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 −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1128 bit
U =
2.Number systemsHard
16 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬¬r → ¬(p ∧ q)¬(p ∧ q) → ¬¬r(p ∧ q) ∧ ¬¬r¬r → (p ∧ q)

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

(p ∧ q) → ¬r
  1. 1¬¬r → ¬(p ∧ q)
  2. 2¬(p ∧ q) → ¬¬r
  3. 3(p ∧ q) ∧ ¬¬r
  4. 4¬r → (p ∧ q)
4.Propositional logic & proofsHard
rq¬qr ∧ q

The premises r ∨ q; ¬r are given. Which conclusion follows from them? Click it.

r ∨ q¬r
  1. 1r
  2. 2q
  3. 3¬q
  4. 4r ∧ q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

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(0, 1, 6, 8, 12, 13)
  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
ABCD000111100001111000011312X41507X61120130150140809111110

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 5, 10, 11, 12) + d(4, 6), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(2, 3, 5, 10, 11, 12) + d(4, 6)
N =
7.Number theory & modular arithmeticMedium

Solve 45x ≡ 1 (mod 49). Give the solution x with 0 < x < 49.

45x ≡ 1 (mod 49)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n479m = cd​ mod n?(n, e) = (1763, 17)d

An RSA key uses p = 41, q = 43 and e = 17. Decrypt the ciphertext c = 479: find the message m.

p = 41q = 43e = 17c = 479
m =
9.Algorithms & complexityMedium

What is the worst-case time complexity of quicksort?

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

A problem is NP-complete if it is:

  1. AOutside NP but harder than all NP problems, such as the halting problem
  2. BSolvable in polynomial time
  3. CIn NP and every NP problem reduces to it in polynomial time
  4. DImpossible to solve
11.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(E)
  2. BO(V)
  3. CO(1)
  4. DO(log V)
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA sorted array updated by shifting
  2. BA deque holding indices in decreasing order of value
  3. CA singly linked list of the window's elements
  4. DA min-heap rebuilt for every window
13.Graph theoryMedium

A connected graph has 7 vertices and 14 edges. How many edges k must be removed to leave a spanning tree?

|V| = 7|E| = 14
k =
14.Graph theoryHard
713415771017ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

start: Around 4
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. ACombine the table with itself using UNION ALL and a WHERE filter
  2. BCreate a second copy of the table on disk
  3. CUse GROUP BY manager_id only
  4. DJoin the employees table with itself using two aliases
16.Databases & SQLHard

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

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

HTTP is stateless. How do websites usually keep a user logged in across requests?

  1. ABy keeping the TCP connection open for days
  2. BBy giving each user a fixed IP address
  3. CWith cookies holding a session identifier or token
  4. DBy storing the password in the URL
18.Computer networksHard

How does fragmentation differ in IPv6 compared with IPv4?

  1. AIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
  2. BIPv6 fragments at the transport layer inside TCP
  3. CIPv6 does not allow packets larger than 576 bytes
  4. DRouters never fragment IPv6 packets; only the source can, using path MTU discovery
19.IPv4 subnettingMedium
255.255.240.011111111111111111111000000000000network bitshost bits

A subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?

255.255.240.0
H =
20.IPv4 subnettingHard

The network 192.168.108.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 21 hosts, C: 47 hosts, D: 8 hosts, E: 4 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?

A: 10B: 21C: 47D: 8E: 4
x =

3Solutions

1.Number systems

Answer: U = 144

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

112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000

Read it as unsigned

1001 0000₂ = 256 − 112 = 144

Answer

U = 144
2.Number systems

Answer: M = 1536 KiB

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.

Bits and bytes

1024·768·16 = 12582912 bit = 1572864 B

Convert to KiB

1572864/1024 = 1536 KiB

Answer

M = 1536 KiB
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: q

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.

Rows with all premises true

q = 1, r = 0: r = 0, q = 1, ¬q = 0, r ∧ q = 0

Conclusion

q (disjunctive syllogism)

Answer

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

Answer: AB = 01, CD = 10

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.

m6: m7, m4, m2, m14 → 0
ABCD0001111000011110111111

The isolated 1

m6 = A′BCD′ → AB = 01, CD = 10

Answer

AB = 01, CD = 10
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.

B′C, BC′D′, A′CD′, A′BC′
ABCD000111100001111000011312X41507X61120130150140809111110

Minimal cover

F = B′C + BC′D′ + A′BC′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 12

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).

Inverse of 45

r = 49, t = 0
r = 45, q = 1, t = 1
r = 4, q = 11, t = −1
r = 1, q = 4, t = 12
45−1 ≡ 12 (mod 49)

Solve

x ≡ 12·1 = 12 ≡ 12 (mod 49)
check: 45·12 = 540 ≡ 1 (mod 49)

Answer

x = 12
8.Number theory & modular arithmetic

Answer: m = 380

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 = 1763, φ(n) = 1680
d = 17−1 mod 1680 = 593

Decrypt by repeated squaring

593 = 512 + 64 + 16 + 1
4791 ≡ 479 (mod 1763)
4792 ≡ 251 (mod 1763)
4794 ≡ 1296 (mod 1763)
4798 ≡ 1240 (mod 1763)
47916 ≡ 264 (mod 1763)
47932 ≡ 939 (mod 1763)
47964 ≡ 221 (mod 1763)
479128 ≡ 1240 (mod 1763)
479256 ≡ 264 (mod 1763)
479512 ≡ 939 (mod 1763)
479593 ≡ 479·264·221·939 ≡ 380 (mod 1763)

Message

m = 380
check: 38017 mod 1763 = 479

Answer

m = 380
9.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).

Answer

C · O(n²)
10.Algorithms & complexity

Answer: C · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

C · In NP and every NP problem reduces to it in polynomial time
11.Data structures

Answer: C · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

C · O(1)
12.Data structures

Answer: B · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

B · A deque holding indices in decreasing order of value
13.Graph theory

Answer: k = 8

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.

Edges of a spanning tree

7 − 1 = 6

Edges to remove

k = 14 − 6 = 8

Answer

k = 8
14.Graph theory

Answer: G

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

1. A (0): B: 7, C: 13
2. B (7): C: 11, D: 22
3. C (11): G: 12
4. G (12): E: 19
5. E (19): F: 29
6. D (22)
7. F (29)

round 4

G

Answer

G
15.Databases & SQL

Answer: D · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

D · Join the employees table with itself using two aliases
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: C · With cookies holding a session identifier or token

Why this is the answer

The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.

Answer

C · With cookies holding a session identifier or token
18.Computer networks

Answer: D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery

Why this is the answer

An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.

Answer

D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
19.IPv4 subnetting

Answer: H = 4094

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

255.255.240.0 = /20

Usable hosts

212 − 2 = 4094

Answer

H = 4094
20.IPv4 subnetting

Answer: x = 112

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

C: 47 hosts → 26 = 64 (/26)
B: 21 hosts → 25 = 32 (/27)
A: 10 hosts → 24 = 16 (/28)
D: 8 hosts → 24 = 16 (/28)
E: 4 hosts → 23 = 8 (/29)

Allocation, largest first

C: .0 – .63
B: .64 – .95
A: .96 – .111
D: .112 – .127
E: .128 – .135

Answer

x = 112

LAN D: 192.168.108.112/28.

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