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 −63 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −638 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1111 1000 AND 1010 0011. What is the result's value R in decimal?

1111 1000AND1010 0011
R =
3.Propositional logic & proofsMedium
¬¬q → ¬(r ∧ p)¬¬q ∧ (r ∧ p)¬¬q ∧ ¬(r ∧ p)¬q ∧ ¬(r ∧ p)

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

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

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

(p → r) ∧ (¬(s ∨ q) → ¬(s → p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010001131214050706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(2, 3, 4). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(2, 3, 4)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKMOD-5

A synchronous mod-5 counter counts 0, 1, …, 4 and then returns to 0. It starts at 4. Which count Q (in decimal) does it show after 12 clock pulses?

Q0 = 412N = 5
Q =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(1573), the number of integers from 1 to 1573 that are coprime to 1573.

n = 1573 = 112·13
φ(1573) =
8.Number theory & modular arithmeticHard
264c = me​ mod n?m = cd​ mod nm(n, e) = (559, 5)d

Encrypt the message m = 264 with the RSA public key (n, e) = (559, 5). Find the ciphertext c.

n = 559e = 5m = 264
c =
9.Algorithms & complexityMedium

Why does dynamic programming give no real speed-up for merge sort?

  1. AIts subproblems do not overlap, so there is nothing to reuse
  2. BMerge sort is already O(n)
  3. CMerge sort has no recursive structure
  4. DDynamic programming only works on graph problems, not on arrays
10.Algorithms & complexityHard

What does the recurrence T(n) = 2T(n/2) + n log n solve to?

  1. AΘ(n log² n)
  2. BΘ(n²)
  3. CΘ(n)
  4. DΘ(n log n)
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(1)
  2. BO(log n)
  3. CO(n)
  4. DO(n²)
12.Data structuresHard

A segment tree is typically used to:

  1. AStore a graph's edges
  2. BSort strings alphabetically by storing each prefix in a separate tree node
  3. CAnswer range queries (such as range sums) and updates in O(log n)
  4. DImplement a stack
13.Graph theoryMedium

A connected graph has 12 vertices and 13 edges. How many edges k must be removed to leave a spanning tree?

|V| = 12|E| = 13
k =
14.Graph theoryHard
1261017132191891614ABCDEFG

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

|V| = 7|E| = 11
W =
15.Databases & SQLMedium

What does a common table expression (WITH name AS (...)) provide?

  1. AAn index on a column
  2. BA permanent table that stays stored on disk until it is dropped
  3. CA new database user
  4. DA named temporary result that can be used within one statement
16.Databases & SQLHard

Why might a query with WHERE YEAR(order_date) = 2025 not use an index on order_date?

  1. AYEAR is not valid SQL
  2. BAn index on order_date is only used for ORDER BY clauses, never for WHERE
  3. CApplying a function to the column prevents a simple index range lookup
  4. DIndexes never work with dates
17.Computer networksMedium

What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?

  1. A/32
  2. B/8
  3. C/24
  4. D/16
18.Computer networksHard

What does TCP fast retransmit do?

  1. ASkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
  2. BResends every segment twice to be safe
  3. CResends a segment after three duplicate ACKs, without waiting for the timeout
  4. DDoubles the window after every loss
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.241.0/24 is divided into /26 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 4?

192.168.241.0/24/26subnet 4 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 184112806413211618040201networkhost

A host has the address 10.20.117.184/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.117.184/25
x =

3Solutions

1.Number systems

Answer: U = 193

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

63 = 0011 1111
inverted: 1100 0000
+1 → 1100 0001

Read it as unsigned

1100 0001₂ = 256 − 63 = 193

Answer

U = 193
2.Number systems

Answer: R = 160

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

AND gives 1 only where both bits are 1.

1111 1000
AND 1010 0011
= 1010 0000

Value of the result

1010 0000₂ = 160

Answer

R = 160
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).

Parts of the implication

a = ¬q, b = r ∧ p

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = ¬q ∧ ¬(r ∧ p)

Answer

¬q ∧ ¬(r ∧ p)

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

4.Propositional logic & proofs

Answer: N = 9

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

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

Split on p

p = 1: 3
p = 0: 6
N = 3 + 6 = 9

Answer

N = 9
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 5

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.

ABC00011110010001131214050706

Group the 1s

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

A′B
AB′C′

Minimal SOP form

F = A′B + AB′C′
L = 5

Answer

L = 5
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 1

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.

Full cycles

Every 5 pulses bring the counter back to the same state.

12 = 2·5 + 2

Count after the remaining pulses

Q = (4 + 2) mod 5 = 1
Q = 001₂

Answer

Q = 1
7.Number theory & modular arithmetic

Answer: φ(1573) = 1320

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

1573 = 112·13

Totient

φ(1573) = φ(121)·φ(13) = 110·12 = 1320

Answer

φ(1573) = 1320
8.Number theory & modular arithmetic

Answer: c = 36

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
2641 ≡ 264 (mod 559)
2642 ≡ 380 (mod 559)
2644 ≡ 178 (mod 559)
2645 ≡ 264·178 ≡ 36 (mod 559)

Ciphertext

c = 36

Answer

c = 36
9.Algorithms & complexity

Answer: A · Its subproblems do not overlap, so there is nothing to reuse

Why this is the answer

Merge sort's halves are disjoint and each is solved once; DP helps only when subproblems recur, as in Fibonacci.

Answer

A · Its subproblems do not overlap, so there is nothing to reuse
10.Algorithms & complexity

Answer: A · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

A · Θ(n log² n)
11.Data structures

Answer: A · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

A · O(1)
12.Data structures

Answer: C · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

C · Answer range queries (such as range sums) and updates in O(log n)
13.Graph theory

Answer: k = 2

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.

Edges of a spanning tree

12 − 1 = 11

Edges to remove

k = 13 − 11 = 2

Answer

k = 2
14.Graph theory

Answer: W = 54

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

DF (2): take
AC (6): take
GC (9): take
BC (10): take
AB (12): skip, closes a cycle
DE (13): take
GE (14): take

Total weight

W = 2 + 6 + 9 + 10 + 13 + 14 = 54

Answer

W = 54
15.Databases & SQL

Answer: D · A named temporary result that can be used within one statement

Why this is the answer

A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.

Answer

D · A named temporary result that can be used within one statement
16.Databases & SQL

Answer: C · Applying a function to the column prevents a simple index range lookup

Why this is the answer

The expression is not "sargable"; using order_date >= '2025-01-01' AND order_date < '2026-01-01' allows an index range scan.

Answer

C · Applying a function to the column prevents a simple index range lookup
17.Computer networks

Answer: C · /24

Why this is the answer

Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.

Answer

C · /24
18.Computer networks

Answer: C · Resends a segment after three duplicate ACKs, without waiting for the timeout

Why this is the answer

Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.

Answer

C · Resends a segment after three duplicate ACKs, without waiting for the timeout
19.IPv4 subnetting

Answer: x = 192

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

26 = 64

Subnet number 4

network: (4 − 1)·64 = 192
broadcast: 192 + 64 − 1 = 255

Answer

x = 192
20.IPv4 subnetting

Answer: x = 255

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: ⌊184/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 255

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