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
place values in two's complement0−12806403201618141211

The 8-bit pattern 0000 1111 stores a signed integer in two's complement. What is its value N in decimal?

0000 11118 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 1110 XOR 1011 1000. What is the result's value R in decimal?

0101 1110XOR1011 1000
R =
3.Propositional logic & proofsMedium
¬r ↔ ¬qr ∧ ¬qr ↔ ¬q¬r ∨ ¬q

Which formula is logically equivalent to ¬(r ↔ q)? Click it.

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

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

((p ∨ ¬s) → q) → (q ∧ (r ↔ p))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​0I1​0I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C′, I₁ = 0, I₂ = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 0I₂ = 0I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100001X31204151706012X13115114X809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(2, 5, 7, 10, 14, 15) + d(3, 8, 13), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(2, 5, 7, 10, 14, 15) + d(3, 8, 13)
L =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 4) and x ≡ 3 (mod 13).

x ≡ 1 (mod 4)x ≡ 3 (mod 13)
x =
8.Number theory & modular arithmeticHard

47 is prime. Use Fermat’s little theorem to find 963677 mod 47.

p = 47k = 63677
963677 mod 47 =
9.Algorithms & complexityMedium

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

  1. AMerge sort has no recursive structure
  2. BIts subproblems do not overlap, so there is nothing to reuse
  3. CDynamic programming only works on graph problems, not on arrays
  4. DMerge sort is already O(n)
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(n² 2ⁿ)
  2. BO(n³)
  3. CO(2n²)
  4. DO(n!)
11.Data structuresMedium

Why is traversing an array often much faster in practice than traversing a linked list of the same length?

  1. AArray elements are contiguous, so they make good use of the CPU cache
  2. BArrays use fewer instructions to compare values
  3. CLinked lists always store elements on disk
  4. DArray traversal is O(1) in total, while linked list traversal is O(n)
12.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(1) time and O(1) extra space
  2. BO(n) time and O(1) extra space
  3. CO(n) time and O(n) extra space
  4. DO(n log n) time and O(1) extra space
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Which query lists customers who have never placed an order?

  1. ASELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id = NULL
  2. BSELECT c.name FROM customers c INNER JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
  3. CSELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NOT NULL
  4. DSELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
16.Databases & SQLHard

Given the functional dependencies A → B, B → C and C → D on R(A, B, C, D), what is the closure A+?

  1. A{A, B, C}
  2. B{A, B}
  3. C{A}
  4. D{A, B, C, D}
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ABusiness policy
  2. BMeasured delay
  3. CLink bandwidth
  4. DHop count
18.Computer networksHard

A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?

  1. AAbout 5 Mbit/s
  2. BAbout 640 Mbit/s
  3. CAbout 50 Mbit/s
  4. DAbout 0.5 Mbit/s
19.IPv4 subnettingMedium
/20 → /2811111111111111111111111111110000network bitshost bits

The network 10.54.16.0/20 is divided into equal subnets with prefix /28. How many subnets N are there?

10.54.16.0/20/28
N =
20.IPv4 subnettingHard

The network 192.168.30.0/24 is split with VLSM for these LANs: A: 9 hosts, B: 112 hosts, C: 29 hosts, D: 39 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: 9B: 112C: 29D: 39
x =

3Solutions

1.Number systems

Answer: N = 15

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.

Add the place values

The other bits keep their normal positive values.

8 + 4 + 2 + 1 = 15

Answer

N = 15
2.Number systems

Answer: R = 230

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

XOR gives 1 where the two bits differ.

0101 1110
XOR 1011 1000
= 1110 0110

Value of the result

1110 0110₂ = 230

Answer

R = 230
3.Propositional logic & proofs

Answer: 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).

The law

¬(a ↔ b) is 1 exactly when a and b differ, which is a ↔ ¬b.
¬(r ↔ q) ≡ r ↔ ¬q

Check with the truth table

q = 1, r = 1: 0 | 1 0 0 0
q = 1, r = 0: 1 | 0 0 1 1
q = 0, r = 1: 1 | 0 1 1 1
q = 0, r = 0: 0 | 1 0 0 1

Answer

r ↔ ¬q

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

4.Propositional logic & proofs

Answer: N = 10

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

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

Split on p

p = 1: 6
p = 0: 4
N = 6 + 4 = 10

Answer

N = 10
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C′ + ABC′

Rows with F = 1

ABC = 000, 110
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: L = 8

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.

BD, B′CD′, A′CD, ACD′, AB′D′, A′B′C, ABC
ABCD00011110000111100001X31204151706012X13115114X809011110

Minimal cover

F = BD + B′CD′ + ACD′
L = 8

Answer

L = 8

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

7.Number theory & modular arithmetic

Answer: x = 29

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 = 1 + 4k

Second congruence

1 + 4k ≡ 3 (mod 13) → 4k ≡ 2 (mod 13)
k ≡ 10·2 ≡ 7 (mod 13)

Result

x = 1 + 4·7 = 29
x = 29
check: 29 mod 4 = 1, 29 mod 13 = 3

Answer

x = 29
8.Number theory & modular arithmetic

Answer: 963677 mod 47 = 27

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

63677 mod 46 = 13
963677 ≡ 913 (mod 47)

Small power

913 mod 47 = 27

Answer

963677 mod 47 = 27
9.Algorithms & complexity

Answer: B · 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

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

Answer: A · 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

A · O(n² 2ⁿ)
11.Data structures

Answer: A · Array elements are contiguous, so they make good use of the CPU cache

Why this is the answer

Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.

Answer

A · Array elements are contiguous, so they make good use of the CPU cache
12.Data structures

Answer: B · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

B · O(n) time and O(1) extra space
13.Graph theory

Answer: H = 12

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.

Count

H = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: N = 3

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

Common neighbours

(M²)BB = 0·0 + 0·0 + 1·1 + 1·1 + 1·1 = 3

Answer

N = 3
15.Databases & SQL

Answer: D · SELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL

Why this is the answer

The left join keeps all customers and fills order columns with NULL when there is no match; IS NULL then picks exactly those customers.

Answer

D · SELECT c.name FROM customers c LEFT JOIN orders o ON o.customer_id = c.id WHERE o.id IS NULL
16.Databases & SQL

Answer: D · {A, B, C, D}

Why this is the answer

Start with A, add B (A → B), then C (B → C), then D (C → D); since A+ contains all attributes, A is a key.

Answer

D · {A, B, C, D}
17.Computer networks

Answer: D · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

D · Hop count
18.Computer networks

Answer: A · About 5 Mbit/s

Why this is the answer

Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.

Answer

A · About 5 Mbit/s
19.IPv4 subnetting

Answer: N = 256

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

28 − 20 = 8

Number of subnets

N = 28 = 256

Answer

N = 256

Each /28 subnet has 24 − 2 = 14 usable hosts.

20.IPv4 subnetting

Answer: x = 128

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

B: 112 hosts → 27 = 128 (/25)
D: 39 hosts → 26 = 64 (/26)
C: 29 hosts → 25 = 32 (/27)
A: 9 hosts → 24 = 16 (/28)

Allocation, largest first

B: .0 – .127
D: .128 – .191
C: .192 – .223
A: .224 – .239

Answer

x = 128

LAN D: 192.168.30.128/26.

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