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 −13 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −138 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1100 0001 OR 1110 1000. What is the result's value R in decimal?

1100 0001OR1110 1000
R =
3.Propositional logic & proofsMedium
¬p → ¬qq → pp ∨ ¬q¬p ∨ q

Which formula is logically equivalent to p → q? Click it.

p → q
  1. 1¬p → ¬q
  2. 2q → p
  3. 3p ∨ ¬q
  4. 4¬p ∨ q
4.Propositional logic & proofsHard
prr → p¬r¬p11100100

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

r → p¬r∴ ¬p
  1. 1p = 1, r = 1
  2. 2p = 1, r = 0
  3. 3p = 0, r = 1
  4. 4p = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010001131214150706

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(2, 3, 4, 5). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(2, 3, 4, 5)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011031204050716012X131151140819X11010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 2, 6, 9, 14, 15) + d(11, 13), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 6, 9, 14, 15) + d(11, 13)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 6 modulo 17: the number x with 0 < x < 17 and 6·x ≡ 1 (mod 17).

a = 6n = 17
6−1 mod 17 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (215, 11)d = ?

An RSA key uses the primes p = 5 and q = 43 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 5q = 43e = 11
d =
9.Algorithms & complexityMedium

A greedy algorithm:

  1. AAlways gives the optimal solution
  2. BTries every possible solution
  3. CMakes the locally best choice at each step
  4. DUses only recursion
10.Algorithms & complexityHard

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

  1. AΘ(n²)
  2. BΘ(n log n)
  3. CΘ(n log² n)
  4. DΘ(n)
11.Data structuresMedium

What balance condition does an AVL tree maintain at every node?

  1. ABoth subtrees of every node contain exactly the same number of nodes
  2. BEvery node is coloured red or black
  3. CThe heights of the left and right subtrees differ by at most 1
  4. DAll leaves are at exactly the same depth
12.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(1)
  2. BO(log n)
  3. CO(n log n)
  4. DO(n)
13.Graph theoryMedium

A tree has 3 vertices of degree 4, 2 vertices of degree 5, 4 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 42 × deg 54 × deg 2
L =
14.Graph theoryHard
910238520176ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 2.

start: Ak = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BD
  4. 4edge DE
  5. 5edge DF
  6. 6edge EF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GE
15.Databases & SQLMedium

Why are indexes created on columns?

  1. ATo make inserts and updates faster on those columns
  2. BTo prevent NULL values
  3. CTo save disk space
  4. DTo speed up searches and joins on those columns
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. BWrite-ahead logging
  3. CStrict two-phase locking with row locks
  4. DDatabase sharding
17.Computer networksMedium

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

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

A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?

  1. AGive the PC a different MAC address so the router can find it from outside
  2. BConfigure port forwarding on the router to the PC's private address
  3. CChange the PC's subnet mask to /32
  4. DTurn off the PC's DNS client
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2412 subnets
H =
20.IPv4 subnettingHard
last octet = 166112806413201608141201networkhost

A host has the address 10.20.67.166/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.67.166/27
x =

3Solutions

1.Number systems

Answer: U = 243

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

13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011

Read it as unsigned

1111 0011₂ = 256 − 13 = 243

Answer

U = 243
2.Number systems

Answer: R = 233

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.

1100 0001
OR 1110 1000
= 1110 1001

Value of the result

1110 1001₂ = 233

Answer

R = 233
3.Propositional logic & proofs

Answer: ¬p ∨ q

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

a → b is 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
p → q ≡ ¬p ∨ q

Check with the truth table

p = 1, q = 1: 1 | 1 1 1 1
p = 1, q = 0: 0 | 1 1 1 0
p = 0, q = 1: 1 | 0 0 0 1
p = 0, q = 0: 1 | 1 1 1 1

Answer

¬p ∨ q

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

4.Propositional logic & proofs

Answer: p = 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

p = 1, r = 1: 1 0 | 0
p = 1, r = 0: 1 1 | 0
p = 0, r = 1: 0 0 | 1
p = 0, r = 0: 1 1 | 1

Counterexample

p = 1, r = 0

Answer

p = 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 = 2

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.

ABC00011110010001131214150706

Group the 0s

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

(A + B)
(A′ + B′)

Minimal POS form

F = (A + B)(A′ + B′)
N = 2

Answer

N = 2
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.

B′C′D, BCD′, AD, A′CD′, A′B′D′, A′B′C′, ABC
ABCD00011110000111101011031204050716012X131151140819X11010

Minimal cover

F = B′C′D + BCD′ + AD + A′B′D′
N = 4

Answer

N = 4

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

7.Number theory & modular arithmetic

Answer: 6−1 mod 17 = 3

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·6 (mod 17).

r = 17, t = 0
r = 6, q = 2, t = 1
r = 5, q = 1, t = −2
r = 1, q = 5, t = 3

Inverse

1 ≡ 3·6 (mod 17)
x = 3
check: 6·3 = 18 = 1·17 + 1

Answer

6−1 mod 17 = 3
8.Number theory & modular arithmetic

Answer: d = 107

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.

Modulus and totient

n = 5·43 = 215
φ(n) = 4·42 = 168

Extended Euclidean algorithm

r = 168, t = 0
r = 11, q = 15, t = 1
r = 3, q = 3, t = −15
r = 2, q = 1, t = 46
r = 1, q = 2, t = −61

Private exponent

d = 107
check: 11·107 = 1177 = 7·168 + 1

Answer

d = 107
9.Algorithms & complexity

Answer: C · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

C · Makes the locally best choice at each step
10.Algorithms & complexity

Answer: C · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

C · Θ(n log² n)
11.Data structures

Answer: C · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

C · The heights of the left and right subtrees differ by at most 1
12.Data structures

Answer: B · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

B · O(log n)
13.Graph theory

Answer: L = 14

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.

Handshake lemma for a tree

L + 30 = 2(L + 9) − 2
L = 14

Answer

L = 14

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: edge BD

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. BD (2)
3. DE (3)
4. EF (5)
5. GE (6)
6. AC (10)

edge number 2

BD

Answer

edge BD
15.Databases & SQL

Answer: D · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

D · To speed up searches and joins on those columns
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 · 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

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

Answer: B · Configure port forwarding on the router to the PC's private address

Why this is the answer

NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.

Answer

B · Configure port forwarding on the router to the PC's private address
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).

Bits to borrow

24 = 16 ≥ 12 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 191

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

25 = 32

Subnet of this host

network: ⌊166/32⌋·32 = 160
broadcast: 160 + 32 − 1 = 191
usable hosts: 161 … 190

Answer

x = 191

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