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 −118 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1188 bit
U =
2.Number systemsHard
20-bit two's complement, N = ??sign???????????????????

What is the most negative integer N that can be stored in 20-bit two's complement?

20 bit
N =
3.Propositional logic & proofsMedium
¬(r ∧ p) ∧ ¬q(r ∧ p) ∧ ¬q¬(r ∧ p) ∧ q¬(r ∧ p) → ¬q

You want to prove (r ∧ p) → q by contradiction. Which statement do you assume at the start? Click it.

(r ∧ p) → q
  1. 1¬(r ∧ p) ∧ ¬q
  2. 2(r ∧ p) ∧ ¬q
  3. 3¬(r ∧ p) ∧ q
  4. 4¬(r ∧ p) → ¬q
4.Propositional logic & proofsHard
∃y ∀x ¬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∃y ∀x ¬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
ABCD000111100001111010011302041517160120131151141819111010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 3, 5, 6, 7, 8, 9, 11, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 3, 5, 6, 7, 8, 9, 11, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 3 (decimal, Q = 0011). Which count Q (in decimal) does it show after 41 clock pulses?

Q0 = 341n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 3) and x ≡ 6 (mod 8).

x ≡ 1 (mod 3)x ≡ 6 (mod 8)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (119, 13)d = ?

An RSA key uses the primes p = 7 and q = 17 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).

p = 7q = 17e = 13
d =
9.Algorithms & complexityMedium

What is the worst-case time complexity of quicksort?

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

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

  1. AΘ(n)
  2. BΘ(n log n)
  3. CΘ(n log² n)
  4. DΘ(n²)
11.Data structuresMedium

A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?

  1. AA single sorted array of vertices
  2. BA binary heap of edges
  3. CAdjacency matrix
  4. DAdjacency list
12.Data structuresHard

In the BST with root 50, left subtree 30 (children 20, 40) and right subtree 70 (children 60, 80), the root is deleted using the in-order successor. Which key becomes the new root?

  1. A60
  2. B70
  3. C80
  4. D40
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
ABCDE

How many walks of length 3 go from E to D? A walk may repeat vertices and edges. Call the number N.

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

Which statement about HAVING is correct?

  1. AIt can use aggregate functions to filter groups
  2. BIt must come before WHERE in the SELECT statement
  3. CIt cannot be used with GROUP BY
  4. DIt sorts the results
16.Databases & SQLHard

Transaction T1 locks row A and waits for row B, while T2 locks row B and waits for row A. What does a typical DBMS do?

  1. AMerges the two transactions into one
  2. BDetects the deadlock and aborts one transaction as the victim
  3. CLets both transactions wait until one of them commits normally
  4. DCommits both transactions at once
17.Computer networksMedium

What does an MX record in DNS specify?

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

Why does TCP choose an unpredictable initial sequence number for each connection?

  1. ATo compress the data more efficiently
  2. BTo encrypt the payload of each segment
  3. CTo select the best route to the destination for the segments of that connection
  4. DTo avoid confusion with old segments and make spoofed segments harder to inject
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2412 subnets
H =
20.IPv4 subnettingHard
last octet = 19012806403211608041211networkhost

A host has the address 10.20.187.19/30. What is the fourth octet x of its network address?

10.20.187.19/30
x =

3Solutions

1.Number systems

Answer: U = 138

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

118 = 0111 0110
inverted: 1000 1001
+1 → 1000 1010

Read it as unsigned

1000 1010₂ = 256 − 118 = 138

Answer

U = 138
2.Number systems

Answer: N = −524288

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

−219 = −524288

Answer

N = −524288
3.Propositional logic & proofs

Answer: (r ∧ p) ∧ ¬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).

Parts of the implication

a = r ∧ p, b = q

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = (r ∧ p) ∧ ¬q

Answer

(r ∧ p) ∧ ¬q

If the assumption leads to a contradiction, the implication must be true.

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: 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.

ABCD000111100001111010011302041517160120131151141819111010

Group the 0s

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

(B + C′ + D)
(B′ + C + D)
(A′ + B′ + C)
(A + B + C + D′)

Minimal POS form

F = (B + C′ + D)(B′ + C + D)(A′ + B′ + C)(A + B + C + D′)
N = 4

Answer

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

Answer: Q = 10

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 16 pulses bring the counter back to the same state.

41 = 2·16 + 9

Count after the remaining pulses

Q = (3 − 9) mod 16 = 10
Q = 1010₂

Answer

Q = 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 + 3k

Second congruence

1 + 3k ≡ 6 (mod 8) → 3k ≡ 5 (mod 8)
k ≡ 3·5 ≡ 7 (mod 8)

Result

x = 1 + 3·7 = 22
x = 22
check: 22 mod 3 = 1, 22 mod 8 = 6

Answer

x = 22
8.Number theory & modular arithmetic

Answer: d = 37

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 = 7·17 = 119
φ(n) = 6·16 = 96

Extended Euclidean algorithm

r = 96, t = 0
r = 13, q = 7, t = 1
r = 5, q = 2, t = −7
r = 3, q = 1, t = 15
r = 2, q = 1, t = −22
r = 1, q = 2, t = 37

Private exponent

d = 37
check: 13·37 = 481 = 5·96 + 1

Answer

d = 37
9.Algorithms & complexity

Answer: B · O(n²)

Why this is the answer

With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).

Answer

B · O(n²)
10.Algorithms & complexity

Answer: C · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

C · Θ(n log² n)
11.Data structures

Answer: D · Adjacency list

Why this is the answer

A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.

Answer

D · Adjacency list
12.Data structures

Answer: A · 60

Why this is the answer

The leftmost node of the right subtree is 60, so 60 replaces 50 and its old position is removed; using the predecessor would give 40.

Answer

A · 60
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: N = 5

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 0 1 0 0
B: 0 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)ED = 1·0 + 2·1 + 1·0 + 1·0 + 3·1 = 5

Answer

N = 5
15.Databases & SQL

Answer: A · It can use aggregate functions to filter groups

Why this is the answer

HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.

Answer

A · It can use aggregate functions to filter groups
16.Databases & SQL

Answer: B · Detects the deadlock and aborts one transaction as the victim

Why this is the answer

A cycle in the wait-for graph is a deadlock; the DBMS rolls back one transaction so the other can continue, and the victim may be retried.

Answer

B · Detects the deadlock and aborts one transaction as the victim
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: D · To avoid confusion with old segments and make spoofed segments harder to inject

Why this is the answer

Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.

Answer

D · To avoid confusion with old segments and make spoofed segments harder to inject
19.IPv4 subnetting

Answer: H = 14

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

24 = 16 ≥ 12 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 16

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

22 = 4

Subnet of this host

network: ⌊19/4⌋·4 = 16
broadcast: 16 + 4 − 1 = 19
usable hosts: 17 … 18

Answer

x = 16

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