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 complement0−12816413211618041201

The 8-bit pattern 0111 1010 stores a signed integer in two's complement. What is its value N in decimal?

0111 10108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 1001 XOR 0010 0010. What is the result's value R in decimal?

1001 1001XOR0010 0010
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (q ∧ (q → r)) → r a tautology, a contradiction or a contingency? Click your answer.

(q ∧ (q → r)) → r
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x (¬F(x) ∨ ¬G(x))∃x (¬F(x) ∧ ¬G(x))∃x (¬F(x) ∨ ¬G(x))∀x (¬F(x) ∧ ¬G(x))

Which statement is equivalent to ¬∀x (F(x) ∨ G(x)), with the negation moved all the way inside? Click it.

¬∀x (F(x) ∨ G(x))
  1. 1∀x (¬F(x) ∨ ¬G(x))
  2. 2∃x (¬F(x) ∧ ¬G(x))
  3. 3∃x (¬F(x) ∨ ¬G(x))
  4. 4∀x (¬F(x) ∧ ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011130214150706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 3, 4, 5). How many product terms N does it have?

F(A, B, C) = Σm(0, 1, 3, 4, 5)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKUP

A 4-bit binary up counter starts at 5 (decimal, Q = 0101). Which count Q (in decimal) does it show after 32 clock pulses?

Q0 = 532n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 3 (mod 5) and x ≡ 5 (mod 13).

x ≡ 3 (mod 5)x ≡ 5 (mod 13)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (799, 3)d = ?

An RSA key uses the primes p = 17 and q = 47 and the public exponent e = 3. Find the private exponent d (0 < d < φ(n)).

p = 17q = 47e = 3
d =
9.Algorithms & complexityMedium

What does it mean that a problem has optimal substructure?

  1. AThe problem can be solved in constant time
  2. BThe problem has exactly one optimal solution, which every subproblem shares
  3. CAn optimal solution can be built from optimal solutions of its subproblems
  4. DEvery subproblem has the same size
10.Algorithms & complexityHard

Is 22n in O(2ⁿ)?

  1. ANo, because 22n grows more slowly than 2ⁿ
  2. BYes, because both are exponential
  3. CNo, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound
  4. DYes, because constant factors in the exponent are always ignored in Big O
11.Data structuresMedium

How long does a search for a word of length L take in a trie with a fixed alphabet?

  1. AO(n), where n is the number of words
  2. BO(L), independent of how many words are stored
  3. CO(log n), where n is the number of words
  4. DO(L × n)
12.Data structuresHard

What is the main advantage of a skip list over a sorted linked list?

  1. AIt stores keys without order
  2. BIt uses no extra memory
  3. CExpected O(log n) search by using several levels of forward pointers
  4. DGuaranteed O(1) search because every node stores a pointer to every other node
13.Graph theoryMedium

A tree has 2 vertices of degree 3, 2 vertices of degree 4, 3 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 32 × deg 43 × deg 5
L =
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

Why are indexes created on columns?

  1. ATo make inserts and updates faster on those columns
  2. BTo speed up searches and joins on those columns
  3. CTo save disk space
  4. DTo prevent NULL values
16.Databases & SQLHard

After customers LEFT JOIN orders, why use COUNT(o.id) rather than COUNT(*) to count orders per customer?

  1. ACOUNT(o.id) counts only distinct customers
  2. BCOUNT(*) is not allowed after a join
  3. CCOUNT(*) ignores rows with NULL values
  4. DCOUNT(*) counts the NULL-filled row as 1 for customers without orders
17.Computer networksMedium

At which OSI layer does a switch normally operate?

  1. ATransport (layer 4)
  2. BData link (layer 2)
  3. CPhysical (layer 1)
  4. DNetwork (layer 3)
18.Computer networksHard

Which of these addresses is a public (not RFC 1918 private) IPv4 address?

  1. A172.32.0.1
  2. B10.255.0.1
  3. C172.31.0.1
  4. D192.168.255.1
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2410 subnets
H =
20.IPv4 subnettingHard

The network 192.168.165.0/24 is split with VLSM for these LANs: A: 6 hosts, B: 27 hosts, C: 23 hosts, D: 65 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 B?

A: 6B: 27C: 23D: 65
x =

3Solutions

1.Number systems

Answer: N = 122

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.

64 + 32 + 16 + 8 + 2 = 122

Answer

N = 122
2.Number systems

Answer: R = 187

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

XOR gives 1 where the two bits differ.

1001 1001
XOR 0010 0010
= 1011 1011

Value of the result

1011 1011₂ = 187

Answer

R = 187
3.Propositional logic & proofs

Answer: tautology

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).

Truth table row by row

q = 1, r = 1: q → r = 1, q ∧ (q → r) = 1, (q ∧ (q → r)) → r = 1
q = 1, r = 0: q → r = 0, q ∧ (q → r) = 0, (q ∧ (q → r)) → r = 1
q = 0, r = 1: q → r = 1, q ∧ (q → r) = 0, (q ∧ (q → r)) → r = 1
q = 0, r = 0: q → r = 1, q ∧ (q → r) = 0, (q ∧ (q → r)) → r = 1

Classify

(q ∧ (q → r)) → r: 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: ∃x (¬F(x) ∧ ¬G(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) ∨ G(x)) ≡ ∃x (¬F(x) ∧ ¬G(x))

Answer

∃x (¬F(x) ∧ ¬G(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: 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.

ABC00011110011011130214150706

Group the 1s

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

B′
A′C

Minimal SOP form

F = B′ + A′C
N = 2

Answer

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

Answer: Q = 5

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.

Full cycles

Every 16 pulses bring the counter back to the same state.

32 = 2·16 + 0

Count after the remaining pulses

Q = (5 + 0) mod 16 = 5
Q = 0101₂

Answer

Q = 5
7.Number theory & modular arithmetic

Answer: x = 18

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

Result

x = 3 + 5·3 = 18
x = 18
check: 18 mod 5 = 3, 18 mod 13 = 5

Answer

x = 18
8.Number theory & modular arithmetic

Answer: d = 491

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 = 17·47 = 799
φ(n) = 16·46 = 736

Extended Euclidean algorithm

r = 736, t = 0
r = 3, q = 245, t = 1
r = 1, q = 3, t = −245

Private exponent

d = 491
check: 3·491 = 1473 = 2·736 + 1

Answer

d = 491
9.Algorithms & complexity

Answer: C · An optimal solution can be built from optimal solutions of its subproblems

Why this is the answer

Optimal substructure is required for both greedy algorithms and dynamic programming to produce correct answers.

Answer

C · An optimal solution can be built from optimal solutions of its subproblems
10.Algorithms & complexity

Answer: C · No, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound

Why this is the answer

A constant added to the exponent gives a constant factor, but a constant multiplying the exponent changes the base and the growth rate.

Answer

C · No, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound
11.Data structures

Answer: B · O(L), independent of how many words are stored

Why this is the answer

Each character selects one child, so a lookup takes L steps no matter how many words the trie contains.

Answer

B · O(L), independent of how many words are stored
12.Data structures

Answer: C · Expected O(log n) search by using several levels of forward pointers

Why this is the answer

Randomised levels allow searches to skip many nodes, giving expected O(log n) search, insert and delete.

Answer

C · Expected O(log n) search by using several levels of forward pointers
13.Graph theory

Answer: L = 17

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.

Handshake lemma for a tree

L + 29 = 2(L + 7) − 2
L = 17

Answer

L = 17

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 0
D: 0 1 0 0 1
E: 0 1 0 1 0

One more step

(M³)DD = 1·0 + 1·1 + 1·0 + 2·0 + 1·1 = 2

Answer

N = 2
15.Databases & SQL

Answer: B · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

B · To speed up searches and joins on those columns
16.Databases & SQL

Answer: D · COUNT(*) counts the NULL-filled row as 1 for customers without orders

Why this is the answer

The left join gives one row with NULL order columns for such customers; COUNT(o.id) skips the NULL and correctly reports 0.

Answer

D · COUNT(*) counts the NULL-filled row as 1 for customers without orders
17.Computer networks

Answer: B · Data link (layer 2)

Why this is the answer

A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.

Answer

B · Data link (layer 2)
18.Computer networks

Answer: A · 172.32.0.1

Why this is the answer

172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.

Answer

A · 172.32.0.1
19.IPv4 subnetting

Answer: H = 14

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

24 = 16 ≥ 10 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 128

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

D: 65 hosts → 27 = 128 (/25)
B: 27 hosts → 25 = 32 (/27)
C: 23 hosts → 25 = 32 (/27)
A: 6 hosts → 23 = 8 (/29)

Allocation, largest first

D: .0 – .127
B: .128 – .159
C: .160 – .191
A: .192 – .199

Answer

x = 128

LAN B: 192.168.165.128/27.

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