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
8-bit two's complement?−128?64?32?16?8?4?2?1

Write −26 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −268 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0100 0111 AND 0101 0110. What is the result's value R in decimal?

0100 0111AND0101 0110
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ¬((c ∧ b) → c) a tautology, a contradiction or a contingency? Click your answer.

¬((c ∧ b) → c)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
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
ABC000111100101324576

Simplify F = B′C′ + BC + AC′ + ABC to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = B′C′ + BC + AC′ + ABC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

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

Find the smallest positive integer x with x ≡ 1 (mod 4) and x ≡ 8 (mod 9).

x ≡ 1 (mod 4)x ≡ 8 (mod 9)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 9169 mod 17.

a = 9k = 169n = 17
9169 mod 17 =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Which statement about P and NP is correct?

  1. AIt has been proven that P ≠ NP
  2. BNP problems cannot be verified quickly
  3. CNP means "not polynomial"
  4. DEvery problem in P is also in NP
11.Data structuresMedium

A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?

  1. AA binary heap of edges
  2. BAdjacency list
  3. CAdjacency matrix
  4. DA single sorted array of vertices
12.Data structuresHard

A segment tree is typically used to:

  1. AAnswer range queries (such as range sums) and updates in O(log n)
  2. BSort strings alphabetically by storing each prefix in a separate tree node
  3. CStore a graph's edges
  4. DImplement a stack
13.Graph theoryMedium

A connected graph has 20 vertices and 20 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 20
k =
14.Graph theoryHard
976132121210ABCDEF

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 GRANT SELECT ON employees TO alice do?

  1. ACopies the employees table to alice
  2. BAllows the user alice to read the employees table
  3. CAllows alice to update and delete rows in the employees table
  4. DMakes alice the owner of the database
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index that covers every column of every table
  2. BAn index that is rebuilt every night
  3. CAn index that contains all columns the query needs, so the table itself need not be read
  4. DAn index on the primary key only
17.Computer networksMedium

What characterises a distributed denial-of-service (DDoS) attack?

  1. AAn attacker reads unencrypted Wi-Fi traffic from a nearby car
  2. BOne attacker guesses a user's password
  3. CA virus deletes files on a single PC
  4. DMany compromised machines flood a target with traffic
18.Computer networksHard

Which routing protocol is a link-state protocol?

  1. ARIP
  2. BBGP
  3. CARP
  4. DOSPF
19.IPv4 subnettingMedium
/20 → /2511111111111111111111111110000000network bitshost bits

The network 10.213.16.0/20 is divided into equal subnets with prefix /25. How many subnets N are there?

10.213.16.0/20/25
N =
20.IPv4 subnettingHard
third octet 174 … 179112806413201618141201commonvaries

A router has routes to these networks: 10.232.174.0/24, 10.232.175.0/24, 10.232.176.0/24, 10.232.177.0/24, 10.232.178.0/24, 10.232.179.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.232.174.0/24 … 10.232.179.0/24
p =

3Solutions

1.Number systems

Answer: U = 230

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

26 = 0001 1010
inverted: 1110 0101
+1 → 1110 0110

Read it as unsigned

1110 0110₂ = 256 − 26 = 230

Answer

U = 230
2.Number systems

Answer: R = 70

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.

0100 0111
AND 0101 0110
= 0100 0110

Value of the result

0100 0110₂ = 70

Answer

R = 70
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

b = 1, c = 1: c ∧ b = 1, (c ∧ b) → c = 1, ¬((c ∧ b) → c) = 0
b = 1, c = 0: c ∧ b = 0, (c ∧ b) → c = 1, ¬((c ∧ b) → c) = 0
b = 0, c = 1: c ∧ b = 0, (c ∧ b) → c = 1, ¬((c ∧ b) → c) = 0
b = 0, c = 0: c ∧ b = 0, (c ∧ b) → c = 1, ¬((c ∧ b) → c) = 0

Classify

¬((c ∧ b) → c): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
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: L = 6

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

B′C′ → m(0, 4)
BC → m(3, 7)
AC′ → m(4, 6)
ABC → m(7)
ABC00011110011001130214051716

Group and read off

F = B′C′ + BC + AC′

Count the literals

L = 6

Answer

L = 6

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

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

Answer: N = 6

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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 0 → Q = 0
2: J = 1, K = 0 → Q = 1
3: J = 1, K = 1 → Q = 0
4: J = 1, K = 1 → Q = 1
5: J = 1, K = 1 → Q = 0
6: J = 1, K = 0 → Q = 1
7: J = 0, K = 1 → Q = 0

Count the changes

0 → 0 → 1 → 0 → 1 → 0 → 1 → 0
N = 6

Answer

N = 6
7.Number theory & modular arithmetic

Answer: x = 17

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 + 4k

Second congruence

1 + 4k ≡ 8 (mod 9) → 4k ≡ 7 (mod 9)
k ≡ 7·7 ≡ 4 (mod 9)

Result

x = 1 + 4·4 = 17
x = 17
check: 17 mod 4 = 1, 17 mod 9 = 8

Answer

x = 17
8.Number theory & modular arithmetic

Answer: 9169 mod 17 = 9

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

169 = 101010012 = 128 + 32 + 8 + 1

Repeated squaring

91 ≡ 9 (mod 17)
92 ≡ 13 (mod 17)
94 ≡ 16 (mod 17)
98 ≡ 1 (mod 17)
916 ≡ 1 (mod 17)
932 ≡ 1 (mod 17)
964 ≡ 1 (mod 17)
9128 ≡ 1 (mod 17)

Multiply the needed powers

9169 ≡ 9·1·1·1 ≡ 9 (mod 17)

Answer

9169 mod 17 = 9
9.Algorithms & complexity

Answer: A · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

A · O(mn)
10.Algorithms & complexity

Answer: D · Every problem in P is also in NP

Why this is the answer

P ⊆ NP: if you can solve a problem quickly you can verify a solution quickly. Whether P = NP is still open.

Answer

D · Every problem in P is also in NP
11.Data structures

Answer: B · Adjacency list

Why this is the answer

A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.

Answer

B · Adjacency list
12.Data structures

Answer: A · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

A · Answer range queries (such as range sums) and updates in O(log n)
13.Graph theory

Answer: k = 1

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.

Edges of a spanning tree

20 − 1 = 19

Edges to remove

k = 20 − 19 = 1

Answer

k = 1
14.Graph theory

Answer: d(F) = 34

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: 9
C (9): B: 16, E: 22
B (16): D: 22
D (22): F: 34
E (22)
F (34)

Shortest path

A → C → B → D → F
d(F) = 34

Answer

d(F) = 34
15.Databases & SQL

Answer: B · Allows the user alice to read the employees table

Why this is the answer

GRANT gives a privilege on an object; SELECT allows reading only, while INSERT, UPDATE and DELETE are separate privileges.

Answer

B · Allows the user alice to read the employees table
16.Databases & SQL

Answer: C · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

C · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: D · Many compromised machines flood a target with traffic

Why this is the answer

In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.

Answer

D · Many compromised machines flood a target with traffic
18.Computer networks

Answer: D · OSPF

Why this is the answer

OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.

Answer

D · OSPF
19.IPv4 subnetting

Answer: N = 32

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

25 − 20 = 5

Number of subnets

N = 25 = 32

Answer

N = 32

Each /25 subnet has 27 − 2 = 126 usable hosts.

20.IPv4 subnetting

Answer: p = 19

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

174 = 10101110
179 = 10110011
common leading bits: 3

Prefix length

p = 16 + 3 = 19

Answer

p = /19

Summary route: 10.232.160.0/19 (it also covers 26 network(s) that are not in the list).

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