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 complement1−12806413201618041211

The 8-bit pattern 1010 1011 stores a signed integer in two's complement. What is its value N in decimal?

1010 10118 bit
N =
2.Number systemsHard
3500 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3500 items its own binary code of the same length. What is the smallest number of bits b needed?

3500
b = bit
3.Propositional logic & proofsMedium
(r ∧ q) ∨ (r ∧ p)(r ∨ q) ∧ (r ∨ p)(r ∨ q) ∧ pr ∧ (q ∨ p)

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

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

The premises (c ∧ b) → a; c; ¬a are given. Which conclusion follows from them? Click it.

(c ∧ b) → ac¬a
  1. 1b ∧ ¬a
  2. 2a
  3. 3¬b
  4. 4b
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​1I1​0I2​1I3​ABS1​S0​F

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

I₀ = C′I₁ = 1I₂ = 0I₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110101113020405X70611211311511418X9111010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 3, 8, 11, 12, 13, 14, 15) + d(7, 9), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 1, 3, 8, 11, 12, 13, 14, 15) + d(7, 9)
L =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 5 (mod 8) and x ≡ 6 (mod 9).

x ≡ 5 (mod 8)x ≡ 6 (mod 9)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 5175 mod 13.

a = 5k = 175n = 13
5175 mod 13 =
9.Algorithms & complexityMedium

Quicksort always uses the first element as pivot. On which input does it show its worst-case behaviour?

  1. AAn array whose elements are all different
  2. BAn array that is already sorted
  3. CAn array of length 2
  4. DAn array in random order
10.Algorithms & complexityHard

Which definition of the class P is correct?

  1. AProblems that need exponential time on every computer
  2. BDecision problems solvable by a deterministic algorithm in polynomial time
  3. CProblems that can be solved in parallel
  4. DProblems whose solutions can be checked but never found in polynomial time
11.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 4
  2. BIndex 6
  3. CIndex 3
  4. DIndex 0
12.Data structuresHard

An AVL node is unbalanced because its left child's right subtree became too tall (left-right case). Which fix is applied?

  1. AA single right rotation at the node
  2. BA single left rotation at the node
  3. CA right rotation at the left child, then a left rotation at the node
  4. DA left rotation at the left child, then a right rotation at the node
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 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?

4 × deg 33 × deg 44 × deg 2
L =
14.Graph theoryHard
214153151510576ABCDEFG

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 does ROLLBACK TO SAVEPOINT s1 do?

  1. ADeletes the savepoint and all tables changed since then
  2. BEnds the transaction and commits everything
  3. CUndoes the whole transaction from the beginning
  4. DUndoes the changes made after savepoint s1, keeping the transaction open
16.Databases & SQLHard

For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?

  1. ABitmap join on the primary key
  2. BHash join
  3. CCross join
  4. DNested loop join with index lookups
17.Computer networksMedium

Which TCP flag signals that a sender has finished sending data?

  1. ASYN
  2. BFIN
  3. CPSH
  4. DURG
18.Computer networksHard

A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?

  1. AName resolution (DNS)
  2. BThe network card driver
  3. CThe physical cable
  4. DThe default gateway
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.95.0/24 is divided into /28 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 12?

192.168.95.0/24/28subnet 12 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 183112806413211608141211networkhost

A host has the address 10.20.13.183/26. What is the fourth octet x of its network address?

10.20.13.183/26
x =

3Solutions

1.Number systems

Answer: N = −85

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 + 8 + 2 + 1 = −85

Answer

N = −85
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 < 3500
212 = 4096 ≥ 3500

Answer

b = 12
3.Propositional logic & proofs

Answer: (r ∨ q) ∧ (r ∨ p)

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).
r ∨ (q ∧ p) ≡ (r ∨ q) ∧ (r ∨ p)

Check with the truth table

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

Answer

(r ∨ q) ∧ (r ∨ p)

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

4.Propositional logic & proofs

Answer: ¬b

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.

Rows with all premises true

a = 0, b = 0, c = 1: b ∧ ¬a = 0, a = 0, ¬b = 1, b = 0

Conclusion

¬b (modus tollens with a conjunction)

Answer

¬b
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

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′B′C′ + A′B + AB

Rows with F = 1

ABC = 000, 010, 011, 110, 111
N = 5

Answer

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

Answer: L = 6

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.

CD, B′D, B′C′, AD, AC′, AB
ABCD0001111000011110101113020405X70611211311511418X9111010

Minimal cover

F = CD + B′C′ + AB
L = 6

Answer

L = 6

Another grouping of the same size exists; it has the same number of terms and literals.

7.Number theory & modular arithmetic

Answer: x = 69

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 + 8k

Second congruence

5 + 8k ≡ 6 (mod 9) → 8k ≡ 1 (mod 9)
k ≡ 8·1 ≡ 8 (mod 9)

Result

x = 5 + 8·8 = 69
x = 69
check: 69 mod 8 = 5, 69 mod 9 = 6

Answer

x = 69
8.Number theory & modular arithmetic

Answer: 5175 mod 13 = 8

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.

Binary exponent

175 = 101011112 = 128 + 32 + 8 + 4 + 2 + 1

Repeated squaring

51 ≡ 5 (mod 13)
52 ≡ 12 (mod 13)
54 ≡ 1 (mod 13)
58 ≡ 1 (mod 13)
516 ≡ 1 (mod 13)
532 ≡ 1 (mod 13)
564 ≡ 1 (mod 13)
5128 ≡ 1 (mod 13)

Multiply the needed powers

5175 ≡ 5·12·1·1·1·1 ≡ 8 (mod 13)

Answer

5175 mod 13 = 8
9.Algorithms & complexity

Answer: B · An array that is already sorted

Why this is the answer

On sorted input the pivot is always the minimum, so one side is empty and the other has n − 1 elements, giving O(n²).

Answer

B · An array that is already sorted
10.Algorithms & complexity

Answer: B · Decision problems solvable by a deterministic algorithm in polynomial time

Why this is the answer

P contains decision problems with a deterministic polynomial-time algorithm, such as checking whether a graph is connected.

Answer

B · Decision problems solvable by a deterministic algorithm in polynomial time
11.Data structures

Answer: A · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

A · Index 4
12.Data structures

Answer: D · A left rotation at the left child, then a right rotation at the node

Why this is the answer

The first rotation converts the left-right case into a left-left case, which a right rotation at the unbalanced node then fixes.

Answer

D · A left rotation at the left child, then a right rotation at the node
13.Graph theory

Answer: L = 12

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 + 32 = 2(L + 11) − 2
L = 12

Answer

L = 12

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

14.Graph theory

Answer: d(F) = 33

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: 2
B (2): C: 16, G: 12
G (12): D: 19, E: 18
C (16)
E (18): F: 33
D (19)
F (33)

Shortest path

A → B → G → E → F
d(F) = 33

Answer

d(F) = 33
15.Databases & SQL

Answer: D · Undoes the changes made after savepoint s1, keeping the transaction open

Why this is the answer

A savepoint marks a point inside a transaction; rolling back to it cancels later work while earlier changes remain pending until COMMIT.

Answer

D · Undoes the changes made after savepoint s1, keeping the transaction open
16.Databases & SQL

Answer: B · Hash join

Why this is the answer

A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.

Answer

B · Hash join
17.Computer networks

Answer: B · FIN

Why this is the answer

Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.

Answer

B · FIN
18.Computer networks

Answer: A · Name resolution (DNS)

Why this is the answer

A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.

Answer

A · Name resolution (DNS)
19.IPv4 subnetting

Answer: x = 176

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

24 = 16

Subnet number 12

network: (12 − 1)·16 = 176
broadcast: 176 + 16 − 1 = 191

Answer

x = 176
20.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 of this host

network: ⌊183/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190

Answer

x = 128

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