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
10-bit two's complement, N = ??sign?????????

What is the most negative integer N that can be stored in 10-bit two's complement?

10 bit
N =
3.Propositional logic & proofsMedium
¬c → ¬b¬b → ¬c¬b ∧ ¬cb → c

Which formula is logically equivalent to c → b? Click it.

c → b
  1. 1¬c → ¬b
  2. 2¬b → ¬c
  3. 3¬b ∧ ¬c
  4. 4b → c
4.Propositional logic & proofsHard
∃x (F(x) → ¬G(x))∀x (¬F(x) ∧ ¬G(x))∀x (F(x) → ¬G(x))∃x (¬F(x) ∧ ¬G(x))

Which statement is equivalent to ¬∃x (F(x) ∧ G(x)), with the negation moved all the way inside? Click it.

¬∃x (F(x) ∧ G(x))
  1. 1∃x (F(x) → ¬G(x))
  2. 2∀x (¬F(x) ∧ ¬G(x))
  3. 3∀x (F(x) → ¬G(x))
  4. 4∃x (¬F(x) ∧ ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010302141507160121130150140819011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(4, 5, 6, 9, 13). How many product terms N does it have?

F(A, B, C, D) = Σm(4, 5, 6, 9, 13)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1001. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 9 clock edges.

Q₃Q₂Q₁Q₀ = 10019
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 29 modulo 73: the number x with 0 < x < 73 and 29·x ≡ 1 (mod 73).

a = 29n = 73
29−1 mod 73 =
8.Number theory & modular arithmeticHard

23 is prime. Use Fermat’s little theorem to find 1863852 mod 23.

p = 23k = 63852
1863852 mod 23 =
9.Algorithms & complexityMedium

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

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

The halting problem shows that:

  1. ALoops should be avoided
  2. BEvery program eventually halts
  3. CSome programs run faster on faster computers, so halting depends on hardware
  4. DNo algorithm can decide for every program and input whether it halts
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 / 2⌋ + 1
  2. BAt index i − 1
  3. CAt index 2i + 1
  4. DAt index ⌊(i − 1) / 2⌋
12.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(1) time and O(1) extra space
  2. BO(n log n) time and O(1) extra space
  3. CO(n) time and O(n) extra space
  4. DO(n) time and O(1) extra space
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
111291110105ABCDEF

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

What does SELECT SUM(amount) FROM orders return when the WHERE clause matches no rows?

  1. ANULL
  2. B0
  3. CAn empty string
  4. DAn error
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

Which IPv4 header field stops packets from circulating forever in a routing loop?

  1. AType of service
  2. BIdentification
  3. CHeader checksum
  4. DTime to live (TTL)
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIPv6 encrypts all addresses
  2. BIPv6 does not use addresses
  3. CIts 128-bit address space is large enough to give every device a global address
  4. DIPv6 only works on local networks, so addresses never need to be shared globally
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 1019 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1019 hosts
p =
20.IPv4 subnettingHard

The network 192.168.160.0/24 is split with VLSM for these LANs: A: 43 hosts, B: 6 hosts, C: 19 hosts, D: 26 hosts, E: 14 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?

A: 43B: 6C: 19D: 26E: 14
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: N = −512

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.

Most negative value

−29 = −512

Answer

N = −512
3.Propositional logic & proofs

Answer: ¬b → ¬c

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

An implication is equivalent to its contrapositive ¬b → ¬a, not to its converse b → a or its inverse ¬a → ¬b.
c → b ≡ ¬b → ¬c

Check with the truth table

b = 1, c = 1: 1 | 1 1 0 1
b = 1, c = 0: 1 | 0 1 0 0
b = 0, c = 1: 0 | 1 0 0 1
b = 0, c = 0: 1 | 1 1 1 1

Answer

¬b → ¬c

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

4.Propositional logic & proofs

Answer: ∀x (F(x) → ¬G(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 (F(x) ∧ G(x)) ≡ ∀x (F(x) → ¬G(x))

Answer

∀x (F(x) → ¬G(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.

ABCD000111100001111000010302141507160121130150140819011010

Group the 1s

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

BC′D
AC′D
A′BD′

Minimal SOP form

F = BC′D + AC′D + A′BD′
N = 3

Answer

N = 3

Another grouping of the same size exists; it has the same number of terms and literals.

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

1001 → 1100 → 0110 → 0011 → 1001 → 1100 → 0110 → 0011 → 1001 → 1100

Decimal value

1100₂ = 12

Answer

Q = 12
7.Number theory & modular arithmetic

Answer: 29−1 mod 73 = 68

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·29 (mod 73).

r = 73, t = 0
r = 29, q = 2, t = 1
r = 15, q = 1, t = −2
r = 14, q = 1, t = 3
r = 1, q = 14, t = −5

Inverse

1 ≡ −5·29 (mod 73)
x = 68
check: 29·68 = 1972 = 27·73 + 1

Answer

29−1 mod 73 = 68
8.Number theory & modular arithmetic

Answer: 1863852 mod 23 = 16

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.

Reduce the exponent

63852 mod 22 = 8
1863852 ≡ 188 (mod 23)

Small power

188 mod 23 = 16

Answer

1863852 mod 23 = 16
9.Algorithms & complexity

Answer: B · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

B · O(log n)
10.Algorithms & complexity

Answer: D · 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

D · No algorithm can decide for every program and input whether it halts
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: D · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

D · O(n) time and O(1) extra space
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: d(F) = 16

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: 1, C: 1
B (1): D: 10, E: 6
C (1)
E (6): F: 16
D (10)
F (16)

Shortest path

A → B → E → F
d(F) = 16

Answer

d(F) = 16
15.Databases & SQL

Answer: A · NULL

Why this is the answer

SUM, AVG, MIN and MAX return NULL on an empty set, while COUNT returns 0; COALESCE(SUM(amount), 0) gives 0 instead.

Answer

A · NULL
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: D · Time to live (TTL)

Why this is the answer

Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.

Answer

D · Time to live (TTL)
18.Computer networks

Answer: C · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

C · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 1019
210 − 2 = 1022 ≥ 1019 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 96

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

A: 43 hosts → 26 = 64 (/26)
D: 26 hosts → 25 = 32 (/27)
C: 19 hosts → 25 = 32 (/27)
E: 14 hosts → 24 = 16 (/28)
B: 6 hosts → 23 = 8 (/29)

Allocation, largest first

A: .0 – .63
D: .64 – .95
C: .96 – .127
E: .128 – .143
B: .144 – .151

Answer

x = 96

LAN C: 192.168.160.96/27.

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