Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 complement0−12816413211618040211

The 8-bit pattern 0111 1001 stores a signed integer in two's complement. What is its value N in decimal?

0111 10018 bit
N =
2.Number systemsHard
24 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px24 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
c ↔ ¬b¬c ∨ ¬bc ∧ ¬b¬c ↔ ¬b

Which formula is logically equivalent to ¬(c ↔ b)? Click it.

¬(c ↔ b)
  1. 1c ↔ ¬b
  2. 2¬c ∨ ¬b
  3. 3c ∧ ¬b
  4. 4¬c ↔ ¬b
4.Propositional logic & proofsHard

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

(r ↔ (p → q)) → (¬s ∧ (¬r → p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​CI1​1I2​CI3​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 = C, I₂ = 1, I₃ = C. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = CI₂ = 1I₃ = C
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​DQ4​Q4​DQ5​Q5​DQ6​Q6​DQ7​Q7​CLKRING

A ring counter is made of 8 D flip-flops with the last output fed back to the first input, and starts with a single 1. How many different states N does it go through?

n = 8
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 10) and x ≡ 6 (mod 13).

x ≡ 1 (mod 10)x ≡ 6 (mod 13)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (667, 23)d = ?

An RSA key uses the primes p = 23 and q = 29 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 23q = 29e = 23
d =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. AConstant
  2. BLogarithmic
  3. CLinear
  4. DExponential
10.Algorithms & complexityHard

What is the amortised cost of incrementing a binary counter, given that one increment may flip many bits?

  1. AO(n) per increment
  2. BO(n log n) per increment
  3. CO(log n) per increment
  4. DO(1) per increment
11.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i + 1) mod n
  2. B2i + 1
  3. Ci + 1
  4. D(i − 1) mod n
12.Data structuresHard

Why are B-trees used in databases and file systems?

  1. AThey store data unsorted
  2. BThey use less memory than any array, so the entire index always fits in main memory
  3. CTheir high branching factor keeps the tree shallow, reducing slow disk reads
  4. DThey need no rebalancing
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
13111421015712ABCDEF

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

What does SELECT COALESCE(phone, 'unknown') FROM customers return for a customer whose phone is NULL?

  1. AAn error
  2. BNULL
  3. CAn empty string
  4. D'unknown'
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AThe Cartesian product, pairing each row of one table with every row of the other
  2. BOnly unmatched rows
  3. CAll rows from both tables, matched where possible and NULL-filled otherwise
  4. DOnly matching rows
17.Computer networksMedium

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

  1. AThe client must log in first
  2. BThe request succeeded
  3. CThe requested resource was not found
  4. DThe server had an internal error
18.Computer networksHard

By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?

  1. A1 Mbit/s
  2. B8 Mbit/s
  3. C2 Mbit/s
  4. D255 Mbit/s
19.IPv4 subnettingMedium
/16 → /2211111111111111111111110000000000network bitshost bits

The network 172.17.0.0/16 is divided into equal subnets with prefix /22. How many subnets N are there?

172.17.0.0/16/22
N =
20.IPv4 subnettingHard
last octet = 56012806413211618040201networkhost

A host has the address 10.20.113.56/30. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.113.56/30
x =

3Solutions

1.Number systems

Answer: N = 121

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.

64 + 32 + 16 + 8 + 1 = 121

Answer

N = 121
2.Number systems

Answer: M = 900 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

640·480·24 = 7372800 bit = 921600 B

Convert to KiB

921600/1024 = 900 KiB

Answer

M = 900 KiB
3.Propositional logic & proofs

Answer: c ↔ ¬b

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) is 1 exactly when a and b differ, which is a ↔ ¬b.
¬(c ↔ b) ≡ c ↔ ¬b

Check with the truth table

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

Answer

c ↔ ¬b

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

4.Propositional logic & proofs

Answer: N = 12

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

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

Split on p

p = 1: 6
p = 0: 6
N = 6 + 6 = 12

Answer

N = 12
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

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′ + A′BC + AB′ + ABC

Rows with F = 1

ABC = 000, 011, 100, 101, 111
N = 5

Answer

N = 5
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 8

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.

Sequence for 8 flip-flops

10000000 → 01000000 → 00100000 → 00010000 → 00001000 → 00000100 → 00000010 → 00000001

Number of states

N = 8

Answer

N = 8
7.Number theory & modular arithmetic

Answer: x = 71

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 = 1 + 10k

Second congruence

1 + 10k ≡ 6 (mod 13) → 10k ≡ 5 (mod 13)
k ≡ 4·5 ≡ 7 (mod 13)

Result

x = 1 + 10·7 = 71
x = 71
check: 71 mod 10 = 1, 71 mod 13 = 6

Answer

x = 71
8.Number theory & modular arithmetic

Answer: d = 375

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 = 23·29 = 667
φ(n) = 22·28 = 616

Extended Euclidean algorithm

r = 616, t = 0
r = 23, q = 26, t = 1
r = 18, q = 1, t = −26
r = 5, q = 3, t = 27
r = 3, q = 1, t = −107
r = 2, q = 1, t = 134
r = 1, q = 2, t = −241

Private exponent

d = 375
check: 23·375 = 8625 = 14·616 + 1

Answer

d = 375
9.Algorithms & complexity

Answer: D · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

D · Exponential
10.Algorithms & complexity

Answer: D · O(1) per increment

Why this is the answer

Over n increments, bit i flips about n / 2ⁱ times, so the total is less than 2n flips, an average of O(1) each.

Answer

D · O(1) per increment
11.Data structures

Answer: A · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

A · (i + 1) mod n
12.Data structures

Answer: C · Their high branching factor keeps the tree shallow, reducing slow disk reads

Why this is the answer

B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.

Answer

C · Their high branching factor keeps the tree shallow, reducing slow disk reads
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: d(F) = 10

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: 13, C: 1
C (1): B: 12, E: 3
E (3): D: 13, F: 10
F (10)
B (12)
D (13)

Shortest path

A → C → E → F
d(F) = 10

Answer

d(F) = 10
15.Databases & SQL

Answer: D · 'unknown'

Why this is the answer

COALESCE checks its arguments from left to right and returns the first that is not NULL, here 'unknown'.

Answer

D · 'unknown'
16.Databases & SQL

Answer: C · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

C · All rows from both tables, matched where possible and NULL-filled otherwise
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: B · 8 Mbit/s

Why this is the answer

C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.

Answer

B · 8 Mbit/s
19.IPv4 subnetting

Answer: N = 64

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).

Borrowed bits

22 − 16 = 6

Number of subnets

N = 26 = 64

Answer

N = 64

Each /22 subnet has 210 − 2 = 1022 usable hosts.

20.IPv4 subnetting

Answer: x = 59

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

22 = 4

Subnet of this host

network: ⌊56/4⌋·4 = 56
broadcast: 56 + 4 − 1 = 59
usable hosts: 57 … 58

Answer

x = 59

ExamProblems · examproblems.com/courses/computer-science-fundamentals