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 complement1−12816413211608041201

The 8-bit pattern 1111 0010 stores a signed integer in two's complement. What is its value N in decimal?

1111 00108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1101 OR 0011 0111. What is the result's value R in decimal?

1101 1101OR0011 0111
R =
3.Propositional logic & proofsMedium
r ∧ ¬pp → r¬p → ¬r¬r → ¬p

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

r → p
  1. 1r ∧ ¬p
  2. 2p → r
  3. 3¬p → ¬r
  4. 4¬r → ¬p
4.Propositional logic & proofsHard

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

¬(r ∧ q) → (¬q → ¬(p → ¬s))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​CI1​C′I2​0I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = C, I₂ = C′, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = CI₂ = C′I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKSHIFT

A 4-bit shift register Q₃Q₂Q₁Q₀ holds 1110. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀, Q₀ is lost) and the serial input enters Q₃. The serial input is 1, 0, 0, 0 at the 4 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.

Q₃Q₂Q₁Q₀ = 11101, 0, 0, 0
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 63 modulo 73: the number x with 0 < x < 73 and 63·x ≡ 1 (mod 73).

a = 63n = 73
63−1 mod 73 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (133, 19)d = ?

An RSA key uses the primes p = 7 and q = 19 and the public exponent e = 19. Find the private exponent d (0 < d < φ(n)).

p = 7q = 19e = 19
d =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the activity that finishes earliest, then repeat with the compatible ones
  2. BPick the shortest activity first
  3. CPick the activity with the fewest overlaps first, then repeat with the compatible ones
  4. DPick the activity that starts earliest
10.Algorithms & complexityHard

Which definition of the class NP is correct?

  1. AProblems that are not polynomial
  2. BDecision problems whose yes-answers can be verified in polynomial time given a certificate
  3. CProblems that cannot be solved in polynomial time
  4. DProblems that no algorithm can solve
11.Data structuresMedium

What balance condition does an AVL tree maintain at every node?

  1. ABoth subtrees of every node contain exactly the same number of nodes
  2. BEvery node is coloured red or black
  3. CAll leaves are at exactly the same depth
  4. DThe heights of the left and right subtrees differ by at most 1
12.Data structuresHard

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(log n)
  2. BO(1 + α), where α = n / m
  3. CO(m)
  4. DO(n)
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
57121310131311ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

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

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

  1. AALTER TABLE
  2. BDELETE FROM with a WHERE clause
  3. CTRUNCATE TABLE
  4. DDROP TABLE
16.Databases & SQLHard

When is a schedule of concurrent transactions called serializable?

  1. AWhen the transactions run on one CPU only
  2. BWhen its effect is equivalent to some serial execution of the same transactions
  3. CWhen every transaction is given a serial number
  4. DWhen no transaction ever reads data
17.Computer networksMedium

What is the network address of the host 192.168.10.77/26?

  1. A192.168.10.64
  2. B192.168.10.128
  3. C192.168.10.0
  4. D192.168.10.76
18.Computer networksHard

In a TCP connection, what does the receive window (rwnd) control?

  1. AThe number of open ports
  2. BThe size of each IP packet header
  3. CHow much unacknowledged data the sender may send, based on the receiver's buffer
  4. DThe retransmission time-out, calculated from the measured round-trip time of each segment
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.47.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 11?

192.168.47.0/24/29subnet 11 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet 58 … 65012806413211618041201commonvaries

A router has routes to these networks: 10.188.58.0/24, 10.188.59.0/24, 10.188.60.0/24, 10.188.61.0/24, 10.188.62.0/24, 10.188.63.0/24, 10.188.64.0/24, 10.188.65.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.188.58.0/24 … 10.188.65.0/24
p =

3Solutions

1.Number systems

Answer: N = −14

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 + 2 = −14

Answer

N = −14
2.Number systems

Answer: R = 255

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

OR gives 1 where at least one bit is 1.

1101 1101
OR 0011 0111
= 1111 1111

Value of the result

1111 1111₂ = 255

Answer

R = 255
3.Propositional logic & proofs

Answer: ¬p → ¬r

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 = r, b = p

Contrapositive

¬b → ¬a = ¬p → ¬r

Answer

¬p → ¬r

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

4.Propositional logic & proofs

Answer: N = 10

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

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

Split on p

p = 1: 6
p = 0: 4
N = 6 + 4 = 10

Answer

N = 10
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′BC + AB′C′

Rows with F = 1

ABC = 000, 001, 011, 100
N = 4

Answer

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

Answer: Q = 1

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

1110 → 1111 → 0111 → 0011 → 0001

Decimal value

0001₂ = 1

Answer

Q = 1
7.Number theory & modular arithmetic

Answer: 63−1 mod 73 = 51

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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·63 (mod 73).

r = 73, t = 0
r = 63, q = 1, t = 1
r = 10, q = 6, t = −1
r = 3, q = 3, t = 7
r = 1, q = 3, t = −22

Inverse

1 ≡ −22·63 (mod 73)
x = 51
check: 63·51 = 3213 = 44·73 + 1

Answer

63−1 mod 73 = 51
8.Number theory & modular arithmetic

Answer: d = 91

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.

Modulus and totient

n = 7·19 = 133
φ(n) = 6·18 = 108

Extended Euclidean algorithm

r = 108, t = 0
r = 19, q = 5, t = 1
r = 13, q = 1, t = −5
r = 6, q = 2, t = 6
r = 1, q = 6, t = −17

Private exponent

d = 91
check: 19·91 = 1729 = 16·108 + 1

Answer

d = 91
9.Algorithms & complexity

Answer: A · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

A · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: B · Decision problems whose yes-answers can be verified in polynomial time given a certificate

Why this is the answer

For an NP problem, a proposed solution (certificate) can be checked quickly, for example verifying a given assignment satisfies a Boolean formula.

Answer

B · Decision problems whose yes-answers can be verified in polynomial time given a certificate
11.Data structures

Answer: D · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

D · The heights of the left and right subtrees differ by at most 1
12.Data structures

Answer: B · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

B · O(1 + α), where α = n / m
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) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: D

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: 5, C: 7
2. B (5): D: 18, E: 16
3. C (7)
4. E (16)
5. D (18): F: 31
6. F (31)

round 5

D

Answer

D
15.Databases & SQL

Answer: C · 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

C · TRUNCATE TABLE
16.Databases & SQL

Answer: B · When its effect is equivalent to some serial execution of the same transactions

Why this is the answer

Serializability is the main correctness criterion for concurrency: the interleaved result must equal the result of some one-at-a-time order.

Answer

B · When its effect is equivalent to some serial execution of the same transactions
17.Computer networks

Answer: A · 192.168.10.64

Why this is the answer

/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.

Answer

A · 192.168.10.64
18.Computer networks

Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer

Why this is the answer

The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.

Answer

C · How much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnetting

Answer: x = 80

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

23 = 8

Subnet number 11

network: (11 − 1)·8 = 80
broadcast: 80 + 8 − 1 = 87

Answer

x = 80
20.IPv4 subnetting

Answer: p = 17

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

58 = 00111010
65 = 01000001
common leading bits: 1

Prefix length

p = 16 + 1 = 17

Answer

p = /17

Summary route: 10.188.0.0/17 (it also covers 120 network(s) that are not in the list).

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