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 complement1−12806413211608141201

The 8-bit pattern 1011 0110 stores a signed integer in two's complement. What is its value N in decimal?

1011 01108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 1010 >> 1. What is the result's value R in decimal?

0101 1010>> 1
R =
3.Propositional logic & proofsMedium
(q → r) → pq → (r → p)(q → p) ∧ (r → p)q ∧ (r → p)

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

(q ∧ r) → p
  1. 1(q → r) → p
  2. 2q → (r → p)
  3. 3(q → p) ∧ (r → p)
  4. 4q ∧ (r → p)
4.Propositional logic & proofsHard
ca ∧ ca¬c

The premises a ∨ c; ¬a are given. Which conclusion follows from them? Click it.

a ∨ c¬a
  1. 1c
  2. 2a ∧ c
  3. 3a
  4. 4¬c
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​1I1​1I2​CI3​ABS1​S0​F

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

I₀ = 1I₁ = 1I₂ = 1I₃ = C
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110100113120405071611201301501408X9111X10

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 2, 3, 6, 11, 12) + d(9, 10), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 2, 3, 6, 11, 12) + d(9, 10)
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(169), the number of integers from 1 to 169 that are coprime to 169.

n = 169 = 132
φ(169) =
8.Number theory & modular arithmeticHard
167c = me​ mod n?m = cd​ mod nm(n, e) = (247, 5)d

Encrypt the message m = 167 with the RSA public key (n, e) = (247, 5). Find the ciphertext c.

n = 247e = 5m = 167
c =
9.Algorithms & complexityMedium

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

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

Which statement about little-o notation is correct?

  1. An = o(n²), but n is not o(n)
  2. Bn = o(n), since every function is little-o of itself
  3. Cn² = o(n)
  4. D2n = o(n), because 2n grows slowly
11.Data structuresMedium

In a hash table using open addressing, what is the maximum possible load factor?

  1. A0.5
  2. B2
  3. CThere is no maximum
  4. D1
12.Data structuresHard

A Fenwick tree (binary indexed tree) over an array of n numbers supports which operations in O(log n)?

  1. AInserting elements at any position
  2. BFinding the median in O(1)
  3. CPrefix-sum queries and single-element updates
  4. DSorting the array and binary search
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
14.Graph theoryHard
1815467819ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

start: Around 5
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

What does EXCEPT (called MINUS in Oracle) return?

  1. AAll rows from both queries without duplicates
  2. BRows that appear in both queries
  3. CRows of the second query that do not appear in the first one
  4. DRows of the first query that do not appear in the second
16.Databases & SQLHard

Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?

  1. A1, 1, 2
  2. B1, 1, 3
  3. C2, 2, 3
  4. D1, 2, 3
17.Computer networksMedium

At which OSI layer does a switch normally operate?

  1. APhysical (layer 1)
  2. BNetwork (layer 3)
  3. CTransport (layer 4)
  4. DData link (layer 2)
18.Computer networksHard

How does fragmentation differ in IPv6 compared with IPv4?

  1. AIPv6 does not allow packets larger than 576 bytes
  2. BIPv6 fragments at the transport layer inside TCP
  3. CRouters never fragment IPv6 packets; only the source can, using path MTU discovery
  4. DIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2420 subnets
H =
20.IPv4 subnettingHard
last octet = 102012816413201608141201networkhost

A host has the address 10.20.56.102/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.56.102/25
x =

3Solutions

1.Number systems

Answer: N = −74

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 + 2 = −74

Answer

N = −74
2.Number systems

Answer: R = 45

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 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0101 1010 >> 1
= 0010 1101

Value of the result

0010 1101₂ = 45

Answer

R = 45
3.Propositional logic & proofs

Answer: 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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(q ∧ r) → p ≡ 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 | 1 1 1 1
p = 1, q = 0, r = 1: 1 | 1 1 1 0
p = 1, q = 0, r = 0: 1 | 1 1 1 0
p = 0, q = 1, r = 1: 0 | 0 0 0 0
p = 0, q = 1, r = 0: 1 | 1 1 0 1
p = 0, q = 0, r = 1: 1 | 0 1 0 0
p = 0, q = 0, r = 0: 1 | 0 1 1 0

Answer

q → (r → p)

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

4.Propositional logic & proofs

Answer: c

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

Conclusion

c (disjunctive syllogism)

Answer

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

Answer: N = 7

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′ + A′B + AB′ + ABC

Rows with F = 1

ABC = 000, 001, 010, 011, 100, 101, 111
N = 7

Answer

N = 7
6.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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′C, A′CD′, A′B′D′, AB′D, ABC′D′
ABCD0001111000011110100113120405071611201301501408X9111X10

Minimal cover

F = B′C + A′CD′ + A′B′D′ + ABC′D′
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: φ(169) = 156

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

169 = 132

Totient

φ(169) = 169 − 13 = 156

Answer

φ(169) = 156
8.Number theory & modular arithmetic

Answer: c = 59

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.

Repeated squaring

5 = 4 + 1
1671 ≡ 167 (mod 247)
1672 ≡ 225 (mod 247)
1674 ≡ 237 (mod 247)
1675 ≡ 167·237 ≡ 59 (mod 247)

Ciphertext

c = 59

Answer

c = 59
9.Algorithms & complexity

Answer: B · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

B · O(n²)
10.Algorithms & complexity

Answer: A · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

A · n = o(n²), but n is not o(n)
11.Data structures

Answer: D · 1

Why this is the answer

With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.

Answer

D · 1
12.Data structures

Answer: C · Prefix-sum queries and single-element updates

Why this is the answer

A Fenwick tree stores partial sums over ranges determined by the low bits of each index, so both prefix sums and point updates take O(log n).

Answer

C · Prefix-sum queries and single-element updates
13.Graph theory

Answer: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
14.Graph theory

Answer: B

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

1. A (0): C: 1
2. C (1): B: 9, E: 5, G: 2
3. G (2)
4. E (5): D: 11
5. B (9)
6. D (11): F: 18
7. F (18)

round 5

B

Answer

B
15.Databases & SQL

Answer: D · Rows of the first query that do not appear in the second

Why this is the answer

A EXCEPT B gives the rows in A that are not in B, so swapping the queries gives a different result.

Answer

D · Rows of the first query that do not appear in the second
16.Databases & SQL

Answer: B · 1, 1, 3

Why this is the answer

RANK gives tied rows the same rank and skips the following numbers (1, 1, 3); DENSE_RANK would give 1, 1, 2.

Answer

B · 1, 1, 3
17.Computer networks

Answer: D · Data link (layer 2)

Why this is the answer

A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.

Answer

D · Data link (layer 2)
18.Computer networks

Answer: C · 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

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

Answer: H = 6

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

25 = 32 ≥ 20 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 127

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

27 = 128

Subnet of this host

network: ⌊102/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126

Answer

x = 127

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