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 −78 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −788 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 0010 OR 0110 1000. What is the result's value R in decimal?

1101 0010OR0110 1000
R =
3.Propositional logic & proofsMedium
(p ∨ r) → q¬(p ∨ r) → ¬q¬q → ¬(p ∨ r)q ∧ ¬(p ∨ r)

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

q → (p ∨ r)
  1. 1(p ∨ r) → q
  2. 2¬(p ∨ r) → ¬q
  3. 3¬q → ¬(p ∨ r)
  4. 4q ∧ ¬(p ∨ r)
4.Propositional logic & proofsHard
∀x (¬F(x) ∧ ¬G(x))∃x (¬F(x) ∧ ¬G(x))∃x (F(x) → ¬G(x))∀x (F(x) → ¬G(x))

Which statement is equivalent to ¬∃x (F(x) ∧ G(x)), with the negation moved all the way inside? Click it.

¬∃x (F(x) ∧ G(x))
  1. 1∀x (¬F(x) ∧ ¬G(x))
  2. 2∃x (¬F(x) ∧ ¬G(x))
  3. 3∃x (F(x) → ¬G(x))
  4. 4∀x (F(x) → ¬G(x))
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, 3, 9, 10, 11, 13)
  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
ABCD0001111000011110101113X214050716X120131151140809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 3, 4, 6, 10, 14, 15) + d(2, 12), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 3, 4, 6, 10, 14, 15) + d(2, 12)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 3) and x ≡ 1 (mod 5).

x ≡ 2 (mod 3)x ≡ 1 (mod 5)
x =
8.Number theory & modular arithmeticHard
76c = me​ mod n?m = cd​ mod nm(n, e) = (235, 13)d

Encrypt the message m = 76 with the RSA public key (n, e) = (235, 13). Find the ciphertext c.

n = 235e = 13m = 76
c =
9.Algorithms & complexityMedium

What does it mean that a problem has optimal substructure?

  1. AAn optimal solution can be built from optimal solutions of its subproblems
  2. BThe problem has exactly one optimal solution, which every subproblem shares
  3. CThe problem can be solved in constant time
  4. DEvery subproblem has the same size
10.Algorithms & complexityHard

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AIt checks whether any edge has a negative weight
  2. BAfter V − 1 rounds, it checks whether any edge can still be relaxed
  3. CIt counts the cycles with depth-first search
  4. DIt runs Dijkstra's algorithm twice and compares the two distance results
11.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA sorted array
  2. BA doubly linked list
  3. CA dynamic array
  4. DA singly linked list
12.Data structuresHard

Which statement about heapsort is correct?

  1. AIt is stable and needs O(n) extra memory
  2. BIt runs in O(n) time on any input
  3. CIt needs a binary search tree as extra storage, so it uses O(n) extra memory
  4. DIt sorts in place in O(n log n) worst-case time but is not stable
13.Graph theoryMedium

A tree has 3 vertices of degree 3, 4 vertices of degree 4, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 34 × deg 42 × deg 53 × deg 2
L =
14.Graph theoryHard
3151219171052144ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 6; skipped edges do not count.

k = 6
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge DE
  6. 6edge DF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GD
  10. 10edge GE
15.Databases & SQLMedium

What does a common table expression (WITH name AS (...)) provide?

  1. AAn index on a column
  2. BA permanent table that stays stored on disk until it is dropped
  3. CA named temporary result that can be used within one statement
  4. DA new database user
16.Databases & SQLHard

What is a phantom read?

  1. AA query that returns no rows
  2. BA repeated query returns new rows inserted by another committed transaction
  3. CReading data that was never written
  4. DReading uncommitted changes
17.Computer networksMedium

What does an MX record in DNS specify?

  1. AThe IPv4 address of a web server
  2. BThe name servers for a zone
  3. CAn alias name for another host
  4. DThe mail servers that accept email for a domain
18.Computer networksHard

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt removes the need for DNS lookups
  2. BIt replaces TCP with a faster transport protocol
  3. CIt increases the bandwidth of the users' own home connections
  4. DIt serves cached content from servers close to the users
19.IPv4 subnettingMedium
/?11111111111111111111111100000000network bitshost bits

A department needs 184 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

184 hosts
p =
20.IPv4 subnettingHard
third octet = 124012816413211618140201networkhost

A host has the address 172.20.124.74/22. What is the third octet x of its network address?

172.20.124.74/22
x =

3Solutions

1.Number systems

Answer: U = 178

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

78 = 0100 1110
inverted: 1011 0001
+1 → 1011 0010

Read it as unsigned

1011 0010₂ = 256 − 78 = 178

Answer

U = 178
2.Number systems

Answer: R = 250

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

OR gives 1 where at least one bit is 1.

1101 0010
OR 0110 1000
= 1111 1010

Value of the result

1111 1010₂ = 250

Answer

R = 250
3.Propositional logic & proofs

Answer: ¬(p ∨ r) → ¬q

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, b = p ∨ r

Contrapositive

¬b → ¬a = ¬(p ∨ r) → ¬q

Answer

¬(p ∨ r) → ¬q

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

4.Propositional logic & proofs

Answer: ∀x (F(x) → ¬G(x))

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 (F(x) ∧ G(x)) ≡ ∀x (F(x) → ¬G(x))

Answer

∀x (F(x) → ¬G(x))

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: N = 4

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

CD′, BD′, A′D′, A′B′, ABC
ABCD0001111000011110101113X214050716X120131151140809011110

Minimal cover

F = CD′ + BD′ + A′B′ + ABC
N = 4

Answer

N = 4

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

7.Number theory & modular arithmetic

Answer: x = 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).

First congruence

x = 2 + 3k

Second congruence

2 + 3k ≡ 1 (mod 5) → 3k ≡ 4 (mod 5)
k ≡ 2·4 ≡ 3 (mod 5)

Result

x = 2 + 3·3 = 11
x = 11
check: 11 mod 3 = 2, 11 mod 5 = 1

Answer

x = 11
8.Number theory & modular arithmetic

Answer: c = 186

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.

Repeated squaring

13 = 8 + 4 + 1
761 ≡ 76 (mod 235)
762 ≡ 136 (mod 235)
764 ≡ 166 (mod 235)
768 ≡ 61 (mod 235)
7613 ≡ 76·166·61 ≡ 186 (mod 235)

Ciphertext

c = 186

Answer

c = 186
9.Algorithms & complexity

Answer: A · An optimal solution can be built from optimal solutions of its subproblems

Why this is the answer

Optimal substructure is required for both greedy algorithms and dynamic programming to produce correct answers.

Answer

A · An optimal solution can be built from optimal solutions of its subproblems
10.Algorithms & complexity

Answer: B · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

B · After V − 1 rounds, it checks whether any edge can still be relaxed
11.Data structures

Answer: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
12.Data structures

Answer: D · It sorts in place in O(n log n) worst-case time but is not stable

Why this is the answer

Heapsort builds a max-heap and extracts the maximum n times within the same array; swaps over long distances can reorder equal keys.

Answer

D · It sorts in place in O(n log n) worst-case time but is not stable
13.Graph theory

Answer: L = 19

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.

Handshake lemma for a tree

L + 41 = 2(L + 12) − 2
L = 19

Answer

L = 19

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: edge GD

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.

Kruskal’s algorithm

1. GC (2)
2. AB (3)
3. GE (4)
4. GB (5)
5. DF (10)
6. GD (14)

edge number 6

GD

Answer

edge GD
15.Databases & SQL

Answer: C · A named temporary result that can be used within one statement

Why this is the answer

A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.

Answer

C · A named temporary result that can be used within one statement
16.Databases & SQL

Answer: B · A repeated query returns new rows inserted by another committed transaction

Why this is the answer

Phantoms occur when another transaction inserts rows that match a range condition; serializable isolation prevents them.

Answer

B · A repeated query returns new rows inserted by another committed transaction
17.Computer networks

Answer: D · The mail servers that accept email for a domain

Why this is the answer

MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.

Answer

D · The mail servers that accept email for a domain
18.Computer networks

Answer: D · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

D · It serves cached content from servers close to the users
19.IPv4 subnetting

Answer: p = 24

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

27 − 2 = 126 < 184
28 − 2 = 254 ≥ 184 → h = 8

Prefix length

p = 32 − 8 = 24

Answer

p = /24
20.IPv4 subnetting

Answer: x = 124

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 in the third octet

22 = 4

Subnet of this host

network: 172.20.124.0
broadcast: 172.20.127.255

Answer

x = 124

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