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 complement0−12806413211608141211

The 8-bit pattern 0011 0111 stores a signed integer in two's complement. What is its value N in decimal?

0011 01118 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0100 0000 >> 1. What is the result's value R in decimal?

0100 0000>> 1
R =
3.Propositional logic & proofsMedium
¬p ∨ q¬p ∨ ¬q¬p ∧ ¬q¬p ∧ q

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

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

How many of the 16 rows of the truth table make ((¬r ∧ q) ∨ (¬s ∧ q)) → (p ∧ (s ∨ p)) true? Call this number N.

((¬r ∧ q) ∨ (¬s ∧ q)) → (p ∧ (s ∨ p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011030214051706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(1, 4, 7). How many product terms N does it have?

F(A, B, C) = Σm(1, 4, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
  8. 8clock edge 8
7.Number theory & modular arithmeticMedium

Find the inverse of 19 modulo 47: the number x with 0 < x < 47 and 19·x ≡ 1 (mod 47).

a = 19n = 47
19−1 mod 47 =
8.Number theory & modular arithmeticHard

How many solutions N does 15x ≡ 12 (mod 21) have in {0, 1, …, 20}?

15x ≡ 12 (mod 21)
N =
9.Algorithms & complexityMedium

What does the recurrence T(n) = 2T(n − 1) + O(1) solve to?

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

Which technique gives an expected O(n log n) running time for quicksort on any input?

  1. AAlways choosing the first element
  2. BChoosing the pivot uniformly at random
  3. CUsing recursion only on the left side
  4. DSorting the input before running quicksort
11.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA singly linked list
  2. BA doubly linked list
  3. CA dynamic array
  4. DA sorted array
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA deque holding indices in decreasing order of value
  2. BA singly linked list of the window's elements
  3. CA min-heap rebuilt for every window
  4. DA sorted array updated by shifting
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| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
128181337111146ABCDEFG

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

First normal form (1NF) requires that:

  1. AEach column contains atomic values and there are no repeating groups
  2. BThere are no foreign keys
  3. CEvery table has exactly one column
  4. DEvery non-key column depends on the whole key and on nothing but the key
16.Databases & SQLHard

Two doctors on call each check that the other is still on call, then both sign off at the same time. Which anomaly can snapshot isolation still allow here?

  1. ANon-repeatable read within one transaction
  2. BWrite skew
  3. CDirty write
  4. DDirty read
17.Computer networksMedium

Using even parity, which parity bit is added to the data bits 1011001?

  1. A2
  2. B1
  3. C4
  4. D0
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. AThe default gateway
  2. BName resolution (DNS)
  3. CThe network card driver
  4. DThe physical cable
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.30.0/24 is divided into /27 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 6?

192.168.30.0/24/27subnet 6 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet 49 … 53012806413211608040211commonvaries

A router has routes to these networks: 10.200.49.0/24, 10.200.50.0/24, 10.200.51.0/24, 10.200.52.0/24, 10.200.53.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.200.49.0/24 … 10.200.53.0/24
p =

3Solutions

1.Number systems

Answer: N = 55

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.

32 + 16 + 4 + 2 + 1 = 55

Answer

N = 55
2.Number systems

Answer: R = 32

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.

0100 0000 >> 1
= 0010 0000

Value of the result

0010 0000₂ = 32

Answer

R = 32
3.Propositional logic & proofs

Answer: ¬p ∨ ¬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

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(p ∧ q) ≡ ¬p ∨ ¬q

Check with the truth table

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

Answer

¬p ∨ ¬q

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

4.Propositional logic & proofs

Answer: N = 13

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

¬r; ¬r ∧ q; ¬s; ¬s ∧ q; (¬r ∧ q) ∨ (¬s ∧ q); s ∨ p; p ∧ (s ∨ p); ((¬r ∧ q) ∨ (¬s ∧ q)) → (p ∧ (s ∨ p))

Split on p

p = 1: 8
p = 0: 5
N = 8 + 5 = 13

Answer

N = 13
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

ABC00011110010011030214051706

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

A′B′C
AB′C′
ABC

Minimal SOP form

F = A′B′C + AB′C′ + ABC
N = 3

Answer

N = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 6

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.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 0 → Q = 0
2: J = 0, K = 0 → Q = 0
3: J = 0, K = 0 → Q = 0
4: J = 1, K = 0 → Q = 1
5: J = 1, K = 0 → Q = 1
6: J = 0, K = 1 → Q = 0
7: J = 0, K = 1 → Q = 0
8: J = 1, K = 0 → Q = 1

First change from 1 to 0

clock edge 6

Answer

clock edge 6
7.Number theory & modular arithmetic

Answer: 19−1 mod 47 = 5

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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·19 (mod 47).

r = 47, t = 0
r = 19, q = 2, t = 1
r = 9, q = 2, t = −2
r = 1, q = 9, t = 5

Inverse

1 ≡ 5·19 (mod 47)
x = 5
check: 19·5 = 95 = 2·47 + 1

Answer

19−1 mod 47 = 5
8.Number theory & modular arithmetic

Answer: N = 3

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).

Divide by the gcd

gcd(15, 21) = 3
5x ≡ 4 (mod 7)

Solve the reduced congruence

5−1 ≡ 3 (mod 7) → x ≡ 5 (mod 7)

Solutions modulo 21

x ∈ {5, 12, 19}
N = 3

Answer

N = 3
9.Algorithms & complexity

Answer: D · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

D · O(2ⁿ)
10.Algorithms & complexity

Answer: B · Choosing the pivot uniformly at random

Why this is the answer

With a random pivot, the expected running time is O(n log n) for every input; bad cases still exist but are very unlikely.

Answer

B · Choosing the pivot uniformly at random
11.Data structures

Answer: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
12.Data structures

Answer: A · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

A · A deque holding indices in decreasing order of value
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) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 2, deg(E) = 3

Odd vertices

C, E → 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: d(F) = 24

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: 12, C: 8
C (8): B: 9, E: 21, G: 12
B (9): D: 17
G (12)
D (17): E: 20, F: 24
E (20)
F (24)

Shortest path

A → C → B → D → F
d(F) = 24

Answer

d(F) = 24
15.Databases & SQL

Answer: A · Each column contains atomic values and there are no repeating groups

Why this is the answer

1NF requires atomic values per cell and a unique row identification, without repeating groups.

Answer

A · Each column contains atomic values and there are no repeating groups
16.Databases & SQL

Answer: B · Write skew

Why this is the answer

Under snapshot isolation both transactions see the old state and change different rows, so no write conflict is detected; serializable isolation prevents this.

Answer

B · Write skew
17.Computer networks

Answer: D · 0

Why this is the answer

1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.

Answer

D · 0
18.Computer networks

Answer: B · 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

B · Name resolution (DNS)
19.IPv4 subnetting

Answer: x = 160

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 number 6

network: (6 − 1)·32 = 160
broadcast: 160 + 32 − 1 = 191

Answer

x = 160
20.IPv4 subnetting

Answer: p = 21

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

49 = 00110001
53 = 00110101
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.200.48.0/21 (it also covers 3 network(s) that are not in the list).

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