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 −77 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −778 bit
U =
2.Number systemsHard
10 bits, unsigned, N = ???????????

What is the largest unsigned integer N that fits in 10 bits?

10 bit
N =
3.Propositional logic & proofsMedium
(q ∧ r) ∨ (q ∧ p)(q ∧ r) ∨ p(q ∨ r) ∧ (q ∨ p)q ∨ (r ∧ p)

Which formula is logically equivalent to q ∧ (r ∨ p)? Click it.

q ∧ (r ∨ p)
  1. 1(q ∧ r) ∨ (q ∧ p)
  2. 2(q ∧ r) ∨ p
  3. 3(q ∨ r) ∧ (q ∨ p)
  4. 4q ∨ (r ∧ p)
4.Propositional logic & proofsHard
pqq → ppq11100100

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

q → pp∴ q
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B + B′C + AC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111010011312141507X6012113115X141819111X10

For F(A, B, C, D) = Σm(0, 2, 3, 4, 5, 8, 9, 11, 13, 15) + d(6, 10, 14), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 2, 3, 4, 5, 8, 9, 11, 13, 15) + d(6, 10, 14)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 9 modulo 29: the number x with 0 < x < 29 and 9·x ≡ 1 (mod 29).

a = 9n = 29
9−1 mod 29 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 32116 mod 77.

a = 32k = 116n = 77
32116 mod 77 =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALogarithmic
  2. BLinear
  3. CConstant
  4. DExponential
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AEvery element must be compared at least once
  2. BBinary search is the only possible algorithm, and it needs log₂ n comparisons
  3. CComparisons take logarithmic time each
  4. DA decision tree with n possible outcomes must have depth at least log₂ n
11.Data structuresMedium

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

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

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CBecause halving at one half would lose elements
  4. DTo keep the array sorted
13.Graph theoryMedium

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

4 × deg 33 × deg 43 × deg 2
L =
14.Graph theoryHard
432151310157ABCDEFG

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

Why does SELECT price * 2 AS double_price FROM t WHERE double_price > 10 fail in most databases?

  1. AAliases may not contain an underscore
  2. BArithmetic is not allowed in the SELECT list
  3. CWHERE can only compare text columns
  4. DWHERE is evaluated before the SELECT list, so the alias does not exist yet
16.Databases & SQLHard

For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?

  1. ACross join
  2. BHash join
  3. CNested loop join with index lookups
  4. DBitmap join on the primary key
17.Computer networksMedium

At which OSI layer does a switch normally operate?

  1. AData link (layer 2)
  2. BTransport (layer 4)
  3. CPhysical (layer 1)
  4. DNetwork (layer 3)
18.Computer networksHard

How does a content delivery network (CDN) mainly improve web performance?

  1. AIt increases the bandwidth of the users' own home connections
  2. BIt serves cached content from servers close to the users
  3. CIt removes the need for DNS lookups
  4. DIt replaces TCP with a faster transport protocol
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
third octet = 48012806413211608040201networkhost

A host has the address 172.20.48.197/19. What is the third octet x of its subnet's broadcast address?

172.20.48.197/19
x =

3Solutions

1.Number systems

Answer: U = 179

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

77 = 0100 1101
inverted: 1011 0010
+1 → 1011 0011

Read it as unsigned

1011 0011₂ = 256 − 77 = 179

Answer

U = 179
2.Number systems

Answer: N = 1023

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

210 − 1 = 1023

Answer

N = 1023
3.Propositional logic & proofs

Answer: (q ∧ r) ∨ (q ∧ p)

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

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
q ∧ (r ∨ p) ≡ (q ∧ r) ∨ (q ∧ p)

Check with the truth table

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

Answer

(q ∧ r) ∨ (q ∧ p)

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

4.Propositional logic & proofs

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

Counterexample

p = 1, q = 0

Answer

p = 1, q = 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: L = 3

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

A′B → m(2, 3)
B′C → m(1, 5)
AC → m(5, 7)
ABC00011110010011131204151706

Group and read off

F = C + A′B

Count the literals

L = 3

Answer

L = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 1

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′, B′D′, B′C, BC′D, A′D′, AD, AC, AB′, A′BC′
ABCD000111100001111010011312141507X6012113115X141819111X10

Essential prime implicants

B′C
N = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: 9−1 mod 29 = 13

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·9 (mod 29).

r = 29, t = 0
r = 9, q = 3, t = 1
r = 2, q = 4, t = −3
r = 1, q = 2, t = 13

Inverse

1 ≡ 13·9 (mod 29)
x = 13
check: 9·13 = 117 = 4·29 + 1

Answer

9−1 mod 29 = 13
8.Number theory & modular arithmetic

Answer: 32116 mod 77 = 23

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

116 = 11101002 = 64 + 32 + 16 + 4

Repeated squaring

321 ≡ 32 (mod 77)
322 ≡ 23 (mod 77)
324 ≡ 67 (mod 77)
328 ≡ 23 (mod 77)
3216 ≡ 67 (mod 77)
3232 ≡ 23 (mod 77)
3264 ≡ 67 (mod 77)

Multiply the needed powers

32116 ≡ 67·67·23·67 ≡ 23 (mod 77)

Answer

32116 mod 77 = 23
9.Algorithms & complexity

Answer: D · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

D · Exponential
10.Algorithms & complexity

Answer: D · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

D · A decision tree with n possible outcomes must have depth at least log₂ n
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: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
13.Graph theory

Answer: L = 12

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 + 10) − 2
L = 12

Answer

L = 12

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

14.Graph theory

Answer: d(F) = 29

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: 3
C (3): G: 8
B (4): D: 19, G: 5
G (5): E: 12
E (12)
D (19): F: 29
F (29)

Shortest path

A → B → D → F
d(F) = 29

Answer

d(F) = 29
15.Databases & SQL

Answer: D · WHERE is evaluated before the SELECT list, so the alias does not exist yet

Why this is the answer

Logically FROM and WHERE run before SELECT, so repeat the expression in WHERE or use a subquery; ORDER BY can use the alias because it runs after SELECT.

Answer

D · WHERE is evaluated before the SELECT list, so the alias does not exist yet
16.Databases & SQL

Answer: B · Hash join

Why this is the answer

A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.

Answer

B · Hash join
17.Computer networks

Answer: A · Data link (layer 2)

Why this is the answer

A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.

Answer

A · Data link (layer 2)
18.Computer networks

Answer: B · It serves cached content from servers close to the users

Why this is the answer

CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.

Answer

B · It serves cached content from servers close to the users
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 = 63

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 in the third octet

25 = 32

Subnet of this host

network: 172.20.32.0
broadcast: 172.20.63.255

Answer

x = 63

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