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 −11 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −118 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 0110 << 1. What is the result's value R in decimal?

0011 0110<< 1
R =
3.Propositional logic & proofsMedium
r ∧ q(r → q) ∧ (q → r)(r → q) ∨ (q → r)(r → q) ∧ (¬q → ¬r)

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

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

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

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
  2. 2∃x ∃y (R(x, y) ∧ ¬R(y, x))
  3. 3∃x ∃y (R(x, y) → ¬R(y, x))
  4. 4∀x ∀y (R(x, y) ∧ ¬R(y, x))
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(2, 3, 4, 6, 10, 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
ABCD0001111000011110101103X214050716112113X1501418X9011010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 4, 6, 8, 12, 13) + d(2, 9, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 4, 6, 8, 12, 13) + d(2, 9, 15)
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(513), the number of integers from 1 to 513 that are coprime to 513.

n = 513 = 33·19
φ(513) =
8.Number theory & modular arithmeticHard
mc = me​ mod n279m = cd​ mod n?(n, e) = (731, 23)d

An RSA key uses p = 17, q = 43 and e = 23. Decrypt the ciphertext c = 279: find the message m.

p = 17q = 43e = 23c = 279
m =
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. CDynamic programming only works on graph problems, not on arrays
  4. DMerge sort has no recursive structure
10.Algorithms & complexityHard

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

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

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

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

Which statement about heapsort is correct?

  1. AIt runs in O(n) time on any input
  2. BIt is stable and needs O(n) extra memory
  3. CIt sorts in place in O(n log n) worst-case time but is not stable
  4. DIt needs a binary search tree as extra storage, so it uses O(n) extra memory
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
12811159155413ABCDEFG

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

The orders table has 5 rows, each referring to an existing customer. The customers table has 3 rows. How many rows does an INNER JOIN on customer_id return?

  1. A3
  2. B15
  3. C5
  4. D8
16.Databases & SQLHard

An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?

  1. AWrite-ahead logging
  2. BStrict two-phase locking with row locks
  3. CDatabase sharding
  4. DOptimistic concurrency control
17.Computer networksMedium

Which medium access method did classic shared (half-duplex) Ethernet use?

  1. ATDMA
  2. BToken passing
  3. CCSMA/CD
  4. DCSMA/CA
18.Computer networksHard

DNS queries normally use UDP. When does DNS typically use TCP instead?

  1. AFor responses too large for UDP and for zone transfers
  2. BOnly for reverse lookups of IP addresses
  3. COnly when the domain name is longer than ten characters
  4. DFor every query sent from a web browser
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/243 subnets
H =
20.IPv4 subnettingHard
last octet = 178112806413211608041201networkhost

A host has the address 10.20.11.178/28. What is the fourth octet x of the last usable host address in its subnet?

10.20.11.178/28
x =

3Solutions

1.Number systems

Answer: U = 245

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

11 = 0000 1011
inverted: 1111 0100
+1 → 1111 0101

Read it as unsigned

1111 0101₂ = 256 − 11 = 245

Answer

U = 245
2.Number systems

Answer: R = 108

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 left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0011 0110 << 1
= 0110 1100

Value of the result

0110 1100₂ = 108

Answer

R = 108
3.Propositional logic & proofs

Answer: (r → q) ∧ (q → 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 ↔ q ≡ (r → q) ∧ (q → r)

Check with the truth table

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

Answer

(r → q) ∧ (q → r)

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

4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, 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 ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, 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 = 11, CD = 11

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.

m15: m14, m13, m11, m7 → 0
ABCD0001111000011110111111

The isolated 1

m15 = ABCD → AB = 11, CD = 11

Answer

AB = 11, CD = 11
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′, B′C′, A′D′, AC′, ABD
ABCD0001111000011110101103X214050716112113X1501418X9011010

Minimal cover

F = B′C′ + A′D′ + AC′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: φ(513) = 324

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).

Prime factors

513 = 33·19

Totient

φ(513) = φ(27)·φ(19) = 18·18 = 324

Answer

φ(513) = 324
8.Number theory & modular arithmetic

Answer: m = 250

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 = 731, φ(n) = 672
d = 23−1 mod 672 = 263

Decrypt by repeated squaring

263 = 256 + 4 + 2 + 1
2791 ≡ 279 (mod 731)
2792 ≡ 355 (mod 731)
2794 ≡ 293 (mod 731)
2798 ≡ 322 (mod 731)
27916 ≡ 613 (mod 731)
27932 ≡ 35 (mod 731)
27964 ≡ 494 (mod 731)
279128 ≡ 613 (mod 731)
279256 ≡ 35 (mod 731)
279263 ≡ 279·355·293·35 ≡ 250 (mod 731)

Message

m = 250
check: 25023 mod 731 = 279

Answer

m = 250
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: B · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

B · O(n log n)
11.Data structures

Answer: D · 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

D · The call stack
12.Data structures

Answer: C · It sorts in place in O(n log n) worst-case time but is not stable

Why this is the answer

Heapsort builds a max-heap and extracts the maximum n times within the same array; swaps over long distances can reorder equal keys.

Answer

C · It sorts in place in O(n log n) worst-case time but is not stable
13.Graph theory

Answer: Euler path, no 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) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 2

Odd vertices

A, B → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: d(F) = 32

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: 12
B (12): C: 20, D: 23, G: 27
C (20): E: 35, G: 25
D (23): F: 32
G (25)
F (32)
E (35)

Shortest path

A → B → D → F
d(F) = 32

Answer

d(F) = 32
15.Databases & SQL

Answer: C · 5

Why this is the answer

Every order finds exactly one matching customer, so the join returns one row per order: 5 rows.

Answer

C · 5
16.Databases & SQL

Answer: D · Optimistic concurrency control

Why this is the answer

Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.

Answer

D · Optimistic concurrency control
17.Computer networks

Answer: C · CSMA/CD

Why this is the answer

Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.

Answer

C · CSMA/CD
18.Computer networks

Answer: A · For responses too large for UDP and for zone transfers

Why this is the answer

If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.

Answer

A · For responses too large for UDP and for zone transfers
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).

Bits to borrow

22 = 4 ≥ 3 → b = 2, /26

Hosts per subnet

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: x = 190

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

24 = 16

Subnet of this host

network: ⌊178/16⌋·16 = 176
broadcast: 176 + 16 − 1 = 191
usable hosts: 177 … 190

Answer

x = 190

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