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 complement?−128?64?32?16?8?4?2?1

Write −59 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −598 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 1011 AND 1111 1001. What is the result's value R in decimal?

1000 1011AND1111 1001
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ¬(p ∨ ¬p) a tautology, a contradiction or a contingency? Click your answer.

¬(p ∨ ¬p)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∃x ∀y ¬R(x, y)∀x ∃y ¬R(x, y)∀x ∀y ¬R(x, y)∃y ∀x ¬R(x, y)

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

¬∀x ∃y R(x, y)
  1. 1∃x ∀y ¬R(x, y)
  2. 2∀x ∃y ¬R(x, y)
  3. 3∀x ∀y ¬R(x, y)
  4. 4∃y ∀x ¬R(x, y)
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, 2, 7, 9, 10, 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
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

Solve 13x ≡ 9 (mod 23). Give the solution x with 0 < x < 23.

13x ≡ 9 (mod 23)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 15181 mod 28.

a = 15k = 181n = 28
15181 mod 28 =
9.Algorithms & complexityMedium

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

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

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. A2-SAT, where each clause has at most two literals
  2. BCircuit satisfiability
  3. CGeneral SAT on formulas in conjunctive normal form
  4. D3-SAT, where each clause has exactly three literals
11.Data structuresMedium

An adjacency matrix for a graph with V vertices uses how much memory?

  1. AO(V + E)
  2. BO(log V)
  3. CO(E)
  4. DO(V²)
12.Data structuresHard

Which property keeps every B-tree perfectly balanced?

  1. AThe keys are stored only in the root
  2. BEvery node has exactly two children
  3. CAll leaves are at the same depth
  4. DEach node stores one key
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
4145152215159ABCDEF

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

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. ACombine the table with itself using UNION ALL and a WHERE filter
  2. BUse GROUP BY manager_id only
  3. CCreate a second copy of the table on disk
  4. DJoin the employees table with itself using two aliases
16.Databases & SQLHard

An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?

  1. AOptimistic concurrency control
  2. BStrict two-phase locking with row locks
  3. CDatabase sharding
  4. DWrite-ahead logging
17.Computer networksMedium

Which range of TCP/UDP port numbers is reserved as "well-known" ports?

  1. A0–65535
  2. B49152–65535
  3. C0–1023
  4. D1024–49151
18.Computer networksHard

By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?

  1. A2 Mbit/s
  2. B255 Mbit/s
  3. C8 Mbit/s
  4. D1 Mbit/s
19.IPv4 subnettingMedium
255.255.255.24011111111111111111111111111110000network bitshost bits

A subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?

255.255.255.240
H =
20.IPv4 subnettingHard
last octet = 83012816403211608041211networkhost

A host has the address 10.20.32.83/30. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.32.83/30
x =

3Solutions

1.Number systems

Answer: U = 197

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.

Invert and add one

59 = 0011 1011
inverted: 1100 0100
+1 → 1100 0101

Read it as unsigned

1100 0101₂ = 256 − 59 = 197

Answer

U = 197
2.Number systems

Answer: R = 137

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.

1000 1011
AND 1111 1001
= 1000 1001

Value of the result

1000 1001₂ = 137

Answer

R = 137
3.Propositional logic & proofs

Answer: contradiction

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).

Truth table row by row

p = 1: ¬p = 0, p ∨ ¬p = 1, ¬(p ∨ ¬p) = 0
p = 0: ¬p = 1, p ∨ ¬p = 1, ¬(p ∨ ¬p) = 0

Classify

¬(p ∨ ¬p): 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ∃x ∀y ¬R(x, y)

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 ∃y R(x, y) ≡ ∃x ∀y ¬R(x, y)

Answer

∃x ∀y ¬R(x, y)

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: AB = 10, CD = 01

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.

m9: m8, m11, m13, m1 → 0
ABCD0001111000011110111111

The isolated 1

m9 = AB′C′D → AB = 10, CD = 01

Answer

AB = 10, CD = 01
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 4

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

First change from 1 to 0

clock edge 4

Answer

clock edge 4
7.Number theory & modular arithmetic

Answer: x = 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).

Inverse of 13

r = 23, t = 0
r = 13, q = 1, t = 1
r = 10, q = 1, t = −1
r = 3, q = 3, t = 2
r = 1, q = 3, t = −7
13−1 ≡ 16 (mod 23)

Solve

x ≡ 16·9 = 144 ≡ 6 (mod 23)
check: 13·6 = 78 ≡ 9 (mod 23)

Answer

x = 6
8.Number theory & modular arithmetic

Answer: 15181 mod 28 = 15

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.

Binary exponent

181 = 101101012 = 128 + 32 + 16 + 4 + 1

Repeated squaring

151 ≡ 15 (mod 28)
152 ≡ 1 (mod 28)
154 ≡ 1 (mod 28)
158 ≡ 1 (mod 28)
1516 ≡ 1 (mod 28)
1532 ≡ 1 (mod 28)
1564 ≡ 1 (mod 28)
15128 ≡ 1 (mod 28)

Multiply the needed powers

15181 ≡ 15·1·1·1·1 ≡ 15 (mod 28)

Answer

15181 mod 28 = 15
9.Algorithms & complexity

Answer: C · 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

C · O(1)
10.Algorithms & complexity

Answer: A · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

A · 2-SAT, where each clause has at most two literals
11.Data structures

Answer: D · O(V²)

Why this is the answer

An adjacency matrix stores V × V entries, efficient for dense graphs; adjacency lists use O(V + E), better for sparse graphs.

Answer

D · O(V²)
12.Data structures

Answer: C · All leaves are at the same depth

Why this is the answer

Nodes split upwards when full and the tree only gets taller when the root splits, so all leaves always stay on the same level.

Answer

C · All leaves are at the same depth
13.Graph theory

Answer: Euler 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) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(F) = 26

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: 4, C: 14
B (4): C: 9, D: 19, E: 13
C (9): E: 11
E (11): D: 13, F: 26
D (13)
F (26)

Shortest path

A → B → C → E → F
d(F) = 26

Answer

d(F) = 26
15.Databases & SQL

Answer: D · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

D · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: A · Optimistic concurrency control

Why this is the answer

Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.

Answer

A · Optimistic concurrency control
17.Computer networks

Answer: C · 0–1023

Why this is the answer

Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.

Answer

C · 0–1023
18.Computer networks

Answer: C · 8 Mbit/s

Why this is the answer

C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.

Answer

C · 8 Mbit/s
19.IPv4 subnetting

Answer: H = 14

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).

Prefix length

255.255.255.240 = /28

Usable hosts

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 83

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

22 = 4

Subnet of this host

network: ⌊83/4⌋·4 = 80
broadcast: 80 + 4 − 1 = 83
usable hosts: 81 … 82

Answer

x = 83

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