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 −15 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −158 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 0001 OR 1001 0011. What is the result's value R in decimal?

0101 0001OR1001 0011
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is c → (c ∨ a) a tautology, a contradiction or a contingency? Click your answer.

c → (c ∨ a)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x (P(x) → ¬Q(x))∃x (P(x) → ¬Q(x))∃x (¬P(x) ∧ ¬Q(x))∀x (¬P(x) ∧ ¬Q(x))

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

¬∃x (P(x) ∧ Q(x))
  1. 1∀x (P(x) → ¬Q(x))
  2. 2∃x (P(x) → ¬Q(x))
  3. 3∃x (¬P(x) ∧ ¬Q(x))
  4. 4∀x (¬P(x) ∧ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010312041517060121130150140809011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(2, 5, 7, 10, 13). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(2, 5, 7, 10, 13)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111000011302X40507X61121130151141809111010

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 8, 11, 12, 13, 14) + d(4, 6), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(3, 8, 11, 12, 13, 14) + d(4, 6)
N =
7.Number theory & modular arithmeticMedium

Solve 8x ≡ 30 (mod 31). Give the solution x with 0 < x < 31.

8x ≡ 30 (mod 31)
x =
8.Number theory & modular arithmeticHard
166c = me​ mod n?m = cd​ mod nm(n, e) = (407, 7)d

Encrypt the message m = 166 with the RSA public key (n, e) = (407, 7). Find the ciphertext c.

n = 407e = 7m = 166
c =
9.Algorithms & complexityMedium

How much stack space does a simple recursive factorial function fact(n) = n × fact(n − 1) use?

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

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AAfter V − 1 rounds, it checks whether any edge can still be relaxed
  2. BIt checks whether any edge has a negative weight
  3. CIt runs Dijkstra's algorithm twice and compares the two distance results
  4. DIt counts the cycles with depth-first search
11.Data structuresMedium

Why is traversing an array often much faster in practice than traversing a linked list of the same length?

  1. AArray traversal is O(1) in total, while linked list traversal is O(n)
  2. BArray elements are contiguous, so they make good use of the CPU cache
  3. CLinked lists always store elements on disk
  4. DArrays use fewer instructions to compare values
12.Data structuresHard

Why are B-trees used in databases and file systems?

  1. AThey store data unsorted
  2. BThey need no rebalancing
  3. CThey use less memory than any array, so the entire index always fits in main memory
  4. DTheir high branching factor keeps the tree shallow, reducing slow disk reads
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| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Which statement about HAVING is correct?

  1. AIt can use aggregate functions to filter groups
  2. BIt sorts the results
  3. CIt must come before WHERE in the SELECT statement
  4. DIt cannot be used with GROUP BY
16.Databases & SQLHard

Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?

  1. AA recursive common table expression (WITH RECURSIVE)
  2. BA GROUP BY on manager_id
  3. CA single self-join of the employees table on manager_id
  4. DA UNIQUE constraint on manager_id
17.Computer networksMedium

What is the main difference between the HTTP methods GET and POST?

  1. AGET uses UDP, POST uses TCP
  2. BGET retrieves a resource, POST submits data in the request body
  3. CGET deletes a resource, POST reads it from the server's database
  4. DGET is encrypted, POST is not
18.Computer networksHard

A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?

  1. A50 kB
  2. B500 kB
  3. C2.5 MB
  4. D4 kB
19.IPv4 subnettingMedium
255.255.240.011111111111111111111000000000000network bitshost bits

A subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?

255.255.240.0
H =
20.IPv4 subnettingHard
third octet 107 … 109012816413201618041211commonvaries

A router has routes to these networks: 10.104.107.0/24, 10.104.108.0/24, 10.104.109.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.104.107.0/24 … 10.104.109.0/24
p =

3Solutions

1.Number systems

Answer: U = 241

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

15 = 0000 1111
inverted: 1111 0000
+1 → 1111 0001

Read it as unsigned

1111 0001₂ = 256 − 15 = 241

Answer

U = 241
2.Number systems

Answer: R = 211

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.

0101 0001
OR 1001 0011
= 1101 0011

Value of the result

1101 0011₂ = 211

Answer

R = 211
3.Propositional logic & proofs

Answer: tautology

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

a = 1, c = 1: c ∨ a = 1, c → (c ∨ a) = 1
a = 1, c = 0: c ∨ a = 1, c → (c ∨ a) = 1
a = 0, c = 1: c ∨ a = 1, c → (c ∨ a) = 1
a = 0, c = 0: c ∨ a = 0, c → (c ∨ a) = 1

Classify

c → (c ∨ a): 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: ∀x (P(x) → ¬Q(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 (P(x) ∧ Q(x)) ≡ ∀x (P(x) → ¬Q(x))

Answer

∀x (P(x) → ¬Q(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: L = 9

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.

ABCD000111100001111000010312041517060121130150140809011110

Group the 1s

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

B′CD′
BC′D
A′BD

Minimal SOP form

F = B′CD′ + BC′D + A′BD
L = 9

Answer

L = 9
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

BD′, B′CD, AC′D′, ABC′
ABCD000111100001111000011302X40507X61121130151141809111010

Minimal cover

F = BD′ + B′CD + AC′D′ + ABC′
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 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).

Inverse of 8

r = 31, t = 0
r = 8, q = 3, t = 1
r = 7, q = 1, t = −3
r = 1, q = 7, t = 4
8−1 ≡ 4 (mod 31)

Solve

x ≡ 4·30 = 120 ≡ 27 (mod 31)
check: 8·27 = 216 ≡ 30 (mod 31)

Answer

x = 27
8.Number theory & modular arithmetic

Answer: c = 309

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

7 = 4 + 2 + 1
1661 ≡ 166 (mod 407)
1662 ≡ 287 (mod 407)
1664 ≡ 155 (mod 407)
1667 ≡ 166·287·155 ≡ 309 (mod 407)

Ciphertext

c = 309

Answer

c = 309
9.Algorithms & complexity

Answer: A · O(n)

Why this is the answer

The recursion is n calls deep before any returns, and each call needs a stack frame, so it uses O(n) space.

Answer

A · O(n)
10.Algorithms & complexity

Answer: A · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

A · After V − 1 rounds, it checks whether any edge can still be relaxed
11.Data structures

Answer: B · Array elements are contiguous, so they make good use of the CPU cache

Why this is the answer

Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.

Answer

B · Array elements are contiguous, so they make good use of the CPU cache
12.Data structures

Answer: D · Their high branching factor keeps the tree shallow, reducing slow disk reads

Why this is the answer

B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.

Answer

D · Their high branching factor keeps the tree shallow, reducing slow disk reads
13.Graph theory

Answer: neither

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) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 3

Odd vertices

A, B, D, E → 4
neither

Answer

neither
14.Graph theory

Answer: N = 3

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 0
E: 0 1 1 0 0

Common neighbours

(M²)CC = 1·1 + 1·1 + 0·0 + 0·0 + 1·1 = 3

Answer

N = 3
15.Databases & SQL

Answer: A · It can use aggregate functions to filter groups

Why this is the answer

HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.

Answer

A · It can use aggregate functions to filter groups
16.Databases & SQL

Answer: A · A recursive common table expression (WITH RECURSIVE)

Why this is the answer

A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.

Answer

A · A recursive common table expression (WITH RECURSIVE)
17.Computer networks

Answer: B · GET retrieves a resource, POST submits data in the request body

Why this is the answer

GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.

Answer

B · GET retrieves a resource, POST submits data in the request body
18.Computer networks

Answer: B · 500 kB

Why this is the answer

The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.

Answer

B · 500 kB
19.IPv4 subnetting

Answer: H = 4094

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

255.255.240.0 = /20

Usable hosts

212 − 2 = 4094

Answer

H = 4094
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

107 = 01101011
109 = 01101101
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.104.104.0/21 (it also covers 5 network(s) that are not in the list).

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