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
place values in two's complement1−12806413211608140211

The 8-bit pattern 1011 0101 stores a signed integer in two's complement. What is its value N in decimal?

1011 01018 bit
N =
2.Number systemsHard
24 bits, unsigned, N = ?????????????????????????

What is the largest unsigned integer N that fits in 24 bits?

24 bit
N =
3.Propositional logic & proofsMedium
¬p ∨ ¬r¬p ∧ r¬p ∧ ¬rp ∨ ¬r

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

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

How many of the 16 rows of the truth table make (d → (b ↔ ¬a)) ∧ ((a ↔ ¬b) ∨ (d ∧ c)) true? Call this number N.

(d → (b ↔ ¬a)) ∧ ((a ↔ ¬b) ∨ (d ∧ c))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302041517060120130151140819011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 5, 7, 9, 10, 14). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 5, 7, 9, 10, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​FFQ4​Q4​FFQ5​Q5​CLKMOD-60

A 20 kHz clock drives a mod-60 counter. Find the frequency fout of its most significant output, which completes one cycle per counting cycle.

fclk = 20 kHzN = 60
fout = kHz
7.Number theory & modular arithmeticMedium

Find the inverse of 15 modulo 19: the number x with 0 < x < 19 and 15·x ≡ 1 (mod 19).

a = 15n = 19
15−1 mod 19 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (85, 11)d = ?

An RSA key uses the primes p = 5 and q = 17 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 5q = 17e = 11
d =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?

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

How do n! and 2ⁿ compare for large n?

  1. AThey grow at the same rate
  2. Bn! is O(n²)
  3. C2ⁿ grows faster than n!
  4. Dn! grows faster than 2ⁿ
11.Data structuresMedium

How does extract-min work in a binary min-heap?

  1. ARemove the last element of the array
  2. BSearch all leaves for the new minimum
  3. CRemove the root and shift all elements one place left
  4. DMove the last element to the root and sift it down
12.Data structuresHard

In a circular buffer, front == rear can mean either empty or full. What is a common way to tell them apart?

  1. AMake the buffer size a power of two
  2. BAlways clear the buffer whenever front equals rear, so the state is empty
  3. CStore the front and rear in the same variable
  4. DKeep a separate element count, or always leave one slot unused
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
1876195293ABCDEF

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

|V| = 6|E| = 8
W =
15.Databases & SQLMedium

Second normal form (2NF) removes:

  1. AForeign keys
  2. BTransitive dependencies between non-key columns via another non-key column
  3. CPartial dependencies of non-key columns on part of a composite key
  4. DAll NULL values
16.Databases & SQLHard

An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?

  1. ADatabase sharding
  2. BOptimistic concurrency control
  3. CStrict two-phase locking with row locks
  4. DWrite-ahead logging
17.Computer networksMedium

Which medium access method did classic shared (half-duplex) Ethernet use?

  1. ACSMA/CA
  2. BTDMA
  3. CCSMA/CD
  4. DToken passing
18.Computer networksHard

How does fragmentation differ in IPv6 compared with IPv4?

  1. ARouters never fragment IPv6 packets; only the source can, using path MTU discovery
  2. BIPv6 does not allow packets larger than 576 bytes
  3. CIPv6 fragments at the transport layer inside TCP
  4. DIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/243 subnets
H =
20.IPv4 subnettingHard
third octet 198 … 202112816403201608141201commonvaries

A router has routes to these networks: 10.71.198.0/24, 10.71.199.0/24, 10.71.200.0/24, 10.71.201.0/24, 10.71.202.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.71.198.0/24 … 10.71.202.0/24
p =

3Solutions

1.Number systems

Answer: N = −75

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 + 32 + 16 + 4 + 1 = −75

Answer

N = −75
2.Number systems

Answer: N = 1.678 × 10⁷

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.

Largest value

224 − 1 = 16777215

Answer

N = 16777215
3.Propositional logic & proofs

Answer: ¬p ∧ ¬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

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(p ∨ r) ≡ ¬p ∧ ¬r

Check with the truth table

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

Answer

¬p ∧ ¬r

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

4.Propositional logic & proofs

Answer: N = 8

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

¬a; b ↔ ¬a; d → (b ↔ ¬a); ¬b; a ↔ ¬b; d ∧ c; (a ↔ ¬b) ∨ (d ∧ c); (d → (b ↔ ¬a)) ∧ ((a ↔ ¬b) ∨ (d ∧ c))

Split on a

a = 1: 4
a = 0: 4
N = 4 + 4 = 8

Answer

N = 8
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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.

ABCD000111100001111010010302041517060120130151140819011110

Group the 1s

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

ACD′
A′BD
A′B′C′D′
AB′C′D

Minimal SOP form

F = ACD′ + A′BD + A′B′C′D′ + AB′C′D
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: fout = 0.3333 kHz

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.

Frequency division

fout = 20 kHz / 60 = 0.3333 kHz

Answer

fout = 0.3333 kHz
7.Number theory & modular arithmetic

Answer: 15−1 mod 19 = 14

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·15 (mod 19).

r = 19, t = 0
r = 15, q = 1, t = 1
r = 4, q = 3, t = −1
r = 3, q = 1, t = 4
r = 1, q = 3, t = −5

Inverse

1 ≡ −5·15 (mod 19)
x = 14
check: 15·14 = 210 = 11·19 + 1

Answer

15−1 mod 19 = 14
8.Number theory & modular arithmetic

Answer: d = 35

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.

Modulus and totient

n = 5·17 = 85
φ(n) = 4·16 = 64

Extended Euclidean algorithm

r = 64, t = 0
r = 11, q = 5, t = 1
r = 9, q = 1, t = −5
r = 2, q = 4, t = 6
r = 1, q = 2, t = −29

Private exponent

d = 35
check: 11·35 = 385 = 6·64 + 1

Answer

d = 35
9.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).

Answer

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

Answer: D · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

D · n! grows faster than 2ⁿ
11.Data structures

Answer: D · Move the last element to the root and sift it down

Why this is the answer

The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).

Answer

D · Move the last element to the root and sift it down
12.Data structures

Answer: D · Keep a separate element count, or always leave one slot unused

Why this is the answer

With a count, empty means count = 0 and full means count = capacity; with one empty slot, full is defined as (rear + 1) mod n == front.

Answer

D · Keep a separate element count, or always leave one slot unused
13.Graph theory

Answer: neither

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) = 3, deg(C) = 3, deg(D) = 3, deg(E) = 3, deg(F) = 2

Odd vertices

B, C, D, E → 4
neither

Answer

neither
14.Graph theory

Answer: W = 26

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

DE (2): take
BE (3): take
CE (5): take
BC (6): skip, closes a cycle
AC (7): take
DF (9): take

Total weight

W = 2 + 3 + 5 + 7 + 9 = 26

Answer

W = 26
15.Databases & SQL

Answer: C · Partial dependencies of non-key columns on part of a composite key

Why this is the answer

2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.

Answer

C · Partial dependencies of non-key columns on part of a composite key
16.Databases & SQL

Answer: B · Optimistic concurrency control

Why this is the answer

Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.

Answer

B · Optimistic concurrency control
17.Computer networks

Answer: C · CSMA/CD

Why this is the answer

Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.

Answer

C · CSMA/CD
18.Computer networks

Answer: A · Routers never fragment IPv6 packets; only the source can, using path MTU discovery

Why this is the answer

An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.

Answer

A · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
19.IPv4 subnetting

Answer: H = 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).

Bits to borrow

22 = 4 ≥ 3 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: p = 20

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

198 = 11000110
202 = 11001010
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.71.192.0/20 (it also covers 11 network(s) that are not in the list).

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