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 complement1−12816413211608040201

The 8-bit pattern 1111 0000 stores a signed integer in two's complement. What is its value N in decimal?

1111 00008 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 0111 AND 1110 1101. What is the result's value R in decimal?

1011 0111AND1110 1101
R =
3.Propositional logic & proofsMedium
pqrp → q(p → q) → r(r → q) ∧ ((p → q) → r)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (r → q) ∧ ((p → q) → r) true? Call this number N.

(r → q) ∧ ((p → q) → r)
N =
4.Propositional logic & proofsHard
c ∧ ac ∨ ab ∨ a¬c

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

b ∨ c¬b ∨ a
  1. 1c ∧ a
  2. 2c ∨ a
  3. 3b ∨ a
  4. 4¬c
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(0, 5, 7, 11, 14, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-10

A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 0. Which count Q (in decimal) does it show after 19 clock pulses?

Q0 = 019N = 10
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 20 modulo 31: the number x with 0 < x < 31 and 20·x ≡ 1 (mod 31).

a = 20n = 31
20−1 mod 31 =
8.Number theory & modular arithmeticHard
mc = me​ mod n26m = cd​ mod n?(n, e) = (95, 11)d

An RSA key uses p = 5, q = 19 and e = 11. Decrypt the ciphertext c = 26: find the message m.

p = 5q = 19e = 11c = 26
m =
9.Algorithms & complexityMedium

What is the amortised time complexity of appending to a dynamic array that doubles its capacity when full?

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

Which statement about selecting the k-th smallest element of an unsorted array is correct?

  1. ASelection is only possible in O(n²)
  2. BQuickselect runs in O(log n) worst case, because it recurses on only one side of the pivot
  3. CSelection always needs Ω(n log n) because it requires sorting
  4. DQuickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
11.Data structuresMedium

What is true of the adjacency matrix of an undirected graph?

  1. AIt is always upper triangular
  2. BIt is symmetric about the main diagonal
  3. CIts main diagonal always contains ones
  4. DEach row has exactly one 1
12.Data structuresHard

A segment tree is typically used to:

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

A connected graph has 20 vertices and 31 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 31
k =
14.Graph theoryHard
ABCDE

How many walks of length 3 go from A to D? A walk may repeat vertices and edges. Call the number N.

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

First normal form (1NF) requires that:

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

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

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

At which OSI layer does a switch normally operate?

  1. AData link (layer 2)
  2. BTransport (layer 4)
  3. CNetwork (layer 3)
  4. DPhysical (layer 1)
18.Computer networksHard

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

  1. AThe number of open ports
  2. BThe size of each IP packet header
  3. CThe retransmission time-out, calculated from the measured round-trip time of each segment
  4. DHow much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnettingMedium
/?11111111111111111111111100000000network bitshost bits

A department needs 221 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

221 hosts
p =
20.IPv4 subnettingHard
third octet 16 … 23012806403211608040201commonvaries

A router has routes to these networks: 10.158.16.0/24, 10.158.17.0/24, 10.158.18.0/24, 10.158.19.0/24, 10.158.20.0/24, 10.158.21.0/24, 10.158.22.0/24, 10.158.23.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.158.16.0/24 … 10.158.23.0/24
p =

3Solutions

1.Number systems

Answer: N = −16

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 + 64 + 32 + 16 = −16

Answer

N = −16
2.Number systems

Answer: R = 165

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.

1011 0111
AND 1110 1101
= 1010 0101

Value of the result

1010 0101₂ = 165

Answer

R = 165
3.Propositional logic & proofs

Answer: N = 3

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.

Truth table row by row

p = 1, q = 1, r = 1: r → q = 1, p → q = 1, (p → q) → r = 1, (r → q) ∧ ((p → q) → r) = 1
p = 1, q = 1, r = 0: r → q = 1, p → q = 1, (p → q) → r = 0, (r → q) ∧ ((p → q) → r) = 0
p = 1, q = 0, r = 1: r → q = 0, p → q = 0, (p → q) → r = 1, (r → q) ∧ ((p → q) → r) = 0
p = 1, q = 0, r = 0: r → q = 1, p → q = 0, (p → q) → r = 1, (r → q) ∧ ((p → q) → r) = 1
p = 0, q = 1, r = 1: r → q = 1, p → q = 1, (p → q) → r = 1, (r → q) ∧ ((p → q) → r) = 1
p = 0, q = 1, r = 0: r → q = 1, p → q = 1, (p → q) → r = 0, (r → q) ∧ ((p → q) → r) = 0
p = 0, q = 0, r = 1: r → q = 0, p → q = 1, (p → q) → r = 1, (r → q) ∧ ((p → q) → r) = 0
p = 0, q = 0, r = 0: r → q = 1, p → q = 1, (p → q) → r = 0, (r → q) ∧ ((p → q) → r) = 0

Count the rows with value 1

(r → q) ∧ ((p → q) → r): 1, 0, 0, 1, 1, 0, 0, 0
N = 3

Answer

N = 3
4.Propositional logic & proofs

Answer: c ∨ a

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

Conclusion

c ∨ a (resolution)

Answer

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

Answer: AB = 00, CD = 00

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m0: m1, m2, m4, m8 → 0
ABCD0001111000011110111111

The isolated 1

m0 = A′B′C′D′ → AB = 00, CD = 00

Answer

AB = 00, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 9

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 10 pulses bring the counter back to the same state.

19 = 1·10 + 9

Count after the remaining pulses

Q = (0 + 9) mod 10 = 9
Q = 1001₂

Answer

Q = 9
7.Number theory & modular arithmetic

Answer: 20−1 mod 31 = 14

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·20 (mod 31).

r = 31, t = 0
r = 20, q = 1, t = 1
r = 11, q = 1, t = −1
r = 9, q = 1, t = 2
r = 2, q = 4, t = −3
r = 1, q = 2, t = 14

Inverse

1 ≡ 14·20 (mod 31)
x = 14
check: 20·14 = 280 = 9·31 + 1

Answer

20−1 mod 31 = 14
8.Number theory & modular arithmetic

Answer: m = 11

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.

Private exponent

n = 95, φ(n) = 72
d = 11−1 mod 72 = 59

Decrypt by repeated squaring

59 = 32 + 16 + 8 + 2 + 1
261 ≡ 26 (mod 95)
262 ≡ 11 (mod 95)
264 ≡ 26 (mod 95)
268 ≡ 11 (mod 95)
2616 ≡ 26 (mod 95)
2632 ≡ 11 (mod 95)
2659 ≡ 26·11·11·26·11 ≡ 11 (mod 95)

Message

m = 11
check: 1111 mod 95 = 26

Answer

m = 11
9.Algorithms & complexity

Answer: A · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

A · O(1)
10.Algorithms & complexity

Answer: D · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case

Why this is the answer

Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.

Answer

D · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
11.Data structures

Answer: B · It is symmetric about the main diagonal

Why this is the answer

In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.

Answer

B · It is symmetric about the main diagonal
12.Data structures

Answer: B · 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

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

Answer: k = 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.

Edges of a spanning tree

20 − 1 = 19

Edges to remove

k = 31 − 19 = 12

Answer

k = 12
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 0
D: 0 1 0 0 1
E: 0 1 0 1 0

One more step

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

Answer

N = 2
15.Databases & SQL

Answer: C · 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

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

Answer: B · 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

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

Answer: A · 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

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

Answer: D · 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

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

Answer: p = 24

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).

Host bits needed

27 − 2 = 126 < 221
28 − 2 = 254 ≥ 221 → h = 8

Prefix length

p = 32 − 8 = 24

Answer

p = /24
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

16 = 00010000
23 = 00010111
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.158.16.0/21

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