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 −120 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1208 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

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

1010 0100AND0100 1110
R =
3.Propositional logic & proofsMedium
¬¬p → ¬(q ∨ r)(q ∨ r) ∧ ¬¬p¬(q ∨ r) → ¬¬p¬p → (q ∨ r)

You want to prove (q ∨ r) → ¬p by contrapositive. Which implication do you prove instead? Click it.

(q ∨ r) → ¬p
  1. 1¬¬p → ¬(q ∨ r)
  2. 2(q ∨ r) ∧ ¬¬p
  3. 3¬(q ∨ r) → ¬¬p
  4. 4¬p → (q ∨ r)
4.Propositional logic & proofsHard
∃x (A(x) → ∀y ¬R(x, y))∃x (A(x) ∧ ∀y ¬R(x, y))∃x (¬A(x) ∧ ∃y R(x, y))∀x (A(x) ∧ ∀y ¬R(x, y))

Which statement is equivalent to ¬∀x (A(x) → ∃y R(x, y)), with the negation moved all the way inside? Click it.

¬∀x (A(x) → ∃y R(x, y))
  1. 1∃x (A(x) → ∀y ¬R(x, y))
  2. 2∃x (A(x) ∧ ∀y ¬R(x, y))
  3. 3∃x (¬A(x) ∧ ∃y R(x, y))
  4. 4∀x (A(x) ∧ ∀y ¬R(x, y))
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(0, 11, 12, 13, 14, 15)
  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
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 11 clock edges.

Q₃Q₂Q₁Q₀ = 000011
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 51 modulo 56: the number x with 0 < x < 56 and 51·x ≡ 1 (mod 56).

a = 51n = 56
51−1 mod 56 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 24142 mod 41.

a = 24k = 142n = 41
24142 mod 41 =
9.Algorithms & complexityMedium

With coins of 1, 3 and 4 and an amount of 6, how many coins does the greedy "largest coin first" method use, compared to the optimum?

  1. AGreedy uses 2 coins; the optimum is 2
  2. BGreedy uses 3 coins; the optimum is 3
  3. CGreedy uses 3 coins; the optimum is 2
  4. DGreedy uses 4 coins; the optimum is 3
10.Algorithms & complexityHard

Which definition of the class NP is correct?

  1. AProblems that no algorithm can solve
  2. BProblems that are not polynomial
  3. CProblems that cannot be solved in polynomial time
  4. DDecision problems whose yes-answers can be verified in polynomial time given a certificate
11.Data structuresMedium

Which abstract data type does Dijkstra's algorithm use to choose the next vertex to process?

  1. AA FIFO queue
  2. BA min-priority queue
  3. CA set with no ordering
  4. DA stack
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA deque holding indices in decreasing order of value
  2. BA sorted array updated by shifting
  3. CA min-heap rebuilt for every window
  4. DA singly linked list of the window's elements
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1142101193131ABCDEFG

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

In relational algebra, the projection operator (π) corresponds to:

  1. AChoosing which columns appear in the result
  2. BChoosing which rows satisfy a given condition
  3. CSorting the result
  4. DCombining two tables on a key
16.Databases & SQLHard

With multiversion concurrency control (MVCC), how can readers avoid blocking writers?

  1. AReaders see a consistent snapshot of committed row versions while writers create new versions
  2. BReaders read uncommitted data directly
  3. CReaders always wait until all writers have committed
  4. DWriters are not allowed while any reader is active
17.Computer networksMedium

Which DNS record type maps a host name to an IPv6 address?

  1. AAAAA
  2. BMX
  3. CCNAME
  4. DA
18.Computer networksHard

A DNS cache poisoning attack aims to:

  1. ADelete DNS servers
  2. BSteal Wi-Fi passwords
  3. CMake a resolver store a false IP address for a domain
  4. DOverload a server with traffic so that it stops answering queries
19.IPv4 subnettingMedium
255.255.255.25211111111111111111111111111111100network bitshost bits

A subnet uses the mask 255.255.255.252. How many usable host addresses H does it have?

255.255.255.252
H =
20.IPv4 subnettingHard

The network 192.168.4.0/24 is split with VLSM for these LANs: A: 14 hosts, B: 16 hosts, C: 47 hosts, D: 8 hosts, E: 5 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 D?

A: 14B: 16C: 47D: 8E: 5
x =

3Solutions

1.Number systems

Answer: U = 136

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

120 = 0111 1000
inverted: 1000 0111
+1 → 1000 1000

Read it as unsigned

1000 1000₂ = 256 − 120 = 136

Answer

U = 136
2.Number systems

Answer: R = 4

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.

1010 0100
AND 0100 1110
= 0000 0100

Value of the result

0000 0100₂ = 4

Answer

R = 4
3.Propositional logic & proofs

Answer: ¬¬p → ¬(q ∨ 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).

Parts of the implication

a = q ∨ r, b = ¬p

Contrapositive

¬b → ¬a = ¬¬p → ¬(q ∨ r)

Answer

¬¬p → ¬(q ∨ r)

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: ∃x (A(x) ∧ ∀y ¬R(x, y))

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 (A(x) → ∃y R(x, y)) ≡ ∃x (A(x) ∧ ∀y ¬R(x, y))

Answer

∃x (A(x) ∧ ∀y ¬R(x, y))

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: AB = 00, 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.

m0: m1, m2, m4, m8 → 0
ABCD0001111000011110111111

The isolated 1

m0 = A′B′C′D′ → AB = 00, CD = 00

Answer

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

Answer: Q = 14

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 → 1000 → 1100 → 1110

Decimal value

1110₂ = 14

Answer

Q = 14
7.Number theory & modular arithmetic

Answer: 51−1 mod 56 = 11

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·51 (mod 56).

r = 56, t = 0
r = 51, q = 1, t = 1
r = 5, q = 10, t = −1
r = 1, q = 5, t = 11

Inverse

1 ≡ 11·51 (mod 56)
x = 11
check: 51·11 = 561 = 10·56 + 1

Answer

51−1 mod 56 = 11
8.Number theory & modular arithmetic

Answer: 24142 mod 41 = 39

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.

Binary exponent

142 = 100011102 = 128 + 8 + 4 + 2

Repeated squaring

241 ≡ 24 (mod 41)
242 ≡ 2 (mod 41)
244 ≡ 4 (mod 41)
248 ≡ 16 (mod 41)
2416 ≡ 10 (mod 41)
2432 ≡ 18 (mod 41)
2464 ≡ 37 (mod 41)
24128 ≡ 16 (mod 41)

Multiply the needed powers

24142 ≡ 2·4·16·16 ≡ 39 (mod 41)

Answer

24142 mod 41 = 39
9.Algorithms & complexity

Answer: C · Greedy uses 3 coins; the optimum is 2

Why this is the answer

Greedy takes 4 + 1 + 1 = 3 coins, but 3 + 3 uses only 2, so greedy is not optimal for this coin system; dynamic programming is.

Answer

C · Greedy uses 3 coins; the optimum is 2
10.Algorithms & complexity

Answer: D · Decision problems whose yes-answers can be verified in polynomial time given a certificate

Why this is the answer

For an NP problem, a proposed solution (certificate) can be checked quickly, for example verifying a given assignment satisfies a Boolean formula.

Answer

D · Decision problems whose yes-answers can be verified in polynomial time given a certificate
11.Data structures

Answer: B · A min-priority queue

Why this is the answer

A min-priority queue, usually a binary heap, returns the unvisited vertex with the smallest distance efficiently.

Answer

B · A min-priority queue
12.Data structures

Answer: A · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

A · A deque holding indices in decreasing order of value
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) = 3, deg(E) = 2, deg(F) = 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: d(F) = 28

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: 11
C (11): B: 15, G: 14
G (14): D: 27, E: 15
B (15): D: 17
E (15)
D (17): F: 28
F (28)

Shortest path

A → C → B → D → F
d(F) = 28

Answer

d(F) = 28
15.Databases & SQL

Answer: A · Choosing which columns appear in the result

Why this is the answer

Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).

Answer

A · Choosing which columns appear in the result
16.Databases & SQL

Answer: A · Readers see a consistent snapshot of committed row versions while writers create new versions

Why this is the answer

MVCC keeps old row versions so a reader can see the data as of its snapshot, while a writer creates a new version instead of overwriting in place.

Answer

A · Readers see a consistent snapshot of committed row versions while writers create new versions
17.Computer networks

Answer: A · AAAA

Why this is the answer

An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.

Answer

A · AAAA
18.Computer networks

Answer: C · Make a resolver store a false IP address for a domain

Why this is the answer

Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.

Answer

C · Make a resolver store a false IP address for a domain
19.IPv4 subnetting

Answer: H = 2

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).

Prefix length

255.255.255.252 = /30

Usable hosts

22 − 2 = 2

Answer

H = 2
20.IPv4 subnetting

Answer: x = 112

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

C: 47 hosts → 26 = 64 (/26)
B: 16 hosts → 25 = 32 (/27)
A: 14 hosts → 24 = 16 (/28)
D: 8 hosts → 24 = 16 (/28)
E: 5 hosts → 23 = 8 (/29)

Allocation, largest first

C: .0 – .63
B: .64 – .95
A: .96 – .111
D: .112 – .127
E: .128 – .135

Answer

x = 112

LAN D: 192.168.4.112/28.

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