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 complement1−12816403211608040211

The 8-bit pattern 1101 0001 stores a signed integer in two's complement. What is its value N in decimal?

1101 00018 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0110 0111 AND 1101 0110. What is the result's value R in decimal?

0110 0111AND1101 0110
R =
3.Propositional logic & proofsMedium
r ∨ (q ∧ p)(r ∧ q) ∨ (r ∧ p)(r ∨ q) ∧ (r ∨ p)(r ∧ q) ∨ p

Which formula is logically equivalent to r ∧ (q ∨ p)? Click it.

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

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

¬((s → q) ∨ r) ↔ ((r ∨ p) ∧ p)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001030214051706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 4, 7). How many product terms N does it have?

F(A, B, C) = Σm(0, 4, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100001131214050706X121131150141809011X10

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 4, 8, 13, 15) + d(10, 12), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(2, 3, 4, 8, 13, 15) + d(10, 12)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 9 (mod 11) and x ≡ 6 (mod 13).

x ≡ 9 (mod 11)x ≡ 6 (mod 13)
x =
8.Number theory & modular arithmeticHard

61 is prime. Use Fermat’s little theorem to find 530087 mod 61.

p = 61k = 30087
530087 mod 61 =
9.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(n log n)
  2. BΩ(n)
  3. CΩ(log n)
  4. DΩ(n²)
10.Algorithms & complexityHard

Is 22n in O(2ⁿ)?

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

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

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

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(log n)
  2. BO(n)
  3. CO(n log n)
  4. DO(1)
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| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
121761319318516ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex G. Click the edge it adds as edge number 3.

start: Gk = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge DE
  6. 6edge DF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GD
  10. 10edge GE
15.Databases & SQLMedium

Which query returns the number of orders per customer?

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

What is a known drawback of decomposing tables into BCNF compared with 3NF?

  1. AThe decomposition always loses data when joined back
  2. BBCNF tables cannot have primary keys
  3. CBCNF allows more redundancy than 3NF
  4. DSome functional dependencies may no longer be checkable within a single table
17.Computer networksMedium

What characterises a distributed denial-of-service (DDoS) attack?

  1. AOne attacker guesses a user's password
  2. BA virus deletes files on a single PC
  3. CMany compromised machines flood a target with traffic
  4. DAn attacker reads unencrypted Wi-Fi traffic from a nearby car
18.Computer networksHard

Why does a browser warn when a site presents a self-signed certificate?

  1. ANo trusted certificate authority vouches that the key belongs to that site
  2. BSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
  3. CThe certificate is always expired
  4. DThe site must be using HTTP instead of HTTPS
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 634 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

634 hosts
p =
20.IPv4 subnettingHard

The network 192.168.249.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 24 hosts, C: 6 hosts, D: 13 hosts, E: 30 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: 5B: 24C: 6D: 13E: 30
x =

3Solutions

1.Number systems

Answer: N = −47

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 + 16 + 1 = −47

Answer

N = −47
2.Number systems

Answer: R = 70

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.

0110 0111
AND 1101 0110
= 0100 0110

Value of the result

0100 0110₂ = 70

Answer

R = 70
3.Propositional logic & proofs

Answer: (r ∧ q) ∨ (r ∧ p)

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

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
r ∧ (q ∨ p) ≡ (r ∧ q) ∨ (r ∧ p)

Check with the truth table

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

Answer

(r ∧ q) ∨ (r ∧ p)

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

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

Split on p

p = 1: 1
p = 0: 7
N = 1 + 7 = 8

Answer

N = 8
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.

ABC00011110011001030214051706

Group the 1s

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

B′C′
ABC

Minimal SOP form

F = B′C′ + ABC
N = 2

Answer

N = 2
6.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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′CD′, BC′D′, AC′D′, AB′D′, ABD, A′B′C, ABC′
ABCD00011110000111100001131214050706X121131150141809011X10

Minimal cover

F = BC′D′ + AC′D′ + ABD + A′B′C
N = 4

Answer

N = 4

Another grouping of the same size exists; it has the same number of terms and literals.

7.Number theory & modular arithmetic

Answer: x = 97

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 = 9 + 11k

Second congruence

9 + 11k ≡ 6 (mod 13) → 11k ≡ 10 (mod 13)
k ≡ 6·10 ≡ 8 (mod 13)

Result

x = 9 + 11·8 = 97
x = 97
check: 97 mod 11 = 9, 97 mod 13 = 6

Answer

x = 97
8.Number theory & modular arithmetic

Answer: 530087 mod 61 = 41

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.

Reduce the exponent

30087 mod 60 = 27
530087 ≡ 527 (mod 61)

Small power

527 mod 61 = 41

Answer

530087 mod 61 = 41
9.Algorithms & complexity

Answer: A · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

A · Ω(n log n)
10.Algorithms & complexity

Answer: A · 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

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

Answer: D · 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

D · O(V²)
12.Data structures

Answer: A · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

A · O(log n)
13.Graph theory

Answer: Euler path, no 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) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 2

Odd vertices

B, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: edge BC

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. GB (3)
2. GD (5)
3. BC (7)
4. AC (1)
5. DE (13)
6. DF (19)

edge number 3

BC

Answer

edge BC
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: D · Some functional dependencies may no longer be checkable within a single table

Why this is the answer

A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.

Answer

D · Some functional dependencies may no longer be checkable within a single table
17.Computer networks

Answer: C · Many compromised machines flood a target with traffic

Why this is the answer

In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.

Answer

C · Many compromised machines flood a target with traffic
18.Computer networks

Answer: A · No trusted certificate authority vouches that the key belongs to that site

Why this is the answer

Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.

Answer

A · No trusted certificate authority vouches that the key belongs to that site
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 634
210 − 2 = 1022 ≥ 634 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 32

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: 30 hosts → 25 = 32 (/27)
B: 24 hosts → 25 = 32 (/27)
D: 13 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
A: 5 hosts → 23 = 8 (/29)

Allocation, largest first

E: .0 – .31
B: .32 – .63
D: .64 – .79
C: .80 – .87
A: .88 – .95

Answer

x = 32

LAN B: 192.168.249.32/27.

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