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 complement1−12806403211608141201

The 8-bit pattern 1001 0110 stores a signed integer in two's complement. What is its value N in decimal?

1001 01108 bit
N =
2.Number systemsHard
8 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬¬r → ¬¬q¬¬q → ¬¬r¬q → ¬r¬r ∧ ¬¬q

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

¬r → ¬q
  1. 1¬¬r → ¬¬q
  2. 2¬¬q → ¬¬r
  3. 3¬q → ¬r
  4. 4¬r ∧ ¬¬q
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
ABCD000111100001111000110312040517161120130150140819011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 2, 6, 7, 9, 12). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(1, 2, 6, 7, 9, 12)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X001X3X2X41507061121130150140819X11010

For F(A, B, C, D) = Σm(5, 9, 12, 13) + d(0, 2, 3, 4, 11), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(5, 9, 12, 13) + d(0, 2, 3, 4, 11)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 14 modulo 23: the number x with 0 < x < 23 and 14·x ≡ 1 (mod 23).

a = 14n = 23
14−1 mod 23 =
8.Number theory & modular arithmeticHard

37 is prime. Use Fermat’s little theorem to find 1856515 mod 37.

p = 37k = 56515
1856515 mod 37 =
9.Algorithms & complexityMedium

A recurrence T(n) = 2T(n/2) + O(n) solves to:

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

What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?

  1. AO(log n) time and O(n) space
  2. BO(n) time and O(1) space
  3. CO(n) time and O(n) space
  4. DO(2ⁿ) time and O(1) space
11.Data structuresMedium

What balance condition does an AVL tree maintain at every node?

  1. AAll leaves are at exactly the same depth
  2. BBoth subtrees of every node contain exactly the same number of nodes
  3. CThe heights of the left and right subtrees differ by at most 1
  4. DEvery node is coloured red or black
12.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. APrimary clustering, where long runs of occupied slots form
  2. BThe need for a hash function
  3. CThe O(n) cost of reading a key's value
  4. DThe need to store keys in the table
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
18120713412ABCDE

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

Why can a table have only one clustered index?

  1. AClustered indexes can only be on the primary key
  2. BThe SQL standard forbids more than one index on a table
  3. CThe rows can be stored physically in only one order
  4. DClustered indexes use too much memory
16.Databases & SQLHard

What is a covering index for a query?

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

A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?

  1. ABoth routes, splitting the traffic equally
  2. B10.0.0.0/8, because it was added first
  3. C10.0.0.0/8, because it covers more addresses than the /16
  4. D10.1.0.0/16, because it is the longest matching prefix
18.Computer networksHard

What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?

  1. AResponses can arrive in any order on one stream, so slow replies do not block others
  2. BThe server can push updates over UDP
  3. CThe connection stays encrypted without TLS
  4. DSeveral requests reuse one TCP connection, avoiding repeated handshakes
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1282 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1282 hosts
p =
20.IPv4 subnettingHard

The network 192.168.184.0/24 is split with VLSM for these LANs: A: 23 hosts, B: 27 hosts, C: 5 hosts, D: 126 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN B?

A: 23B: 27C: 5D: 126
x =

3Solutions

1.Number systems

Answer: N = −106

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.

−128 + 16 + 4 + 2 = −106

Answer

N = −106
2.Number systems

Answer: M = 256 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

512·512·8 = 2097152 bit = 262144 B

Convert to KiB

262144/1024 = 256 KiB

Answer

M = 256 KiB
3.Propositional logic & proofs

Answer: ¬¬q → ¬¬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 = ¬q

Contrapositive

¬b → ¬a = ¬¬q → ¬¬r

Answer

¬¬q → ¬¬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: L = 13

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.

ABCD000111100001111000110312040517161120130150140819011010

Group the 1s

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

B′C′D
A′CD′
A′BC
ABC′D′

Minimal SOP form

F = B′C′D + A′CD′ + A′BC + ABC′D′
L = 13

Answer

L = 13
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 1

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.

BC′, AC′D, AB′D
ABCD0001111000011110X001X3X2X41507061121130150140819X11010

Essential prime implicants

BC′
N = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: 14−1 mod 23 = 5

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·14 (mod 23).

r = 23, t = 0
r = 14, q = 1, t = 1
r = 9, q = 1, t = −1
r = 5, q = 1, t = 2
r = 4, q = 1, t = −3
r = 1, q = 4, t = 5

Inverse

1 ≡ 5·14 (mod 23)
x = 5
check: 14·5 = 70 = 3·23 + 1

Answer

14−1 mod 23 = 5
8.Number theory & modular arithmetic

Answer: 1856515 mod 37 = 5

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.

Reduce the exponent

56515 mod 36 = 31
1856515 ≡ 1831 (mod 37)

Small power

1831 mod 37 = 5

Answer

1856515 mod 37 = 5
9.Algorithms & complexity

Answer: D · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

D · O(n log n)
10.Algorithms & complexity

Answer: B · O(n) time and O(1) space

Why this is the answer

Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.

Answer

B · O(n) time and O(1) space
11.Data structures

Answer: C · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

C · The heights of the left and right subtrees differ by at most 1
12.Data structures

Answer: A · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

A · Primary clustering, where long runs of occupied slots form
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) = 4, deg(F) = 1

Odd vertices

C, F → 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: W = 25

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

AC (1): take
DE (4): take
BD (7): take
BE (12): skip, closes a cycle
CE (13): take

Total weight

W = 1 + 4 + 7 + 13 = 25

Answer

W = 25
15.Databases & SQL

Answer: C · The rows can be stored physically in only one order

Why this is the answer

A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.

Answer

C · The rows can be stored physically in only one order
16.Databases & SQL

Answer: D · 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

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

Answer: D · 10.1.0.0/16, because it is the longest matching prefix

Why this is the answer

IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.

Answer

D · 10.1.0.0/16, because it is the longest matching prefix
18.Computer networks

Answer: D · Several requests reuse one TCP connection, avoiding repeated handshakes

Why this is the answer

Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.

Answer

D · Several requests reuse one TCP connection, avoiding repeated handshakes
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 < 1282
211 − 2 = 2046 ≥ 1282 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 128

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

D: 126 hosts → 27 = 128 (/25)
B: 27 hosts → 25 = 32 (/27)
A: 23 hosts → 25 = 32 (/27)
C: 5 hosts → 23 = 8 (/29)

Allocation, largest first

D: .0 – .127
B: .128 – .159
A: .160 – .191
C: .192 – .199

Answer

x = 128

LAN B: 192.168.184.128/27.

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