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 −71 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −718 bit
U =
2.Number systemsHard
8 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(p → q) ∧ (¬q → ¬p)(p → q) ∨ (q → p)p ∧ q(p → q) ∧ (q → p)

Which formula is logically equivalent to p ↔ q? Click it.

p ↔ q
  1. 1(p → q) ∧ (¬q → ¬p)
  2. 2(p → q) ∨ (q → p)
  3. 3p ∧ q
  4. 4(p → q) ∧ (q → 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
ABCD000111100001111010111302040517160120131150141809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 1, 3, 6, 7, 8, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 1, 3, 6, 7, 8, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​DQ4​Q4​DQ5​Q5​DQ6​Q6​DQ7​Q7​CLKJOHNSON

A Johnson (twisted-ring) counter is made of 8 D flip-flops, with the inverted last output fed back to the first input, and starts at all zeros. How many different states N does it go through?

n = 8
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(325), the number of integers from 1 to 325 that are coprime to 325.

n = 325 = 52·13
φ(325) =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 3374 mod 59.

a = 33k = 74n = 59
3374 mod 59 =
9.Algorithms & complexityMedium

What does amortised analysis measure?

  1. AThe expected cost over random inputs
  2. BThe cost of an algorithm on the best input
  3. CThe average cost per operation over a worst-case sequence of operations
  4. DThe cost of the single most expensive operation anywhere in the sequence
10.Algorithms & complexityHard

Taking both endpoints of every edge in a maximal matching gives a vertex cover. How good is this cover?

  1. AAt most log n times the minimum
  2. BAt most twice the size of a minimum vertex cover
  3. COf unknown quality compared to the optimum
  4. DAlways a minimum vertex cover
11.Data structuresMedium

Which data structure is used by depth-first search (iterative version)?

  1. AA heap
  2. BA stack
  3. CA hash table only
  4. DA queue
12.Data structuresHard

Why are B-trees used in databases and file systems?

  1. AThey store data unsorted
  2. BTheir high branching factor keeps the tree shallow, reducing slow disk reads
  3. CThey use less memory than any array, so the entire index always fits in main memory
  4. DThey need no rebalancing
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
1581452113ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.

k = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

What is a dirty read?

  1. AReading data from a corrupted disk
  2. BReading the same row twice and getting the same value
  3. CReading data written by another transaction that has not yet committed
  4. DReading data without a WHERE clause
16.Databases & SQLHard

Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?

  1. A2, 2, 3
  2. B1, 1, 2
  3. C1, 1, 3
  4. D1, 2, 3
17.Computer networksMedium

What is the IPv6 loopback address?

  1. A::1
  2. Bff02::1
  3. Cfe80::1
  4. D::
18.Computer networksHard

By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?

  1. A8 Mbit/s
  2. B2 Mbit/s
  3. C1 Mbit/s
  4. D255 Mbit/s
19.IPv4 subnettingMedium
255.255.240.011111111111111111111000000000000network bitshost bits

A subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?

255.255.240.0
H =
20.IPv4 subnettingHard
third octet = 195112816403201608041211networkhost

A host has the address 172.20.195.50/21. What is the third octet x of its network address?

172.20.195.50/21
x =

3Solutions

1.Number systems

Answer: U = 185

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

71 = 0100 0111
inverted: 1011 1000
+1 → 1011 1001

Read it as unsigned

1011 1001₂ = 256 − 71 = 185

Answer

U = 185
2.Number systems

Answer: M = 256 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

512·512·8 = 2097152 bit = 262144 B

Convert to KiB

262144/1024 = 256 KiB

Answer

M = 256 KiB
3.Propositional logic & proofs

Answer: (p → q) ∧ (q → 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).

The law

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

Check with the truth table

p = 1, q = 1: 1 | 1 1 1 1
p = 1, q = 0: 0 | 0 1 0 0
p = 0, q = 1: 0 | 1 1 0 0
p = 0, q = 0: 1 | 1 1 0 1

Answer

(p → q) ∧ (q → p)

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

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: L = 12

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.

ABCD000111100001111010111302040517160120131150141809011010

Group the 1s

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

B′C′D′
BCD
A′B′D
A′BC

Minimal SOP form

F = B′C′D′ + BCD + A′B′D + A′BC
L = 12

Answer

L = 12
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 16

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.

Sequence for 8 flip-flops

00000000 → 10000000 → 11000000 → 11100000 → 11110000 → 11111000 → 11111100 → 11111110 → 11111111 → 01111111 → 00111111 → 00011111 → 00001111 → 00000111 → 00000011 → 00000001

Number of states

N = 16

Answer

N = 16
7.Number theory & modular arithmetic

Answer: φ(325) = 240

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

325 = 52·13

Totient

φ(325) = φ(13)·φ(25) = 12·20 = 240

Answer

φ(325) = 240
8.Number theory & modular arithmetic

Answer: 3374 mod 59 = 17

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

74 = 10010102 = 64 + 8 + 2

Repeated squaring

331 ≡ 33 (mod 59)
332 ≡ 27 (mod 59)
334 ≡ 21 (mod 59)
338 ≡ 28 (mod 59)
3316 ≡ 17 (mod 59)
3332 ≡ 53 (mod 59)
3364 ≡ 36 (mod 59)

Multiply the needed powers

3374 ≡ 27·28·36 ≡ 17 (mod 59)

Answer

3374 mod 59 = 17
9.Algorithms & complexity

Answer: C · The average cost per operation over a worst-case sequence of operations

Why this is the answer

Amortised analysis spreads the cost of rare expensive operations over many cheap ones, giving a guaranteed average for any sequence.

Answer

C · The average cost per operation over a worst-case sequence of operations
10.Algorithms & complexity

Answer: B · At most twice the size of a minimum vertex cover

Why this is the answer

The matching edges share no vertices, so the optimum needs at least one vertex per edge, while this method takes two: a 2-approximation.

Answer

B · At most twice the size of a minimum vertex cover
11.Data structures

Answer: B · A stack

Why this is the answer

DFS explores as deep as possible, naturally using a stack (or recursion's call stack); BFS uses a queue.

Answer

B · A stack
12.Data structures

Answer: B · Their high branching factor keeps the tree shallow, reducing slow disk reads

Why this is the answer

B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.

Answer

B · Their high branching factor keeps the tree shallow, reducing slow disk reads
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) = 1, deg(B) = 2, deg(C) = 2, deg(D) = 3, deg(E) = 3, deg(F) = 1

Odd vertices

A, D, E, F → 4
neither

Answer

neither
14.Graph theory

Answer: edge AC

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

1. CE (2)
2. BE (3)
3. BD (5)
4. AC (8)

edge number 4

AC

Answer

edge AC
15.Databases & SQL

Answer: C · Reading data written by another transaction that has not yet committed

Why this is the answer

A dirty read sees uncommitted changes that might be undone; it is possible only at the read uncommitted level.

Answer

C · Reading data written by another transaction that has not yet committed
16.Databases & SQL

Answer: C · 1, 1, 3

Why this is the answer

RANK gives tied rows the same rank and skips the following numbers (1, 1, 3); DENSE_RANK would give 1, 1, 2.

Answer

C · 1, 1, 3
17.Computer networks

Answer: A · ::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

A · ::1
18.Computer networks

Answer: A · 8 Mbit/s

Why this is the answer

C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.

Answer

A · 8 Mbit/s
19.IPv4 subnetting

Answer: H = 4094

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

255.255.240.0 = /20

Usable hosts

212 − 2 = 4094

Answer

H = 4094
20.IPv4 subnetting

Answer: x = 192

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

23 = 8

Subnet of this host

network: 172.20.192.0
broadcast: 172.20.199.255

Answer

x = 192

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