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 −91 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −918 bit
U =
2.Number systemsHard
3324 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3324 items its own binary code of the same length. What is the smallest number of bits b needed?

3324
b = bit
3.Propositional logic & proofsMedium
(q ∨ p) ∧ (q ∨ r)(q ∧ p) ∨ rq ∨ (p ∧ r)(q ∧ p) ∨ (q ∧ r)

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

q ∧ (p ∨ r)
  1. 1(q ∨ p) ∧ (q ∨ r)
  2. 2(q ∧ p) ∨ r
  3. 3q ∨ (p ∧ r)
  4. 4(q ∧ p) ∨ (q ∧ r)
4.Propositional logic & proofsHard
pqrr → qp → qr → p111110101100011010001000

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

r → qp → q∴ r → p
  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
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(3, 4, 5, 8, 12, 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
ABCD000111100001111010110312X41507160120131150140809111X10

For F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 11, 15) + d(4, 10), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 11, 15) + d(4, 10)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 43 modulo 99: the number x with 0 < x < 99 and 43·x ≡ 1 (mod 99).

a = 43n = 99
43−1 mod 99 =
8.Number theory & modular arithmeticHard

Solve 9x ≡ 3 (mod 12). Give the smallest positive solution x.

9x ≡ 3 (mod 12)
x =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n/2) + O(1) solve to?

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

The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?

  1. ABecause n is always small
  2. BBecause it only works for W = 0
  3. CBecause the algorithm is randomised and its running time varies with each input
  4. DW is a number, so the time is exponential in the number of bits used to write W
11.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AThat keys are stored in insertion order
  2. BConstant-time search in every case
  3. CO(log n) height through rebalancing
  4. DThat no node has two children
12.Data structuresHard

In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?

  1. AStoring each vertex's current distance from the source during the search
  2. BChecking whether an edge's two endpoints are already connected
  3. CSorting the edges by weight
  4. DChoosing the starting vertex
13.Graph theoryMedium

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

3 × deg 44 × deg 2
L =
14.Graph theoryHard
619104151175ABCDEF

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 AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge EF
  8. 8edge BE
15.Databases & SQLMedium

A table stores the customer's address on every order row. Changing the address in one row but not in the others is an example of:

  1. AA deletion anomaly
  2. BA phantom read
  3. CAn update anomaly
  4. DAn insertion anomaly
16.Databases & SQLHard

Two transactions both read stock = 10, each subtracts 1 and writes the result, giving 9 instead of 8. Which anomaly is this?

  1. ADeletion anomaly
  2. BPhantom read
  3. CDirty read
  4. DLost update
17.Computer networksMedium

Which IPv4 header field stops packets from circulating forever in a routing loop?

  1. AType of service
  2. BIdentification
  3. CHeader checksum
  4. DTime to live (TTL)
18.Computer networksHard

You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?

  1. A/28
  2. B/25
  3. C/27
  4. D/26
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.63.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 3?

192.168.63.0/24/26subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 240112816413211608040201networkhost

A host has the address 10.20.62.240/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.62.240/29
x =

3Solutions

1.Number systems

Answer: U = 165

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

91 = 0101 1011
inverted: 1010 0100
+1 → 1010 0101

Read it as unsigned

1010 0101₂ = 256 − 91 = 165

Answer

U = 165
2.Number systems

Answer: b = 12 bit

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.

Powers of two

211 = 2048 < 3324
212 = 4096 ≥ 3324

Answer

b = 12
3.Propositional logic & proofs

Answer: (q ∧ p) ∨ (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).

The law

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
q ∧ (p ∨ r) ≡ (q ∧ p) ∨ (q ∧ r)

Check with the truth table

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

Answer

(q ∧ p) ∨ (q ∧ r)

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

4.Propositional logic & proofs

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

Counterexample

p = 0, q = 1, r = 1

Answer

p = 0, q = 1, 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: AB = 00, CD = 11

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.

m3: m2, m1, m7, m11 → 0
ABCD0001111000011110111111

The isolated 1

m3 = A′B′CD → AB = 00, CD = 11

Answer

AB = 00, CD = 11
6.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.

Prime implicants

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

B′CD′, A′D′, A′C′, ACD, AB′C
ABCD000111100001111010110312X41507160120131150140809111X10

Essential prime implicants

A′D′, A′C′, ACD
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 43−1 mod 99 = 76

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·43 (mod 99).

r = 99, t = 0
r = 43, q = 2, t = 1
r = 13, q = 3, t = −2
r = 4, q = 3, t = 7
r = 1, q = 4, t = −23

Inverse

1 ≡ −23·43 (mod 99)
x = 76
check: 43·76 = 3268 = 33·99 + 1

Answer

43−1 mod 99 = 76
8.Number theory & modular arithmetic

Answer: x = 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(9, 12) = 3
3x ≡ 1 (mod 4)

Solve the reduced congruence

3−1 ≡ 3 (mod 4) → x ≡ 3 (mod 4)

Solutions modulo 12

x ∈ {3, 7, 11}
x = 3

Answer

x = 3
9.Algorithms & complexity

Answer: D · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

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

Answer: D · W is a number, so the time is exponential in the number of bits used to write W

Why this is the answer

The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.

Answer

D · W is a number, so the time is exponential in the number of bits used to write W
11.Data structures

Answer: C · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

C · O(log n) height through rebalancing
12.Data structures

Answer: B · Checking whether an edge's two endpoints are already connected

Why this is the answer

Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.

Answer

B · Checking whether an edge's two endpoints are already connected
13.Graph theory

Answer: L = 8

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 + 20 = 2(L + 7) − 2
L = 8

Answer

L = 8

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

14.Graph theory

Answer: edge EF

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. BD (4)
2. BE (5)
3. AB (6)
4. EF (7)
5. BC (10)

edge number 4

EF

Answer

edge EF
15.Databases & SQL

Answer: C · An update anomaly

Why this is the answer

Redundant storage means one change must be made in many rows; missing some leaves the data inconsistent, which normalisation prevents.

Answer

C · An update anomaly
16.Databases & SQL

Answer: D · Lost update

Why this is the answer

In a lost update, both transactions work from the same old value and the second write erases the first; locking (SELECT ... FOR UPDATE) or atomic updates prevent it.

Answer

D · Lost update
17.Computer networks

Answer: D · Time to live (TTL)

Why this is the answer

Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.

Answer

D · Time to live (TTL)
18.Computer networks

Answer: C · /27

Why this is the answer

/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.

Answer

C · /27
19.IPv4 subnetting

Answer: x = 128

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 number 3

network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191

Answer

x = 128
20.IPv4 subnetting

Answer: x = 246

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 of this host

network: ⌊240/8⌋·8 = 240
broadcast: 240 + 8 − 1 = 247
usable hosts: 241 … 246

Answer

x = 246

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