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
16 bit1 px1920 px1080 pxM = ? KiB

An uncompressed image is 1920 × 1080 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1920 × 1080 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
b ∧ ¬c¬b ∨ ¬c¬b ∧ c¬b → ¬c

Which formula is logically equivalent to ¬(b → c)? Click it.

¬(b → c)
  1. 1b ∧ ¬c
  2. 2¬b ∨ ¬c
  3. 3¬b ∧ c
  4. 4¬b → ¬c
4.Propositional logic & proofsHard

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

(d ∧ ¬a) ∨ (d ∧ (b ∨ c))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110312140507061121131151141809011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 2, 4, 8, 12, 13, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 2, 4, 8, 12, 13, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKUP

A 4-bit binary up counter starts at 2 (decimal, Q = 0010). Which count Q (in decimal) does it show after 36 clock pulses?

Q0 = 236n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 14 modulo 73: the number x with 0 < x < 73 and 14·x ≡ 1 (mod 73).

a = 14n = 73
14−1 mod 73 =
8.Number theory & modular arithmeticHard

17 is prime. Use Fermat’s little theorem to find 2173146 mod 17.

p = 17k = 73146
2173146 mod 17 =
9.Algorithms & complexityMedium

Which sorting algorithm is stable?

  1. ASelection sort (typical version)
  2. BHeap sort
  3. CMerge sort
  4. DQuicksort (typical in-place version)
10.Algorithms & complexityHard

How do n! and 2ⁿ compare for large n?

  1. An! is O(n²)
  2. B2ⁿ grows faster than n!
  3. CThey grow at the same rate
  4. Dn! grows faster than 2ⁿ
11.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AIts right child, always
  2. BThe rightmost node of its left subtree
  3. CThe leftmost node of its right subtree
  4. DIts parent, always
12.Data structuresHard

What is the running time of breadth-first search on a graph with V vertices stored as an adjacency matrix?

  1. AO(V²)
  2. BO(V + E)
  3. CO(V)
  4. DO(E log V)
13.Graph theoryMedium

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

4 × deg 32 × deg 43 × deg 5
L =
14.Graph theoryHard
118991197510ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

start: Around 5
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
15.Databases & SQLMedium

The orders table has 5 rows, each referring to an existing customer. The customers table has 3 rows. How many rows does an INNER JOIN on customer_id return?

  1. A3
  2. B8
  3. C15
  4. D5
16.Databases & SQLHard

Two transactions both read stock = 10, each subtracts 1 and writes the result, giving 9 instead of 8. Which anomaly is this?

  1. APhantom read
  2. BDeletion anomaly
  3. CLost update
  4. DDirty read
17.Computer networksMedium

HTTP is stateless. How do websites usually keep a user logged in across requests?

  1. AWith cookies holding a session identifier or token
  2. BBy giving each user a fixed IP address
  3. CBy storing the password in the URL
  4. DBy keeping the TCP connection open for days
18.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A6 kbit/s
  2. B24 kbit/s
  3. C3 kbit/s
  4. D12 kbit/s
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1461 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1461 hosts
p =
20.IPv4 subnettingHard
third octet = 121012816413211618040211networkhost

A host has the address 172.20.121.217/20. What is the third octet x of its network address?

172.20.121.217/20
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: M = 4050 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

1920·1080·16 = 33177600 bit = 4147200 B

Convert to KiB

4147200/1024 = 4050 KiB

Answer

M = 4050 KiB
3.Propositional logic & proofs

Answer: b ∧ ¬c

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) ≡ ¬(¬a ∨ b) ≡ a ∧ ¬b: the only row where an implication fails.
¬(b → c) ≡ b ∧ ¬c

Check with the truth table

b = 1, c = 1: 0 | 0 0 0 1
b = 1, c = 0: 1 | 1 1 0 1
b = 0, c = 1: 0 | 0 1 1 0
b = 0, c = 0: 0 | 0 1 0 1

Answer

b ∧ ¬c

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

4.Propositional logic & proofs

Answer: N = 7

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

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

Split on a

a = 1: 3
a = 0: 4
N = 3 + 4 = 7

Answer

N = 7
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 6

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.

ABCD000111100001111000110312140507061121131151141809011010

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

Minimal POS form

F = (B + C′ + D′)(A + B′ + D′)(A′ + B + D′)(A + B′ + C′)(A′ + B + C′)(A + B + C + D)
N = 6

Answer

N = 6

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

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

Answer: Q = 6

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.

Full cycles

Every 16 pulses bring the counter back to the same state.

36 = 2·16 + 4

Count after the remaining pulses

Q = (2 + 4) mod 16 = 6
Q = 0110₂

Answer

Q = 6
7.Number theory & modular arithmetic

Answer: 14−1 mod 73 = 47

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·14 (mod 73).

r = 73, t = 0
r = 14, q = 5, t = 1
r = 3, q = 4, t = −5
r = 2, q = 1, t = 21
r = 1, q = 2, t = −26

Inverse

1 ≡ −26·14 (mod 73)
x = 47
check: 14·47 = 658 = 9·73 + 1

Answer

14−1 mod 73 = 47
8.Number theory & modular arithmetic

Answer: 2173146 mod 17 = 16

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.

Reduce the exponent

73146 mod 16 = 10
2173146 ≡ 2110 (mod 17)

Small power

2110 mod 17 = 16

Answer

2173146 mod 17 = 16
9.Algorithms & complexity

Answer: C · Merge sort

Why this is the answer

Merge sort is stable when merging takes from the left half first on ties. In-place quicksort, heap sort and selection sort are usually not stable.

Answer

C · Merge sort
10.Algorithms & complexity

Answer: D · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

D · n! grows faster than 2ⁿ
11.Data structures

Answer: C · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

C · The leftmost node of its right subtree
12.Data structures

Answer: A · O(V²)

Why this is the answer

Each of the V vertices scans a row of V entries, giving O(V²); with adjacency lists BFS is O(V + E).

Answer

A · O(V²)
13.Graph theory

Answer: L = 19

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 + 35 = 2(L + 9) − 2
L = 19

Answer

L = 19

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

14.Graph theory

Answer: D

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

1. A (0): B: 11, C: 8
2. C (8): E: 19
3. B (11): D: 20
4. E (19): F: 24
5. D (20)
6. F (24)

round 5

D

Answer

D
15.Databases & SQL

Answer: D · 5

Why this is the answer

Every order finds exactly one matching customer, so the join returns one row per order: 5 rows.

Answer

D · 5
16.Databases & SQL

Answer: C · Lost update

Why this is the answer

In a lost update, both transactions work from the same old value and the second write erases the first; locking (SELECT ... FOR UPDATE) or atomic updates prevent it.

Answer

C · Lost update
17.Computer networks

Answer: A · With cookies holding a session identifier or token

Why this is the answer

The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.

Answer

A · With cookies holding a session identifier or token
18.Computer networks

Answer: D · 12 kbit/s

Why this is the answer

2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.

Answer

D · 12 kbit/s
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 < 1461
211 − 2 = 2046 ≥ 1461 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 112

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

24 = 16

Subnet of this host

network: 172.20.112.0
broadcast: 172.20.127.255

Answer

x = 112

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