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 −51 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −518 bit
U =
2.Number systemsHard
32-bit two's complement, N = ??sign???????????????????????????????

What is the most negative integer N that can be stored in 32-bit two's complement?

32 bit
N =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is ((r → p) ∧ (r → q)) → (p → q) a tautology, a contradiction or a contingency? Click your answer.

((r → p) ∧ (r → q)) → (p → q)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x ∃y ¬R(x, y)∀x ∀y ¬R(x, y)∃x ∃y ¬R(x, y)∃x ∀y ¬R(x, y)

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

¬∃x ∀y R(x, y)
  1. 1∀x ∃y ¬R(x, y)
  2. 2∀x ∀y ¬R(x, y)
  3. 3∃x ∃y ¬R(x, y)
  4. 4∃x ∀y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010302040517060121131151140809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(7, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(7, 13, 14, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011X312X41507160121130150141809111110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 8, 10, 11, 13) + d(3, 4), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 8, 10, 11, 13) + d(3, 4)
L =
7.Number theory & modular arithmeticMedium

Solve 23x ≡ 28 (mod 73). Give the solution x with 0 < x < 73.

23x ≡ 28 (mod 73)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (703, 13)d = ?

An RSA key uses the primes p = 19 and q = 37 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).

p = 19q = 37e = 13
d =
9.Algorithms & complexityMedium

Which statement about a recursive function without a reachable base case is correct?

  1. AThe compiler always turns it into a loop that ends safely
  2. BIt returns 0 automatically
  3. CIt keeps calling itself until the call stack overflows
  4. DIt runs in O(1) time
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AComparisons take logarithmic time each
  2. BA decision tree with n possible outcomes must have depth at least log₂ n
  3. CEvery element must be compared at least once
  4. DBinary search is the only possible algorithm, and it needs log₂ n comparisons
11.Data structuresMedium

What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?

  1. A√n
  2. Blog₂ n
  3. C1
  4. Dn (it becomes a linked list)
12.Data structuresHard

Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?

  1. AO(α(n))
  2. BO(1)
  3. CO(n)
  4. DO(log n)
13.Graph theoryMedium

A connected graph has 8 vertices and 11 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 11
k =
14.Graph theoryHard
24102011812ABCDE

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 5|E| = 7
W =
15.Databases & SQLMedium

What does a common table expression (WITH name AS (...)) provide?

  1. AAn index on a column
  2. BA new database user
  3. CA permanent table that stays stored on disk until it is dropped
  4. DA named temporary result that can be used within one statement
16.Databases & SQLHard

Two-phase locking (2PL) ensures:

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

Which host is in the same /16 subnet as 10.1.2.3?

  1. A10.1.200.5
  2. B11.1.2.3
  3. C10.2.2.3
  4. D10.0.2.3
18.Computer networksHard

In a TCP connection, what does the receive window (rwnd) control?

  1. AThe size of each IP packet header
  2. BHow much unacknowledged data the sender may send, based on the receiver's buffer
  3. CThe retransmission time-out, calculated from the measured round-trip time of each segment
  4. DThe number of open ports
19.IPv4 subnettingMedium
/16 → /2111111111111111111111100000000000network bitshost bits

The network 172.17.0.0/16 is divided into equal subnets with prefix /21. How many subnets N are there?

172.17.0.0/16/21
N =
20.IPv4 subnettingHard

The network 192.168.77.0/24 is split with VLSM for these LANs: A: 7 hosts, B: 6 hosts, C: 23 hosts, D: 19 hosts, E: 3 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: 7B: 6C: 23D: 19E: 3
x =

3Solutions

1.Number systems

Answer: U = 205

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

51 = 0011 0011
inverted: 1100 1100
+1 → 1100 1101

Read it as unsigned

1100 1101₂ = 256 − 51 = 205

Answer

U = 205
2.Number systems

Answer: N = −2.147 × 10⁹

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.

Most negative value

−231 = −2147483648

Answer

N = −2147483648
3.Propositional logic & proofs

Answer: contingency

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).

Truth table row by row

p = 1, q = 1, r = 1: r → p = 1, r → q = 1, (r → p) ∧ (r → q) = 1, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 1, q = 1, r = 0: r → p = 1, r → q = 1, (r → p) ∧ (r → q) = 1, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 1, q = 0, r = 1: r → p = 1, r → q = 0, (r → p) ∧ (r → q) = 0, p → q = 0, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 1, q = 0, r = 0: r → p = 1, r → q = 1, (r → p) ∧ (r → q) = 1, p → q = 0, ((r → p) ∧ (r → q)) → (p → q) = 0
p = 0, q = 1, r = 1: r → p = 0, r → q = 1, (r → p) ∧ (r → q) = 0, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 0, q = 1, r = 0: r → p = 1, r → q = 1, (r → p) ∧ (r → q) = 1, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 0, q = 0, r = 1: r → p = 0, r → q = 0, (r → p) ∧ (r → q) = 0, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1
p = 0, q = 0, r = 0: r → p = 1, r → q = 1, (r → p) ∧ (r → q) = 1, p → q = 1, ((r → p) ∧ (r → q)) → (p → q) = 1

Classify

((r → p) ∧ (r → q)) → (p → q): 1, 1, 1, 0, 1, 1, 1, 1
1 in 7 of the 8 rows, 0 in the others: a contingency.

Answer

contingency
4.Propositional logic & proofs

Answer: ∀x ∃y ¬R(x, y)

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) ≡ ∀x ∃y ¬R(x, y)

Answer

∀x ∃y ¬R(x, y)

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 = 9

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.

ABCD000111100001111000010302040517060121131151140809011010

Group the 1s

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

BCD
ABD
ABC

Minimal SOP form

F = BCD + ABD + ABC
L = 9

Answer

L = 9
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′D′, B′C, BC′D, A′D′, A′C′, A′B′
ABCD00011110000111101011X312X41507160121130150141809111110

Minimal cover

F = B′D′ + B′C + BC′D + A′D′ + A′C′
L = 11

Answer

L = 11

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

7.Number theory & modular arithmetic

Answer: x = 52

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 23

r = 73, t = 0
r = 23, q = 3, t = 1
r = 4, q = 5, t = −3
r = 3, q = 1, t = 16
r = 1, q = 3, t = −19
23−1 ≡ 54 (mod 73)

Solve

x ≡ 54·28 = 1512 ≡ 52 (mod 73)
check: 23·52 = 1196 ≡ 28 (mod 73)

Answer

x = 52
8.Number theory & modular arithmetic

Answer: d = 349

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 = 19·37 = 703
φ(n) = 18·36 = 648

Extended Euclidean algorithm

r = 648, t = 0
r = 13, q = 49, t = 1
r = 11, q = 1, t = −49
r = 2, q = 5, t = 50
r = 1, q = 2, t = −299

Private exponent

d = 349
check: 13·349 = 4537 = 7·648 + 1

Answer

d = 349
9.Algorithms & complexity

Answer: C · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

C · It keeps calling itself until the call stack overflows
10.Algorithms & complexity

Answer: B · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

B · A decision tree with n possible outcomes must have depth at least log₂ n
11.Data structures

Answer: D · n (it becomes a linked list)

Why this is the answer

Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.

Answer

D · n (it becomes a linked list)
12.Data structures

Answer: C · O(n)

Why this is the answer

Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.

Answer

C · O(n)
13.Graph theory

Answer: k = 4

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

8 − 1 = 7

Edges to remove

k = 11 − 7 = 4

Answer

k = 4
14.Graph theory

Answer: W = 25

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.

Kruskal’s algorithm

CE (1): take
AB (2): take
AC (4): take
BC (10): skip, closes a cycle
BE (12): skip, closes a cycle
DE (18): take

Total weight

W = 1 + 2 + 4 + 18 = 25

Answer

W = 25
15.Databases & SQL

Answer: D · A named temporary result that can be used within one statement

Why this is the answer

A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.

Answer

D · A named temporary result that can be used within one statement
16.Databases & SQL

Answer: D · Conflict-serializable schedules

Why this is the answer

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

Answer

D · Conflict-serializable schedules
17.Computer networks

Answer: A · 10.1.200.5

Why this is the answer

With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.

Answer

A · 10.1.200.5
18.Computer networks

Answer: B · How much unacknowledged data the sender may send, based on the receiver's buffer

Why this is the answer

The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.

Answer

B · How much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnetting

Answer: N = 32

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

21 − 16 = 5

Number of subnets

N = 25 = 32

Answer

N = 32

Each /21 subnet has 211 − 2 = 2046 usable hosts.

20.IPv4 subnetting

Answer: x = 32

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: 23 hosts → 25 = 32 (/27)
D: 19 hosts → 25 = 32 (/27)
A: 7 hosts → 24 = 16 (/28)
B: 6 hosts → 23 = 8 (/29)
E: 3 hosts → 23 = 8 (/29)

Allocation, largest first

C: .0 – .31
D: .32 – .63
A: .64 – .79
B: .80 – .87
E: .88 – .95

Answer

x = 32

LAN D: 192.168.77.32/27.

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