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 complement1−12816403201608040201

The 8-bit pattern 1100 0000 stores a signed integer in two's complement. What is its value N in decimal?

1100 00008 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 0010 >> 3. What is the result's value R in decimal?

0011 0010>> 3
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ((c → a) ∧ a) → c a tautology, a contradiction or a contingency? Click your answer.

((c → a) ∧ a) → c
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x (¬P(x) → ¬Q(x))∃x (P(x) ∧ ¬Q(x))∀x (P(x) ∧ ¬Q(x))∀x (P(x) → ¬Q(x))

Which statement is equivalent to ¬∃x (P(x) → Q(x)), with the negation moved all the way inside? Click it.

¬∃x (P(x) → Q(x))
  1. 1∀x (¬P(x) → ¬Q(x))
  2. 2∃x (P(x) ∧ ¬Q(x))
  3. 3∀x (P(x) ∧ ¬Q(x))
  4. 4∀x (P(x) → ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

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(4, 5, 7, 11, 12)
  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
ABCD00011110000111100001131204150706112013015014X8X9011010

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 5, 12) + d(8, 9), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(2, 3, 5, 12) + d(8, 9)
L =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(24), the number of integers from 1 to 24 that are coprime to 24.

n = 24 = 23·3
φ(24) =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (1189, 17)d = ?

An RSA key uses the primes p = 29 and q = 41 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).

p = 29q = 41e = 17
d =
9.Algorithms & complexityMedium

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

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

A problem is NP-complete if it is:

  1. AImpossible to solve
  2. BSolvable in polynomial time
  3. CIn NP and every NP problem reduces to it in polynomial time
  4. DOutside NP but harder than all NP problems, such as the halting problem
11.Data structuresMedium

A trie (prefix tree) is especially efficient for:

  1. AStoring matrices
  2. BPrefix searches such as autocomplete
  3. CSorting numbers
  4. DFinding shortest paths
12.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CBecause halving at one half would lose elements
  4. DTo keep the array sorted
13.Graph theoryMedium

A tree has 4 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?

4 × deg 44 × deg 2
L =
14.Graph theoryHard
131813215115ABCDEF

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 a database trigger?

  1. AA query that a user runs by hand
  2. BA scheduled backup of the database
  3. CCode that runs automatically when a specified event such as INSERT, UPDATE or DELETE occurs on a table
  4. DA type of index that speeds up searches
16.Databases & SQLHard

Which window function assigns 1, 2, 3, ... to rows within each partition without ties?

  1. ARANK()
  2. BROW_NUMBER()
  3. CCOUNT()
  4. DDENSE_RANK()
17.Computer networksMedium

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

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

A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?

  1. ASlow start
  2. BNagle's algorithm
  3. CFast retransmit
  4. DThe three-way handshake
19.IPv4 subnettingMedium
/24 → /3011111111111111111111111111111100network bitshost bits

The network 10.64.67.0/24 is divided into equal subnets with prefix /30. How many subnets N are there?

10.64.67.0/24/30
N =
20.IPv4 subnettingHard

The network 192.168.247.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 23 hosts, C: 41 hosts, D: 43 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 A?

A: 10B: 23C: 41D: 43
x =

3Solutions

1.Number systems

Answer: N = −64

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 + 64 = −64

Answer

N = −64
2.Number systems

Answer: R = 6

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.

Bit by bit

Shifting right by 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0011 0010 >> 3
= 0000 0110

Value of the result

0000 0110₂ = 6

Answer

R = 6
3.Propositional logic & proofs

Answer: contingency

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).

Truth table row by row

a = 1, c = 1: c → a = 1, (c → a) ∧ a = 1, ((c → a) ∧ a) → c = 1
a = 1, c = 0: c → a = 1, (c → a) ∧ a = 1, ((c → a) ∧ a) → c = 0
a = 0, c = 1: c → a = 0, (c → a) ∧ a = 0, ((c → a) ∧ a) → c = 1
a = 0, c = 0: c → a = 1, (c → a) ∧ a = 0, ((c → a) ∧ a) → c = 1

Classify

((c → a) ∧ a) → c: 1, 0, 1, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
4.Propositional logic & proofs

Answer: ∀x (P(x) ∧ ¬Q(x))

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 (P(x) → Q(x)) ≡ ∀x (P(x) ∧ ¬Q(x))

Answer

∀x (P(x) ∧ ¬Q(x))

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: AB = 10, 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.

m11: m10, m9, m15, m3 → 0
ABCD000111100001111011111

The isolated 1

m11 = AB′CD → AB = 10, CD = 11

Answer

AB = 10, CD = 11
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.

AC′D′, A′B′C, A′BC′D
ABCD00011110000111100001131204150706112013015014X8X9011010

Minimal cover

F = AC′D′ + A′B′C + A′BC′D
L = 10

Answer

L = 10
7.Number theory & modular arithmetic

Answer: φ(24) = 8

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).

Prime factors

24 = 23·3

Totient

φ(24) = φ(8)·φ(3) = 4·2 = 8

Answer

φ(24) = 8
8.Number theory & modular arithmetic

Answer: d = 593

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 = 29·41 = 1189
φ(n) = 28·40 = 1120

Extended Euclidean algorithm

r = 1120, t = 0
r = 17, q = 65, t = 1
r = 15, q = 1, t = −65
r = 2, q = 7, t = 66
r = 1, q = 2, t = −527

Private exponent

d = 593
check: 17·593 = 10081 = 9·1120 + 1

Answer

d = 593
9.Algorithms & complexity

Answer: A · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

A · O(n)
10.Algorithms & complexity

Answer: C · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

C · In NP and every NP problem reduces to it in polynomial time
11.Data structures

Answer: B · Prefix searches such as autocomplete

Why this is the answer

Tries store strings by characters along paths, so lookups and prefix queries take time proportional to the key length.

Answer

B · Prefix searches such as autocomplete
12.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
13.Graph theory

Answer: L = 10

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 + 24 = 2(L + 8) − 2
L = 10

Answer

L = 10

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

14.Graph theory

Answer: d(F) = 8

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: 13, C: 1
C (1): B: 9, E: 3
E (3): D: 18, F: 8
F (8)
B (9)
D (18)

Shortest path

A → C → E → F
d(F) = 8

Answer

d(F) = 8
15.Databases & SQL

Answer: C · Code that runs automatically when a specified event such as INSERT, UPDATE or DELETE occurs on a table

Why this is the answer

Triggers execute automatically before or after data changes, for example to write an audit log or enforce a complex rule.

Answer

C · Code that runs automatically when a specified event such as INSERT, UPDATE or DELETE occurs on a table
16.Databases & SQL

Answer: B · ROW_NUMBER()

Why this is the answer

ROW_NUMBER gives unique sequential numbers; RANK leaves gaps after ties; DENSE_RANK does not.

Answer

B · ROW_NUMBER()
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: B · Nagle's algorithm

Why this is the answer

Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.

Answer

B · Nagle's algorithm
19.IPv4 subnetting

Answer: N = 64

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).

Borrowed bits

30 − 24 = 6

Number of subnets

N = 26 = 64

Answer

N = 64

Each /30 subnet has 22 − 2 = 2 usable hosts.

20.IPv4 subnetting

Answer: x = 160

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

D: 43 hosts → 26 = 64 (/26)
C: 41 hosts → 26 = 64 (/26)
B: 23 hosts → 25 = 32 (/27)
A: 10 hosts → 24 = 16 (/28)

Allocation, largest first

D: .0 – .63
C: .64 – .127
B: .128 – .159
A: .160 – .175

Answer

x = 160

LAN A: 192.168.247.160/28.

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