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 −2 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −28 bit
U =
2.Number systemsHard
12-bit two's complement, N = ??sign???????????

What is the most negative integer N that can be stored in 12-bit two's complement?

12 bit
N =
3.Propositional logic & proofsMedium
pqr¬r ↔ (q ∨ ¬p)¬p → r(¬r ↔ (q ∨ ¬p)) → (¬p → r)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (¬r ↔ (q ∨ ¬p)) → (¬p → r) true? Call this number N.

(¬r ↔ (q ∨ ¬p)) → (¬p → r)
N =
4.Propositional logic & proofsHard
qq ∧ ¬p¬qp

The premises (r ∧ q) → p; r; ¬p are given. Which conclusion follows from them? Click it.

(r ∧ q) → pr¬p
  1. 1q
  2. 2q ∧ ¬p
  3. 3¬q
  4. 4p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′BC + BC + A′B + ABC′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111000111312041517161120130150140819X11X10

Find the minimal SOP form of F(A, B, C, D) = Σm(1, 2, 3, 5, 6, 7, 9, 12) + d(10, 11), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(1, 2, 3, 5, 6, 7, 9, 12) + d(10, 11)
L =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 7) and x ≡ 2 (mod 10).

x ≡ 1 (mod 7)x ≡ 2 (mod 10)
x =
8.Number theory & modular arithmeticHard

59 is prime. Use Fermat’s little theorem to find 152504 mod 59.

p = 59k = 2504
152504 mod 59 =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALogarithmic
  2. BExponential
  3. CLinear
  4. DConstant
10.Algorithms & complexityHard

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

  1. AIt counts the cycles with depth-first search
  2. BIt checks whether any edge has a negative weight
  3. CAfter V − 1 rounds, it checks whether any edge can still be relaxed
  4. DIt runs Dijkstra's algorithm twice and compares the two distance results
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. A50
  2. B40
  3. C70
  4. D20
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  2. BEvery path from a node to its descendant leaves has the same number of black nodes
  3. CThe root must be red
  4. DEvery node must be red
13.Graph theoryMedium

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

2 × deg 32 × deg 2
L =
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

What does INTERSECT return when used between two SELECT queries?

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

Boyce–Codd normal form (BCNF) requires that for every non-trivial functional dependency X → Y:

  1. AY is atomic
  2. BX is a superkey
  3. CY is a primary key
  4. DX contains NULL
17.Computer networksMedium

What does an MX record in DNS specify?

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

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

  1. ASeveral requests reuse one TCP connection, avoiding repeated handshakes
  2. BThe connection stays encrypted without TLS
  3. CResponses can arrive in any order on one stream, so slow replies do not block others
  4. DThe server can push updates over UDP
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 941 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

941 hosts
p =
20.IPv4 subnettingHard
last octet = 2012806403201608041201networkhost

A host has the address 10.20.52.2/26. What is the fourth octet x of the last usable host address in its subnet?

10.20.52.2/26
x =

3Solutions

1.Number systems

Answer: U = 254

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

2 = 0000 0010
inverted: 1111 1101
+1 → 1111 1110

Read it as unsigned

1111 1110₂ = 256 − 2 = 254

Answer

U = 254
2.Number systems

Answer: N = −2048

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.

Most negative value

−211 = −2048

Answer

N = −2048
3.Propositional logic & proofs

Answer: N = 6

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: ¬r = 0, ¬p = 0, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 0, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 1, q = 1, r = 0: ¬r = 1, ¬p = 0, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 1, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 1, q = 0, r = 1: ¬r = 0, ¬p = 0, q ∨ ¬p = 0, ¬r ↔ (q ∨ ¬p) = 1, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 1, q = 0, r = 0: ¬r = 1, ¬p = 0, q ∨ ¬p = 0, ¬r ↔ (q ∨ ¬p) = 0, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 0, q = 1, r = 1: ¬r = 0, ¬p = 1, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 0, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 0, q = 1, r = 0: ¬r = 1, ¬p = 1, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 1, ¬p → r = 0, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 0
p = 0, q = 0, r = 1: ¬r = 0, ¬p = 1, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 0, ¬p → r = 1, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 1
p = 0, q = 0, r = 0: ¬r = 1, ¬p = 1, q ∨ ¬p = 1, ¬r ↔ (q ∨ ¬p) = 1, ¬p → r = 0, (¬r ↔ (q ∨ ¬p)) → (¬p → r) = 0

Count the rows with value 1

(¬r ↔ (q ∨ ¬p)) → (¬p → r): 1, 1, 1, 1, 1, 0, 1, 0
N = 6

Answer

N = 6
4.Propositional logic & proofs

Answer: ¬q

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.

Rows with all premises true

p = 0, q = 0, r = 1: q = 0, q ∧ ¬p = 0, ¬q = 1, p = 0

Conclusion

¬q (modus tollens with a conjunction)

Answer

¬q
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 1

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

A′BC → m(3)
BC → m(3, 7)
A′B → m(2, 3)
ABC′ → m(6)
ABC00011110010001131204051716

Group and read off

F = B

Count the literals

L = 1

Answer

L = 1
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 10

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, B′C, A′D, A′C, ABC′D′
ABCD000111100001111000111312041517161120130150140819X11X10

Minimal cover

F = B′D + A′D + A′C + ABC′D′
L = 10

Answer

L = 10
7.Number theory & modular arithmetic

Answer: x = 22

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 + 7k

Second congruence

1 + 7k ≡ 2 (mod 10) → 7k ≡ 1 (mod 10)
k ≡ 3·1 ≡ 3 (mod 10)

Result

x = 1 + 7·3 = 22
x = 22
check: 22 mod 7 = 1, 22 mod 10 = 2

Answer

x = 22
8.Number theory & modular arithmetic

Answer: 152504 mod 59 = 19

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

2504 mod 58 = 10
152504 ≡ 1510 (mod 59)

Small power

1510 mod 59 = 19

Answer

152504 mod 59 = 19
9.Algorithms & complexity

Answer: B · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

B · Exponential
10.Algorithms & complexity

Answer: C · 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

C · 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: B · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

B · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: L = 4

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 + 10 = 2(L + 4) − 2
L = 4

Answer

L = 4

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

14.Graph theory

Answer: N = 4

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)CC = 1·1 + 2·1 + 3·0 + 2·0 + 1·1 = 4

Answer

N = 4
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: B · X is a superkey

Why this is the answer

BCNF demands that every determinant is a candidate key (superkey); 3NF allows an exception when Y is part of a candidate key.

Answer

B · X is a superkey
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: A · 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

A · Several requests reuse one TCP connection, avoiding repeated handshakes
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 941
210 − 2 = 1022 ≥ 941 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 62

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

26 = 64

Subnet of this host

network: ⌊2/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62

Answer

x = 62

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