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 −62 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −628 bit
U =
2.Number systemsHard
16-bit two's complement, N = ??sign???????????????

What is the most negative integer N that can be stored in 16-bit two's complement?

16 bit
N =
3.Propositional logic & proofsMedium
¬q ∨ ¬p¬q ∧ ¬pq ∨ ¬p¬q ∧ p

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

¬(q ∨ p)
  1. 1¬q ∨ ¬p
  2. 2¬q ∧ ¬p
  3. 3q ∨ ¬p
  4. 4¬q ∧ p
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ¬((c → ¬(b ∧ a)) → ¬((d ↔ a) ∨ d)) true? Call this number N.

¬((c → ¬(b ∧ a)) → ¬((d ↔ a) ∨ d))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010111312040517160120130150140809011110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 1, 2, 3, 6, 7, 10). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 1, 2, 3, 6, 7, 10)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKRING

A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 11 clock edges.

Q₃Q₂Q₁Q₀ = 110011
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 39 modulo 73: the number x with 0 < x < 73 and 39·x ≡ 1 (mod 73).

a = 39n = 73
39−1 mod 73 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 6859 mod 77.

a = 68k = 59n = 77
6859 mod 77 =
9.Algorithms & complexityMedium

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

  1. ALogarithmic
  2. BConstant
  3. CExponential
  4. DLinear
10.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n!)
  2. BO(2n²)
  3. CO(n³)
  4. DO(n² 2ⁿ)
11.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the preorder traversal?

  1. A1 2 3 4 5
  2. B4 2 5 1 3
  3. C1 2 4 5 3
  4. D4 5 2 3 1
12.Data structuresHard

Which statement about heapsort is correct?

  1. AIt is stable and needs O(n) extra memory
  2. BIt sorts in place in O(n log n) worst-case time but is not stable
  3. CIt needs a binary search tree as extra storage, so it uses O(n) extra memory
  4. DIt runs in O(n) time on any input
13.Graph theoryMedium

A connected graph has 15 vertices and 19 edges. How many edges k must be removed to leave a spanning tree?

|V| = 15|E| = 19
k =
14.Graph theoryHard
19818313161094ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 9
W =
15.Databases & SQLMedium

What does INTERSECT return when used between two SELECT queries?

  1. ARows in the first result but not the second
  2. BAll rows from both results
  3. CRows that appear in both results
  4. DThe Cartesian product of the results
16.Databases & SQLHard

What does SELECT name, dept, AVG(salary) OVER (PARTITION BY dept) FROM emp return?

  1. AAn error because name is not in a GROUP BY
  2. BOne row per department with its average salary
  3. COnly employees who earn more than their department average
  4. DEvery employee row together with the average salary of their department
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ABusiness policy
  2. BMeasured delay
  3. CHop count
  4. DLink bandwidth
18.Computer networksHard

The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?

  1. ADistance-vector; split horizon and poison reverse
  2. BLink-state; flooding link updates
  3. CPath-vector; longer AS paths
  4. DStatic routing; adding default routes
19.IPv4 subnettingMedium
255.255.248.011111111111111111111100000000000network bitshost bits

A subnet uses the mask 255.255.248.0. How many usable host addresses H does it have?

255.255.248.0
H =
20.IPv4 subnettingHard
last octet = 248112816413211618040201networkhost

A host has the address 10.20.220.248/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.220.248/28
x =

3Solutions

1.Number systems

Answer: U = 194

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

62 = 0011 1110
inverted: 1100 0001
+1 → 1100 0010

Read it as unsigned

1100 0010₂ = 256 − 62 = 194

Answer

U = 194
2.Number systems

Answer: N = −32768

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.

Most negative value

−215 = −32768

Answer

N = −32768
3.Propositional logic & proofs

Answer: ¬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

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(q ∨ p) ≡ ¬q ∧ ¬p

Check with the truth table

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

Answer

¬q ∧ ¬p

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

4.Propositional logic & proofs

Answer: N = 11

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

b ∧ a; ¬(b ∧ a); c → ¬(b ∧ a); d ↔ a; (d ↔ a) ∨ d; ¬((d ↔ a) ∨ d); (c → ¬(b ∧ a)) → ¬((d ↔ a) ∨ d); ¬((c → ¬(b ∧ a)) → ¬((d ↔ a) ∨ d))

Split on a

a = 1: 3
a = 0: 8
N = 3 + 8 = 11

Answer

N = 11
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 7

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.

ABCD000111100001111010111312040517160120130150140809011110

Group the 1s

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

B′CD′
A′C
A′B′

Minimal SOP form

F = B′CD′ + A′C + A′B′
L = 7

Answer

L = 7
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 9

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.

State after every edge

1100 → 0110 → 0011 → 1001 → 1100 → 0110 → 0011 → 1001 → 1100 → 0110 → 0011 → 1001

Decimal value

1001₂ = 9

Answer

Q = 9
7.Number theory & modular arithmetic

Answer: 39−1 mod 73 = 15

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·39 (mod 73).

r = 73, t = 0
r = 39, q = 1, t = 1
r = 34, q = 1, t = −1
r = 5, q = 6, t = 2
r = 4, q = 1, t = −13
r = 1, q = 4, t = 15

Inverse

1 ≡ 15·39 (mod 73)
x = 15
check: 39·15 = 585 = 8·73 + 1

Answer

39−1 mod 73 = 15
8.Number theory & modular arithmetic

Answer: 6859 mod 77 = 17

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

59 = 1110112 = 32 + 16 + 8 + 2 + 1

Repeated squaring

681 ≡ 68 (mod 77)
682 ≡ 4 (mod 77)
684 ≡ 16 (mod 77)
688 ≡ 25 (mod 77)
6816 ≡ 9 (mod 77)
6832 ≡ 4 (mod 77)

Multiply the needed powers

6859 ≡ 68·4·25·9·4 ≡ 17 (mod 77)

Answer

6859 mod 77 = 17
9.Algorithms & complexity

Answer: C · Exponential

Why this is the answer

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

Answer

C · Exponential
10.Algorithms & complexity

Answer: D · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

D · O(n² 2ⁿ)
11.Data structures

Answer: C · 1 2 4 5 3

Why this is the answer

Preorder visits 1, then its left subtree (2, 4, 5), then 3; the other options are inorder, postorder and level-order.

Answer

C · 1 2 4 5 3
12.Data structures

Answer: B · It sorts in place in O(n log n) worst-case time but is not stable

Why this is the answer

Heapsort builds a max-heap and extracts the maximum n times within the same array; swaps over long distances can reorder equal keys.

Answer

B · It sorts in place in O(n log n) worst-case time but is not stable
13.Graph theory

Answer: k = 5

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 = 19 − 14 = 5

Answer

k = 5
14.Graph theory

Answer: W = 54

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.

Kruskal’s algorithm

BD (3): take
GD (4): take
AC (8): take
GB (9): skip, closes a cycle
EF (10): take
CE (13): take
DE (16): take

Total weight

W = 3 + 4 + 8 + 10 + 13 + 16 = 54

Answer

W = 54
15.Databases & SQL

Answer: C · Rows that appear in both results

Why this is the answer

INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.

Answer

C · Rows that appear in both results
16.Databases & SQL

Answer: D · Every employee row together with the average salary of their department

Why this is the answer

Unlike GROUP BY, a window function keeps every input row and adds a value computed over its partition, here the department.

Answer

D · Every employee row together with the average salary of their department
17.Computer networks

Answer: C · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

C · Hop count
18.Computer networks

Answer: A · Distance-vector; split horizon and poison reverse

Why this is the answer

After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.

Answer

A · Distance-vector; split horizon and poison reverse
19.IPv4 subnetting

Answer: H = 2046

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

255.255.248.0 = /21

Usable hosts

211 − 2 = 2046

Answer

H = 2046
20.IPv4 subnetting

Answer: x = 255

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: ⌊248/16⌋·16 = 240
broadcast: 240 + 16 − 1 = 255
usable hosts: 241 … 254

Answer

x = 255

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