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 −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −258 bit
U =
2.Number systemsHard
20 bits, unsigned, N = ?????????????????????

What is the largest unsigned integer N that fits in 20 bits?

20 bit
N =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (q ↔ r) ∧ (q ↔ ¬r) a tautology, a contradiction or a contingency? Click your answer.

(q ↔ r) ∧ (q ↔ ¬r)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
qrr → q¬r¬q11100100

The argument with premises r → q; ¬r and conclusion ¬q is invalid. Click the row of the truth table that is a counterexample.

r → q¬r∴ ¬q
  1. 1q = 1, r = 1
  2. 2q = 1, r = 0
  3. 3q = 0, r = 1
  4. 4q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011011031204151706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 2, 5, 7). How many product terms N does it have?

F(A, B, C) = Σm(0, 1, 2, 5, 7)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKT123456

A positive-edge-triggered T 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 = 0T
  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.Number theory & modular arithmeticMedium

Find the inverse of 10 modulo 17: the number x with 0 < x < 17 and 10·x ≡ 1 (mod 17).

a = 10n = 17
10−1 mod 17 =
8.Number theory & modular arithmeticHard
mc = me​ mod n216m = cd​ mod n?(n, e) = (703, 23)d

An RSA key uses p = 19, q = 37 and e = 23. Decrypt the ciphertext c = 216: find the message m.

p = 19q = 37e = 23c = 216
m =
9.Algorithms & complexityMedium

How many levels does the recursion of merge sort have for an array of n elements?

  1. AAbout log₂ n
  2. BAbout √n
  3. CAbout n
  4. DAbout n / 2
10.Algorithms & complexityHard

To show that a new problem X is NP-hard, what must you do?

  1. AReduce a known NP-hard problem to X in polynomial time
  2. BShow that X can be verified in polynomial time
  3. CReduce X to a known NP-hard problem in polynomial time
  4. DShow that X has no polynomial algorithm for small inputs
11.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AAll keys in a heap are equal
  2. BIt is a complete binary tree, so positions follow simple index formulas
  3. CHeaps never contain more than 100 elements
  4. DArrays automatically keep the keys in sorted order, so no pointers are needed
12.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA stack of all visited edges
  2. BA max-heap ordered by out-degree
  3. CA queue (or any collection) of vertices whose in-degree is 0
  4. DAn adjacency matrix recording which vertices have already been removed
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
91815319281ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 4.

start: Ak = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE
15.Databases & SQLMedium

Third normal form (3NF) removes:

  1. ARedundant indexes on columns that are not part of a key
  2. BAtomic values
  3. CTransitive dependencies between non-key attributes
  4. DComposite keys
16.Databases & SQLHard

What does EXPLAIN (or EXPLAIN ANALYZE in PostgreSQL) show for a query?

  1. AThe execution plan chosen by the optimiser, such as scans and join methods
  2. BThe list of users who ran the query
  3. CThe meaning of each SQL keyword in the query
  4. DA rewrite of the query in relational algebra notation only
17.Computer networksMedium

What is the usual order of messages when a client obtains an address from DHCP?

  1. ADiscover, Request, Offer, Acknowledge
  2. BOffer, Discover, Acknowledge, Request
  3. CRequest, Offer, Discover, Acknowledge
  4. DDiscover, Offer, Request, Acknowledge
18.Computer networksHard

What does TCP fast retransmit do?

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

The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/243 subnets
H =
20.IPv4 subnettingHard
last octet = 56012806413211618040201networkhost

A host has the address 10.20.252.56/29. What is the fourth octet x of its network address?

10.20.252.56/29
x =

3Solutions

1.Number systems

Answer: U = 231

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

25 = 0001 1001
inverted: 1110 0110
+1 → 1110 0111

Read it as unsigned

1110 0111₂ = 256 − 25 = 231

Answer

U = 231
2.Number systems

Answer: N = 1.049 × 10⁶

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.

Largest value

220 − 1 = 1048575

Answer

N = 1048575
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

q = 1, r = 1: q ↔ r = 1, ¬r = 0, q ↔ ¬r = 0, (q ↔ r) ∧ (q ↔ ¬r) = 0
q = 1, r = 0: q ↔ r = 0, ¬r = 1, q ↔ ¬r = 1, (q ↔ r) ∧ (q ↔ ¬r) = 0
q = 0, r = 1: q ↔ r = 0, ¬r = 0, q ↔ ¬r = 1, (q ↔ r) ∧ (q ↔ ¬r) = 0
q = 0, r = 0: q ↔ r = 1, ¬r = 1, q ↔ ¬r = 0, (q ↔ r) ∧ (q ↔ ¬r) = 0

Classify

(q ↔ r) ∧ (q ↔ ¬r): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: q = 1, r = 0

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.

Premises and conclusion row by row

q = 1, r = 1: 1 0 | 0
q = 1, r = 0: 1 1 | 0
q = 0, r = 1: 0 0 | 1
q = 0, r = 0: 1 1 | 1

Counterexample

q = 1, r = 0

Answer

q = 1, r = 0

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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.

ABC00011110011011031204151706

Group the 1s

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

B′C
A′C′
AC

Minimal SOP form

F = B′C + A′C′ + AC
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: clock edge 3

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 T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

1: T = 1 → Q = 1
2: T = 0 → Q = 1
3: T = 1 → Q = 0
4: T = 0 → Q = 0
5: T = 0 → Q = 0
6: T = 0 → Q = 0

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: 10−1 mod 17 = 12

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·10 (mod 17).

r = 17, t = 0
r = 10, q = 1, t = 1
r = 7, q = 1, t = −1
r = 3, q = 2, t = 2
r = 1, q = 3, t = −5

Inverse

1 ≡ −5·10 (mod 17)
x = 12
check: 10·12 = 120 = 7·17 + 1

Answer

10−1 mod 17 = 12
8.Number theory & modular arithmetic

Answer: m = 524

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 = 703, φ(n) = 648
d = 23−1 mod 648 = 479

Decrypt by repeated squaring

479 = 256 + 128 + 64 + 16 + 8 + 4 + 2 + 1
2161 ≡ 216 (mod 703)
2162 ≡ 258 (mod 703)
2164 ≡ 482 (mod 703)
2168 ≡ 334 (mod 703)
21616 ≡ 482 (mod 703)
21632 ≡ 334 (mod 703)
21664 ≡ 482 (mod 703)
216128 ≡ 334 (mod 703)
216256 ≡ 482 (mod 703)
216479 ≡ 216·258·482·334·482·482·334·482 ≡ 524 (mod 703)

Message

m = 524
check: 52423 mod 703 = 216

Answer

m = 524
9.Algorithms & complexity

Answer: A · About log₂ n

Why this is the answer

Halving from n down to 1 takes about log₂ n levels, and each level does O(n) merging work, giving O(n log n).

Answer

A · About log₂ n
10.Algorithms & complexity

Answer: A · Reduce a known NP-hard problem to X in polynomial time

Why this is the answer

If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.

Answer

A · Reduce a known NP-hard problem to X in polynomial time
11.Data structures

Answer: B · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

B · It is a complete binary tree, so positions follow simple index formulas
12.Data structures

Answer: C · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

C · A queue (or any collection) of vertices whose in-degree is 0
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: edge DF

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.

Prim’s algorithm

1. AB (9)
2. BE (1)
3. BD (3)
4. DF (2)
5. BC (15)

edge number 4

DF

Answer

edge DF
15.Databases & SQL

Answer: C · Transitive dependencies between non-key attributes

Why this is the answer

In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.

Answer

C · Transitive dependencies between non-key attributes
16.Databases & SQL

Answer: A · The execution plan chosen by the optimiser, such as scans and join methods

Why this is the answer

EXPLAIN displays the plan (index or sequential scans, join algorithms, estimated costs); EXPLAIN ANALYZE also runs the query and reports actual times.

Answer

A · The execution plan chosen by the optimiser, such as scans and join methods
17.Computer networks

Answer: D · Discover, Offer, Request, Acknowledge

Why this is the answer

The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.

Answer

D · Discover, Offer, Request, Acknowledge
18.Computer networks

Answer: D · 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

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

Answer: H = 62

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).

Bits to borrow

22 = 4 ≥ 3 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: x = 56

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

23 = 8

Subnet of this host

network: ⌊56/8⌋·8 = 56
broadcast: 56 + 8 − 1 = 63
usable hosts: 57 … 62

Answer

x = 56

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