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
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 1011 OR 1001 0011. What is the result's value R in decimal?

0111 1011OR1001 0011
R =
3.Propositional logic & proofsMedium
¬r ∧ ¬qr ∨ ¬q¬r ∧ q¬r ∨ ¬q

Which formula is logically equivalent to ¬(r ∨ q)? Click it.

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

Which statement is equivalent to ¬∀x (A(x) → B(x)), with the negation moved all the way inside? Click it.

¬∀x (A(x) → B(x))
  1. 1∃x (A(x) ∧ ¬B(x))
  2. 2∀x (A(x) ∧ ¬B(x))
  3. 3∀x (A(x) → ¬B(x))
  4. 4∃x (A(x) → ¬B(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010302040507161120131151141809111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(6, 8, 10, 11, 12, 14, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(6, 8, 10, 11, 12, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​FFQ4​Q4​CLKMOD-25

A mod-25 counter is built from the minimum number of flip-flops. How many of the possible states U are never used?

N = 25
U =
7.Number theory & modular arithmeticMedium

Find the inverse of 48 modulo 53: the number x with 0 < x < 53 and 48·x ≡ 1 (mod 53).

a = 48n = 53
48−1 mod 53 =
8.Number theory & modular arithmeticHard
100c = me​ mod n?m = cd​ mod nm(n, e) = (1517, 17)d

Encrypt the message m = 100 with the RSA public key (n, e) = (1517, 17). Find the ciphertext c.

n = 1517e = 17m = 100
c =
9.Algorithms & complexityMedium

Which algorithm finds the shortest paths from one source in a graph with non-negative edge weights?

  1. ADepth-first search
  2. BDijkstra's algorithm
  3. CKruskal's algorithm
  4. DBellman–Ford is required in all cases
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. BIt runs Dijkstra's algorithm twice and compares the two distance results
  3. CIt counts the cycles with depth-first search
  4. DAfter V − 1 rounds, it checks whether any edge can still be relaxed
11.Data structuresMedium

The keys 50, 30, 70, 20, 40 are inserted in this order into an empty binary search tree. What is the left child of 30?

  1. A70
  2. B50
  3. C40
  4. D20
12.Data structuresHard

Why do databases often use B+ trees rather than plain B-trees for indexes?

  1. AAll records are in linked leaves, which makes range scans efficient
  2. BB+ trees store data only in the root
  3. CB+ trees use binary internal nodes, which reduces the number of disk reads
  4. DB+ trees do not need to be balanced
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
121437914997ABCDEFG

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

In relational algebra, the projection operator (π) corresponds to:

  1. AChoosing which columns appear in the result
  2. BChoosing which rows satisfy a given condition
  3. CCombining two tables on a key
  4. DSorting the result
16.Databases & SQLHard

Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?

  1. AA UNIQUE constraint on manager_id
  2. BA GROUP BY on manager_id
  3. CA recursive common table expression (WITH RECURSIVE)
  4. DA single self-join of the employees table on manager_id
17.Computer networksMedium

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

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

A DNS cache poisoning attack aims to:

  1. ADelete DNS servers
  2. BOverload a server with traffic so that it stops answering queries
  3. CMake a resolver store a false IP address for a domain
  4. DSteal Wi-Fi passwords
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1510 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1510 hosts
p =
20.IPv4 subnettingHard
last octet = 201112816403201618040211networkhost

A host has the address 10.20.231.201/30. What is the fourth octet x of its network address?

10.20.231.201/30
x =

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: R = 251

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.

0111 1011
OR 1001 0011
= 1111 1011

Value of the result

1111 1011₂ = 251

Answer

R = 251
3.Propositional logic & proofs

Answer: ¬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).

The law

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(r ∨ q) ≡ ¬r ∧ ¬q

Check with the truth table

q = 1, r = 1: 0 | 0 1 0 0
q = 1, r = 0: 0 | 0 0 1 1
q = 0, r = 1: 0 | 0 1 0 1
q = 0, r = 0: 1 | 1 1 0 1

Answer

¬r ∧ ¬q

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

4.Propositional logic & proofs

Answer: ∃x (A(x) ∧ ¬B(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 (A(x) → B(x)) ≡ ∃x (A(x) ∧ ¬B(x))

Answer

∃x (A(x) ∧ ¬B(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: N = 3

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.

ABCD000111100001111000010302040507161120131151141809111110

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

BCD′
AD′
AC

Minimal SOP form

F = BCD′ + AD′ + AC
N = 3

Answer

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

Answer: U = 7

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.

Number of flip-flops

2ⁿ ≥ 25 → n = 5 (2⁵ = 32)

Unused states

U = 32 − 25 = 7

Answer

U = 7
7.Number theory & modular arithmetic

Answer: 48−1 mod 53 = 21

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·48 (mod 53).

r = 53, t = 0
r = 48, q = 1, t = 1
r = 5, q = 9, t = −1
r = 3, q = 1, t = 10
r = 2, q = 1, t = −11
r = 1, q = 2, t = 21

Inverse

1 ≡ 21·48 (mod 53)
x = 21
check: 48·21 = 1008 = 19·53 + 1

Answer

48−1 mod 53 = 21
8.Number theory & modular arithmetic

Answer: c = 898

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

17 = 16 + 1
1001 ≡ 100 (mod 1517)
1002 ≡ 898 (mod 1517)
1004 ≡ 877 (mod 1517)
1008 ≡ 10 (mod 1517)
10016 ≡ 100 (mod 1517)
10017 ≡ 100·100 ≡ 898 (mod 1517)

Ciphertext

c = 898

Answer

c = 898
9.Algorithms & complexity

Answer: B · Dijkstra's algorithm

Why this is the answer

Dijkstra's algorithm works with non-negative weights; Bellman–Ford handles negative weights; Kruskal builds minimum spanning trees.

Answer

B · Dijkstra's algorithm
10.Algorithms & complexity

Answer: D · 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

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

Answer: D · 20

Why this is the answer

20 is smaller than 50 and smaller than 30, so it becomes the left child of 30; 40 becomes its right child.

Answer

D · 20
12.Data structures

Answer: A · All records are in linked leaves, which makes range scans efficient

Why this is the answer

In a B+ tree internal nodes only guide the search and the leaves are linked in key order, so a range query finds the first key and then walks along the leaves.

Answer

A · All records are in linked leaves, which makes range scans efficient
13.Graph theory

Answer: Euler 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) = 2, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(F) = 40

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: 12
B (12): C: 26, D: 15, G: 26
D (15): G: 24
G (24): E: 31
C (26)
E (31): F: 40
F (40)

Shortest path

A → B → D → G → E → F
d(F) = 40

Answer

d(F) = 40
15.Databases & SQL

Answer: A · Choosing which columns appear in the result

Why this is the answer

Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).

Answer

A · Choosing which columns appear in the result
16.Databases & SQL

Answer: C · A recursive common table expression (WITH RECURSIVE)

Why this is the answer

A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.

Answer

C · A recursive common table expression (WITH RECURSIVE)
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: C · Make a resolver store a false IP address for a domain

Why this is the answer

Poisoning inserts forged records into a resolver's cache. DNSSEC adds signatures so resolvers can verify records.

Answer

C · Make a resolver store a false IP address for a domain
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 < 1510
211 − 2 = 2046 ≥ 1510 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 200

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: ⌊201/4⌋·4 = 200
broadcast: 200 + 4 − 1 = 203
usable hosts: 201 … 202

Answer

x = 200

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