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 −101 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1018 bit
U =
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
p ↔ ¬r¬p ∨ ¬r¬p ↔ ¬rp ∧ ¬r

Which formula is logically equivalent to ¬(p ↔ r)? Click it.

¬(p ↔ r)
  1. 1p ↔ ¬r
  2. 2¬p ∨ ¬r
  3. 3¬p ↔ ¬r
  4. 4p ∧ ¬r
4.Propositional logic & proofsHard

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

(r ∧ (q ∨ ¬r)) ∨ ((¬p ∨ ¬q) → (s ∧ p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(2, 3, 4, 8, 10, 14)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 12 (decimal, Q = 1100). Which count Q (in decimal) does it show after 25 clock pulses?

Q0 = 1225n = 4
Q =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(75), the number of integers from 1 to 75 that are coprime to 75.

n = 75 = 3·52
φ(75) =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (187, 11)d = ?

An RSA key uses the primes p = 11 and q = 17 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 11q = 17e = 11
d =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms are always rounded to integers
  2. BAll logarithms in computer science have base 10, so it is implied
  3. CThe base only matters for small n
  4. DLogarithms with different bases differ only by a constant factor
10.Algorithms & complexityHard

Which definition of the class P is correct?

  1. ADecision problems solvable by a deterministic algorithm in polynomial time
  2. BProblems whose solutions can be checked but never found in polynomial time
  3. CProblems that can be solved in parallel
  4. DProblems that need exponential time on every computer
11.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AOnly the colliding keys are moved to a new overflow list at the end of the table
  2. BThe existing keys are sorted in place
  3. CA larger table is allocated and every key is inserted again
  4. DThe oldest keys are deleted to make room
12.Data structuresHard

What is the running time of breadth-first search on a graph with V vertices stored as an adjacency matrix?

  1. AO(V + E)
  2. BO(V²)
  3. CO(E log V)
  4. DO(V)
13.Graph theoryMedium

A connected graph has 20 vertices and 25 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 25
k =
14.Graph theoryHard
1451281371417ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

Why does SELECT price * 2 AS double_price FROM t WHERE double_price > 10 fail in most databases?

  1. AArithmetic is not allowed in the SELECT list
  2. BWHERE is evaluated before the SELECT list, so the alias does not exist yet
  3. CWHERE can only compare text columns
  4. DAliases may not contain an underscore
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AOnly unmatched rows
  2. BOnly matching rows
  3. CThe Cartesian product, pairing each row of one table with every row of the other
  4. DAll rows from both tables, matched where possible and NULL-filled otherwise
17.Computer networksMedium

A browser receives HTTP status code 404. What does it mean?

  1. AThe request succeeded
  2. BThe server had an internal error
  3. CThe requested resource was not found
  4. DThe client must log in first
18.Computer networksHard

Which routing protocol is a link-state protocol?

  1. AARP
  2. BRIP
  3. CBGP
  4. DOSPF
19.IPv4 subnettingMedium
/20 → /2711111111111111111111111111100000network bitshost bits

The network 10.94.176.0/20 is divided into equal subnets with prefix /27. How many subnets N are there?

10.94.176.0/20/27
N =
20.IPv4 subnettingHard
last octet = 199112816403201608141211networkhost

A host has the address 10.20.22.199/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.22.199/29
x =

3Solutions

1.Number systems

Answer: U = 155

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

101 = 0110 0101
inverted: 1001 1010
+1 → 1001 1011

Read it as unsigned

1001 1011₂ = 256 − 101 = 155

Answer

U = 155
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: 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).

The law

¬(a ↔ b) is 1 exactly when a and b differ, which is a ↔ ¬b.
¬(p ↔ r) ≡ p ↔ ¬r

Check with the truth table

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

Answer

p ↔ ¬r

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

4.Propositional logic & proofs

Answer: N = 8

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

Split on p

p = 1: 6
p = 0: 2
N = 6 + 2 = 8

Answer

N = 8
5.Digital logic: gates, K-maps & flip-flops

Answer: AB = 01, CD = 00

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m4: m5, m6, m0, m12 → 0
ABCD0001111000011110111111

The isolated 1

m4 = A′BC′D′ → AB = 01, CD = 00

Answer

AB = 01, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 3

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.

25 = 1·16 + 9

Count after the remaining pulses

Q = (12 − 9) mod 16 = 3
Q = 0011₂

Answer

Q = 3
7.Number theory & modular arithmetic

Answer: φ(75) = 40

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

75 = 3·52

Totient

φ(75) = φ(3)·φ(25) = 2·20 = 40

Answer

φ(75) = 40
8.Number theory & modular arithmetic

Answer: d = 131

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 = 11·17 = 187
φ(n) = 10·16 = 160

Extended Euclidean algorithm

r = 160, t = 0
r = 11, q = 14, t = 1
r = 6, q = 1, t = −14
r = 5, q = 1, t = 15
r = 1, q = 5, t = −29

Private exponent

d = 131
check: 11·131 = 1441 = 9·160 + 1

Answer

d = 131
9.Algorithms & complexity

Answer: D · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

D · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: A · Decision problems solvable by a deterministic algorithm in polynomial time

Why this is the answer

P contains decision problems with a deterministic polynomial-time algorithm, such as checking whether a graph is connected.

Answer

A · Decision problems solvable by a deterministic algorithm in polynomial time
11.Data structures

Answer: C · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

C · A larger table is allocated and every key is inserted again
12.Data structures

Answer: B · O(V²)

Why this is the answer

Each of the V vertices scans a row of V entries, giving O(V²); with adjacency lists BFS is O(V + E).

Answer

B · O(V²)
13.Graph theory

Answer: k = 6

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.

Edges of a spanning tree

20 − 1 = 19

Edges to remove

k = 25 − 19 = 6

Answer

k = 6
14.Graph theory

Answer: d(F) = 37

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): C: 14
C (14): G: 28
G (28): B: 35, D: 29, E: 35
D (29): B: 34, F: 37
B (34)
E (35)
F (37)

Shortest path

A → C → G → D → F
d(F) = 37

Answer

d(F) = 37
15.Databases & SQL

Answer: B · WHERE is evaluated before the SELECT list, so the alias does not exist yet

Why this is the answer

Logically FROM and WHERE run before SELECT, so repeat the expression in WHERE or use a subquery; ORDER BY can use the alias because it runs after SELECT.

Answer

B · WHERE is evaluated before the SELECT list, so the alias does not exist yet
16.Databases & SQL

Answer: D · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

D · All rows from both tables, matched where possible and NULL-filled otherwise
17.Computer networks

Answer: C · The requested resource was not found

Why this is the answer

404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.

Answer

C · The requested resource was not found
18.Computer networks

Answer: D · OSPF

Why this is the answer

OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.

Answer

D · OSPF
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

27 − 20 = 7

Number of subnets

N = 27 = 128

Answer

N = 128

Each /27 subnet has 25 − 2 = 30 usable hosts.

20.IPv4 subnetting

Answer: x = 198

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 of this host

network: ⌊199/8⌋·8 = 192
broadcast: 192 + 8 − 1 = 199
usable hosts: 193 … 198

Answer

x = 198

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