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 −105 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1058 bit
U =
2.Number systemsHard
8 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬(a ∨ c) → ¬¬b¬(a ∨ c) ∧ ¬¬b¬(a ∨ c) ∧ ¬b(a ∨ c) ∧ ¬¬b

You want to prove (a ∨ c) → ¬b by contradiction. Which statement do you assume at the start? Click it.

(a ∨ c) → ¬b
  1. 1¬(a ∨ c) → ¬¬b
  2. 2¬(a ∨ c) ∧ ¬¬b
  3. 3¬(a ∨ c) ∧ ¬b
  4. 4(a ∨ c) ∧ ¬¬b
4.Propositional logic & proofsHard
qrq ∨ rq¬r11100100

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

q ∨ rq∴ ¬r
  1. 1q = 1, r = 1
  2. 2q = 1, r = 0
  3. 3q = 0, r = 1
  4. 4q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​1I1​C′I2​CI3​ABS1​S0​F

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

I₀ = 0I₁ = 1I₂ = C′I₃ = C
N =
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 10 clock edges.

Q₃Q₂Q₁Q₀ = 000010
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 42 modulo 67: the number x with 0 < x < 67 and 42·x ≡ 1 (mod 67).

a = 42n = 67
42−1 mod 67 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (481, 23)d = ?

An RSA key uses the primes p = 13 and q = 37 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 13q = 37e = 23
d =
9.Algorithms & complexityMedium

Quicksort always uses the first element as pivot. On which input does it show its worst-case behaviour?

  1. AAn array of length 2
  2. BAn array whose elements are all different
  3. CAn array in random order
  4. DAn array that is already sorted
10.Algorithms & complexityHard

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n²)
  2. BΘ(n log n)
  3. CΘ(n³)
  4. DΘ(n² log n)
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(log n)
  2. BO(n²)
  3. CO(n)
  4. DO(1)
12.Data structuresHard

In a circular buffer, front == rear can mean either empty or full. What is a common way to tell them apart?

  1. AMake the buffer size a power of two
  2. BStore the front and rear in the same variable
  3. CAlways clear the buffer whenever front equals rear, so the state is empty
  4. DKeep a separate element count, or always leave one slot unused
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
14.Graph theoryHard
210738154118ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 2.

start: Dk = 2
  1. 1edge AC
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge GB
  9. 9edge GE
15.Databases & SQLMedium

What does a common table expression (WITH name AS (...)) provide?

  1. AAn index on a column
  2. BA new database user
  3. CA named temporary result that can be used within one statement
  4. DA permanent table that stays stored on disk until it is dropped
16.Databases & SQLHard

What is the result of SELECT COUNT(col) FROM t if all values in col are NULL and the table has 10 rows?

  1. A0
  2. B10
  3. CNULL
  4. DAn error
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. AHop count
  2. BLink bandwidth
  3. CBusiness policy
  4. DMeasured delay
18.Computer networksHard

A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?

  1. AThe network card driver
  2. BThe default gateway
  3. CName resolution (DNS)
  4. DThe physical cable
19.IPv4 subnettingMedium
/16 → /2211111111111111111111110000000000network bitshost bits

The network 172.26.0.0/16 is divided into equal subnets with prefix /22. How many subnets N are there?

172.26.0.0/16/22
N =
20.IPv4 subnettingHard

The network 192.168.245.0/24 is split with VLSM for these LANs: A: 16 hosts, B: 9 hosts, C: 12 hosts, D: 6 hosts, E: 41 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?

A: 16B: 9C: 12D: 6E: 41
x =

3Solutions

1.Number systems

Answer: U = 151

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

105 = 0110 1001
inverted: 1001 0110
+1 → 1001 0111

Read it as unsigned

1001 0111₂ = 256 − 105 = 151

Answer

U = 151
2.Number systems

Answer: M = 300 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

640·480·8 = 2457600 bit = 307200 B

Convert to KiB

307200/1024 = 300 KiB

Answer

M = 300 KiB
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).

Parts of the implication

a = a ∨ c, b = ¬b

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = (a ∨ c) ∧ ¬¬b

Answer

(a ∨ c) ∧ ¬¬b

If the assumption leads to a contradiction, the implication must be true.

4.Propositional logic & proofs

Answer: 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

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

Counterexample

q = 1, r = 1

Answer

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 = 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 + AB′C′ + ABC

Rows with F = 1

ABC = 010, 011, 100, 111
N = 4

Answer

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

Answer: Q = 12

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 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: 42−1 mod 67 = 8

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·42 (mod 67).

r = 67, t = 0
r = 42, q = 1, t = 1
r = 25, q = 1, t = −1
r = 17, q = 1, t = 2
r = 8, q = 2, t = −3
r = 1, q = 8, t = 8

Inverse

1 ≡ 8·42 (mod 67)
x = 8
check: 42·8 = 336 = 5·67 + 1

Answer

42−1 mod 67 = 8
8.Number theory & modular arithmetic

Answer: d = 263

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 = 13·37 = 481
φ(n) = 12·36 = 432

Extended Euclidean algorithm

r = 432, t = 0
r = 23, q = 18, t = 1
r = 18, q = 1, t = −18
r = 5, q = 3, t = 19
r = 3, q = 1, t = −75
r = 2, q = 1, t = 94
r = 1, q = 2, t = −169

Private exponent

d = 263
check: 23·263 = 6049 = 14·432 + 1

Answer

d = 263
9.Algorithms & complexity

Answer: D · An array that is already sorted

Why this is the answer

On sorted input the pivot is always the minimum, so one side is empty and the other has n − 1 elements, giving O(n²).

Answer

D · An array that is already sorted
10.Algorithms & complexity

Answer: C · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

C · Θ(n³)
11.Data structures

Answer: D · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

D · O(1)
12.Data structures

Answer: D · Keep a separate element count, or always leave one slot unused

Why this is the answer

With a count, empty means count = 0 and full means count = capacity; with one empty slot, full is defined as (rear + 1) mod n == front.

Answer

D · Keep a separate element count, or always leave one slot unused
13.Graph theory

Answer: H = 3

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.

Count

H = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
14.Graph theory

Answer: edge GB

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.

Prim’s algorithm

1. BD (7)
2. GB (1)
3. DE (8)
4. CE (3)
5. AC (2)
6. EF (4)

edge number 2

GB

Answer

edge GB
15.Databases & SQL

Answer: C · A named temporary result that can be used within one statement

Why this is the answer

A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.

Answer

C · A named temporary result that can be used within one statement
16.Databases & SQL

Answer: A · 0

Why this is the answer

COUNT(col) counts non-NULL values, so it returns 0; COUNT(*) would return 10.

Answer

A · 0
17.Computer networks

Answer: A · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

A · Hop count
18.Computer networks

Answer: C · Name resolution (DNS)

Why this is the answer

A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.

Answer

C · Name resolution (DNS)
19.IPv4 subnetting

Answer: N = 64

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

22 − 16 = 6

Number of subnets

N = 26 = 64

Answer

N = 64

Each /22 subnet has 210 − 2 = 1022 usable hosts.

20.IPv4 subnetting

Answer: x = 96

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

E: 41 hosts → 26 = 64 (/26)
A: 16 hosts → 25 = 32 (/27)
C: 12 hosts → 24 = 16 (/28)
B: 9 hosts → 24 = 16 (/28)
D: 6 hosts → 23 = 8 (/29)

Allocation, largest first

E: .0 – .63
A: .64 – .95
C: .96 – .111
B: .112 – .127
D: .128 – .135

Answer

x = 96

LAN C: 192.168.245.96/28.

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