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 −75 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −758 bit
U =
2.Number systemsHard
8 bit1 px1024 px768 pxM = ? KiB

An uncompressed image is 1024 × 768 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

1024 × 768 px8 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
¬(p ∧ r) → ¬q¬(p ∧ r) ∧ q(p ∧ r) ∧ ¬q¬(p ∧ r) ∧ ¬q

You want to prove (p ∧ r) → q by contradiction. Which statement do you assume at the start? Click it.

(p ∧ r) → q
  1. 1¬(p ∧ r) → ¬q
  2. 2¬(p ∧ r) ∧ q
  3. 3(p ∧ r) ∧ ¬q
  4. 4¬(p ∧ r) ∧ ¬q
4.Propositional logic & proofsHard
¬r¬qrq ∧ ¬r

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

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

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

F = BC + BC′ + ABC′ + A′B′C
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​TQ4​Q4​TQ5​Q5​TQ6​Q6​TQ7​Q7​CLK

An 8-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 12 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 8tpd = 12 ns
fmax = MHz
7.Number theory & modular arithmeticMedium

Find the inverse of 39 modulo 89: the number x with 0 < x < 89 and 39·x ≡ 1 (mod 89).

a = 39n = 89
39−1 mod 89 =
8.Number theory & modular arithmeticHard
mc = me​ mod n1374m = cd​ mod n?(n, e) = (1591, 13)d

An RSA key uses p = 37, q = 43 and e = 13. Decrypt the ciphertext c = 1374: find the message m.

p = 37q = 43e = 13c = 1374
m =
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 3 coins; the optimum is 2
  2. BGreedy uses 4 coins; the optimum is 3
  3. CGreedy uses 2 coins; the optimum is 2
  4. DGreedy uses 3 coins; the optimum is 3
10.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n!)
  2. BO(n³)
  3. CO(2n²)
  4. DO(n² 2ⁿ)
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(n²)
  2. BO(1)
  3. CO(log n)
  4. DO(n)
12.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA stack of all visited edges
  2. BA max-heap ordered by out-degree
  3. CAn adjacency matrix recording which vertices have already been removed
  4. DA queue (or any collection) of vertices whose in-degree is 0
13.Graph theoryMedium

A connected graph has 16 vertices and 17 edges. How many edges k must be removed to leave a spanning tree?

|V| = 16|E| = 17
k =
14.Graph theoryHard
ABCDE

How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

In relational algebra, the selection operator (σ) corresponds to which SQL clause?

  1. ASELECT list
  2. BGROUP BY
  3. CWHERE
  4. DORDER BY
16.Databases & SQLHard

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

  1. A1, 1, 3
  2. B1, 1, 2
  3. C2, 2, 3
  4. D1, 2, 3
17.Computer networksMedium

A browser receives HTTP status code 404. What does it mean?

  1. AThe requested resource was not found
  2. BThe client must log in first
  3. CThe server had an internal error
  4. DThe request succeeded
18.Computer networksHard

Which of these addresses is a public (not RFC 1918 private) IPv4 address?

  1. A172.32.0.1
  2. B172.31.0.1
  3. C10.255.0.1
  4. D192.168.255.1
19.IPv4 subnettingMedium
255.255.252.011111111111111111111110000000000network bitshost bits

A subnet uses the mask 255.255.252.0. How many usable host addresses H does it have?

255.255.252.0
H =
20.IPv4 subnettingHard
third octet = 182112806413211608141201networkhost

A host has the address 172.20.182.87/21. What is the third octet x of its subnet's broadcast address?

172.20.182.87/21
x =

3Solutions

1.Number systems

Answer: U = 181

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

75 = 0100 1011
inverted: 1011 0100
+1 → 1011 0101

Read it as unsigned

1011 0101₂ = 256 − 75 = 181

Answer

U = 181
2.Number systems

Answer: M = 768 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

1024·768·8 = 6291456 bit = 786432 B

Convert to KiB

786432/1024 = 768 KiB

Answer

M = 768 KiB
3.Propositional logic & proofs

Answer: (p ∧ r) ∧ ¬q

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).

Parts of the implication

a = p ∧ r, b = q

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = (p ∧ r) ∧ ¬q

Answer

(p ∧ r) ∧ ¬q

If the assumption leads to a contradiction, the implication must be true.

4.Propositional logic & proofs

Answer: r

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

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

Conclusion

r (modus ponens)

Answer

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

Answer: L = 3

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

BC → m(3, 7)
BC′ → m(2, 6)
ABC′ → m(6)
A′B′C → m(1)
ABC00011110010011131204051716

Group and read off

F = B + A′C

Count the literals

L = 3

Answer

L = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: fmax = 10.42 MHz

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.

Worst-case settling time

T = 8·12 ns = 96 ns

Maximum frequency

fmax = 1/96 ns = 10.42 MHz

Answer

fmax = 10.42 MHz
7.Number theory & modular arithmetic

Answer: 39−1 mod 89 = 16

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·39 (mod 89).

r = 89, t = 0
r = 39, q = 2, t = 1
r = 11, q = 3, t = −2
r = 6, q = 1, t = 7
r = 5, q = 1, t = −9
r = 1, q = 5, t = 16

Inverse

1 ≡ 16·39 (mod 89)
x = 16
check: 39·16 = 624 = 7·89 + 1

Answer

39−1 mod 89 = 16
8.Number theory & modular arithmetic

Answer: m = 537

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 = 1591, φ(n) = 1512
d = 13−1 mod 1512 = 349

Decrypt by repeated squaring

349 = 256 + 64 + 16 + 8 + 4 + 1
13741 ≡ 1374 (mod 1591)
13742 ≡ 950 (mod 1591)
13744 ≡ 403 (mod 1591)
13748 ≡ 127 (mod 1591)
137416 ≡ 219 (mod 1591)
137432 ≡ 231 (mod 1591)
137464 ≡ 858 (mod 1591)
1374128 ≡ 1122 (mod 1591)
1374256 ≡ 403 (mod 1591)
1374349 ≡ 1374·403·127·219·858·403 ≡ 537 (mod 1591)

Message

m = 537
check: 53713 mod 1591 = 1374

Answer

m = 537
9.Algorithms & complexity

Answer: A · 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

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

Answer: D · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

D · O(n² 2ⁿ)
11.Data structures

Answer: B · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

B · O(1)
12.Data structures

Answer: D · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

D · A queue (or any collection) of vertices whose in-degree is 0
13.Graph theory

Answer: k = 2

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

16 − 1 = 15

Edges to remove

k = 17 − 15 = 2

Answer

k = 2
14.Graph theory

Answer: N = 4

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 0 0 0
B: 1 0 1 1 1
C: 0 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)EE = 1·0 + 2·1 + 1·1 + 1·1 + 3·0 = 4

Answer

N = 4
15.Databases & SQL

Answer: C · WHERE

Why this is the answer

σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.

Answer

C · WHERE
16.Databases & SQL

Answer: A · 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

A · 1, 1, 3
17.Computer networks

Answer: A · The requested resource was not found

Why this is the answer

404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.

Answer

A · The requested resource was not found
18.Computer networks

Answer: A · 172.32.0.1

Why this is the answer

172.16.0.0/12 runs from 172.16.0.0 to 172.31.255.255, so 172.32.0.1 lies outside it. The other three are in private ranges.

Answer

A · 172.32.0.1
19.IPv4 subnetting

Answer: H = 1022

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).

Prefix length

255.255.252.0 = /22

Usable hosts

210 − 2 = 1022

Answer

H = 1022
20.IPv4 subnetting

Answer: x = 183

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

23 = 8

Subnet of this host

network: 172.20.176.0
broadcast: 172.20.183.255

Answer

x = 183

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