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 −34 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −348 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 0010 OR 0100 1000. What is the result's value R in decimal?

0111 0010OR0100 1000
R =
3.Propositional logic & proofsMedium
(r ∧ p) ∨ (r ∧ q)(r ∨ p) ∧ (r ∨ q)(r ∧ p) ∨ qr ∨ (p ∧ q)

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

r ∧ (p ∨ q)
  1. 1(r ∧ p) ∨ (r ∧ q)
  2. 2(r ∨ p) ∧ (r ∨ q)
  3. 3(r ∧ p) ∨ q
  4. 4r ∨ (p ∧ q)
4.Propositional logic & proofsHard
¬rr¬r ∧ qq

The premises r → q; ¬q are given. Which conclusion follows from them? Click it.

r → q¬q
  1. 1¬r
  2. 2r
  3. 3¬r ∧ q
  4. 4q
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​CI1​0I2​C′I3​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 = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = CI₂ = 0I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 10 clock edges.

Q₃Q₂Q₁Q₀ = 000010
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 8) and x ≡ 8 (mod 11).

x ≡ 2 (mod 8)x ≡ 8 (mod 11)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 17127 mod 91.

a = 17k = 127n = 91
17127 mod 91 =
9.Algorithms & complexityMedium

About how many comparisons does binary search need in the worst case on a sorted array of 1,000,000 elements?

  1. AAbout 500,000
  2. BAbout 1,000
  3. CAbout 20
  4. DAbout 100
10.Algorithms & complexityHard

The halting problem shows that:

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

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

  1. A2
  2. BThere is no maximum
  3. C1
  4. D0.5
12.Data structuresHard

For a hash table to work correctly, which rule must hash and equality follow?

  1. AEqual keys must have equal hash values
  2. BDifferent keys must have different hash values
  3. CEqual hash values imply equal keys
  4. DHash values must be prime numbers
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
241327102157ABCDEFG

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

Table a has 3 rows and table b has 4 rows. How many rows does SELECT * FROM a CROSS JOIN b return?

  1. A12
  2. B7
  3. C3
  4. D4
16.Databases & SQLHard

R(A, B, C, D) has the functional dependencies A → B and C → D. What is the candidate key?

  1. A{A}
  2. B{B, D}
  3. C{C}
  4. D{A, C}
17.Computer networksMedium

What does an MX record in DNS specify?

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

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

  1. AHow much unacknowledged data the sender may send, based on the receiver's buffer
  2. BThe retransmission time-out, calculated from the measured round-trip time of each segment
  3. CThe size of each IP packet header
  4. DThe number of open ports
19.IPv4 subnettingMedium
255.255.255.12811111111111111111111111110000000network bitshost bits

A subnet uses the mask 255.255.255.128. How many usable host addresses H does it have?

255.255.255.128
H =
20.IPv4 subnettingHard
third octet = 87012816403211608141211networkhost

A host has the address 172.20.87.216/20. What is the third octet x of its subnet's broadcast address?

172.20.87.216/20
x =

3Solutions

1.Number systems

Answer: U = 222

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

34 = 0010 0010
inverted: 1101 1101
+1 → 1101 1110

Read it as unsigned

1101 1110₂ = 256 − 34 = 222

Answer

U = 222
2.Number systems

Answer: R = 122

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

OR gives 1 where at least one bit is 1.

0111 0010
OR 0100 1000
= 0111 1010

Value of the result

0111 1010₂ = 122

Answer

R = 122
3.Propositional logic & proofs

Answer: (r ∧ p) ∨ (r ∧ 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).

The law

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
r ∧ (p ∨ q) ≡ (r ∧ p) ∨ (r ∧ q)

Check with the truth table

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

Answer

(r ∧ p) ∨ (r ∧ q)

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

4.Propositional logic & proofs

Answer: ¬r

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

q = 0, r = 0: ¬r = 1, r = 0, ¬r ∧ q = 0, q = 0

Conclusion

¬r (modus tollens)

Answer

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

Answer: N = 2

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 + ABC′

Rows with F = 1

ABC = 011, 110
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 12

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 → 1000 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: x = 74

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 = 2 + 8k

Second congruence

2 + 8k ≡ 8 (mod 11) → 8k ≡ 6 (mod 11)
k ≡ 7·6 ≡ 9 (mod 11)

Result

x = 2 + 8·9 = 74
x = 74
check: 74 mod 8 = 2, 74 mod 11 = 8

Answer

x = 74
8.Number theory & modular arithmetic

Answer: 17127 mod 91 = 17

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

127 = 11111112 = 64 + 32 + 16 + 8 + 4 + 2 + 1

Repeated squaring

171 ≡ 17 (mod 91)
172 ≡ 16 (mod 91)
174 ≡ 74 (mod 91)
178 ≡ 16 (mod 91)
1716 ≡ 74 (mod 91)
1732 ≡ 16 (mod 91)
1764 ≡ 74 (mod 91)

Multiply the needed powers

17127 ≡ 17·16·74·16·74·16·74 ≡ 17 (mod 91)

Answer

17127 mod 91 = 17
9.Algorithms & complexity

Answer: C · About 20

Why this is the answer

220 ≈ 1,048,576, so after about 20 halvings only one element remains.

Answer

C · About 20
10.Algorithms & complexity

Answer: B · 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

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

Answer: C · 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

C · 1
12.Data structures

Answer: A · Equal keys must have equal hash values

Why this is the answer

If two equal keys hashed differently they would be stored in different buckets and lookups would fail; unequal keys may still share a hash.

Answer

A · Equal keys must have equal hash values
13.Graph theory

Answer: H = 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.

Count

H = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: d(F) = 20

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, C: 4
B (2): D: 4, G: 12
C (4): G: 6
D (4)
G (6): E: 13
E (13): F: 20
F (20)

Shortest path

A → C → G → E → F
d(F) = 20

Answer

d(F) = 20
15.Databases & SQL

Answer: A · 12

Why this is the answer

A cross join gives the Cartesian product, 3 × 4 = 12 rows.

Answer

A · 12
16.Databases & SQL

Answer: D · {A, C}

Why this is the answer

A and C are not determined by anything, so every key must contain both, and {A, C}+ = {A, B, C, D}.

Answer

D · {A, C}
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: A · 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

A · How much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnetting

Answer: H = 126

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).

Prefix length

255.255.255.128 = /25

Usable hosts

27 − 2 = 126

Answer

H = 126
20.IPv4 subnetting

Answer: x = 95

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 in the third octet

24 = 16

Subnet of this host

network: 172.20.80.0
broadcast: 172.20.95.255

Answer

x = 95

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