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 −4 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −48 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0001 1101 >> 2. What is the result's value R in decimal?

0001 1101>> 2
R =
3.Propositional logic & proofsMedium
¬r → ¬pp → r¬r ∨ pr ∨ ¬p

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

r → p
  1. 1¬r → ¬p
  2. 2p → r
  3. 3¬r ∨ p
  4. 4r ∨ ¬p
4.Propositional logic & proofsHard
∀x (F(x) → ¬G(x))∃x (F(x) ∧ ¬G(x))∀x (F(x) ∧ ¬G(x))∃x (F(x) → ¬G(x))

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

¬∀x (F(x) → G(x))
  1. 1∀x (F(x) → ¬G(x))
  2. 2∃x (F(x) ∧ ¬G(x))
  3. 3∀x (F(x) ∧ ¬G(x))
  4. 4∃x (F(x) → ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110011001130214151706

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 3, 4, 5, 7). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 3, 4, 5, 7)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKUP

A 3-bit binary up counter starts at 6 (decimal, Q = 110). Which count Q (in decimal) does it show after 12 clock pulses?

Q0 = 612n = 3
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 8 (mod 10) and x ≡ 6 (mod 13).

x ≡ 8 (mod 10)x ≡ 6 (mod 13)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (301, 17)d = ?

An RSA key uses the primes p = 7 and q = 43 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).

p = 7q = 43e = 17
d =
9.Algorithms & complexityMedium

With coins of 1, 3 and 4 and an amount of 6, how many coins does the greedy "largest coin first" method use, compared to the optimum?

  1. AGreedy uses 4 coins; the optimum is 3
  2. BGreedy uses 2 coins; the optimum is 2
  3. CGreedy uses 3 coins; the optimum is 2
  4. DGreedy uses 3 coins; the optimum is 3
10.Algorithms & complexityHard

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, because insertion sort is not a comparison sort
  2. BYes, but only for arrays shorter than 100
  3. CYes, the lower bound is therefore wrong
  4. DNo, the lower bound applies to the worst case over all inputs, not to every input
11.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AThe leftmost node of its right subtree
  2. BIts parent, always
  3. CThe rightmost node of its left subtree
  4. DIts right child, always
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA hash map combined with a doubly linked list
  2. BA binary search tree only
  3. CA single sorted array
  4. DA stack only
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
411111253143ABCDEFG

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

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

Which query returns the number of orders per customer?

  1. ASELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
  2. BSELECT DISTINCT COUNT(*) FROM orders
  3. CSELECT customer_id FROM orders WHERE COUNT(*) > 0
  4. DSELECT customer_id, COUNT(*) FROM orders ORDER BY customer_id
16.Databases & SQLHard

Two-phase locking (2PL) ensures:

  1. AThat deadlocks can never happen
  2. BThat no locks are needed
  3. CConflict-serializable schedules
  4. DThat transactions never wait
17.Computer networksMedium

Which TCP flag signals that a sender has finished sending data?

  1. ASYN
  2. BFIN
  3. CPSH
  4. DURG
18.Computer networksHard

What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?

  1. AThe connection stays encrypted without TLS
  2. BResponses can arrive in any order on one stream, so slow replies do not block others
  3. CThe server can push updates over UDP
  4. DSeveral requests reuse one TCP connection, avoiding repeated handshakes
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1776 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1776 hosts
p =
20.IPv4 subnettingHard
third octet = 228112816413201608140201networkhost

A host has the address 172.20.228.62/23. What is the third octet x of its network address?

172.20.228.62/23
x =

3Solutions

1.Number systems

Answer: U = 252

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

4 = 0000 0100
inverted: 1111 1011
+1 → 1111 1100

Read it as unsigned

1111 1100₂ = 256 − 4 = 252

Answer

U = 252
2.Number systems

Answer: R = 7

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 right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0001 1101 >> 2
= 0000 0111

Value of the result

0000 0111₂ = 7

Answer

R = 7
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 is 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
r → p ≡ ¬r ∨ p

Check with the truth table

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

Answer

¬r ∨ p

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

4.Propositional logic & proofs

Answer: ∃x (F(x) ∧ ¬G(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 (F(x) → G(x)) ≡ ∃x (F(x) ∧ ¬G(x))

Answer

∃x (F(x) ∧ ¬G(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 = 6

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.

ABC00011110011001130214151706

Group the 1s

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

B′C′
BC
AC

Minimal SOP form

F = B′C′ + BC + AC
L = 6

Answer

L = 6

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: Q = 2

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.

Full cycles

Every 8 pulses bring the counter back to the same state.

12 = 1·8 + 4

Count after the remaining pulses

Q = (6 + 4) mod 8 = 2
Q = 010₂

Answer

Q = 2
7.Number theory & modular arithmetic

Answer: x = 58

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 = 8 + 10k

Second congruence

8 + 10k ≡ 6 (mod 13) → 10k ≡ 11 (mod 13)
k ≡ 4·11 ≡ 5 (mod 13)

Result

x = 8 + 10·5 = 58
x = 58
check: 58 mod 10 = 8, 58 mod 13 = 6

Answer

x = 58
8.Number theory & modular arithmetic

Answer: d = 89

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.

Modulus and totient

n = 7·43 = 301
φ(n) = 6·42 = 252

Extended Euclidean algorithm

r = 252, t = 0
r = 17, q = 14, t = 1
r = 14, q = 1, t = −14
r = 3, q = 4, t = 15
r = 2, q = 1, t = −74
r = 1, q = 2, t = 89

Private exponent

d = 89
check: 17·89 = 1513 = 6·252 + 1

Answer

d = 89
9.Algorithms & complexity

Answer: C · Greedy uses 3 coins; the optimum is 2

Why this is the answer

Greedy takes 4 + 1 + 1 = 3 coins, but 3 + 3 uses only 2, so greedy is not optimal for this coin system; dynamic programming is.

Answer

C · Greedy uses 3 coins; the optimum is 2
10.Algorithms & complexity

Answer: D · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

D · No, the lower bound applies to the worst case over all inputs, not to every input
11.Data structures

Answer: A · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

A · The leftmost node of its right subtree
12.Data structures

Answer: A · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

A · A hash map combined with a doubly linked list
13.Graph theory

Answer: Euler 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) = 2, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(D) = 5

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: 4, C: 1
C (1): E: 13, G: 15
B (4): D: 5
D (5): E: 10, G: 8
G (8)
E (10): F: 13
F (13)

Shortest path

A → B → D
d(D) = 5

Answer

d(D) = 5
15.Databases & SQL

Answer: A · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id

Why this is the answer

GROUP BY customer_id makes one group per customer, and COUNT(*) counts the rows in each group.

Answer

A · SELECT customer_id, COUNT(*) FROM orders GROUP BY customer_id
16.Databases & SQL

Answer: C · Conflict-serializable schedules

Why this is the answer

2PL guarantees serializability, but deadlocks remain possible and must be detected or prevented.

Answer

C · Conflict-serializable schedules
17.Computer networks

Answer: B · FIN

Why this is the answer

Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.

Answer

B · FIN
18.Computer networks

Answer: D · Several requests reuse one TCP connection, avoiding repeated handshakes

Why this is the answer

Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.

Answer

D · Several requests reuse one TCP connection, avoiding repeated handshakes
19.IPv4 subnetting

Answer: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1776
211 − 2 = 2046 ≥ 1776 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 228

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 in the third octet

21 = 2

Subnet of this host

network: 172.20.228.0
broadcast: 172.20.229.255

Answer

x = 228

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