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 −90 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −908 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1111 << 3. What is the result's value R in decimal?

1000 1111<< 3
R =
3.Propositional logic & proofsMedium
¬b ∨ c¬b ∧ c¬b ∨ ¬c¬b ∧ ¬c

Which formula is logically equivalent to ¬(b ∧ c)? Click it.

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

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

((q → p) ↔ r) ∧ (¬(s → ¬p) → s)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

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(3, 6, 7, 8, 10, 11, 13)
  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
CLKD1234567

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0D
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 4 (mod 8) and x ≡ 7 (mod 13).

x ≡ 4 (mod 8)x ≡ 7 (mod 13)
x =
8.Number theory & modular arithmeticHard
483c = me​ mod n?m = cd​ mod nm(n, e) = (697, 7)d

Encrypt the message m = 483 with the RSA public key (n, e) = (697, 7). Find the ciphertext c.

n = 697e = 7m = 483
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

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

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA sorted array
  2. BA doubly linked list
  3. CA dynamic array
  4. DA singly linked list
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery path from a node to its descendant leaves has the same number of black nodes
  2. BThe root must be red
  3. CEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  4. DEvery node must be red
13.Graph theoryMedium

A connected graph has 12 vertices and 20 edges. How many edges k must be removed to leave a spanning tree?

|V| = 12|E| = 20
k =
14.Graph theoryHard
4148123255ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

start: Around 4
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

Which query returns the number of different cities in the customers table?

  1. ASELECT COUNT(city) FROM customers GROUP BY city
  2. BSELECT COUNT(DISTINCT city) FROM customers
  3. CSELECT SUM(DISTINCT city) FROM customers
  4. DSELECT DISTINCT COUNT(city) FROM customers
16.Databases & SQLHard

Two-phase locking (2PL) ensures:

  1. AThat transactions never wait
  2. BThat no locks are needed
  3. CThat deadlocks can never happen
  4. DConflict-serializable schedules
17.Computer networksMedium

What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?

  1. A/32
  2. B/8
  3. C/24
  4. D/16
18.Computer networksHard

In a SYN flood attack, what resource is the attacker trying to exhaust?

  1. AThe switch's MAC address table
  2. BThe server's disk space
  3. CThe DNS cache of the client
  4. DThe server's table of half-open connections
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/245 subnets
H =
20.IPv4 subnettingHard
last octet = 111012816413201618141211networkhost

A host has the address 10.20.238.111/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.238.111/28
x =

3Solutions

1.Number systems

Answer: U = 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.

Invert and add one

90 = 0101 1010
inverted: 1010 0101
+1 → 1010 0110

Read it as unsigned

1010 0110₂ = 256 − 90 = 166

Answer

U = 166
2.Number systems

Answer: R = 120

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 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.

1000 1111 << 3
= 0111 1000

Value of the result

0111 1000₂ = 120

Answer

R = 120
3.Propositional logic & proofs

Answer: ¬b ∨ ¬c

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

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(b ∧ c) ≡ ¬b ∨ ¬c

Check with the truth table

b = 1, c = 1: 0 | 1 0 0 0
b = 1, c = 0: 1 | 0 0 1 0
b = 0, c = 1: 1 | 1 1 1 0
b = 0, c = 0: 1 | 1 0 1 1

Answer

¬b ∨ ¬c

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

4.Propositional logic & proofs

Answer: N = 8

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

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

Split on p

p = 1: 4
p = 0: 4
N = 4 + 4 = 8

Answer

N = 8
5.Digital logic: gates, K-maps & flip-flops

Answer: AB = 11, 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.

m13: m12, m15, m9, m5 → 0
ABCD00011110000111101111111

The isolated 1

m13 = ABC′D → AB = 11, CD = 01

Answer

AB = 11, CD = 01
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 1 → Q = 1
3: D = 0 → Q = 0
4: D = 0 → Q = 0
5: D = 1 → Q = 1
6: D = 0 → Q = 0
7: D = 1 → Q = 1

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: x = 20

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 = 4 + 8k

Second congruence

4 + 8k ≡ 7 (mod 13) → 8k ≡ 3 (mod 13)
k ≡ 5·3 ≡ 2 (mod 13)

Result

x = 4 + 8·2 = 20
x = 20
check: 20 mod 8 = 4, 20 mod 13 = 7

Answer

x = 20
8.Number theory & modular arithmetic

Answer: c = 624

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

7 = 4 + 2 + 1
4831 ≡ 483 (mod 697)
4832 ≡ 491 (mod 697)
4834 ≡ 616 (mod 697)
4837 ≡ 483·491·616 ≡ 624 (mod 697)

Ciphertext

c = 624

Answer

c = 624
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: 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: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
12.Data structures

Answer: A · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

A · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: k = 9

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.

Edges of a spanning tree

12 − 1 = 11

Edges to remove

k = 20 − 11 = 9

Answer

k = 9
14.Graph theory

Answer: G

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: 4, C: 1
2. C (1): E: 9
3. B (4): G: 6
4. G (6): D: 11
5. E (9)
6. D (11): F: 14
7. F (14)

round 4

G

Answer

G
15.Databases & SQL

Answer: B · SELECT COUNT(DISTINCT city) FROM customers

Why this is the answer

COUNT(DISTINCT city) counts unique non-NULL cities; DISTINCT outside COUNT only removes duplicate result rows.

Answer

B · SELECT COUNT(DISTINCT city) FROM customers
16.Databases & SQL

Answer: D · Conflict-serializable schedules

Why this is the answer

2PL guarantees serializability, but deadlocks remain possible and must be detected or prevented.

Answer

D · Conflict-serializable schedules
17.Computer networks

Answer: C · /24

Why this is the answer

Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.

Answer

C · /24
18.Computer networks

Answer: D · The server's table of half-open connections

Why this is the answer

Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.

Answer

D · The server's table of half-open connections
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 ≥ 5 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 111

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

24 = 16

Subnet of this host

network: ⌊111/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110

Answer

x = 111

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