Isolation
16.Databases & SQL

Answer: D · Students with a grade above the average grade

Why this is the answer

The scalar subquery computes the average, and the outer query keeps students above it.

Answer

D · Students with a grade above the average grade
17.Computer networks

Answer: C · 192.168.0.0/23

Why this is the answer

0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.

Answer

C · 192.168.0.0/23
18.Computer networks

Answer: C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack

Why this is the answer

ARP spoofing poisons hosts' ARP caches with the attacker's MAC for the gateway IP. The attacker can then read, change or drop traffic; dynamic ARP inspection helps prevent it.

Answer

C · Victims send traffic for the gateway to the attacker, enabling a man-in-the-middle attack
19.IPv4 subnetting

Answer: H = 62

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

255.255.255.192 = /26

Usable hosts

26 − 2 = 62

Answer

H = 62
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

96 = 01100000
102 = 01100110
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.158.96.0/21 (it also covers 1 network(s) that are not in the list).

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