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 −99 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −998 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 1001 OR 1111 0110. What is the result's value R in decimal?

1011 1001OR1111 0110
R =
3.Propositional logic & proofsMedium
(p → q) ∧ (r → q)(p ∧ r) → qp ∨ (r → q)(p → q) ∨ (r → q)

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

(p ∨ r) → q
  1. 1(p → q) ∧ (r → q)
  2. 2(p ∧ r) → q
  3. 3p ∨ (r → q)
  4. 4(p → q) ∨ (r → q)
4.Propositional logic & proofsHard
∃x A(x)∃x ¬A(x)∀x A(x)∀x ¬A(x)

Which statement is equivalent to ¬∀x A(x), with the negation moved all the way inside? Click it.

¬∀x A(x)
  1. 1∃x A(x)
  2. 2∃x ¬A(x)
  3. 3∀x A(x)
  4. 4∀x ¬A(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000111302140517161121131151140819111110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15)
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 32: the number x with 0 < x < 32 and 19·x ≡ 1 (mod 32).

a = 19n = 32
19−1 mod 32 =
8.Number theory & modular arithmeticHard

How many solutions N does 18x ≡ 12 (mod 30) have in {0, 1, …, 29}?

18x ≡ 12 (mod 30)
N =
9.Algorithms & complexityMedium

Breadth-first search on an unweighted graph finds:

  1. AShortest paths in graphs that contain negative edge weights
  2. BShortest paths in number of edges from the start node
  3. CA topological order in all graphs
  4. DThe minimum spanning tree
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AComparisons take logarithmic time each
  2. BBinary search is the only possible algorithm, and it needs log₂ n comparisons
  3. CA decision tree with n possible outcomes must have depth at least log₂ n
  4. DEvery element must be compared at least once
11.Data structuresMedium

In a binary heap stored in an array starting at index 0, where is the parent of the node at index i?

  1. AAt index i − 1
  2. BAt index 2i + 1
  3. CAt index ⌊i / 2⌋ + 1
  4. DAt index ⌊(i − 1) / 2⌋
12.Data structuresHard

For each element of an array you want the next greater element to its right, in O(n) total time. Which structure helps?

  1. AA stack that keeps elements waiting for their answer
  2. BA queue that holds every pair of elements
  3. CA hash table of all elements
  4. DA binary search tree rebuilt for each element
13.Graph theoryMedium

A tree has 2 vertices of degree 4, 2 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 42 × deg 5
L =
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at A? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

What does the functional dependency A → B mean?

  1. AEach value of B is associated with exactly one value of A
  2. BA and B are both primary keys
  3. CEach value of A is associated with exactly one value of B
  4. DA is a foreign key referencing B
16.Databases & SQLHard

What is a known drawback of decomposing tables into BCNF compared with 3NF?

  1. AThe decomposition always loses data when joined back
  2. BBCNF allows more redundancy than 3NF
  3. CSome functional dependencies may no longer be checkable within a single table
  4. DBCNF tables cannot have primary keys
17.Computer networksMedium

What is the main purpose of a VLAN?

  1. ATo split one physical switched network into separate broadcast domains
  2. BTo give each device a public IP address
  3. CTo increase the speed of every switch port
  4. DTo encrypt all traffic between switches so other departments cannot read it
18.Computer networksHard

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt increases the bandwidth of the users' own home connections
  2. BIt serves cached content from servers close to the users
  3. CIt replaces TCP with a faster transport protocol
  4. DIt removes the need for DNS lookups
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 = 230112816413201608141201networkhost

A host has the address 10.20.16.230/25. What is the fourth octet x of the last usable host address in its subnet?

10.20.16.230/25
x =

3Solutions

1.Number systems

Answer: U = 157

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

99 = 0110 0011
inverted: 1001 1100
+1 → 1001 1101

Read it as unsigned

1001 1101₂ = 256 − 99 = 157

Answer

U = 157
2.Number systems

Answer: R = 255

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.

1011 1001
OR 1111 0110
= 1111 1111

Value of the result

1111 1111₂ = 255

Answer

R = 255
3.Propositional logic & proofs

Answer: (p → q) ∧ (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

(a ∨ b) → c needs c whenever either a or b holds: (a → c) ∧ (b → c).
(p ∨ r) → q ≡ (p → q) ∧ (r → q)

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

Answer

(p → q) ∧ (r → q)

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

4.Propositional logic & proofs

Answer: ∃x ¬A(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 A(x) ≡ ∃x ¬A(x)

Answer

∃x ¬A(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 = 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.

ABCD000111100001111000111302140517161121131151140819111110

Group the 0s

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

(B + C + D)
(A + B + D)
(A + B′ + C + D′)

Minimal POS form

F = (B + C + D)(A + B + D)(A + B′ + C + D′)
N = 3

Answer

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

Answer: clock edge 3

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

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: 19−1 mod 32 = 27

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 32).

r = 32, t = 0
r = 19, q = 1, t = 1
r = 13, q = 1, t = −1
r = 6, q = 2, t = 2
r = 1, q = 6, t = −5

Inverse

1 ≡ −5·19 (mod 32)
x = 27
check: 19·27 = 513 = 16·32 + 1

Answer

19−1 mod 32 = 27
8.Number theory & modular arithmetic

Answer: N = 6

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(18, 30) = 6
3x ≡ 2 (mod 5)

Solve the reduced congruence

3−1 ≡ 2 (mod 5) → x ≡ 4 (mod 5)

Solutions modulo 30

x ∈ {4, 9, 14, 19, 24, 29}
N = 6

Answer

N = 6
9.Algorithms & complexity

Answer: B · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

B · Shortest paths in number of edges from the start node
10.Algorithms & complexity

Answer: C · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

C · A decision tree with n possible outcomes must have depth at least log₂ n
11.Data structures

Answer: D · At index ⌊(i − 1) / 2⌋

Why this is the answer

Children of i are at 2i + 1 and 2i + 2, so the parent of i is ⌊(i − 1) / 2⌋; with 1-based indexing it is ⌊i / 2⌋.

Answer

D · At index ⌊(i − 1) / 2⌋
12.Data structures

Answer: A · A stack that keeps elements waiting for their answer

Why this is the answer

Scanning left to right, pop every stacked element smaller than the current one (the current element is their answer), then push it; each element moves at most twice.

Answer

A · A stack that keeps elements waiting for their answer
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 + 18 = 2(L + 4) − 2
L = 12

Answer

L = 12

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

14.Graph theory

Answer: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)AA = 2·0 + 1·1 + 1·1 + 1·0 + 2·0 = 2

Answer

N = 2
15.Databases & SQL

Answer: C · Each value of A is associated with exactly one value of B

Why this is the answer

A → B means any two rows with the same A value must have the same B value; A determines B.

Answer

C · Each value of A is associated with exactly one value of B
16.Databases & SQL

Answer: C · Some functional dependencies may no longer be checkable within a single table

Why this is the answer

A lossless BCNF decomposition always exists, but it may not preserve dependencies; 3NF can always be both lossless and dependency-preserving.

Answer

C · Some functional dependencies may no longer be checkable within a single table
17.Computer networks

Answer: A · To split one physical switched network into separate broadcast domains

Why this is the answer

VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.

Answer

A · To split one physical switched network into separate broadcast domains
18.Computer networks

Answer: B · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

B · It serves cached content from servers close to the users
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 = 254

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: ⌊230/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 254

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