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 complement0−12816403201618040211

The 8-bit pattern 0100 1001 stores a signed integer in two's complement. What is its value N in decimal?

0100 10018 bit
N =
2.Number systemsHard
16 bit1 px1920 px1080 pxM = ? KiB

An uncompressed image is 1920 × 1080 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1920 × 1080 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
r ∧ ¬p¬p → ¬r¬r → ¬pp → r

You want to prove r → p by contrapositive. Which implication do you prove instead? Click it.

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

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

¬∃x ∀y R(x, y)
  1. 1∀x ∀y ¬R(x, y)
  2. 2∃x ∃y ¬R(x, y)
  3. 3∀x ∃y ¬R(x, y)
  4. 4∃x ∀y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​1I1​1I2​1I3​ABS1​S0​F

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

I₀ = 0I₁ = 1I₂ = 1I₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​FFQ4​Q4​FFQ5​Q5​CLKMOD-50

A 5 kHz clock drives a mod-50 counter. Find the frequency fout of its most significant output, which completes one cycle per counting cycle.

fclk = 5 kHzN = 50
fout = kHz
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(147), the number of integers from 1 to 147 that are coprime to 147.

n = 147 = 3·72
φ(147) =
8.Number theory & modular arithmeticHard
mc = me​ mod n831m = cd​ mod n?(n, e) = (1739, 19)d

An RSA key uses p = 37, q = 47 and e = 19. Decrypt the ciphertext c = 831: find the message m.

p = 37q = 47e = 19c = 831
m =
9.Algorithms & complexityMedium

In a sorted array, you look for two elements that sum to a target by moving one pointer from each end inwards. What is the time complexity?

  1. AO(n log n)
  2. BO(n²)
  3. CO(n)
  4. DO(log n)
10.Algorithms & complexityHard

What is the time complexity of building a binary heap from n unsorted elements with the bottom-up method?

  1. AO(log n)
  2. BO(n)
  3. CO(n²)
  4. DO(n log n)
11.Data structuresMedium

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(V²)
  2. BO(V × E)
  3. CO(E²)
  4. DO(V + E)
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA binary search tree only
  2. BA stack only
  3. CA single sorted array
  4. DA hash map combined with a doubly linked list
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
ABCDE

How many walks of length 2 go from C to D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

A table has rows where col is NULL. Which rows does WHERE col = NULL return?

  1. ANo rows, because the comparison is unknown
  2. BOnly the rows where col is NULL, as with IS NULL
  3. CAll rows of the table
  4. DAn error in every database
16.Databases & SQLHard

For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?

  1. ACross join
  2. BNested loop join with index lookups
  3. CBitmap join on the primary key
  4. DHash join
17.Computer networksMedium

What is a cyclic redundancy check (CRC) used for in Ethernet frames?

  1. AChoosing the best route
  2. BEncrypting the frame
  3. CDetecting transmission errors
  4. DCorrecting every bit error automatically
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. A255 Mbit/s
  2. B8 Mbit/s
  3. C1 Mbit/s
  4. D2 Mbit/s
19.IPv4 subnettingMedium
/16 → /2011111111111111111111000000000000network bitshost bits

The network 172.22.0.0/16 is divided into equal subnets with prefix /20. How many subnets N are there?

172.22.0.0/16/20
N =
20.IPv4 subnettingHard
last octet = 184112806413211618040201networkhost

A host has the address 10.20.202.184/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.202.184/25
x =

3Solutions

1.Number systems

Answer: N = 73

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 + 8 + 1 = 73

Answer

N = 73
2.Number systems

Answer: M = 4050 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

1920·1080·16 = 33177600 bit = 4147200 B

Convert to KiB

4147200/1024 = 4050 KiB

Answer

M = 4050 KiB
3.Propositional logic & proofs

Answer: ¬p → ¬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 = r, b = p

Contrapositive

¬b → ¬a = ¬p → ¬r

Answer

¬p → ¬r

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

4.Propositional logic & proofs

Answer: ∀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 ∀y R(x, y) ≡ ∀x ∃y ¬R(x, y)

Answer

∀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: N = 6

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 + AB′ + AB

Rows with F = 1

ABC = 010, 011, 100, 101, 110, 111
N = 6

Answer

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

Answer: fout = 0.1 kHz

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.

Frequency division

fout = 5 kHz / 50 = 0.1 kHz

Answer

fout = 0.1 kHz
7.Number theory & modular arithmetic

Answer: φ(147) = 84

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).

Prime factors

147 = 3·72

Totient

φ(147) = φ(49)·φ(3) = 42·2 = 84

Answer

φ(147) = 84
8.Number theory & modular arithmetic

Answer: m = 723

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.

Private exponent

n = 1739, φ(n) = 1656
d = 19−1 mod 1656 = 523

Decrypt by repeated squaring

523 = 512 + 8 + 2 + 1
8311 ≡ 831 (mod 1739)
8312 ≡ 178 (mod 1739)
8314 ≡ 382 (mod 1739)
8318 ≡ 1587 (mod 1739)
83116 ≡ 497 (mod 1739)
83132 ≡ 71 (mod 1739)
83164 ≡ 1563 (mod 1739)
831128 ≡ 1413 (mod 1739)
831256 ≡ 197 (mod 1739)
831512 ≡ 551 (mod 1739)
831523 ≡ 831·178·1587·551 ≡ 723 (mod 1739)

Message

m = 723
check: 72319 mod 1739 = 831

Answer

m = 723
9.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

If the sum is too small the left pointer moves right, if too large the right pointer moves left, so there are at most n steps.

Answer

C · O(n)
10.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

Summing sift-down costs over the levels gives a convergent series, so heapify runs in linear time.

Answer

B · O(n)
11.Data structures

Answer: D · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

D · O(V + E)
12.Data structures

Answer: D · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

D · A hash map combined with a doubly linked list
13.Graph theory

Answer: H = 2520

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 = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 0
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 0 1 1 0

Common neighbours

(M²)CD = 1·0 + 1·1 + 0·0 + 0·0 + 1·1 = 2

Answer

N = 2
15.Databases & SQL

Answer: A · No rows, because the comparison is unknown

Why this is the answer

Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.

Answer

A · No rows, because the comparison is unknown
16.Databases & SQL

Answer: D · Hash join

Why this is the answer

A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.

Answer

D · Hash join
17.Computer networks

Answer: C · Detecting transmission errors

Why this is the answer

The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.

Answer

C · Detecting transmission errors
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 = 16

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

20 − 16 = 4

Number of subnets

N = 24 = 16

Answer

N = 16

Each /20 subnet has 212 − 2 = 4094 usable hosts.

20.IPv4 subnetting

Answer: x = 255

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

27 = 128

Subnet of this host

network: ⌊184/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 255

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