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 −17 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −178 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0110 1100 << 2. What is the result's value R in decimal?

0110 1100<< 2
R =
3.Propositional logic & proofsMedium
¬(q ∨ p) ∧ ¬¬r(q ∨ p) ∧ ¬¬r¬(q ∨ p) ∧ ¬r¬(q ∨ p) → ¬¬r

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

(q ∨ p) → ¬r
  1. 1¬(q ∨ p) ∧ ¬¬r
  2. 2(q ∨ p) ∧ ¬¬r
  3. 3¬(q ∨ p) ∧ ¬r
  4. 4¬(q ∨ p) → ¬¬r
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
MUX0I0​CI1​CI2​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 = 0, I₁ = C, I₂ = C, I₃ = 1. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = CI₂ = CI₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011030214050716X12013015X141819011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 4, 6, 8, 9, 10) + d(12, 14), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 1, 4, 6, 8, 9, 10) + d(12, 14)
L =
7.Number theory & modular arithmeticMedium

Solve 41x ≡ 2 (mod 70). Give the solution x with 0 < x < 70.

41x ≡ 2 (mod 70)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 2741 mod 68.

a = 27k = 41n = 68
2741 mod 68 =
9.Algorithms & complexityMedium

What does it mean that a problem has optimal substructure?

  1. AThe problem can be solved in constant time
  2. BAn optimal solution can be built from optimal solutions of its subproblems
  3. CEvery subproblem has the same size
  4. DThe problem has exactly one optimal solution, which every subproblem shares
10.Algorithms & complexityHard

The halting problem shows that:

  1. ANo algorithm can decide for every program and input whether it halts
  2. BSome programs run faster on faster computers, so halting depends on hardware
  3. CEvery program eventually halts
  4. DLoops should be avoided
11.Data structuresMedium

What balance condition does an AVL tree maintain at every node?

  1. AEvery node is coloured red or black
  2. BAll leaves are at exactly the same depth
  3. CBoth subtrees of every node contain exactly the same number of nodes
  4. DThe heights of the left and right subtrees differ by at most 1
12.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(α(n)), the inverse Ackermann function, which is practically constant
  2. BO(n²)
  3. CO(n)
  4. DO(log n) exactly
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
2711517209ABCDE

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

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

A table has rows where col is NULL. Which rows does WHERE col = NULL return?

  1. AAn error in every database
  2. BNo rows, because the comparison is unknown
  3. CAll rows of the table
  4. DOnly the rows where col is NULL, as with IS NULL
16.Databases & SQLHard

Table enrolment(student_id, course_id, student_name, grade) has the key (student_id, course_id). Which normal form does it violate?

  1. AFirst normal form, because the grade column does not hold atomic values
  2. BNone, it is in BCNF
  3. CThird normal form only, because grade depends on student_name
  4. DSecond normal form, because student_name depends on student_id only
17.Computer networksMedium

How does TLS typically use asymmetric and symmetric cryptography?

  1. ASymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
  2. BAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
  3. COnly asymmetric cryptography is used throughout
  4. DOnly symmetric cryptography is used, with a fixed shared password
18.Computer networksHard

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

  1. AThe size of each IP packet header
  2. BThe retransmission time-out, calculated from the measured round-trip time of each segment
  3. CHow much unacknowledged data the sender may send, based on the receiver's buffer
  4. DThe number of open ports
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/248 subnets
H =
20.IPv4 subnettingHard
last octet = 131112806403201608041211networkhost

A host has the address 10.20.84.131/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.84.131/27
x =

3Solutions

1.Number systems

Answer: U = 239

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

17 = 0001 0001
inverted: 1110 1110
+1 → 1110 1111

Read it as unsigned

1110 1111₂ = 256 − 17 = 239

Answer

U = 239
2.Number systems

Answer: R = 176

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 left by 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0110 1100 << 2
= 1011 0000

Value of the result

1011 0000₂ = 176

Answer

R = 176
3.Propositional logic & proofs

Answer: (q ∨ 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).

Parts of the implication

a = q ∨ p, b = ¬r

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = (q ∨ p) ∧ ¬¬r

Answer

(q ∨ p) ∧ ¬¬r

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

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: N = 4

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′C + AB

Rows with F = 1

ABC = 011, 101, 110, 111
N = 4

Answer

N = 4
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.

C′D′, BD′, B′C′, AD′
ABCD00011110000111101011030214050716X12013015X141819011110

Minimal cover

F = BD′ + B′C′ + AD′
L = 6

Answer

L = 6
7.Number theory & modular arithmetic

Answer: x = 12

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).

Inverse of 41

r = 70, t = 0
r = 41, q = 1, t = 1
r = 29, q = 1, t = −1
r = 12, q = 2, t = 2
r = 5, q = 2, t = −5
r = 2, q = 2, t = 12
r = 1, q = 2, t = −29
41−1 ≡ 41 (mod 70)

Solve

x ≡ 41·2 = 82 ≡ 12 (mod 70)
check: 41·12 = 492 ≡ 2 (mod 70)

Answer

x = 12
8.Number theory & modular arithmetic

Answer: 2741 mod 68 = 7

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

41 = 1010012 = 32 + 8 + 1

Repeated squaring

271 ≡ 27 (mod 68)
272 ≡ 49 (mod 68)
274 ≡ 21 (mod 68)
278 ≡ 33 (mod 68)
2716 ≡ 1 (mod 68)
2732 ≡ 1 (mod 68)

Multiply the needed powers

2741 ≡ 27·33·1 ≡ 7 (mod 68)

Answer

2741 mod 68 = 7
9.Algorithms & complexity

Answer: B · An optimal solution can be built from optimal solutions of its subproblems

Why this is the answer

Optimal substructure is required for both greedy algorithms and dynamic programming to produce correct answers.

Answer

B · An optimal solution can be built from optimal solutions of its subproblems
10.Algorithms & complexity

Answer: A · No algorithm can decide for every program and input whether it halts

Why this is the answer

Turing showed the halting problem is undecidable: assuming a decider exists leads to a contradiction with a program that does the opposite.

Answer

A · No algorithm can decide for every program and input whether it halts
11.Data structures

Answer: D · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

D · The heights of the left and right subtrees differ by at most 1
12.Data structures

Answer: A · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

A · O(α(n)), the inverse Ackermann function, which is practically constant
13.Graph theory

Answer: Euler path, no circuit

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) = 1, deg(B) = 4, deg(C) = 2, deg(D) = 1, deg(E) = 2

Odd vertices

A, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: W = 27

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

BC (1): take
AB (2): take
AC (7): skip, closes a cycle
BE (9): take
BD (15): take

Total weight

W = 1 + 2 + 9 + 15 = 27

Answer

W = 27
15.Databases & SQL

Answer: B · No rows, because the comparison is unknown

Why this is the answer

Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.

Answer

B · No rows, because the comparison is unknown
16.Databases & SQL

Answer: D · Second normal form, because student_name depends on student_id only

Why this is the answer

student_name depends on only part of the key, which is a partial dependency; moving it to a students table fixes the 2NF violation.

Answer

D · Second normal form, because student_name depends on student_id only
17.Computer networks

Answer: B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data

Why this is the answer

Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.

Answer

B · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
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: H = 30

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

23 = 8 ≥ 8 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 159

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

25 = 32

Subnet of this host

network: ⌊131/32⌋·32 = 128
broadcast: 128 + 32 − 1 = 159
usable hosts: 129 … 158

Answer

x = 159

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