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
place values in two's complement1−12806413201608041211

The 8-bit pattern 1010 0011 stores a signed integer in two's complement. What is its value N in decimal?

1010 00118 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1110 0100 >> 2. What is the result's value R in decimal?

1110 0100>> 2
R =
3.Propositional logic & proofsMedium
a ∨ (b → c)(a ∧ b) → c(a → c) ∨ (b → c)(a → c) ∧ (b → c)

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

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

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

(r ↔ p) → ((r ∧ s) ∧ q)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B′C + AC + AB
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKMOD-7

A synchronous mod-7 counter counts 0, 1, …, 6 and then returns to 0. It starts at 5. Which count Q (in decimal) does it show after 19 clock pulses?

Q0 = 519N = 7
Q =
7.Number theory & modular arithmeticMedium

Solve 49x ≡ 38 (mod 59). Give the solution x with 0 < x < 59.

49x ≡ 38 (mod 59)
x =
8.Number theory & modular arithmeticHard

Solve 26x ≡ 2 (mod 30). Give the smallest positive solution x.

26x ≡ 2 (mod 30)
x =
9.Algorithms & complexityMedium

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms with different bases differ only by a constant factor
  2. BLogarithms are always rounded to integers
  3. CThe base only matters for small n
  4. DAll logarithms in computer science have base 10, so it is implied
10.Algorithms & complexityHard

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AAfter V − 1 rounds, it checks whether any edge can still be relaxed
  2. BIt checks whether any edge has a negative weight
  3. CIt counts the cycles with depth-first search
  4. DIt runs Dijkstra's algorithm twice and compares the two distance results
11.Data structuresMedium

What is the value of the postfix expression 3 4 + 2 * ?

  1. A11
  2. B24
  3. C14
  4. D10
12.Data structuresHard

A segment tree is typically used to:

  1. AAnswer range queries (such as range sums) and updates in O(log n)
  2. BImplement a stack
  3. CSort strings alphabetically by storing each prefix in a separate tree node
  4. DStore a graph's edges
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
17187101616ABCDE

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

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

Which ACID property guarantees that committed changes survive a system crash?

  1. AConsistency
  2. BIsolation
  3. CAtomicity
  4. DDurability
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index on the primary key only
  2. BAn index that is rebuilt every night
  3. CAn index that covers every column of every table
  4. DAn index that contains all columns the query needs, so the table itself need not be read
17.Computer networksMedium

What does ARP do?

  1. ARoutes packets on the internet
  2. BResolves domain names
  3. CFinds the MAC address that belongs to an IP address on the local network
  4. DAssigns IP addresses automatically to hosts when they join the local network
18.Computer networksHard

Which of these addresses is a public (not RFC 1918 private) IPv4 address?

  1. A172.32.0.1
  2. B10.255.0.1
  3. C192.168.255.1
  4. D172.31.0.1
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.126.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 9?

192.168.126.0/24/29subnet 9 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet = 157112806403211618140211networkhost

A host has the address 172.20.157.51/23. What is the third octet x of its network address?

172.20.157.51/23
x =

3Solutions

1.Number systems

Answer: N = −93

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.

Add the place values

The other bits keep their normal positive values.

−128 + 32 + 2 + 1 = −93

Answer

N = −93
2.Number systems

Answer: R = 57

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

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1110 0100 >> 2
= 0011 1001

Value of the result

0011 1001₂ = 57

Answer

R = 57
3.Propositional logic & proofs

Answer: (a → c) ∧ (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) → c needs c whenever either a or b holds: (a → c) ∧ (b → c).
(a ∨ b) → c ≡ (a → c) ∧ (b → c)

Check with the truth table

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

Answer

(a → c) ∧ (b → c)

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

4.Propositional logic & proofs

Answer: N = 9

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

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

Split on p

p = 1: 5
p = 0: 4
N = 5 + 4 = 9

Answer

N = 9
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 4

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′C → m(1)
AC → m(5, 7)
AB → m(6, 7)
ABC00011110010011030204151716

Group and read off

F = B′C + AB

Count the literals

L = 4

Answer

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

Answer: Q = 3

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 7 pulses bring the counter back to the same state.

19 = 2·7 + 5

Count after the remaining pulses

Q = (5 + 5) mod 7 = 3
Q = 011₂

Answer

Q = 3
7.Number theory & modular arithmetic

Answer: x = 8

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 49

r = 59, t = 0
r = 49, q = 1, t = 1
r = 10, q = 4, t = −1
r = 9, q = 1, t = 5
r = 1, q = 9, t = −6
49−1 ≡ 53 (mod 59)

Solve

x ≡ 53·38 = 2014 ≡ 8 (mod 59)
check: 49·8 = 392 ≡ 38 (mod 59)

Answer

x = 8
8.Number theory & modular arithmetic

Answer: x = 7

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).

Divide by the gcd

gcd(26, 30) = 2
13x ≡ 1 (mod 15)

Solve the reduced congruence

13−1 ≡ 7 (mod 15) → x ≡ 7 (mod 15)

Solutions modulo 30

x ∈ {7, 22}
x = 7

Answer

x = 7
9.Algorithms & complexity

Answer: A · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

A · Logarithms with different bases differ only by a constant factor
10.Algorithms & complexity

Answer: A · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

A · After V − 1 rounds, it checks whether any edge can still be relaxed
11.Data structures

Answer: C · 14

Why this is the answer

Push 3 and 4, + gives 7; push 2, * gives 7 × 2 = 14.

Answer

C · 14
12.Data structures

Answer: A · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

A · Answer range queries (such as range sums) and updates in O(log n)
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) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: W = 31

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

DE (1): take
BE (6): take
BC (7): take
BD (10): skip, closes a cycle
CE (16): skip, closes a cycle
AB (17): take

Total weight

W = 1 + 6 + 7 + 17 = 31

Answer

W = 31
15.Databases & SQL

Answer: D · Durability

Why this is the answer

Durability means that once a transaction commits, its effects are kept, usually by writing a log to stable storage.

Answer

D · Durability
16.Databases & SQL

Answer: D · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

D · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: C · Finds the MAC address that belongs to an IP address on the local network

Why this is the answer

ARP broadcasts "who has this IP?" and the owner replies with its MAC address, so frames can be delivered locally.

Answer

C · Finds the MAC address that belongs to an IP address on the local network
18.Computer networks

Answer: A · 172.32.0.1

Why this is the answer

172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.

Answer

A · 172.32.0.1
19.IPv4 subnetting

Answer: x = 71

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

23 = 8

Subnet number 9

network: (9 − 1)·8 = 64
broadcast: 64 + 8 − 1 = 71

Answer

x = 71
20.IPv4 subnetting

Answer: x = 156

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: 172.20.156.0
broadcast: 172.20.157.255

Answer

x = 156

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