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 complement?−128?64?32?16?8?4?2?1

Write −82 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −828 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1010 1101 AND 0010 0111. What is the result's value R in decimal?

1010 1101AND0010 0111
R =
3.Propositional logic & proofsMedium
¬(p ∨ q) → ¬¬r¬r → (p ∨ q)¬¬r → ¬(p ∨ q)(p ∨ q) ∧ ¬¬r

You want to prove (p ∨ q) → ¬r by contrapositive. Which implication do you prove instead? Click it.

(p ∨ q) → ¬r
  1. 1¬(p ∨ q) → ¬¬r
  2. 2¬r → (p ∨ q)
  3. 3¬¬r → ¬(p ∨ q)
  4. 4(p ∨ q) ∧ ¬¬r
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
ABCD0001111000011110111111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(1, 4, 7, 8, 9, 10, 13, 14, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X011031214X507061120130150141809011010

Find the minimal SOP form of F(A, B, C, D) = Σm(1, 2, 4, 8, 12) + d(0, 5), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(1, 2, 4, 8, 12) + d(0, 5)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 3) and x ≡ 3 (mod 13).

x ≡ 1 (mod 3)x ≡ 3 (mod 13)
x =
8.Number theory & modular arithmeticHard
235c = me​ mod n?m = cd​ mod nm(n, e) = (287, 11)d

Encrypt the message m = 235 with the RSA public key (n, e) = (287, 11). Find the ciphertext c.

n = 287e = 11m = 235
c =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Which definition of the class P is correct?

  1. ADecision problems solvable by a deterministic algorithm in polynomial time
  2. BProblems whose solutions can be checked but never found in polynomial time
  3. CProblems that can be solved in parallel
  4. DProblems that need exponential time on every computer
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(n²)
  2. BO(log n)
  3. CO(1)
  4. DO(n)
12.Data structuresHard

What is the running time of breadth-first search on a graph with V vertices stored as an adjacency matrix?

  1. AO(V + E)
  2. BO(E log V)
  3. CO(V)
  4. DO(V²)
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
6122101352011ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 5.

start: Dk = 5
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE
15.Databases & SQLMedium

A table has rows where col is NULL. Which rows does WHERE col = NULL return?

  1. AOnly the rows where col is NULL, as with IS NULL
  2. BAn error in every database
  3. CAll rows of the table
  4. DNo rows, because the comparison is unknown
16.Databases & SQLHard

When is decomposing R into R1 and R2 guaranteed to be lossless?

  1. AWhen the common attributes of R1 and R2 form a superkey of R1 or of R2
  2. BWhen R1 and R2 have no attributes in common
  3. CWhen every attribute of R appears in both R1 and R2
  4. DWhen R1 and R2 have the same number of columns
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. AHop count
  2. BBusiness policy
  3. CLink bandwidth
  4. DMeasured delay
18.Computer networksHard

A server's private key is stolen a year from now. Which TLS property keeps today's recorded sessions safe?

  1. AUsing a longer domain name
  2. BCompressing the data before encryption
  3. CForward secrecy from ephemeral Diffie-Hellman key exchange
  4. DSending the certificate twice
19.IPv4 subnettingMedium
255.255.255.19211111111111111111111111111000000network bitshost bits

A subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?

255.255.255.192
H =
20.IPv4 subnettingHard
last octet = 47012806413201618141211networkhost

A host has the address 10.20.80.47/30. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.80.47/30
x =

3Solutions

1.Number systems

Answer: U = 174

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.

Invert and add one

82 = 0101 0010
inverted: 1010 1101
+1 → 1010 1110

Read it as unsigned

1010 1110₂ = 256 − 82 = 174

Answer

U = 174
2.Number systems

Answer: R = 37

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

AND gives 1 only where both bits are 1.

1010 1101
AND 0010 0111
= 0010 0101

Value of the result

0010 0101₂ = 37

Answer

R = 37
3.Propositional logic & proofs

Answer: ¬¬r → ¬(p ∨ q)

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).

Parts of the implication

a = p ∨ q, b = ¬r

Contrapositive

¬b → ¬a = ¬¬r → ¬(p ∨ q)

Answer

¬¬r → ¬(p ∨ q)

The contrapositive is equivalent to the original statement, so proving it proves the statement.

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: AB = 01, CD = 00

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m4: m5, m6, m0, m12 → 0
ABCD0001111000011110111111111

The isolated 1

m4 = A′BC′D′ → AB = 01, CD = 00

Answer

AB = 01, CD = 00
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.

C′D′, A′C′, A′B′D′
ABCD0001111000011110X011031214X507061120130150141809011010

Minimal cover

F = C′D′ + A′C′ + A′B′D′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 16

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 = 1 + 3k

Second congruence

1 + 3k ≡ 3 (mod 13) → 3k ≡ 2 (mod 13)
k ≡ 9·2 ≡ 5 (mod 13)

Result

x = 1 + 3·5 = 16
x = 16
check: 16 mod 3 = 1, 16 mod 13 = 3

Answer

x = 16
8.Number theory & modular arithmetic

Answer: c = 65

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.

Repeated squaring

11 = 8 + 2 + 1
2351 ≡ 235 (mod 287)
2352 ≡ 121 (mod 287)
2354 ≡ 4 (mod 287)
2358 ≡ 16 (mod 287)
23511 ≡ 235·121·16 ≡ 65 (mod 287)

Ciphertext

c = 65

Answer

c = 65
9.Algorithms & complexity

Answer: A · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

A · O(mn)
10.Algorithms & complexity

Answer: A · Decision problems solvable by a deterministic algorithm in polynomial time

Why this is the answer

P contains decision problems with a deterministic polynomial-time algorithm, such as checking whether a graph is connected.

Answer

A · Decision problems solvable by a deterministic algorithm in polynomial time
11.Data structures

Answer: C · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

C · O(1)
12.Data structures

Answer: D · O(V²)

Why this is the answer

Each of the V vertices scans a row of V entries, giving O(V²); with adjacency lists BFS is O(V + E).

Answer

D · O(V²)
13.Graph theory

Answer: neither

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) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 3

Odd vertices

A, B, D, E → 4
neither

Answer

neither
14.Graph theory

Answer: edge AB

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. DF (5)
2. DE (13)
3. CE (10)
4. BC (2)
5. AB (6)

edge number 5

AB

Answer

edge AB
15.Databases & SQL

Answer: D · No rows, because the comparison is unknown

Why this is the answer

Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.

Answer

D · No rows, because the comparison is unknown
16.Databases & SQL

Answer: A · When the common attributes of R1 and R2 form a superkey of R1 or of R2

Why this is the answer

If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.

Answer

A · When the common attributes of R1 and R2 form a superkey of R1 or of R2
17.Computer networks

Answer: A · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

A · Hop count
18.Computer networks

Answer: C · Forward secrecy from ephemeral Diffie-Hellman key exchange

Why this is the answer

With ephemeral (EC)DHE, each session key is derived from temporary values that are discarded. The long-term key only signs the exchange, so stealing it later does not decrypt old traffic.

Answer

C · Forward secrecy from ephemeral Diffie-Hellman key exchange
19.IPv4 subnetting

Answer: H = 62

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

255.255.255.192 = /26

Usable hosts

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: x = 47

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: ⌊47/4⌋·4 = 44
broadcast: 44 + 4 − 1 = 47
usable hosts: 45 … 46

Answer

x = 47

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