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 −93 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −938 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 1000 XOR 1110 0010. What is the result's value R in decimal?

0111 1000XOR1110 0010
R =
3.Propositional logic & proofsMedium
¬q ∧ p¬q ∨ ¬pq ∧ ¬p¬q → ¬p

Which formula is logically equivalent to ¬(q → p)? Click it.

¬(q → p)
  1. 1¬q ∧ p
  2. 2¬q ∨ ¬p
  3. 3q ∧ ¬p
  4. 4¬q → ¬p
4.Propositional logic & proofsHard
pqq ∨ pq¬p11100100

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

q ∨ pq∴ ¬p
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010110312040507061120131151140809011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 10, 12, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 10, 12, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD12345678

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0D
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 10 modulo 19: the number x with 0 < x < 19 and 10·x ≡ 1 (mod 19).

a = 10n = 19
10−1 mod 19 =
8.Number theory & modular arithmeticHard
287c = me​ mod n?m = cd​ mod nm(n, e) = (731, 19)d

Encrypt the message m = 287 with the RSA public key (n, e) = (731, 19). Find the ciphertext c.

n = 731e = 19m = 287
c =
9.Algorithms & complexityMedium

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

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

What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?

  1. AO(2ⁿ) time and O(1) space
  2. BO(log n) time and O(n) space
  3. CO(n) time and O(1) space
  4. DO(n) time and O(n) space
11.Data structuresMedium

A trie (prefix tree) is especially efficient for:

  1. AFinding shortest paths
  2. BPrefix searches such as autocomplete
  3. CStoring matrices
  4. DSorting numbers
12.Data structuresHard

For each element of an array you want the next greater element to its right, in O(n) total time. Which structure helps?

  1. AA stack that keeps elements waiting for their answer
  2. BA hash table of all elements
  3. CA queue that holds every pair of elements
  4. DA binary search tree rebuilt for each element
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
6121314426139104ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 6; A itself is round 1.

start: Around 6
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

Table a has 3 rows and table b has 4 rows. How many rows does SELECT * FROM a CROSS JOIN b return?

  1. A3
  2. B12
  3. C4
  4. D7
16.Databases & SQLHard

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

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

What uniquely identifies one end of a TCP connection on the internet?

  1. AA subnet mask and gateway
  2. BAn IP address together with a port number
  3. CA MAC address alone
  4. DA domain name alone
18.Computer networksHard

A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?

  1. ADNS needs ICMP to resolve names
  2. BPath MTU discovery fails because "fragmentation needed" messages are blocked
  3. CTCP cannot open connections without ICMP
  4. DTLS certificates are sent over ICMP
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/244 subnets
H =
20.IPv4 subnettingHard
third octet 201 … 204112816403201618040211commonvaries

A router has routes to these networks: 10.30.201.0/24, 10.30.202.0/24, 10.30.203.0/24, 10.30.204.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.30.201.0/24 … 10.30.204.0/24
p =

3Solutions

1.Number systems

Answer: U = 163

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

93 = 0101 1101
inverted: 1010 0010
+1 → 1010 0011

Read it as unsigned

1010 0011₂ = 256 − 93 = 163

Answer

U = 163
2.Number systems

Answer: R = 154

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.

0111 1000
XOR 1110 0010
= 1001 1010

Value of the result

1001 1010₂ = 154

Answer

R = 154
3.Propositional logic & proofs

Answer: q ∧ ¬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

¬(a → b) ≡ ¬(¬a ∨ b) ≡ a ∧ ¬b: the only row where an implication fails.
¬(q → p) ≡ q ∧ ¬p

Check with the truth table

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

Answer

q ∧ ¬p

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

4.Propositional logic & proofs

Answer: p = 1, q = 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

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

Counterexample

p = 1, q = 1

Answer

p = 1, q = 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

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.

ABCD000111100001111010110312040507061120131151140809011110

Group the 0s

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

(B + C′ + D′)
(B′ + C + D′)
(A + B′)
(A′ + B + C)

Minimal POS form

F = (B + C′ + D′)(B′ + C + D′)(A + B′)(A′ + B + C)
N = 4

Answer

N = 4

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

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

Answer: N = 4

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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 1 → Q = 1
3: D = 0 → Q = 0
4: D = 1 → Q = 1
5: D = 0 → Q = 0
6: D = 0 → Q = 0
7: D = 0 → Q = 0
8: D = 0 → Q = 0

Count the changes

0 → 1 → 1 → 0 → 1 → 0 → 0 → 0 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: 10−1 mod 19 = 2

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·10 (mod 19).

r = 19, t = 0
r = 10, q = 1, t = 1
r = 9, q = 1, t = −1
r = 1, q = 9, t = 2

Inverse

1 ≡ 2·10 (mod 19)
x = 2
check: 10·2 = 20 = 1·19 + 1

Answer

10−1 mod 19 = 2
8.Number theory & modular arithmetic

Answer: c = 77

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.

Repeated squaring

19 = 16 + 2 + 1
2871 ≡ 287 (mod 731)
2872 ≡ 497 (mod 731)
2874 ≡ 662 (mod 731)
2878 ≡ 375 (mod 731)
28716 ≡ 273 (mod 731)
28719 ≡ 287·497·273 ≡ 77 (mod 731)

Ciphertext

c = 77

Answer

c = 77
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: C · O(n) time and O(1) space

Why this is the answer

Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.

Answer

C · O(n) time and O(1) space
11.Data structures

Answer: B · Prefix searches such as autocomplete

Why this is the answer

Tries store strings by characters along paths, so lookups and prefix queries take time proportional to the key length.

Answer

B · Prefix searches such as autocomplete
12.Data structures

Answer: A · A stack that keeps elements waiting for their answer

Why this is the answer

Scanning left to right, pop every stacked element smaller than the current one (the current element is their answer), then push it; each element moves at most twice.

Answer

A · A stack that keeps elements waiting for their answer
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: G

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

1. A (0): B: 6, C: 12
2. B (6): D: 20
3. C (12): E: 16, G: 21
4. E (16): D: 18, F: 29, G: 20
5. D (18): F: 24
6. G (20)
7. F (24)

round 6

G

Answer

G
15.Databases & SQL

Answer: B · 12

Why this is the answer

A cross join gives the Cartesian product, 3 × 4 = 12 rows.

Answer

B · 12
16.Databases & SQL

Answer: B · 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

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

Answer: B · An IP address together with a port number

Why this is the answer

A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.

Answer

B · An IP address together with a port number
18.Computer networks

Answer: B · Path MTU discovery fails because "fragmentation needed" messages are blocked

Why this is the answer

PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.

Answer

B · Path MTU discovery fails because "fragmentation needed" messages are blocked
19.IPv4 subnetting

Answer: H = 62

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

22 = 4 ≥ 4 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: p = 21

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

201 = 11001001
204 = 11001100
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.30.200.0/21 (it also covers 4 network(s) that are not in the list).

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