Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 −91 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −918 bit
U =
2.Number systemsHard
32 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (r → q) ∨ (q → r) a tautology, a contradiction or a contingency? Click your answer.

(r → q) ∨ (q → r)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make (s ∧ q) ↔ ((¬q ∨ s) → (r ↔ p)) true? Call this number N.

(s ∧ q) ↔ ((¬q ∨ s) → (r ↔ p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000111312141507161120130151141819011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 2, 3, 4, 5, 6, 8, 9, 12, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 2, 3, 4, 5, 6, 8, 9, 12, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Solve 70x ≡ 77 (mod 83). Give the solution x with 0 < x < 83.

70x ≡ 77 (mod 83)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n105m = cd​ mod n?(n, e) = (731, 17)d

An RSA key uses p = 17, q = 43 and e = 17. Decrypt the ciphertext c = 105: find the message m.

p = 17q = 43e = 17c = 105
m =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n/2) + O(1) solve to?

  1. AO(n)
  2. BO(n log n)
  3. CO(log n)
  4. DO(1)
10.Algorithms & complexityHard

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

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

In a binary heap stored in an array starting at index 0, where is the parent of the node at index i?

  1. AAt index i − 1
  2. BAt index ⌊(i − 1) / 2⌋
  3. CAt index 2i + 1
  4. DAt index ⌊i / 2⌋ + 1
12.Data structuresHard

An AVL node is unbalanced because its left child's right subtree became too tall (left-right case). Which fix is applied?

  1. AA single right rotation at the node
  2. BA left rotation at the left child, then a right rotation at the node
  3. CA right rotation at the left child, then a left rotation at the node
  4. DA single left rotation at the node
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
ABCDEF

How many walks of length 2 go from C to B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

What does a common table expression (WITH name AS (...)) provide?

  1. AA new database user
  2. BAn index on a column
  3. CA permanent table that stays stored on disk until it is dropped
  4. DA named temporary result that can be used within one statement
16.Databases & SQLHard

A correlated subquery is one that:

  1. AUses only aggregate functions
  2. BReferences columns of the outer query and is evaluated per outer row
  3. CRuns only once before the outer query and caches its result for every row
  4. DJoins two tables with a foreign key
17.Computer networksMedium

How many usable host addresses does a /26 IPv4 subnet have?

  1. A30
  2. B126
  3. C62
  4. D64
18.Computer networksHard

In the TCP/IP model, which layer adds port numbers?

  1. ATransport layer
  2. BInternet layer
  3. CApplication layer
  4. DLink layer
19.IPv4 subnettingMedium
/16 → /2311111111111111111111111000000000network bitshost bits

The network 172.22.0.0/16 is divided into equal subnets with prefix /23. How many subnets N are there?

172.22.0.0/16/23
N =
20.IPv4 subnettingHard
third octet = 205112816403201618140211networkhost

A host has the address 172.20.205.29/23. What is the third octet x of its subnet's broadcast address?

172.20.205.29/23
x =

3Solutions

1.Number systems

Answer: U = 165

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

91 = 0101 1011
inverted: 1010 0100
+1 → 1010 0101

Read it as unsigned

1010 0101₂ = 256 − 91 = 165

Answer

U = 165
2.Number systems

Answer: M = 1200 KiB

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.

Bits and bytes

640·480·32 = 9830400 bit = 1228800 B

Convert to KiB

1228800/1024 = 1200 KiB

Answer

M = 1200 KiB
3.Propositional logic & proofs

Answer: tautology

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: r → q = 1, q → r = 1, (r → q) ∨ (q → r) = 1
q = 1, r = 0: r → q = 1, q → r = 0, (r → q) ∨ (q → r) = 1
q = 0, r = 1: r → q = 0, q → r = 1, (r → q) ∨ (q → r) = 1
q = 0, r = 0: r → q = 1, q → r = 1, (r → q) ∨ (q → r) = 1

Classify

(r → q) ∨ (q → r): 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: N = 6

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

s ∧ q; ¬q; ¬q ∨ s; r ↔ p; (¬q ∨ s) → (r ↔ p); (s ∧ q) ↔ ((¬q ∨ s) → (r ↔ p))

Split on p

p = 1: 3
p = 0: 3
N = 3 + 3 = 6

Answer

N = 6
5.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.

ABCD000111100001111000111312141507161120130151141819011010

Group the 0s

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

(B′ + C′ + D′)
(A′ + B′ + D′)
(A′ + B + C′)
(A + B + C + D)

Minimal POS form

F = (B′ + C′ + D′)(A′ + B′ + D′)(A′ + B + C′)(A + B + C + D)
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 1 → Q = 0
2: J = 1, K = 1 → Q = 1
3: J = 0, K = 1 → Q = 0
4: J = 1, K = 0 → Q = 1
5: J = 0, K = 0 → Q = 1
6: J = 0, K = 0 → Q = 1
7: J = 1, K = 0 → Q = 1
8: J = 1, K = 1 → Q = 0

Count the changes

0 → 0 → 1 → 0 → 1 → 1 → 1 → 1 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 26

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).

Inverse of 70

r = 83, t = 0
r = 70, q = 1, t = 1
r = 13, q = 5, t = −1
r = 5, q = 2, t = 6
r = 3, q = 1, t = −13
r = 2, q = 1, t = 19
r = 1, q = 2, t = −32
70−1 ≡ 51 (mod 83)

Solve

x ≡ 51·77 = 3927 ≡ 26 (mod 83)
check: 70·26 = 1820 ≡ 77 (mod 83)

Answer

x = 26
8.Number theory & modular arithmetic

Answer: m = 632

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 = 731, φ(n) = 672
d = 17−1 mod 672 = 593

Decrypt by repeated squaring

593 = 512 + 64 + 16 + 1
1051 ≡ 105 (mod 731)
1052 ≡ 60 (mod 731)
1054 ≡ 676 (mod 731)
1058 ≡ 101 (mod 731)
10516 ≡ 698 (mod 731)
10532 ≡ 358 (mod 731)
10564 ≡ 239 (mod 731)
105128 ≡ 103 (mod 731)
105256 ≡ 375 (mod 731)
105512 ≡ 273 (mod 731)
105593 ≡ 105·698·239·273 ≡ 632 (mod 731)

Message

m = 632
check: 63217 mod 731 = 105

Answer

m = 632
9.Algorithms & complexity

Answer: C · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

C · O(log n)
10.Algorithms & complexity

Answer: B · 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

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

Answer: B · At index ⌊(i − 1) / 2⌋

Why this is the answer

Children of i are at 2i + 1 and 2i + 2, so the parent of i is ⌊(i − 1) / 2⌋; with 1-based indexing it is ⌊i / 2⌋.

Answer

B · At index ⌊(i − 1) / 2⌋
12.Data structures

Answer: B · A left rotation at the left child, then a right rotation at the node

Why this is the answer

The first rotation converts the left-right case into a left-left case, which a right rotation at the unbalanced node then fixes.

Answer

B · A left rotation at the left child, then a right rotation at the node
13.Graph theory

Answer: H = 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.

Count

H = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0 0
B: 1 0 0 1 1 0
C: 1 0 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 0
F: 0 0 0 1 0 0

Common neighbours

(M²)CB = 1·1 + 0·0 + 0·0 + 0·1 + 1·1 + 0·0 = 2

Answer

N = 2
15.Databases & SQL

Answer: D · A named temporary result that can be used within one statement

Why this is the answer

A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.

Answer

D · A named temporary result that can be used within one statement
16.Databases & SQL

Answer: B · References columns of the outer query and is evaluated per outer row

Why this is the answer

Correlated subqueries depend on the current outer row, so they are logically evaluated per row (optimisers may rewrite them as joins).

Answer

B · References columns of the outer query and is evaluated per outer row
17.Computer networks

Answer: C · 62

Why this is the answer

A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.

Answer

C · 62
18.Computer networks

Answer: A · Transport layer

Why this is the answer

TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.

Answer

A · Transport layer
19.IPv4 subnetting

Answer: N = 128

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).

Borrowed bits

23 − 16 = 7

Number of subnets

N = 27 = 128

Answer

N = 128

Each /23 subnet has 29 − 2 = 510 usable hosts.

20.IPv4 subnetting

Answer: x = 205

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

21 = 2

Subnet of this host

network: 172.20.204.0
broadcast: 172.20.205.255

Answer

x = 205

ExamProblems · examproblems.com/courses/computer-science-fundamentals