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 complement1−12806413201618041211

The 8-bit pattern 1010 1011 stores a signed integer in two's complement. What is its value N in decimal?

1010 10118 bit
N =
2.Number systemsHard
16 bits, unsigned, N = ?????????????????

What is the largest unsigned integer N that fits in 16 bits?

16 bit
N =
3.Propositional logic & proofsMedium
a ∧ (c → b)a → (c → b)(a → b) ∧ (c → b)(a → c) → b

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

(a ∧ c) → b
  1. 1a ∧ (c → b)
  2. 2a → (c → b)
  3. 3(a → b) ∧ (c → b)
  4. 4(a → c) → b
4.Propositional logic & proofsHard
pqrr → (p ∨ q)rp111110101100011010001000

The argument with premises r → (p ∨ q); r and conclusion p is invalid. Click the row of the truth table that is a counterexample.

r → (p ∨ q)r∴ p
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011130214151706

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(1, 3, 4, 5, 7). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(1, 3, 4, 5, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100001131214151706012013X15014X809111110

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 4, 5, 7, 10, 11) + d(8, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(2, 3, 4, 5, 7, 10, 11) + d(8, 15)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 3 (mod 5) and x ≡ 9 (mod 12).

x ≡ 3 (mod 5)x ≡ 9 (mod 12)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 1742 mod 30.

a = 17k = 42n = 30
1742 mod 30 =
9.Algorithms & complexityMedium

About how many comparisons does binary search need in the worst case on a sorted array of 1,000,000 elements?

  1. AAbout 20
  2. BAbout 100
  3. CAbout 1,000
  4. DAbout 500,000
10.Algorithms & complexityHard

Which statement about little-o notation is correct?

  1. An = o(n), since every function is little-o of itself
  2. Bn = o(n²), but n is not o(n)
  3. C2n = o(n), because 2n grows slowly
  4. Dn² = o(n)
11.Data structuresMedium

How does extract-min work in a binary min-heap?

  1. ARemove the last element of the array
  2. BRemove the root and shift all elements one place left
  3. CMove the last element to the root and sift it down
  4. DSearch all leaves for the new minimum
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA binary search tree only
  2. BA single sorted array
  3. CA stack only
  4. DA hash map combined with a doubly linked list
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
51615112317ABCDEF

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

|V| = 6|E| = 8
W =
15.Databases & SQLMedium

Which statement removes all rows of a table, keeps its structure and cannot take a WHERE clause?

  1. ATRUNCATE TABLE
  2. BDROP TABLE
  3. CALTER TABLE
  4. DDELETE FROM with a WHERE clause
16.Databases & SQLHard

What is a phantom read?

  1. AA repeated query returns new rows inserted by another committed transaction
  2. BReading uncommitted changes
  3. CReading data that was never written
  4. DA query that returns no rows
17.Computer networksMedium

A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?

  1. A10.1.0.0/16, because it is the longest matching prefix
  2. B10.0.0.0/8, because it was added first
  3. C10.0.0.0/8, because it covers more addresses than the /16
  4. DBoth routes, splitting the traffic equally
18.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A12 kbit/s
  2. B24 kbit/s
  3. C3 kbit/s
  4. D6 kbit/s
19.IPv4 subnettingMedium
/22 → /2911111111111111111111111111111000network bitshost bits

The network 10.50.84.0/22 is divided into equal subnets with prefix /29. How many subnets N are there?

10.50.84.0/22/29
N =
20.IPv4 subnettingHard
last octet = 247112816413211608141211networkhost

A host has the address 10.20.232.247/27. What is the fourth octet x of the last usable host address in its subnet?

10.20.232.247/27
x =

3Solutions

1.Number systems

Answer: N = −85

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 + 32 + 8 + 2 + 1 = −85

Answer

N = −85
2.Number systems

Answer: N = 65535

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.

Largest value

216 − 1 = 65535

Answer

N = 65535
3.Propositional logic & proofs

Answer: a → (c → b)

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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(a ∧ c) → b ≡ a → (c → b)

Check with the truth table

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

Answer

a → (c → b)

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

4.Propositional logic & proofs

Answer: p = 0, q = 1, r = 1

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.

Premises and conclusion row by row

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

Counterexample

p = 0, q = 1, r = 1

Answer

p = 0, q = 1, r = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: N = 2

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.

ABC00011110010011130214151706

Group the 0s

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

(B′ + C)
(A + C)

Minimal POS form

F = (B′ + C)(A + C)
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

CD, B′C, A′BD, AB′D′, A′BC′
ABCD00011110000111100001131214151706012013X15014X809111110

Minimal cover

F = CD + B′C + A′BC′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 33

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 = 3 + 5k

Second congruence

3 + 5k ≡ 9 (mod 12) → 5k ≡ 6 (mod 12)
k ≡ 5·6 ≡ 6 (mod 12)

Result

x = 3 + 5·6 = 33
x = 33
check: 33 mod 5 = 3, 33 mod 12 = 9

Answer

x = 33
8.Number theory & modular arithmetic

Answer: 1742 mod 30 = 19

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

42 = 1010102 = 32 + 8 + 2

Repeated squaring

171 ≡ 17 (mod 30)
172 ≡ 19 (mod 30)
174 ≡ 1 (mod 30)
178 ≡ 1 (mod 30)
1716 ≡ 1 (mod 30)
1732 ≡ 1 (mod 30)

Multiply the needed powers

1742 ≡ 19·1·1 ≡ 19 (mod 30)

Answer

1742 mod 30 = 19
9.Algorithms & complexity

Answer: A · About 20

Why this is the answer

220 ≈ 1,048,576, so after about 20 halvings only one element remains.

Answer

A · About 20
10.Algorithms & complexity

Answer: B · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

B · n = o(n²), but n is not o(n)
11.Data structures

Answer: C · Move the last element to the root and sift it down

Why this is the answer

The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).

Answer

C · Move the last element to the root and sift it down
12.Data structures

Answer: D · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

D · A hash map combined with a doubly linked list
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) = 2, deg(B) = 3, deg(C) = 3, deg(D) = 1, deg(E) = 3

Odd vertices

B, C, D, E → 4
neither

Answer

neither
14.Graph theory

Answer: W = 26

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

EF (1): take
DE (2): take
DF (3): skip, closes a cycle
AB (5): take
BE (7): take
CE (11): take

Total weight

W = 1 + 2 + 5 + 7 + 11 = 26

Answer

W = 26
15.Databases & SQL

Answer: A · TRUNCATE TABLE

Why this is the answer

TRUNCATE quickly empties a table and keeps its definition; DELETE can remove selected rows, and DROP removes the table itself.

Answer

A · TRUNCATE TABLE
16.Databases & SQL

Answer: A · A repeated query returns new rows inserted by another committed transaction

Why this is the answer

Phantoms occur when another transaction inserts rows that match a range condition; serializable isolation prevents them.

Answer

A · A repeated query returns new rows inserted by another committed transaction
17.Computer networks

Answer: A · 10.1.0.0/16, because it is the longest matching prefix

Why this is the answer

IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.

Answer

A · 10.1.0.0/16, because it is the longest matching prefix
18.Computer networks

Answer: A · 12 kbit/s

Why this is the answer

2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.

Answer

A · 12 kbit/s
19.IPv4 subnetting

Answer: N = 128

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).

Borrowed bits

29 − 22 = 7

Number of subnets

N = 27 = 128

Answer

N = 128

Each /29 subnet has 23 − 2 = 6 usable hosts.

20.IPv4 subnetting

Answer: x = 254

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

25 = 32

Subnet of this host

network: ⌊247/32⌋·32 = 224
broadcast: 224 + 32 − 1 = 255
usable hosts: 225 … 254

Answer

x = 254

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