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 −32 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −328 bit
U =
2.Number systemsHard
3359 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3359 items its own binary code of the same length. What is the smallest number of bits b needed?

3359
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is b ↔ ¬b a tautology, a contradiction or a contingency? Click your answer.

b ↔ ¬b
  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)∀x ∀y ¬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∀x ∀y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = AB′C′ + A′B′ + A′B′C to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = AB′C′ + A′B′ + A′B′C
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101001031204X507X60121130151140809X11110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 2, 10, 13, 14) + d(5, 6, 11), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 2, 10, 13, 14) + d(5, 6, 11)
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(289), the number of integers from 1 to 289 that are coprime to 289.

n = 289 = 172
φ(289) =
8.Number theory & modular arithmeticHard
mc = me​ mod n615m = cd​ mod n?(n, e) = (713, 23)d

An RSA key uses p = 23, q = 31 and e = 23. Decrypt the ciphertext c = 615: find the message m.

p = 23q = 31e = 23c = 615
m =
9.Algorithms & complexityMedium

Why does dynamic programming give no real speed-up for merge sort?

  1. AMerge sort is already O(n)
  2. BDynamic programming only works on graph problems, not on arrays
  3. CMerge sort has no recursive structure
  4. DIts subproblems do not overlap, so there is nothing to reuse
10.Algorithms & complexityHard

Karatsuba multiplication satisfies T(n) = 3T(n/2) + O(n). What does this solve to?

  1. AΘ(n3)
  2. BΘ(n²)
  3. CΘ(n log n)
  4. DΘ(nlog₂ 3), about n1.585
11.Data structuresMedium

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA binary search tree
  2. BA queue
  3. CA hash table
  4. DA stack
12.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(log n) exactly
  2. BO(n²)
  3. CO(α(n)), the inverse Ackermann function, which is practically constant
  4. DO(n)
13.Graph theoryMedium

A connected graph has 15 vertices and 15 edges. How many edges k must be removed to leave a spanning tree?

|V| = 15|E| = 15
k =
14.Graph theoryHard
5121761423161398ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 3.

start: Bk = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge EF
  8. 8edge GB
  9. 9edge GC
  10. 10edge GD
  11. 11edge GE
15.Databases & SQLMedium

What does the "A" in ACID stand for?

  1. AAtomicity
  2. BAccuracy
  3. CAvailability
  4. DAggregation
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AThe Cartesian product, pairing each row of one table with every row of the other
  2. BAll rows from both tables, matched where possible and NULL-filled otherwise
  3. COnly matching rows
  4. DOnly unmatched rows
17.Computer networksMedium

What is the IPv6 loopback address?

  1. Aff02::1
  2. Bfe80::1
  3. C::
  4. D::1
18.Computer networksHard

Which of these addresses is a public (not RFC 1918 private) IPv4 address?

  1. A10.255.0.1
  2. B172.32.0.1
  3. C172.31.0.1
  4. D192.168.255.1
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1299 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1299 hosts
p =
20.IPv4 subnettingHard
last octet = 149112806403211608140211networkhost

A host has the address 10.20.48.149/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.48.149/28
x =

3Solutions

1.Number systems

Answer: U = 224

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

32 = 0010 0000
inverted: 1101 1111
+1 → 1110 0000

Read it as unsigned

1110 0000₂ = 256 − 32 = 224

Answer

U = 224
2.Number systems

Answer: b = 12 bit

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.

Powers of two

211 = 2048 < 3359
212 = 4096 ≥ 3359

Answer

b = 12
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

b = 1: ¬b = 0, b ↔ ¬b = 0
b = 0: ¬b = 1, b ↔ ¬b = 0

Classify

b ↔ ¬b: 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: L = 4

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

AB′C′ → m(4)
A′B′ → m(0, 1)
A′B′C → m(1)
ABC00011110011011030214050706

Group and read off

F = B′C′ + A′B′

Count the literals

L = 4

Answer

L = 4
6.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.

Prime implicants

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

CD′, BC′D, A′B′D′, AB′C
ABCD00011110000111101001031204X507X60121130151140809X11110

Minimal cover

F = CD′ + BC′D + A′B′D′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: φ(289) = 272

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).

Prime factors

289 = 172

Totient

φ(289) = 289 − 17 = 272

Answer

φ(289) = 272
8.Number theory & modular arithmetic

Answer: m = 316

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 = 713, φ(n) = 660
d = 23−1 mod 660 = 287

Decrypt by repeated squaring

287 = 256 + 16 + 8 + 4 + 2 + 1
6151 ≡ 615 (mod 713)
6152 ≡ 335 (mod 713)
6154 ≡ 284 (mod 713)
6158 ≡ 87 (mod 713)
61516 ≡ 439 (mod 713)
61532 ≡ 211 (mod 713)
61564 ≡ 315 (mod 713)
615128 ≡ 118 (mod 713)
615256 ≡ 377 (mod 713)
615287 ≡ 615·335·284·87·439·377 ≡ 316 (mod 713)

Message

m = 316
check: 31623 mod 713 = 615

Answer

m = 316
9.Algorithms & complexity

Answer: D · Its subproblems do not overlap, so there is nothing to reuse

Why this is the answer

Merge sort's halves are disjoint and each is solved once; DP helps only when subproblems recur, as in Fibonacci.

Answer

D · Its subproblems do not overlap, so there is nothing to reuse
10.Algorithms & complexity

Answer: D · Θ(nlog₂ 3), about n1.585

Why this is the answer

Since log₂ 3 ≈ 1.585 > 1, the leaves dominate by the master theorem, beating the O(n²) schoolbook method.

Answer

D · Θ(nlog₂ 3), about n1.585
11.Data structures

Answer: D · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

D · A stack
12.Data structures

Answer: C · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

C · O(α(n)), the inverse Ackermann function, which is practically constant
13.Graph theory

Answer: k = 1

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.

Edges of a spanning tree

15 − 1 = 14

Edges to remove

k = 15 − 14 = 1

Answer

k = 1
14.Graph theory

Answer: edge DE

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 (5)
2. BD (6)
3. DE (2)
4. EF (3)
5. GE (8)
6. AC (12)

edge number 3

DE

Answer

edge DE
15.Databases & SQL

Answer: A · Atomicity

Why this is the answer

ACID: atomicity, consistency, isolation, durability. Atomicity means a transaction is applied completely or not at all.

Answer

A · Atomicity
16.Databases & SQL

Answer: B · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

B · All rows from both tables, matched where possible and NULL-filled otherwise
17.Computer networks

Answer: D · ::1

Why this is the answer

::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.

Answer

D · ::1
18.Computer networks

Answer: B · 172.32.0.1

Why this is the answer

172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.

Answer

B · 172.32.0.1
19.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1299
211 − 2 = 2046 ≥ 1299 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 159

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

24 = 16

Subnet of this host

network: ⌊149/16⌋·16 = 144
broadcast: 144 + 16 − 1 = 159
usable hosts: 145 … 158

Answer

x = 159

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