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 complement?−128?64?32?16?8?4?2?1

Write −113 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1138 bit
U =
2.Number systemsHard
340 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 340 items its own binary code of the same length. What is the smallest number of bits b needed?

340
b = bit
3.Propositional logic & proofsMedium
(p → q) ∧ (r → q)p ∨ (r → q)(p ∧ r) → q(p → q) ∨ (r → q)

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

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

How many of the 16 rows of the truth table make ((r ∧ ¬q) ↔ s) → ((r ∨ ¬p) ∧ (¬q → s)) true? Call this number N.

((r ∧ ¬q) ↔ s) → ((r ∨ ¬p) ∧ (¬q → s))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​C′I1​1I2​0I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 0, I₁ = C′, I₂ = 1, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = C′I₂ = 1I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-12

A synchronous mod-12 counter counts 0, 1, …, 11 and then returns to 0. It starts at 2. Which count Q (in decimal) does it show after 35 clock pulses?

Q0 = 235N = 12
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 5 (mod 11) and x ≡ 4 (mod 13).

x ≡ 5 (mod 11)x ≡ 4 (mod 13)
x =
8.Number theory & modular arithmeticHard

79 is prime. Use Fermat’s little theorem to find 1599159 mod 79.

p = 79k = 99159
1599159 mod 79 =
9.Algorithms & complexityMedium

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

  1. ABellman–Ford is required in all cases
  2. BDepth-first search
  3. CDijkstra's algorithm
  4. DKruskal's algorithm
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. BW is a number, so the time is exponential in the number of bits used to write W
  3. CBecause it only works for W = 0
  4. DBecause the algorithm is randomised and its running time varies with each input
11.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AHeaps never contain more than 100 elements
  2. BArrays automatically keep the keys in sorted order, so no pointers are needed
  3. CIt is a complete binary tree, so positions follow simple index formulas
  4. DAll keys in a heap are equal
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. BSorting the edges by weight
  3. CChecking whether an edge's two endpoints are already connected
  4. DChoosing the starting vertex
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
14.Graph theoryHard
55722310410515ABCDEFG

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

What is SQL injection?

  1. AAn attack where user input changes the structure of a SQL query
  2. BA database backup technique
  3. CA technique for inserting many rows quickly with a single INSERT statement
  4. DA way to add indexes
16.Databases & SQLHard

When is a schedule of concurrent transactions called serializable?

  1. AWhen every transaction is given a serial number
  2. BWhen the transactions run on one CPU only
  3. CWhen its effect is equivalent to some serial execution of the same transactions
  4. DWhen no transaction ever reads data
17.Computer networksMedium

TLS provides which security properties for HTTPS?

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

In a TCP connection, what does the receive window (rwnd) control?

  1. AThe number of open ports
  2. BThe size of each IP packet header
  3. CHow much unacknowledged data the sender may send, based on the receiver's buffer
  4. DThe retransmission time-out, calculated from the measured round-trip time of each segment
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.53.0/24 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?

192.168.53.0/24/26subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard

The network 192.168.240.0/24 is split with VLSM for these LANs: A: 66 hosts, B: 5 hosts, C: 16 hosts, D: 19 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 D?

A: 66B: 5C: 16D: 19
x =

3Solutions

1.Number systems

Answer: U = 143

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.

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: b = 9 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

28 = 256 < 340
29 = 512 ≥ 340

Answer

b = 9
3.Propositional logic & proofs

Answer: (p → q) ∧ (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

(a ∨ b) → c needs c whenever either a or b holds: (a → c) ∧ (b → c).
(p ∨ r) → q ≡ (p → q) ∧ (r → q)

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

Answer

(p → q) ∧ (r → q)

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

4.Propositional logic & proofs

Answer: N = 13

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.

Columns

¬q; r ∧ ¬q; (r ∧ ¬q) ↔ s; ¬p; r ∨ ¬p; ¬q → s; (r ∨ ¬p) ∧ (¬q → s); ((r ∧ ¬q) ↔ s) → ((r ∨ ¬p) ∧ (¬q → s))

Split on p

p = 1: 6
p = 0: 7
N = 6 + 7 = 13

Answer

N = 13
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′BC′ + AB′

Rows with F = 1

ABC = 010, 100, 101
N = 3

Answer

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

Answer: Q = 1

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.

Full cycles

Every 12 pulses bring the counter back to the same state.

35 = 2·12 + 11

Count after the remaining pulses

Q = (2 + 11) mod 12 = 1
Q = 0001₂

Answer

Q = 1
7.Number theory & modular arithmetic

Answer: x = 82

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 = 5 + 11k

Second congruence

5 + 11k ≡ 4 (mod 13) → 11k ≡ 12 (mod 13)
k ≡ 6·12 ≡ 7 (mod 13)

Result

x = 5 + 11·7 = 82
x = 82
check: 82 mod 11 = 5, 82 mod 13 = 4

Answer

x = 82
8.Number theory & modular arithmetic

Answer: 1599159 mod 79 = 41

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

99159 mod 78 = 21
1599159 ≡ 1521 (mod 79)

Small power

1521 mod 79 = 41

Answer

1599159 mod 79 = 41
9.Algorithms & complexity

Answer: C · 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

C · Dijkstra's algorithm
10.Algorithms & complexity

Answer: B · 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

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

Answer: C · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

C · It is a complete binary tree, so positions follow simple index formulas
12.Data structures

Answer: C · 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

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

Answer: H = 3

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.

Count

H = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
14.Graph theory

Answer: d(F) = 11

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: 5, C: 5
B (5): D: 7, G: 15
C (5): E: 7
D (7): F: 17, G: 12
E (7): F: 11
F (11)
G (12)

Shortest path

A → C → E → F
d(F) = 11

Answer

d(F) = 11
15.Databases & SQL

Answer: A · An attack where user input changes the structure of a SQL query

Why this is the answer

When user input is concatenated into SQL, attackers can inject commands. Prepared statements with parameters keep data separate from code.

Answer

A · An attack where user input changes the structure of a SQL query
16.Databases & SQL

Answer: C · When its effect is equivalent to some serial execution of the same transactions

Why this is the answer

Serializability is the main correctness criterion for concurrency: the interleaved result must equal the result of some one-at-a-time order.

Answer

C · When its effect is equivalent to some serial execution of the same transactions
17.Computer networks

Answer: B · 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

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

Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer

Why this is the answer

The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.

Answer

C · How much unacknowledged data the sender may send, based on the receiver's buffer
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 = 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

A: 66 hosts → 27 = 128 (/25)
D: 19 hosts → 25 = 32 (/27)
C: 16 hosts → 25 = 32 (/27)
B: 5 hosts → 23 = 8 (/29)

Allocation, largest first

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

Answer

x = 128

LAN D: 192.168.240.128/27.

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