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 complement1−12816413211608041211

The 8-bit pattern 1111 0011 stores a signed integer in two's complement. What is its value N in decimal?

1111 00118 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1110 1100 AND 0100 1010. What is the result's value R in decimal?

1110 1100AND0100 1010
R =
3.Propositional logic & proofsMedium
abcb ↔ ca ∧ (b ↔ c)¬(a ∧ c) ↔ (a ∧ (b ↔ c))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) true? Call this number N.

¬(a ∧ c) ↔ (a ∧ (b ↔ c))
N =
4.Propositional logic & proofsHard
¬pr ∨ qp ∧ qp ∨ q

The premises r ∨ p; ¬r ∨ q are given. Which conclusion follows from them? Click it.

r ∨ p¬r ∨ q
  1. 1¬p
  2. 2r ∨ q
  3. 3p ∧ q
  4. 4p ∨ q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010312140507160121130151141809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 2, 4, 6, 8, 13, 14). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 2, 4, 6, 8, 13, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-10

A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 6. Which count Q (in decimal) does it show after 16 clock pulses?

Q0 = 616N = 10
Q =
7.Number theory & modular arithmeticMedium

Solve 26x ≡ 48 (mod 67). Give the solution x with 0 < x < 67.

26x ≡ 48 (mod 67)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (629, 5)d = ?

An RSA key uses the primes p = 17 and q = 37 and the public exponent e = 5. Find the private exponent d (0 < d < φ(n)).

p = 17q = 37e = 5
d =
9.Algorithms & complexityMedium

Dynamic programming is useful when a problem has:

  1. AOnly random inputs
  2. BNo subproblems at all, so it can be solved in one step
  3. CA single loop
  4. DOverlapping subproblems and optimal substructure
10.Algorithms & complexityHard

Topological sorting is possible only for:

  1. ADirected acyclic graphs
  2. BGraphs with negative weights
  3. CComplete graphs
  4. DUndirected graphs
11.Data structuresMedium

An adjacency matrix for a graph with V vertices uses how much memory?

  1. AO(log V)
  2. BO(V²)
  3. CO(E)
  4. DO(V + E)
12.Data structuresHard

Which property keeps every B-tree perfectly balanced?

  1. AAll leaves are at the same depth
  2. BEach node stores one key
  3. CEvery node has exactly two children
  4. DThe keys are stored only in the root
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 10
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
221361310107515ABCDEFG

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

Which query returns the number of orders per customer?

  1. ASELECT customer_id FROM orders WHERE COUNT(*) > 0
  2. BSELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
  3. CSELECT customer_id, COUNT(*) FROM orders ORDER BY customer_id
  4. DSELECT DISTINCT COUNT(*) FROM orders
16.Databases & SQLHard

When is decomposing R into R1 and R2 guaranteed to be lossless?

  1. AWhen the common attributes of R1 and R2 form a superkey of R1 or of R2
  2. BWhen every attribute of R appears in both R1 and R2
  3. CWhen R1 and R2 have the same number of columns
  4. DWhen R1 and R2 have no attributes in common
17.Computer networksMedium

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

  1. AThe server had an internal error
  2. BThe request succeeded
  3. CThe requested resource was not found
  4. DThe client must log in first
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. C6 kbit/s
  4. D3 kbit/s
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1689 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1689 hosts
p =
20.IPv4 subnettingHard
last octet = 112012816413211608040201networkhost

A host has the address 10.20.213.112/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.213.112/29
x =

3Solutions

1.Number systems

Answer: N = −13

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 + 1 = −13

Answer

N = −13
2.Number systems

Answer: R = 72

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

AND gives 1 only where both bits are 1.

1110 1100
AND 0100 1010
= 0100 1000

Value of the result

0100 1000₂ = 72

Answer

R = 72
3.Propositional logic & proofs

Answer: N = 2

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.

Truth table row by row

a = 1, b = 1, c = 1: a ∧ c = 1, ¬(a ∧ c) = 0, b ↔ c = 1, a ∧ (b ↔ c) = 1, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0
a = 1, b = 1, c = 0: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 0, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0
a = 1, b = 0, c = 1: a ∧ c = 1, ¬(a ∧ c) = 0, b ↔ c = 0, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 1
a = 1, b = 0, c = 0: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 1, a ∧ (b ↔ c) = 1, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 1
a = 0, b = 1, c = 1: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 1, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0
a = 0, b = 1, c = 0: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 0, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0
a = 0, b = 0, c = 1: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 0, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0
a = 0, b = 0, c = 0: a ∧ c = 0, ¬(a ∧ c) = 1, b ↔ c = 1, a ∧ (b ↔ c) = 0, ¬(a ∧ c) ↔ (a ∧ (b ↔ c)) = 0

Count the rows with value 1

¬(a ∧ c) ↔ (a ∧ (b ↔ c)): 0, 0, 1, 1, 0, 0, 0, 0
N = 2

Answer

N = 2
4.Propositional logic & proofs

Answer: p ∨ q

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.

Rows with all premises true

p = 1, q = 1, r = 1: ¬p = 0, r ∨ q = 1, p ∧ q = 1, p ∨ q = 1
p = 1, q = 1, r = 0: ¬p = 0, r ∨ q = 1, p ∧ q = 1, p ∨ q = 1
p = 1, q = 0, r = 0: ¬p = 0, r ∨ q = 0, p ∧ q = 0, p ∨ q = 1
p = 0, q = 1, r = 1: ¬p = 1, r ∨ q = 1, p ∧ q = 0, p ∨ q = 1

Conclusion

p ∨ q (resolution)

Answer

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

Answer: N = 4

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.

ABCD000111100001111010010312140507160121130151141809011010

Group the 1s

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

B′C′D′
BCD′
A′D′
ABC′D

Minimal SOP form

F = B′C′D′ + BCD′ + A′D′ + ABC′D
N = 4

Answer

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

Answer: Q = 2

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 10 pulses bring the counter back to the same state.

16 = 1·10 + 6

Count after the remaining pulses

Q = (6 + 6) mod 10 = 2
Q = 0010₂

Answer

Q = 2
7.Number theory & modular arithmetic

Answer: x = 7

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).

Inverse of 26

r = 67, t = 0
r = 26, q = 2, t = 1
r = 15, q = 1, t = −2
r = 11, q = 1, t = 3
r = 4, q = 2, t = −5
r = 3, q = 1, t = 13
r = 1, q = 3, t = −18
26−1 ≡ 49 (mod 67)

Solve

x ≡ 49·48 = 2352 ≡ 7 (mod 67)
check: 26·7 = 182 ≡ 48 (mod 67)

Answer

x = 7
8.Number theory & modular arithmetic

Answer: d = 461

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 = 17·37 = 629
φ(n) = 16·36 = 576

Extended Euclidean algorithm

r = 576, t = 0
r = 5, q = 115, t = 1
r = 1, q = 5, t = −115

Private exponent

d = 461
check: 5·461 = 2305 = 4·576 + 1

Answer

d = 461
9.Algorithms & complexity

Answer: D · Overlapping subproblems and optimal substructure

Why this is the answer

Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.

Answer

D · Overlapping subproblems and optimal substructure
10.Algorithms & complexity

Answer: A · Directed acyclic graphs

Why this is the answer

A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.

Answer

A · Directed acyclic graphs
11.Data structures

Answer: B · O(V²)

Why this is the answer

An adjacency matrix stores V × V entries, efficient for dense graphs; adjacency lists use O(V + E), better for sparse graphs.

Answer

B · O(V²)
12.Data structures

Answer: A · All leaves are at the same depth

Why this is the answer

Nodes split upwards when full and the tree only gets taller when the root splits, so all leaves always stay on the same level.

Answer

A · All leaves are at the same depth
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) = 1, deg(B) = 3, deg(C) = 3, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 3

Odd vertices

A, B, C, G → 4
neither

Answer

neither
14.Graph theory

Answer: d(F) = 18

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: 2
C (2): B: 4, E: 8, G: 7
B (4): D: 17
G (7)
E (8): F: 18
D (17)
F (18)

Shortest path

A → C → E → F
d(F) = 18

Answer

d(F) = 18
15.Databases & SQL

Answer: B · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id

Why this is the answer

GROUP BY customer_id makes one group per customer, and COUNT(*) counts the rows in each group.

Answer

B · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
16.Databases & SQL

Answer: A · When the common attributes of R1 and R2 form a superkey of R1 or of R2

Why this is the answer

If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.

Answer

A · When the common attributes of R1 and R2 form a superkey of R1 or of R2
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: 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: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1689
211 − 2 = 2046 ≥ 1689 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 119

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: ⌊112/8⌋·8 = 112
broadcast: 112 + 8 − 1 = 119
usable hosts: 113 … 118

Answer

x = 119

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