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 complement?−128?64?32?16?8?4?2?1

Write −84 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −848 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1111 0101 >> 2. What is the result's value R in decimal?

1111 0101>> 2
R =
3.Propositional logic & proofsMedium
pqrp → (q ∧ p)(r → ¬q) → (p → (q ∧ p))¬((r → ¬q) → (p → (q ∧ p)))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is ¬((r → ¬q) → (p → (q ∧ p))) true? Call this number N.

¬((r → ¬q) → (p → (q ∧ p)))
N =
4.Propositional logic & proofsHard
¬qqpp ∧ r

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

p ∨ rp → qr → q
  1. 1¬q
  2. 2q
  3. 3p
  4. 4p ∧ r
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′B′ + ABC′ + ABC to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′B′ + ABC′ + ABC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011130214050716012113115X1408X9111X10

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 3, 4, 6, 11, 13, 15) + d(9, 10, 14), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 3, 4, 6, 11, 13, 15) + d(9, 10, 14)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 58 modulo 61: the number x with 0 < x < 61 and 58·x ≡ 1 (mod 61).

a = 58n = 61
58−1 mod 61 =
8.Number theory & modular arithmeticHard
mc = me​ mod n1223m = cd​ mod n?(n, e) = (1763, 19)d

An RSA key uses p = 41, q = 43 and e = 19. Decrypt the ciphertext c = 1223: find the message m.

p = 41q = 43e = 19c = 1223
m =
9.Algorithms & complexityMedium

How many levels does the recursion of merge sort have for an array of n elements?

  1. AAbout n / 2
  2. BAbout n
  3. CAbout log₂ n
  4. DAbout √n
10.Algorithms & complexityHard

Topological sorting is possible only for:

  1. AComplete graphs
  2. BGraphs with negative weights
  3. CDirected acyclic graphs
  4. DUndirected graphs
11.Data structuresMedium

How long does a search for a word of length L take in a trie with a fixed alphabet?

  1. AO(log n), where n is the number of words
  2. BO(L × n)
  3. CO(L), independent of how many words are stored
  4. DO(n), where n is the number of words
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  2. BEvery node must be red
  3. CEvery path from a node to its descendant leaves has the same number of black nodes
  4. DThe root must be red
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1037212871013ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

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

Why are indexes created on columns?

  1. ATo save disk space
  2. BTo prevent NULL values
  3. CTo speed up searches and joins on those columns
  4. DTo make inserts and updates faster on those columns
16.Databases & SQLHard

Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?

  1. A1, 2, 3
  2. B1, 1, 3
  3. C1, 1, 2
  4. D2, 2, 3
17.Computer networksMedium

How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?

  1. A8 ms
  2. B1 ms
  3. C12 ms
  4. D0.125 ms
18.Computer networksHard

A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?

  1. AConfigure port forwarding on the router to the PC's private address
  2. BTurn off the PC's DNS client
  3. CGive the PC a different MAC address so the router can find it from outside
  4. DChange the PC's subnet mask to /32
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 950 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

950 hosts
p =
20.IPv4 subnettingHard
last octet = 38012806413201608141201networkhost

A host has the address 10.20.128.38/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.128.38/27
x =

3Solutions

1.Number systems

Answer: U = 172

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.

Invert and add one

84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100

Read it as unsigned

1010 1100₂ = 256 − 84 = 172

Answer

U = 172
2.Number systems

Answer: R = 61

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.

1111 0101 >> 2
= 0011 1101

Value of the result

0011 1101₂ = 61

Answer

R = 61
3.Propositional logic & proofs

Answer: N = 2

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.

Truth table row by row

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

Count the rows with value 1

¬((r → ¬q) → (p → (q ∧ p))): 0, 0, 1, 1, 0, 0, 0, 0
N = 2

Answer

N = 2
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

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

Conclusion

q (proof by cases)

Answer

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

Answer: L = 4

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′B′ → m(0, 1)
ABC′ → m(6)
ABC → m(7)
ABC00011110011011030204051716

Group and read off

F = A′B′ + AB

Count the literals

L = 4

Answer

L = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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, BCD′, AD, AC, A′C′D′, A′BD′, A′B′C′
ABCD00011110000111101011130214050716012113115X1408X9111X10

Minimal cover

F = B′D + BCD′ + AD + A′C′D′
N = 4

Answer

N = 4

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

7.Number theory & modular arithmetic

Answer: 58−1 mod 61 = 20

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·58 (mod 61).

r = 61, t = 0
r = 58, q = 1, t = 1
r = 3, q = 19, t = −1
r = 1, q = 3, t = 20

Inverse

1 ≡ 20·58 (mod 61)
x = 20
check: 58·20 = 1160 = 19·61 + 1

Answer

58−1 mod 61 = 20
8.Number theory & modular arithmetic

Answer: m = 949

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 = 19−1 mod 1680 = 619

Decrypt by repeated squaring

619 = 512 + 64 + 32 + 8 + 2 + 1
12231 ≡ 1223 (mod 1763)
12232 ≡ 705 (mod 1763)
12234 ≡ 1622 (mod 1763)
12238 ≡ 488 (mod 1763)
122316 ≡ 139 (mod 1763)
122332 ≡ 1691 (mod 1763)
122364 ≡ 1658 (mod 1763)
1223128 ≡ 447 (mod 1763)
1223256 ≡ 590 (mod 1763)
1223512 ≡ 789 (mod 1763)
1223619 ≡ 1223·705·488·1691·1658·789 ≡ 949 (mod 1763)

Message

m = 949
check: 94919 mod 1763 = 1223

Answer

m = 949
9.Algorithms & complexity

Answer: C · About log₂ n

Why this is the answer

Halving from n down to 1 takes about log₂ n levels, and each level does O(n) merging work, giving O(n log n).

Answer

C · About log₂ n
10.Algorithms & complexity

Answer: C · Directed acyclic graphs

Why this is the answer

A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.

Answer

C · Directed acyclic graphs
11.Data structures

Answer: C · O(L), independent of how many words are stored

Why this is the answer

Each character selects one child, so a lookup takes L steps no matter how many words the trie contains.

Answer

C · O(L), independent of how many words are stored
12.Data structures

Answer: C · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

C · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: neither

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) = 3, deg(F) = 1

Odd vertices

A, C, E, F → 4
neither

Answer

neither
14.Graph theory

Answer: B

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: 10, C: 3
2. C (3): E: 15, G: 13
3. B (10): D: 12
4. D (12): F: 19
5. G (13)
6. E (15)
7. F (19)

round 3

B

Answer

B
15.Databases & SQL

Answer: C · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

C · To speed up searches and joins on those columns
16.Databases & SQL

Answer: B · 1, 1, 3

Why this is the answer

RANK gives tied rows the same rank and skips the following numbers (1, 1, 3); DENSE_RANK would give 1, 1, 2.

Answer

B · 1, 1, 3
17.Computer networks

Answer: B · 1 ms

Why this is the answer

1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.

Answer

B · 1 ms
18.Computer networks

Answer: A · Configure port forwarding on the router to the PC's private address

Why this is the answer

NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.

Answer

A · Configure port forwarding on the router to the PC's private address
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 950
210 − 2 = 1022 ≥ 950 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 63

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

25 = 32

Subnet of this host

network: ⌊38/32⌋·32 = 32
broadcast: 32 + 32 − 1 = 63
usable hosts: 33 … 62

Answer

x = 63

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