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 −115 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1158 bit
U =
2.Number systemsHard
8 bit1 px640 px480 pxM = ? KiB

An uncompressed image is 640 × 480 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

640 × 480 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬q ∧ p¬q ∨ ¬p¬q ∧ ¬pq ∨ ¬p

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

¬(q ∨ p)
  1. 1¬q ∧ p
  2. 2¬q ∨ ¬p
  3. 3¬q ∧ ¬p
  4. 4q ∨ ¬p
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((p ∨ ¬q) ↔ (p ∨ r)) ∨ (s ↔ q) true? Call this number N.

((p ∨ ¬q) ↔ (p ∨ r)) ∨ (s ↔ q)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011302040507161121130151141809111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 3, 6, 8, 11, 12, 13, 14). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 3, 6, 8, 11, 12, 13, 14)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 8 modulo 23: the number x with 0 < x < 23 and 8·x ≡ 1 (mod 23).

a = 8n = 23
8−1 mod 23 =
8.Number theory & modular arithmeticHard

19 is prime. Use Fermat’s little theorem to find 479441 mod 19.

p = 19k = 79441
479441 mod 19 =
9.Algorithms & complexityMedium

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

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

Why can Dijkstra's algorithm give wrong results when some edge weights are negative?

  1. AIt goes into an infinite loop on every negative edge
  2. BIt only works on undirected graphs
  3. CA vertex is finalised when extracted, but a later negative edge could still give it a shorter path
  4. DNegative numbers cannot be stored in a priority queue
11.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AA larger table is allocated and every key is inserted again
  2. BOnly the colliding keys are moved to a new overflow list at the end of the table
  3. CThe existing keys are sorted in place
  4. DThe oldest keys are deleted to make room
12.Data structuresHard

Why do databases often use B+ trees rather than plain B-trees for indexes?

  1. AB+ trees do not need to be balanced
  2. BAll records are in linked leaves, which makes range scans efficient
  3. CB+ trees use binary internal nodes, which reduces the number of disk reads
  4. DB+ trees store data only in the root
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
51014115491381ABCDEFG

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

Which query returns the highest salary in each department?

  1. ASELECT dept, MAX(salary) FROM emp GROUP BY dept
  2. BSELECT dept, MAX(salary) FROM emp ORDER BY dept
  3. CSELECT MAX(dept), salary FROM emp GROUP BY salary
  4. DSELECT dept, salary FROM emp WHERE salary = MAX(salary)
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index that contains all columns the query needs, so the table itself need not be read
  2. BAn index that is rebuilt every night
  3. CAn index that covers every column of every table
  4. DAn index on the primary key only
17.Computer networksMedium

What is the role of a default gateway on a host?

  1. AIt is the switch port the host is connected to
  2. BIt is the DNS server
  3. CIt is the router used for traffic to other networks
  4. DIt assigns MAC addresses
18.Computer networksHard

A remote-terminal application feels sluggish because small keystroke packets are held back. Which TCP mechanism is often responsible?

  1. ASlow start
  2. BNagle's algorithm
  3. CFast retransmit
  4. DThe three-way handshake
19.IPv4 subnettingMedium
/16 → /1911111111111111111110000000000000network bitshost bits

The network 172.17.0.0/16 is divided into equal subnets with prefix /19. How many subnets N are there?

172.17.0.0/16/19
N =
20.IPv4 subnettingHard
last octet = 32012806413201608040201networkhost

A host has the address 10.20.116.32/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.116.32/29
x =

3Solutions

1.Number systems

Answer: U = 141

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

115 = 0111 0011
inverted: 1000 1100
+1 → 1000 1101

Read it as unsigned

1000 1101₂ = 256 − 115 = 141

Answer

U = 141
2.Number systems

Answer: M = 300 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

640·480·8 = 2457600 bit = 307200 B

Convert to KiB

307200/1024 = 300 KiB

Answer

M = 300 KiB
3.Propositional logic & proofs

Answer: ¬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

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(q ∨ p) ≡ ¬q ∧ ¬p

Check with the truth table

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

Answer

¬q ∧ ¬p

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

4.Propositional logic & proofs

Answer: N = 14

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

¬q; p ∨ ¬q; p ∨ r; (p ∨ ¬q) ↔ (p ∨ r); s ↔ q; ((p ∨ ¬q) ↔ (p ∨ r)) ∨ (s ↔ q)

Split on p

p = 1: 8
p = 0: 6
N = 8 + 6 = 14

Answer

N = 14
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 12

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.

ABCD000111100001111010011302040507161121130151141809111010

Group the 1s

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

B′C′D′
B′CD
BCD′
ABC′

Minimal SOP form

F = B′C′D′ + B′CD + BCD′ + ABC′
L = 12

Answer

L = 12
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 0 → Q = 1
2: J = 1, K = 0 → Q = 1
3: J = 0, K = 1 → Q = 0
4: J = 1, K = 0 → Q = 1
5: J = 1, K = 1 → Q = 0
6: J = 0, K = 0 → Q = 0
7: J = 0, K = 0 → Q = 0

Count the changes

0 → 1 → 1 → 0 → 1 → 0 → 0 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: 8−1 mod 23 = 3

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·8 (mod 23).

r = 23, t = 0
r = 8, q = 2, t = 1
r = 7, q = 1, t = −2
r = 1, q = 7, t = 3

Inverse

1 ≡ 3·8 (mod 23)
x = 3
check: 8·3 = 24 = 1·23 + 1

Answer

8−1 mod 23 = 3
8.Number theory & modular arithmetic

Answer: 479441 mod 19 = 6

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.

Reduce the exponent

79441 mod 18 = 7
479441 ≡ 47 (mod 19)

Small power

47 mod 19 = 6

Answer

479441 mod 19 = 6
9.Algorithms & complexity

Answer: C · 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

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

Answer: C · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path

Why this is the answer

The greedy step relies on non-negative weights so that extracted distances are final; Bellman–Ford is used when weights may be negative.

Answer

C · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path
11.Data structures

Answer: A · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

A · A larger table is allocated and every key is inserted again
12.Data structures

Answer: B · All records are in linked leaves, which makes range scans efficient

Why this is the answer

In a B+ tree internal nodes only guide the search and the leaves are linked in key order, so a range query finds the first key and then walks along the leaves.

Answer

B · All records are in linked leaves, which makes range scans efficient
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) = 2, deg(C) = 3, deg(D) = 2, deg(E) = 4, deg(F) = 2

Odd vertices

A, C → 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) = 28

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: 5, C: 10
B (5): G: 18
C (10): E: 21
G (18): D: 26, E: 19
E (19): D: 24, F: 28
D (24)
F (28)

Shortest path

A → B → G → E → F
d(F) = 28

Answer

d(F) = 28
15.Databases & SQL

Answer: A · SELECT dept, MAX(salary) FROM emp GROUP BY dept

Why this is the answer

GROUP BY dept creates a group per department and MAX(salary) picks the highest value in each.

Answer

A · SELECT dept, MAX(salary) FROM emp GROUP BY dept
16.Databases & SQL

Answer: A · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

A · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: C · It is the router used for traffic to other networks

Why this is the answer

The host sends packets for non-local destinations to the default gateway, usually its local router.

Answer

C · It is the router used for traffic to other networks
18.Computer networks

Answer: B · Nagle's algorithm

Why this is the answer

Nagle's algorithm waits to bundle small segments while earlier data is unacknowledged. This saves bandwidth but adds delay for interactive traffic.

Answer

B · Nagle's algorithm
19.IPv4 subnetting

Answer: N = 8

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).

Borrowed bits

19 − 16 = 3

Number of subnets

N = 23 = 8

Answer

N = 8

Each /19 subnet has 213 − 2 = 8190 usable hosts.

20.IPv4 subnetting

Answer: x = 39

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

23 = 8

Subnet of this host

network: ⌊32/8⌋·8 = 32
broadcast: 32 + 8 − 1 = 39
usable hosts: 33 … 38

Answer

x = 39

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