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 −117 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1178 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0001 1111 OR 1101 0111. What is the result's value R in decimal?

0001 1111OR1101 0111
R =
3.Propositional logic & proofsMedium
p → q¬q → ¬p¬p → ¬q¬p ∧ ¬q

Which formula is logically equivalent to q → p? Click it.

q → p
  1. 1p → q
  2. 2¬q → ¬p
  3. 3¬p → ¬q
  4. 4¬p ∧ ¬q
4.Propositional logic & proofsHard
∃x (¬A(x) ∨ ¬B(x))∃x (¬A(x) ∧ ¬B(x))∀x (¬A(x) ∨ ¬B(x))∀x (¬A(x) ∧ ¬B(x))

Which statement is equivalent to ¬∀x (A(x) ∨ B(x)), with the negation moved all the way inside? Click it.

¬∀x (A(x) ∨ B(x))
  1. 1∃x (¬A(x) ∨ ¬B(x))
  2. 2∃x (¬A(x) ∧ ¬B(x))
  3. 3∀x (¬A(x) ∨ ¬B(x))
  4. 4∀x (¬A(x) ∧ ¬B(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​CI1​1I2​1I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = C, I₂ = 1, I₃ = 1. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = CI₂ = 1I₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X00113X214050716012013X15X141819X11010

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 4, 6, 8, 9) + d(0, 2, 11, 14, 15), where d marks the don't-care conditions. How many literals L does it contain?

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

Find the smallest positive integer x with x ≡ 1 (mod 5) and x ≡ 3 (mod 6).

x ≡ 1 (mod 5)x ≡ 3 (mod 6)
x =
8.Number theory & modular arithmeticHard

17 is prime. Use Fermat’s little theorem to find 2986638 mod 17.

p = 17k = 86638
2986638 mod 17 =
9.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(log n)
  2. BΩ(n log n)
  3. CΩ(n)
  4. DΩ(n²)
10.Algorithms & complexityHard

What did the Cook–Levin theorem establish?

  1. AP is not equal to NP
  2. BThe halting problem is undecidable
  3. CThe Boolean satisfiability problem (SAT) is NP-complete
  4. DEvery NP problem can be solved in exponential time
11.Data structuresMedium

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

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

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. AA single right rotation at the unbalanced node
  2. BRecolouring the node and its child
  3. CA left rotation at the child, then a right rotation at the node
  4. DA single left rotation at the unbalanced node
13.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
16961281710ABCDE

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

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

Which query lists customers who have never placed an order?

  1. ASELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NOT NULL
  2. BSELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
  3. CSELECT c.name FROM customers c INNER JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
  4. DSELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id = NULL
16.Databases & SQLHard

Given the functional dependencies A → B, B → C and C → D on R(A, B, C, D), what is the closure A+?

  1. A{A, B, C}
  2. B{A, B}
  3. C{A, B, C, D}
  4. D{A}
17.Computer networksMedium

Using even parity, which parity bit is added to the data bits 1011001?

  1. A4
  2. B2
  3. C1
  4. D0
18.Computer networksHard

Why does TCP choose an unpredictable initial sequence number for each connection?

  1. ATo encrypt the payload of each segment
  2. BTo avoid confusion with old segments and make spoofed segments harder to inject
  3. CTo select the best route to the destination for the segments of that connection
  4. DTo compress the data more efficiently
19.IPv4 subnettingMedium
255.255.255.24011111111111111111111111111110000network bitshost bits

A subnet uses the mask 255.255.255.240. How many usable host addresses H does it have?

255.255.255.240
H =
20.IPv4 subnettingHard
last octet = 119012816413211608141211networkhost

A host has the address 10.20.148.119/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.148.119/28
x =

3Solutions

1.Number systems

Answer: U = 139

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

117 = 0111 0101
inverted: 1000 1010
+1 → 1000 1011

Read it as unsigned

1000 1011₂ = 256 − 117 = 139

Answer

U = 139
2.Number systems

Answer: R = 223

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.

Bit by bit

OR gives 1 where at least one bit is 1.

0001 1111
OR 1101 0111
= 1101 1111

Value of the result

1101 1111₂ = 223

Answer

R = 223
3.Propositional logic & proofs

Answer: ¬p → ¬q

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

An implication is equivalent to its contrapositive ¬b → ¬a, not to its converse b → a or its inverse ¬a → ¬b.
q → p ≡ ¬p → ¬q

Check with the truth table

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

Answer

¬p → ¬q

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

4.Propositional logic & proofs

Answer: ∃x (¬A(x) ∧ ¬B(x))

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 (A(x) ∨ B(x)) ≡ ∃x (¬A(x) ∧ ¬B(x))

Answer

∃x (¬A(x) ∧ ¬B(x))

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: N = 7

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′BC + AB′ + AB

Rows with F = 1

ABC = 000, 001, 011, 100, 101, 110, 111
N = 7

Answer

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

Answer: L = 8

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.

B′C′D′, B′CD, BCD′, A′D′, AB′D, A′B′C, AB′C′
ABCD0001111000011110X00113X214050716012013X15X141819X11010

Minimal cover

F = B′CD + A′D′ + AB′C′
L = 8

Answer

L = 8

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

7.Number theory & modular arithmetic

Answer: x = 21

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).

First congruence

x = 1 + 5k

Second congruence

1 + 5k ≡ 3 (mod 6) → 5k ≡ 2 (mod 6)
k ≡ 5·2 ≡ 4 (mod 6)

Result

x = 1 + 5·4 = 21
x = 21
check: 21 mod 5 = 1, 21 mod 6 = 3

Answer

x = 21
8.Number theory & modular arithmetic

Answer: 2986638 mod 17 = 15

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

86638 mod 16 = 14
2986638 ≡ 2914 (mod 17)

Small power

2914 mod 17 = 15

Answer

2986638 mod 17 = 15
9.Algorithms & complexity

Answer: B · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

B · Ω(n log n)
10.Algorithms & complexity

Answer: C · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

C · The Boolean satisfiability problem (SAT) is NP-complete
11.Data structures

Answer: C · 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

C · A stack
12.Data structures

Answer: A · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

A · A single right rotation at the unbalanced node
13.Graph theory

Answer: Euler circuit

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.

Degrees

deg(A) = 2, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2, deg(F) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: W = 35

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

BC (6): take
CE (8): take
AC (9): take
BE (10): skip, closes a cycle
BD (12): take

Total weight

W = 6 + 8 + 9 + 12 = 35

Answer

W = 35
15.Databases & SQL

Answer: B · SELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL

Why this is the answer

The left join keeps all customers and fills order columns with NULL when there is no match; IS NULL then picks exactly those customers.

Answer

B · SELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
16.Databases & SQL

Answer: C · {A, B, C, D}

Why this is the answer

Start with A, add B (A → B), then C (B → C), then D (C → D); since A+ contains all attributes, A is a key.

Answer

C · {A, B, C, D}
17.Computer networks

Answer: D · 0

Why this is the answer

1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.

Answer

D · 0
18.Computer networks

Answer: B · To avoid confusion with old segments and make spoofed segments harder to inject

Why this is the answer

Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.

Answer

B · To avoid confusion with old segments and make spoofed segments harder to inject
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).

Prefix length

255.255.255.240 = /28

Usable hosts

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 127

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: ⌊119/16⌋·16 = 112
broadcast: 112 + 16 − 1 = 127
usable hosts: 113 … 126

Answer

x = 127

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