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 −108 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1088 bit
U =
2.Number systemsHard
1274 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 1274 items its own binary code of the same length. What is the smallest number of bits b needed?

1274
b = bit
3.Propositional logic & proofsMedium
¬r ∧ p¬r → ¬pr ∧ ¬p¬r ∨ ¬p

Which formula is logically equivalent to ¬(r → p)? Click it.

¬(r → p)
  1. 1¬r ∧ p
  2. 2¬r → ¬p
  3. 3r ∧ ¬p
  4. 4¬r ∨ ¬p
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
ABC00011110011001131214051716

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 2, 3, 4, 6, 7). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 2, 3, 4, 6, 7)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111000010302X4151706X120131151141809011X10

For F(A, B, C, D) = Σm(5, 7, 8, 14, 15) + d(4, 10, 12), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(5, 7, 8, 14, 15) + d(4, 10, 12)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 4 (mod 6) and x ≡ 1 (mod 11).

x ≡ 4 (mod 6)x ≡ 1 (mod 11)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n587m = cd​ mod n?(n, e) = (713, 19)d

An RSA key uses p = 23, q = 31 and e = 19. Decrypt the ciphertext c = 587: find the message m.

p = 23q = 31e = 19c = 587
m =
9.Algorithms & complexityMedium

You sort an array of n elements once and then answer k membership queries with binary search. What is the total cost?

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

Which statement about P and NP is correct?

  1. ANP problems cannot be verified quickly
  2. BNP means "not polynomial"
  3. CIt has been proven that P ≠ NP
  4. DEvery problem in P is also in NP
11.Data structuresMedium

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

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

The keys 5, 3, 8, 1 are inserted in this order into an empty array-based binary min-heap. What is the final array?

  1. A[3, 1, 8, 5]
  2. B[1, 3, 8, 5]
  3. C[1, 5, 8, 3]
  4. D[1, 3, 5, 8]
13.Graph theoryMedium

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

3 × deg 3
L =
14.Graph theoryHard
1121215528911ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

What does EXCEPT (called MINUS in Oracle) return?

  1. ARows of the first query that do not appear in the second
  2. BRows of the second query that do not appear in the first one
  3. CRows that appear in both queries
  4. DAll rows from both queries without duplicates
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AAll rows from both tables, matched where possible and NULL-filled otherwise
  2. BThe Cartesian product, pairing each row of one table with every row of the other
  3. COnly matching rows
  4. DOnly unmatched rows
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

In a SYN flood attack, what resource is the attacker trying to exhaust?

  1. AThe server's disk space
  2. BThe switch's MAC address table
  3. CThe server's table of half-open connections
  4. DThe DNS cache of the client
19.IPv4 subnettingMedium
255.255.255.22411111111111111111111111111100000network bitshost bits

A subnet uses the mask 255.255.255.224. How many usable host addresses H does it have?

255.255.255.224
H =
20.IPv4 subnettingHard
last octet = 18012806403211608041201networkhost

A host has the address 10.20.48.18/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.48.18/27
x =

3Solutions

1.Number systems

Answer: U = 148

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

108 = 0110 1100
inverted: 1001 0011
+1 → 1001 0100

Read it as unsigned

1001 0100₂ = 256 − 108 = 148

Answer

U = 148
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 < 1274
211 = 2048 ≥ 1274

Answer

b = 11
3.Propositional logic & proofs

Answer: r ∧ ¬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

¬(a → b) ≡ ¬(¬a ∨ b) ≡ a ∧ ¬b: the only row where an implication fails.
¬(r → p) ≡ r ∧ ¬p

Check with the truth table

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

Answer

r ∧ ¬p

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: L = 2

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.

ABC00011110011001131214051716

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

C′
B

Minimal SOP form

F = C′ + B
L = 2

Answer

L = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 1

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.

BCD, AD′, A′BD, A′BC′, ABC
ABCD000111100001111000010302X4151706X120131151141809011X10

Essential prime implicants

AD′
N = 1

Answer

N = 1
7.Number theory & modular arithmetic

Answer: x = 34

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 + 6k

Second congruence

4 + 6k ≡ 1 (mod 11) → 6k ≡ 8 (mod 11)
k ≡ 2·8 ≡ 5 (mod 11)

Result

x = 4 + 6·5 = 34
x = 34
check: 34 mod 6 = 4, 34 mod 11 = 1

Answer

x = 34
8.Number theory & modular arithmetic

Answer: m = 108

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.

Private exponent

n = 713, φ(n) = 660
d = 19−1 mod 660 = 139

Decrypt by repeated squaring

139 = 128 + 8 + 2 + 1
5871 ≡ 587 (mod 713)
5872 ≡ 190 (mod 713)
5874 ≡ 450 (mod 713)
5878 ≡ 8 (mod 713)
58716 ≡ 64 (mod 713)
58732 ≡ 531 (mod 713)
58764 ≡ 326 (mod 713)
587128 ≡ 39 (mod 713)
587139 ≡ 587·190·8·39 ≡ 108 (mod 713)

Message

m = 108
check: 10819 mod 713 = 587

Answer

m = 108
9.Algorithms & complexity

Answer: D · O(n log n + k log n)

Why this is the answer

Sorting costs O(n log n) and each query O(log n); this beats k linear scans (O(nk)) when k is large.

Answer

D · O(n log n + k log n)
10.Algorithms & complexity

Answer: D · Every problem in P is also in NP

Why this is the answer

P ⊆ NP: if you can solve a problem quickly you can verify a solution quickly. Whether P = NP is still open.

Answer

D · Every problem in P is also in NP
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: B · [1, 3, 8, 5]

Why this is the answer

After 5, 3, 8 the array is [3, 5, 8]; 1 goes to index 3, swaps with its parent 5 and then with 3, giving [1, 3, 8, 5].

Answer

B · [1, 3, 8, 5]
13.Graph theory

Answer: L = 5

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 + 9 = 2(L + 3) − 2
L = 5

Answer

L = 5

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

14.Graph theory

Answer: d(F) = 21

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.

Dijkstra round by round

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 1, C: 12
B (1): D: 16, E: 12
C (12)
E (12): D: 14, F: 21
D (14)
F (21)

Shortest path

A → B → E → F
d(F) = 21

Answer

d(F) = 21
15.Databases & SQL

Answer: A · Rows of the first query that do not appear in the second

Why this is the answer

A EXCEPT B gives the rows in A that are not in B, so swapping the queries gives a different result.

Answer

A · Rows of the first query that do not appear in the second
16.Databases & SQL

Answer: A · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

A · All rows from both tables, matched where possible and NULL-filled otherwise
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: C · The server's table of half-open connections

Why this is the answer

Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.

Answer

C · The server's table of half-open connections
19.IPv4 subnetting

Answer: H = 30

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

255.255.255.224 = /27

Usable hosts

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 31

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 of this host

network: ⌊18/32⌋·32 = 0
broadcast: 0 + 32 − 1 = 31
usable hosts: 1 … 30

Answer

x = 31

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