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 −74 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −748 bit
U =
2.Number systemsHard
24 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ¬(r ∨ ¬r) a tautology, a contradiction or a contingency? Click your answer.

¬(r ∨ ¬r)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
abcb → (a ∨ c)ba111110101100011010001000

The argument with premises b → (a ∨ c); b and conclusion a is invalid. Click the row of the truth table that is a counterexample.

b → (a ∨ c)b∴ a
  1. 1a = 1, b = 1, c = 1
  2. 2a = 1, b = 1, c = 0
  3. 3a = 1, b = 0, c = 1
  4. 4a = 1, b = 0, c = 0
  5. 5a = 0, b = 1, c = 1
  6. 6a = 0, b = 1, c = 0
  7. 7a = 0, b = 0, c = 1
  8. 8a = 0, b = 0, c = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​0I1​C′I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = C′, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 0I₂ = C′I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKDOWN

A 3-bit binary down counter starts at 1 (decimal, Q = 001). Which count Q (in decimal) does it show after 16 clock pulses?

Q0 = 116n = 3
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 13 modulo 17: the number x with 0 < x < 17 and 13·x ≡ 1 (mod 17).

a = 13n = 17
13−1 mod 17 =
8.Number theory & modular arithmeticHard

61 is prime. Use Fermat’s little theorem to find 1850318 mod 61.

p = 61k = 50318
1850318 mod 61 =
9.Algorithms & complexityMedium

Dynamic programming is useful when a problem has:

  1. AA single loop
  2. BOnly random inputs
  3. COverlapping subproblems and optimal substructure
  4. DNo subproblems at all, so it can be solved in one step
10.Algorithms & complexityHard

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AKeys must be sorted first
  2. BArrays have O(n) access
  3. CMany keys may collide into the same bucket
  4. DHash functions are slow
11.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AHeaps never contain more than 100 elements
  2. BAll keys in a heap are equal
  3. CArrays automatically keep the keys in sorted order, so no pointers are needed
  4. DIt is a complete binary tree, so positions follow simple index formulas
12.Data structuresHard

What does a splay tree do after each access?

  1. AMoves the accessed node to a leaf
  2. BRecolours the nodes on the search path
  3. CRebuilds the whole tree in balanced form
  4. DRotates the accessed node up to 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| = 11
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1531168436ABCDEF

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. AWHERE can only compare text columns
  2. BArithmetic is not allowed in the SELECT list
  3. CAliases may not contain an underscore
  4. DWHERE is evaluated before the SELECT list, so the alias does not exist yet
16.Databases & SQLHard

What is the result of SELECT COUNT(col) FROM t if all values in col are NULL and the table has 10 rows?

  1. A10
  2. BAn error
  3. CNULL
  4. D0
17.Computer networksMedium

Which TCP flag signals that a sender has finished sending data?

  1. APSH
  2. BFIN
  3. CURG
  4. DSYN
18.Computer networksHard

Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?

  1. AOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
  2. BBy routing A's frames through C
  3. CBy making A and C use collision detection so they stop sending when frames overlap
  4. DBy increasing the TTL of the frames
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1509 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1509 hosts
p =
20.IPv4 subnettingHard
third octet 119 … 123012816413211608141211commonvaries

A router has routes to these networks: 10.187.119.0/24, 10.187.120.0/24, 10.187.121.0/24, 10.187.122.0/24, 10.187.123.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.187.119.0/24 … 10.187.123.0/24
p =

3Solutions

1.Number systems

Answer: U = 182

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

74 = 0100 1010
inverted: 1011 0101
+1 → 1011 0110

Read it as unsigned

1011 0110₂ = 256 − 74 = 182

Answer

U = 182
2.Number systems

Answer: M = 768 KiB

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.

Bits and bytes

512·512·24 = 6291456 bit = 786432 B

Convert to KiB

786432/1024 = 768 KiB

Answer

M = 768 KiB
3.Propositional logic & proofs

Answer: contradiction

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).

Truth table row by row

r = 1: ¬r = 0, r ∨ ¬r = 1, ¬(r ∨ ¬r) = 0
r = 0: ¬r = 1, r ∨ ¬r = 1, ¬(r ∨ ¬r) = 0

Classify

¬(r ∨ ¬r): 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: a = 0, b = 1, c = 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

a = 1, b = 1, c = 1: 1 1 | 1
a = 1, b = 1, c = 0: 1 1 | 1
a = 1, b = 0, c = 1: 1 0 | 1
a = 1, b = 0, c = 0: 1 0 | 1
a = 0, b = 1, c = 1: 1 1 | 0
a = 0, b = 1, c = 0: 0 1 | 0
a = 0, b = 0, c = 1: 1 0 | 0
a = 0, b = 0, c = 0: 1 0 | 0

Counterexample

a = 0, b = 1, c = 1

Answer

a = 0, b = 1, c = 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 = 3

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C′ + AB′C′ + ABC′

Rows with F = 1

ABC = 000, 100, 110
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 1

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 8 pulses bring the counter back to the same state.

16 = 2·8 + 0

Count after the remaining pulses

Q = (1 − 0) mod 8 = 1
Q = 001₂

Answer

Q = 1
7.Number theory & modular arithmetic

Answer: 13−1 mod 17 = 4

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·13 (mod 17).

r = 17, t = 0
r = 13, q = 1, t = 1
r = 4, q = 3, t = −1
r = 1, q = 4, t = 4

Inverse

1 ≡ 4·13 (mod 17)
x = 4
check: 13·4 = 52 = 3·17 + 1

Answer

13−1 mod 17 = 4
8.Number theory & modular arithmetic

Answer: 1850318 mod 61 = 36

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

50318 mod 60 = 38
1850318 ≡ 1838 (mod 61)

Small power

1838 mod 61 = 36

Answer

1850318 mod 61 = 36
9.Algorithms & complexity

Answer: C · 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

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

Answer: C · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

C · Many keys may collide into the same bucket
11.Data structures

Answer: D · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

D · It is a complete binary tree, so positions follow simple index formulas
12.Data structures

Answer: D · Rotates the accessed node up to the root

Why this is the answer

Splaying moves the accessed node to the root through rotations; a single operation can cost O(n), but the amortised cost is O(log n).

Answer

D · Rotates the accessed node up to the root
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) = 1, deg(B) = 4, deg(C) = 3, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 4

Odd vertices

A, C → 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) = 24

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): B: 15
B (15): C: 18, D: 26, E: 21
C (18)
E (21): F: 24
F (24)
D (26)

Shortest path

A → B → E → F
d(F) = 24

Answer

d(F) = 24
15.Databases & SQL

Answer: D · 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

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

Answer: D · 0

Why this is the answer

COUNT(col) counts non-NULL values, so it returns 0; COUNT(*) would return 10.

Answer

D · 0
17.Computer networks

Answer: B · FIN

Why this is the answer

Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.

Answer

B · FIN
18.Computer networks

Answer: A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait

Why this is the answer

A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.

Answer

A · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
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 < 1509
211 − 2 = 2046 ≥ 1509 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: p = 20

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

119 = 01110111
123 = 01111011
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.187.112.0/20 (it also covers 11 network(s) that are not in the list).

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