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 complement0−12806403201608040201

Set the bits so they store −113 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1138 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 0101 OR 1001 1101. What is the result's value R in decimal?

0101 0101OR1001 1101
R =
3.Propositional logic & proofsMedium
¬c → ¬a¬c ∨ ac ∨ ¬aa → c

Which formula is logically equivalent to c → a? Click it.

c → a
  1. 1¬c → ¬a
  2. 2¬c ∨ a
  3. 3c ∨ ¬a
  4. 4a → c
4.Propositional logic & proofsHard
pqrr → (q ∨ p)rq111110101100011010001000

The argument with premises r → (q ∨ p); r and conclusion q is invalid. Click the row of the truth table that is a counterexample.

r → (q ∨ p)r∴ q
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010302140507061121131151141809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(4, 8, 12, 13, 14, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(4, 8, 12, 13, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111010110312040507X6012X131150141809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 2, 8, 10, 15) + d(6, 13), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 1, 2, 8, 10, 15) + d(6, 13)
L =
7.Number theory & modular arithmeticMedium

Find the inverse of 20 modulo 91: the number x with 0 < x < 91 and 20·x ≡ 1 (mod 91).

a = 20n = 91
20−1 mod 91 =
8.Number theory & modular arithmeticHard

How many solutions N does 12x ≡ 15 (mod 21) have in {0, 1, …, 20}?

12x ≡ 15 (mod 21)
N =
9.Algorithms & complexityMedium

Which statement about a recursive function without a reachable base case is correct?

  1. AIt runs in O(1) time
  2. BThe compiler always turns it into a loop that ends safely
  3. CIt keeps calling itself until the call stack overflows
  4. DIt returns 0 automatically
10.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is always solvable in O(V + E)
  2. BIt can be solved by Dijkstra's algorithm with negated weights
  3. CIt is undecidable
  4. DIt is NP-hard
11.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 3
  2. BIndex 0
  3. CIndex 4
  4. DIndex 6
12.Data structuresHard

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

  1. AThey store data unsorted
  2. BThey need no rebalancing
  3. CTheir high branching factor keeps the tree shallow, reducing slow disk reads
  4. DThey use less memory than any array, so the entire index always fits in main memory
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| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
2012191315134965ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 4.

start: Bk = 4
  1. 1edge AC
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge GB
  9. 9edge GC
  10. 10edge GD
  11. 11edge GE
15.Databases & SQLMedium

What does INTERSECT return when used between two SELECT queries?

  1. ARows in the first result but not the second
  2. BAll rows from both results
  3. CRows that appear in both results
  4. DThe Cartesian product of the results
16.Databases & SQLHard

What is a known drawback of decomposing tables into BCNF compared with 3NF?

  1. ABCNF tables cannot have primary keys
  2. BBCNF allows more redundancy than 3NF
  3. CSome functional dependencies may no longer be checkable within a single table
  4. DThe decomposition always loses data when joined back
17.Computer networksMedium

TLS provides which security properties for HTTPS?

  1. AAnonymity of the user
  2. BOnly faster loading
  3. CProtection against all malware and phishing on the visited site
  4. DConfidentiality, integrity and server authentication
18.Computer networksHard

A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?

  1. AA missing or wrong default gateway
  2. BThe TCP window is too small
  3. CA duplicate MAC address on the internet
  4. DA faulty DNS server setting
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.30.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 3?

192.168.30.0/24/29subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 251112816413211618041211networkhost

A host has the address 10.20.137.251/27. What is the fourth octet x of the last usable host address in its subnet?

10.20.137.251/27
x =

3Solutions

1.Number systems

Answer: U = 143

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

113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111

Read it as unsigned

1000 1111₂ = 256 − 113 = 143

Answer

U = 143
2.Number systems

Answer: R = 221

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.

0101 0101
OR 1001 1101
= 1101 1101

Value of the result

1101 1101₂ = 221

Answer

R = 221
3.Propositional logic & proofs

Answer: ¬c ∨ a

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 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
c → a ≡ ¬c ∨ a

Check with the truth table

a = 1, c = 1: 1 | 1 1 1 1
a = 1, c = 0: 1 | 0 1 0 0
a = 0, c = 1: 0 | 1 0 1 1
a = 0, c = 0: 1 | 1 1 1 1

Answer

¬c ∨ a

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

4.Propositional logic & proofs

Answer: p = 1, q = 0, r = 1

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.

Premises and conclusion row by row

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

Counterexample

p = 1, q = 0, r = 1

Answer

p = 1, q = 0, r = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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.

ABCD000111100001111000010302140507061121131151141809011010

Group the 1s

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

BC′D′
AC′D′
AB

Minimal SOP form

F = BC′D′ + AC′D′ + AB
N = 3

Answer

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

Answer: L = 8

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.

B′D′, A′CD′, ABD, A′B′C′
ABCD000111100001111010110312040507X6012X131150141809011110

Minimal cover

F = B′D′ + ABD + A′B′C′
L = 8

Answer

L = 8
7.Number theory & modular arithmetic

Answer: 20−1 mod 91 = 41

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·20 (mod 91).

r = 91, t = 0
r = 20, q = 4, t = 1
r = 11, q = 1, t = −4
r = 9, q = 1, t = 5
r = 2, q = 4, t = −9
r = 1, q = 2, t = 41

Inverse

1 ≡ 41·20 (mod 91)
x = 41
check: 20·41 = 820 = 9·91 + 1

Answer

20−1 mod 91 = 41
8.Number theory & modular arithmetic

Answer: N = 3

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).

Divide by the gcd

gcd(12, 21) = 3
4x ≡ 5 (mod 7)

Solve the reduced congruence

4−1 ≡ 2 (mod 7) → x ≡ 3 (mod 7)

Solutions modulo 21

x ∈ {3, 10, 17}
N = 3

Answer

N = 3
9.Algorithms & complexity

Answer: C · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

C · It keeps calling itself until the call stack overflows
10.Algorithms & complexity

Answer: D · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

D · It is NP-hard
11.Data structures

Answer: C · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

C · Index 4
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: Euler path, no 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) = 3, deg(D) = 2, deg(E) = 3

Odd vertices

C, E → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: edge DF

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.

Prim’s algorithm

1. GB (4)
2. GE (5)
3. EF (3)
4. DF (1)
5. GC (9)
6. AC (20)

edge number 4

DF

Answer

edge DF
15.Databases & SQL

Answer: C · Rows that appear in both results

Why this is the answer

INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.

Answer

C · Rows that appear in both results
16.Databases & SQL

Answer: C · Some functional dependencies may no longer be checkable within a single table

Why this is the answer

A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.

Answer

C · Some functional dependencies may no longer be checkable within a single table
17.Computer networks

Answer: D · Confidentiality, integrity and server authentication

Why this is the answer

TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).

Answer

D · Confidentiality, integrity and server authentication
18.Computer networks

Answer: A · A missing or wrong default gateway

Why this is the answer

Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.

Answer

A · A missing or wrong default gateway
19.IPv4 subnetting

Answer: x = 23

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

23 = 8

Subnet number 3

network: (3 − 1)·8 = 16
broadcast: 16 + 8 − 1 = 23

Answer

x = 23
20.IPv4 subnetting

Answer: x = 254

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

25 = 32

Subnet of this host

network: ⌊251/32⌋·32 = 224
broadcast: 224 + 32 − 1 = 255
usable hosts: 225 … 254

Answer

x = 254

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