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 complement1−12816403211608041201

The 8-bit pattern 1101 0010 stores a signed integer in two's complement. What is its value N in decimal?

1101 00108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0110 0011 >> 2. What is the result's value R in decimal?

0110 0011>> 2
R =
3.Propositional logic & proofsMedium
(q ∨ r) ∧ (q ∨ p)q ∧ (r ∨ p)(q ∨ r) ∧ p(q ∧ r) ∨ (q ∧ p)

Which formula is logically equivalent to q ∨ (r ∧ p)? Click it.

q ∨ (r ∧ p)
  1. 1(q ∨ r) ∧ (q ∨ p)
  2. 2q ∧ (r ∨ p)
  3. 3(q ∨ r) ∧ p
  4. 4(q ∧ r) ∨ (q ∧ p)
4.Propositional logic & proofsHard
abb → a¬b¬a11100100

The argument with premises b → a; ¬b and conclusion ¬a is invalid. Click the row of the truth table that is a counterexample.

b → a¬b∴ ¬a
  1. 1a = 1, b = 1
  2. 2a = 1, b = 0
  3. 3a = 0, b = 1
  4. 4a = 0, b = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = B′C + A′B + A′B′C + ABC′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101001X31214150716112013015114X819011110

For F(A, B, C, D) = Σm(0, 2, 4, 5, 6, 9, 10, 12, 14) + d(3, 8), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 2, 4, 5, 6, 9, 10, 12, 14) + d(3, 8)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 39 modulo 43: the number x with 0 < x < 43 and 39·x ≡ 1 (mod 43).

a = 39n = 43
39−1 mod 43 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 44121 mod 53.

a = 44k = 121n = 53
44121 mod 53 =
9.Algorithms & complexityMedium

Why does dynamic programming give no real speed-up for merge sort?

  1. AMerge sort is already O(n)
  2. BIts subproblems do not overlap, so there is nothing to reuse
  3. CMerge sort has no recursive structure
  4. DDynamic programming only works on graph problems, not on arrays
10.Algorithms & complexityHard

What is the time complexity of building a binary heap from n unsorted elements with the bottom-up method?

  1. AO(n)
  2. BO(log n)
  3. CO(n log n)
  4. DO(n²)
11.Data structuresMedium

An adjacency matrix for a graph with V vertices uses how much memory?

  1. AO(log V)
  2. BO(V + E)
  3. CO(E)
  4. DO(V²)
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 left rotation at the child, then a right rotation at the node
  2. BRecolouring the node and its child
  3. CA single left rotation at the unbalanced node
  4. DA single right rotation at the unbalanced node
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
162031914815ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex E. Click the edge it adds as edge number 2.

start: Ek = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

In relational algebra, the selection operator (σ) corresponds to which SQL clause?

  1. AWHERE
  2. BORDER BY
  3. CSELECT list
  4. DGROUP BY
16.Databases & SQLHard

A correlated subquery is one that:

  1. AUses only aggregate functions
  2. BJoins two tables with a foreign key
  3. CReferences columns of the outer query and is evaluated per outer row
  4. DRuns only once before the outer query and caches its result for every row
17.Computer networksMedium

A browser receives HTTP status code 404. What does it mean?

  1. AThe requested resource was not found
  2. BThe server had an internal error
  3. CThe request succeeded
  4. DThe client must log in first
18.Computer networksHard

A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?

  1. AGive the PC a different MAC address so the router can find it from outside
  2. BConfigure port forwarding on the router to the PC's private address
  3. CTurn off the PC's DNS client
  4. DChange the PC's subnet mask to /32
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.78.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 7?

192.168.78.0/24/27subnet 7 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet = 120012816413211618040201networkhost

A host has the address 172.20.120.251/20. What is the third octet x of its network address?

172.20.120.251/20
x =

3Solutions

1.Number systems

Answer: N = −46

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 + 64 + 16 + 2 = −46

Answer

N = −46
2.Number systems

Answer: R = 24

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.

0110 0011 >> 2
= 0001 1000

Value of the result

0001 1000₂ = 24

Answer

R = 24
3.Propositional logic & proofs

Answer: (q ∨ r) ∧ (q ∨ p)

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

∨ distributes over ∧: a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
q ∨ (r ∧ p) ≡ (q ∨ r) ∧ (q ∨ p)

Check with the truth table

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

Answer

(q ∨ r) ∧ (q ∨ p)

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

4.Propositional logic & proofs

Answer: a = 1, b = 0

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.

Premises and conclusion row by row

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

Counterexample

a = 1, b = 0

Answer

a = 1, b = 0

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: L = 6

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

B′C → m(1, 5)
A′B → m(2, 3)
A′B′C → m(1)
ABC′ → m(6)
ABC00011110010011131204150716

Group and read off

F = B′C + BC′ + A′C

Count the literals

L = 6

Answer

L = 6

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

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

Answer: N = 3

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.

D′, A′B′C, A′BC′, AB′C′
ABCD00011110000111101001X31214150716112013015114X819011110

Essential prime implicants

D′, A′BC′, AB′C′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 39−1 mod 43 = 32

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·39 (mod 43).

r = 43, t = 0
r = 39, q = 1, t = 1
r = 4, q = 9, t = −1
r = 3, q = 1, t = 10
r = 1, q = 3, t = −11

Inverse

1 ≡ −11·39 (mod 43)
x = 32
check: 39·32 = 1248 = 29·43 + 1

Answer

39−1 mod 43 = 32
8.Number theory & modular arithmetic

Answer: 44121 mod 53 = 42

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.

Binary exponent

121 = 11110012 = 64 + 32 + 16 + 8 + 1

Repeated squaring

441 ≡ 44 (mod 53)
442 ≡ 28 (mod 53)
444 ≡ 42 (mod 53)
448 ≡ 15 (mod 53)
4416 ≡ 13 (mod 53)
4432 ≡ 10 (mod 53)
4464 ≡ 47 (mod 53)

Multiply the needed powers

44121 ≡ 44·15·13·10·47 ≡ 42 (mod 53)

Answer

44121 mod 53 = 42
9.Algorithms & complexity

Answer: B · Its subproblems do not overlap, so there is nothing to reuse

Why this is the answer

Merge sort's halves are disjoint and each is solved once; DP helps only when subproblems recur, as in Fibonacci.

Answer

B · Its subproblems do not overlap, so there is nothing to reuse
10.Algorithms & complexity

Answer: A · O(n)

Why this is the answer

Summing sift-down costs over the levels gives a convergent series, so heapify runs in linear time.

Answer

A · O(n)
11.Data structures

Answer: D · O(V²)

Why this is the answer

An adjacency matrix stores V × V entries, efficient for dense graphs; adjacency lists use O(V + E), better for sparse graphs.

Answer

D · O(V²)
12.Data structures

Answer: D · 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

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

Answer: H = 360

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.

Count

H = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: edge CE

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.

Prim’s algorithm

1. DE (8)
2. CE (14)
3. BC (3)
4. AB (16)

edge number 2

CE

Answer

edge CE
15.Databases & SQL

Answer: A · WHERE

Why this is the answer

σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.

Answer

A · WHERE
16.Databases & SQL

Answer: C · References columns of the outer query and is evaluated per outer row

Why this is the answer

Correlated subqueries depend on the current outer row, so they are logically evaluated per row (optimisers may rewrite them as joins).

Answer

C · References columns of the outer query and is evaluated per outer row
17.Computer networks

Answer: A · The requested resource was not found

Why this is the answer

404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.

Answer

A · The requested resource was not found
18.Computer networks

Answer: B · Configure port forwarding on the router to the PC's private address

Why this is the answer

NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.

Answer

B · Configure port forwarding on the router to the PC's private address
19.IPv4 subnetting

Answer: x = 192

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

25 = 32

Subnet number 7

network: (7 − 1)·32 = 192
broadcast: 192 + 32 − 1 = 223

Answer

x = 192
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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