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 complement0−12806403201608040201

Set the bits so they store −78 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −788 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0110 0001 XOR 1110 1101. What is the result's value R in decimal?

0110 0001XOR1110 1101
R =
3.Propositional logic & proofsMedium
r ∧ p(r → p) ∧ (¬p → ¬r)(r → p) ∨ (p → r)(r → p) ∧ (p → r)

Which formula is logically equivalent to r ↔ p? Click it.

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

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

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

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

F = AB′C + ABC′ + ABC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD123456

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 6 rising clock edges. How many times N does Q change state?

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

Find the inverse of 53 modulo 97: the number x with 0 < x < 97 and 53·x ≡ 1 (mod 97).

a = 53n = 97
53−1 mod 97 =
8.Number theory & modular arithmeticHard

Solve 6x ≡ 60 (mod 81). Give the smallest positive solution x.

6x ≡ 60 (mod 81)
x =
9.Algorithms & complexityMedium

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

  1. AO(n log n)
  2. BO(n)
  3. CO(log n)
  4. DO(n²)
10.Algorithms & complexityHard

Why does binary search not give O(log n) time on a singly linked list?

  1. AComparisons on linked list nodes are slower
  2. BBinary search needs at least twice the memory
  3. CLinked lists cannot be sorted
  4. DReaching the middle element takes O(n) steps, since there is no direct index access
11.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ASeparate chaining
  2. BDouble hashing
  3. CLinear probing
  4. DQuadratic probing
12.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. AThe need for a hash function
  2. BThe O(n) cost of reading a key's value
  3. CThe need to store keys in the table
  4. DPrimary clustering, where long runs of occupied slots form
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
31110122078417ABCDEF

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

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

A table stores the customer's address on every order row. Changing the address in one row but not in the others is an example of:

  1. AAn insertion anomaly
  2. BA phantom read
  3. CA deletion anomaly
  4. DAn update anomaly
16.Databases & SQLHard

Boyce–Codd normal form (BCNF) requires that for every non-trivial functional dependency X → Y:

  1. AY is a primary key
  2. BY is atomic
  3. CX contains NULL
  4. DX is a superkey
17.Computer networksMedium

How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?

  1. A0.125 ms
  2. B1 ms
  3. C12 ms
  4. D8 ms
18.Computer networksHard

Why does a browser warn when a site presents a self-signed certificate?

  1. AThe certificate is always expired
  2. BThe site must be using HTTP instead of HTTPS
  3. CNo trusted certificate authority vouches that the key belongs to that site
  4. DSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1891 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1891 hosts
p =
20.IPv4 subnettingHard
last octet = 39012806413201608141211networkhost

A host has the address 10.20.160.39/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.160.39/29
x =

3Solutions

1.Number systems

Answer: U = 178

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.

Invert and add one

78 = 0100 1110
inverted: 1011 0001
+1 → 1011 0010

Read it as unsigned

1011 0010₂ = 256 − 78 = 178

Answer

U = 178
2.Number systems

Answer: R = 140

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

XOR gives 1 where the two bits differ.

0110 0001
XOR 1110 1101
= 1000 1100

Value of the result

1000 1100₂ = 140

Answer

R = 140
3.Propositional logic & proofs

Answer: (r → p) ∧ (p → r)

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 means the implication holds in both directions: (a → b) ∧ (b → a).
r ↔ p ≡ (r → p) ∧ (p → r)

Check with the truth table

p = 1, r = 1: 1 | 1 1 1 1
p = 1, r = 0: 0 | 0 1 1 0
p = 0, r = 1: 0 | 0 0 1 0
p = 0, r = 0: 1 | 0 1 1 1

Answer

(r → p) ∧ (p → r)

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

4.Propositional logic & proofs

Answer: ∀x ∃y ¬R(x, y)

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) ≡ ∀x ∃y ¬R(x, y)

Answer

∀x ∃y ¬R(x, y)

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

AB′C → m(5)
ABC′ → m(6)
ABC → m(7)
ABC00011110010001030204151716

Group and read off

F = AC + AB

Count the literals

L = 4

Answer

L = 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 D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 0 → Q = 0
3: D = 0 → Q = 0
4: D = 0 → Q = 0
5: D = 1 → Q = 1
6: D = 0 → Q = 0

Count the changes

0 → 1 → 0 → 0 → 0 → 1 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: 53−1 mod 97 = 11

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·53 (mod 97).

r = 97, t = 0
r = 53, q = 1, t = 1
r = 44, q = 1, t = −1
r = 9, q = 4, t = 2
r = 8, q = 1, t = −9
r = 1, q = 8, t = 11

Inverse

1 ≡ 11·53 (mod 97)
x = 11
check: 53·11 = 583 = 6·97 + 1

Answer

53−1 mod 97 = 11
8.Number theory & modular arithmetic

Answer: x = 10

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(6, 81) = 3
2x ≡ 20 (mod 27)

Solve the reduced congruence

2−1 ≡ 14 (mod 27) → x ≡ 10 (mod 27)

Solutions modulo 81

x ∈ {10, 37, 64}
x = 10

Answer

x = 10
9.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

By the master theorem nlog₂ 2 = n dominates the constant work, so T(n) = Θ(n), as when finding a maximum by splitting in halves.

Answer

B · O(n)
10.Algorithms & complexity

Answer: D · Reaching the middle element takes O(n) steps, since there is no direct index access

Why this is the answer

Binary search needs O(1) access to the middle; on a linked list walking there costs linear time, so the total is O(n).

Answer

D · Reaching the middle element takes O(n) steps, since there is no direct index access
11.Data structures

Answer: A · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

A · Separate chaining
12.Data structures

Answer: D · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

D · Primary clustering, where long runs of occupied slots form
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

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: W = 36

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 (3): take
EF (4): take
DE (7): take
DF (8): skip, closes a cycle
BC (10): take
AC (11): skip, closes a cycle
BD (12): take

Total weight

W = 3 + 4 + 7 + 10 + 12 = 36

Answer

W = 36
15.Databases & SQL

Answer: D · An update anomaly

Why this is the answer

Redundant storage means one change must be made in many rows; missing some leaves the data inconsistent, which normalisation prevents.

Answer

D · An update anomaly
16.Databases & SQL

Answer: D · X is a superkey

Why this is the answer

BCNF demands that every determinant is a candidate key (superkey); 3NF allows an exception when Y is part of a candidate key.

Answer

D · X is a superkey
17.Computer networks

Answer: B · 1 ms

Why this is the answer

1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.

Answer

B · 1 ms
18.Computer networks

Answer: C · No trusted certificate authority vouches that the key belongs to that site

Why this is the answer

Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.

Answer

C · No trusted certificate authority vouches that the key belongs to that site
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 < 1891
211 − 2 = 2046 ≥ 1891 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 38

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 of this host

network: ⌊39/8⌋·8 = 32
broadcast: 32 + 8 − 1 = 39
usable hosts: 33 … 38

Answer

x = 38

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