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 complement1−12806403201608040211

The 8-bit pattern 1000 0001 stores a signed integer in two's complement. What is its value N in decimal?

1000 00018 bit
N =
2.Number systemsHard
1060 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 1060 items its own binary code of the same length. What is the smallest number of bits b needed?

1060
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

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

(r → p) ↔ (p → r)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∃x ∃y (R(x, y) → ¬R(y, x))∀x ∀y (R(x, y) ∧ ¬R(y, x))∃x ∃y (R(x, y) ∧ ¬R(y, x))∃x ∃y (¬R(x, y) ∧ ¬R(y, x))

Which statement is equivalent to ¬∀x ∀y (R(x, y) → R(y, x)), with the negation moved all the way inside? Click it.

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∃x ∃y (R(x, y) → ¬R(y, x))
  2. 2∀x ∀y (R(x, y) ∧ ¬R(y, x))
  3. 3∃x ∃y (R(x, y) ∧ ¬R(y, x))
  4. 4∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B′ + A′B′C + ABC′ + A′BC′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKDOWN

A 4-bit binary down counter starts at 12 (decimal, Q = 1100). Which count Q (in decimal) does it show after 39 clock pulses?

Q0 = 1239n = 4
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 4 (mod 7) and x ≡ 8 (mod 9).

x ≡ 4 (mod 7)x ≡ 8 (mod 9)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (943, 23)d = ?

An RSA key uses the primes p = 23 and q = 41 and the public exponent e = 23. Find the private exponent d (0 < d < φ(n)).

p = 23q = 41e = 23
d =
9.Algorithms & complexityMedium

Which statement about a recursive function without a reachable base case is correct?

  1. AIt runs in O(1) time
  2. BIt returns 0 automatically
  3. CThe compiler always turns it into a loop that ends safely
  4. DIt keeps calling itself until the call stack overflows
10.Algorithms & complexityHard

Which definition of the class NP is correct?

  1. ADecision problems whose yes-answers can be verified in polynomial time given a certificate
  2. BProblems that are not polynomial
  3. CProblems that no algorithm can solve
  4. DProblems that cannot be solved in polynomial time
11.Data structuresMedium

Which data structure does a running program use to keep track of active function calls and their local variables?

  1. AA hash table of functions
  2. BThe call stack
  3. CThe heap memory queue
  4. DA binary search tree of frames
12.Data structuresHard

Why might a server choose its hash function at random from a universal family?

  1. ATo remove the need to resize the table
  2. BTo guarantee that no two keys can ever collide, whatever keys an attacker sends
  3. CTo make it very unlikely that an attacker can pick many keys that all collide
  4. DTo make the hash table sorted
13.Graph theoryMedium

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

4 × deg 34 × deg 42 × deg 5
L =
14.Graph theoryHard
11141815168192ABCDEF

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

|V| = 6|E| = 9
W =
15.Databases & SQLMedium

First normal form (1NF) requires that:

  1. AEvery table has exactly one column
  2. BEvery non-key column depends on the whole key and on nothing but the key
  3. CEach column contains atomic values and there are no repeating groups
  4. DThere are no foreign keys
16.Databases & SQLHard

What does the following return: SELECT name FROM students WHERE grade > (SELECT AVG(grade) FROM students)?

  1. AStudents with a grade above the average grade
  2. BAn error because subqueries are not allowed in WHERE
  3. CThe average grade of all students
  4. DStudents with the highest grade only
17.Computer networksMedium

Which range of TCP/UDP port numbers is reserved as "well-known" ports?

  1. A1024–49151
  2. B0–1023
  3. C0–65535
  4. D49152–65535
18.Computer networksHard

After repeated collisions, classic Ethernet stations choose their waiting time from a range that doubles each time. Why?

  1. ATo synchronise all stations to one clock
  2. BTo make the cable carry signals further
  3. CTo give the station with the highest MAC address priority on the shared cable
  4. DTo lower the chance of colliding again when many stations compete
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.227.0/24 is divided into /25 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 2?

192.168.227.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 120012816413211618040201networkhost

A host has the address 10.20.225.120/30. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.225.120/30
x =

3Solutions

1.Number systems

Answer: N = −127

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 + 1 = −127

Answer

N = −127
2.Number systems

Answer: b = 11 bit

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.

Powers of two

210 = 1024 < 1060
211 = 2048 ≥ 1060

Answer

b = 11
3.Propositional logic & proofs

Answer: contingency

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

p = 1, r = 1: r → p = 1, p → r = 1, (r → p) ↔ (p → r) = 1
p = 1, r = 0: r → p = 1, p → r = 0, (r → p) ↔ (p → r) = 0
p = 0, r = 1: r → p = 0, p → r = 1, (r → p) ↔ (p → r) = 0
p = 0, r = 0: r → p = 1, p → r = 1, (r → p) ↔ (p → r) = 1

Classify

(r → p) ↔ (p → r): 1, 0, 0, 1
1 in 2 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, 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 ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, 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: 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′ → m(0, 1)
A′B′C → m(1)
ABC′ → m(6)
A′BC′ → m(2)
ABC00011110011011031204050716

Group and read off

F = BC′ + A′B′

Count the literals

L = 4

Answer

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

Answer: Q = 5

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.

39 = 2·16 + 7

Count after the remaining pulses

Q = (12 − 7) mod 16 = 5
Q = 0101₂

Answer

Q = 5
7.Number theory & modular arithmetic

Answer: x = 53

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 = 4 + 7k

Second congruence

4 + 7k ≡ 8 (mod 9) → 7k ≡ 4 (mod 9)
k ≡ 4·4 ≡ 7 (mod 9)

Result

x = 4 + 7·7 = 53
x = 53
check: 53 mod 7 = 4, 53 mod 9 = 8

Answer

x = 53
8.Number theory & modular arithmetic

Answer: d = 727

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.

Modulus and totient

n = 23·41 = 943
φ(n) = 22·40 = 880

Extended Euclidean algorithm

r = 880, t = 0
r = 23, q = 38, t = 1
r = 6, q = 3, t = −38
r = 5, q = 1, t = 115
r = 1, q = 5, t = −153

Private exponent

d = 727
check: 23·727 = 16721 = 19·880 + 1

Answer

d = 727
9.Algorithms & complexity

Answer: D · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

D · It keeps calling itself until the call stack overflows
10.Algorithms & complexity

Answer: A · Decision problems whose yes-answers can be verified in polynomial time given a certificate

Why this is the answer

For an NP problem, a proposed solution (certificate) can be checked quickly, for example verifying a given assignment satisfies a Boolean formula.

Answer

A · Decision problems whose yes-answers can be verified in polynomial time given a certificate
11.Data structures

Answer: B · The call stack

Why this is the answer

Each call pushes a stack frame and each return pops it, which is why the most recent call finishes first.

Answer

B · The call stack
12.Data structures

Answer: C · To make it very unlikely that an attacker can pick many keys that all collide

Why this is the answer

With universal hashing the expected number of collisions is small for any set of keys, since the attacker does not know which function was chosen.

Answer

C · To make it very unlikely that an attacker can pick many keys that all collide
13.Graph theory

Answer: L = 20

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 + 38 = 2(L + 10) − 2
L = 20

Answer

L = 20

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

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

AB (1): take
BE (2): take
BC (4): take
DF (8): take
AC (11): skip, closes a cycle
CE (15): skip, closes a cycle
DE (16): take

Total weight

W = 1 + 2 + 4 + 8 + 16 = 31

Answer

W = 31
15.Databases & SQL

Answer: C · Each column contains atomic values and there are no repeating groups

Why this is the answer

1NF requires atomic values per cell and a unique row identification, without repeating groups.

Answer

C · Each column contains atomic values and there are no repeating groups
16.Databases & SQL

Answer: A · Students with a grade above the average grade

Why this is the answer

The scalar subquery computes the average, and the outer query keeps students above it.

Answer

A · Students with a grade above the average grade
17.Computer networks

Answer: B · 0–1023

Why this is the answer

Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.

Answer

B · 0–1023
18.Computer networks

Answer: D · To lower the chance of colliding again when many stations compete

Why this is the answer

Each further collision suggests more contention, so the random range grows and stations are more likely to pick different slots. It adapts to load without central control.

Answer

D · To lower the chance of colliding again when many stations compete
19.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

27 = 128

Subnet number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 255
20.IPv4 subnetting

Answer: x = 123

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

22 = 4

Subnet of this host

network: ⌊120/4⌋·4 = 120
broadcast: 120 + 4 − 1 = 123
usable hosts: 121 … 122

Answer

x = 123

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