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 complement1−12816403201608040201

The 8-bit pattern 1100 0000 stores a signed integer in two's complement. What is its value N in decimal?

1100 00008 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 1010 AND 0111 1101. What is the result's value R in decimal?

1011 1010AND0111 1101
R =
3.Propositional logic & proofsMedium
¬¬q → ¬p¬p → ¬¬q¬q → pp ∧ ¬¬q

You want to prove p → ¬q by contrapositive. Which implication do you prove instead? Click it.

p → ¬q
  1. 1¬¬q → ¬p
  2. 2¬p → ¬¬q
  3. 3¬q → p
  4. 4p ∧ ¬¬q
4.Propositional logic & proofsHard

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

((¬r → q) ↔ ¬p) ∨ (s ∨ ¬q)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXC′I0​1I1​0I2​0I3​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 = 1, I₂ = 0, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = C′I₁ = 1I₂ = 0I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110101103X214151716X12013015X140819011010

For F(A, B, C, D) = Σm(0, 1, 4, 5, 6, 7, 9) + d(2, 12, 14), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 4, 5, 6, 7, 9) + d(2, 12, 14)
N =
7.Number theory & modular arithmeticMedium

Solve 11x ≡ 4 (mod 41). Give the solution x with 0 < x < 41.

11x ≡ 4 (mod 41)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (319, 11)d = ?

An RSA key uses the primes p = 11 and q = 29 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 11q = 29e = 11
d =
9.Algorithms & complexityMedium

Which sorting algorithm is stable?

  1. AQuicksort (typical in-place version)
  2. BSelection sort (typical version)
  3. CHeap sort
  4. DMerge sort
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

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo make the list sorted
  2. BTo simplify code by removing special cases at the head or tail
  3. CTo prevent the list from growing beyond a fixed maximum number of nodes
  4. DTo store the number of elements
12.Data structuresHard

Why might a server choose its hash function at random from a universal family?

  1. ATo guarantee that no two keys can ever collide, whatever keys an attacker sends
  2. BTo remove the need to resize the table
  3. CTo make it very unlikely that an attacker can pick many keys that all collide
  4. DTo make the hash table sorted
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
ABCDEF

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

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

What does ROLLBACK TO SAVEPOINT s1 do?

  1. AUndoes the whole transaction from the beginning
  2. BEnds the transaction and commits everything
  3. CDeletes the savepoint and all tables changed since then
  4. DUndoes the changes made after savepoint s1, keeping the transaction open
16.Databases & SQLHard

Transaction T1 locks row A and waits for row B, while T2 locks row B and waits for row A. What does a typical DBMS do?

  1. ALets both transactions wait until one of them commits normally
  2. BMerges the two transactions into one
  3. CDetects the deadlock and aborts one transaction as the victim
  4. DCommits both transactions at once
17.Computer networksMedium

What characterises a distributed denial-of-service (DDoS) attack?

  1. AOne attacker guesses a user's password
  2. BA virus deletes files on a single PC
  3. CMany compromised machines flood a target with traffic
  4. DAn attacker reads unencrypted Wi-Fi traffic from a nearby car
18.Computer networksHard

Which routing protocol is a link-state protocol?

  1. AARP
  2. BBGP
  3. COSPF
  4. DRIP
19.IPv4 subnettingMedium
255.255.255.19211111111111111111111111111000000network bitshost bits

A subnet uses the mask 255.255.255.192. How many usable host addresses H does it have?

255.255.255.192
H =
20.IPv4 subnettingHard
third octet 3 … 8012806403201608041211commonvaries

A router has routes to these networks: 10.119.3.0/24, 10.119.4.0/24, 10.119.5.0/24, 10.119.6.0/24, 10.119.7.0/24, 10.119.8.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.119.3.0/24 … 10.119.8.0/24
p =

3Solutions

1.Number systems

Answer: N = −64

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.

−128 + 64 = −64

Answer

N = −64
2.Number systems

Answer: R = 56

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

AND gives 1 only where both bits are 1.

1011 1010
AND 0111 1101
= 0011 1000

Value of the result

0011 1000₂ = 56

Answer

R = 56
3.Propositional logic & proofs

Answer: ¬¬q → ¬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).

Parts of the implication

a = p, b = ¬q

Contrapositive

¬b → ¬a = ¬¬q → ¬p

Answer

¬¬q → ¬p

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: N = 14

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

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

Split on p

p = 1: 6
p = 0: 8
N = 6 + 8 = 14

Answer

N = 14
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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′ + A′B

Rows with F = 1

ABC = 000, 010, 011
N = 3

Answer

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

Answer: N = 2

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′C′D, A′D′, A′C′, A′B
ABCD0001111000011110101103X214151716X12013015X140819011010

Essential prime implicants

B′C′D, A′B
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: x = 19

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 11

r = 41, t = 0
r = 11, q = 3, t = 1
r = 8, q = 1, t = −3
r = 3, q = 2, t = 4
r = 2, q = 1, t = −11
r = 1, q = 2, t = 15
11−1 ≡ 15 (mod 41)

Solve

x ≡ 15·4 = 60 ≡ 19 (mod 41)
check: 11·19 = 209 ≡ 4 (mod 41)

Answer

x = 19
8.Number theory & modular arithmetic

Answer: d = 51

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 = 11·29 = 319
φ(n) = 10·28 = 280

Extended Euclidean algorithm

r = 280, t = 0
r = 11, q = 25, t = 1
r = 5, q = 2, t = −25
r = 1, q = 5, t = 51

Private exponent

d = 51
check: 11·51 = 561 = 2·280 + 1

Answer

d = 51
9.Algorithms & complexity

Answer: D · Merge sort

Why this is the answer

Merge sort is stable when merging takes from the left half first on ties. In-place quicksort, heap sort and selection sort are usually not stable.

Answer

D · Merge sort
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: B · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

B · To simplify code by removing special cases at the head or tail
12.Data structures

Answer: C · To make it very unlikely that an attacker can pick many keys that all collide

Why this is the answer

With universal hashing the expected number of collisions is small for any set of keys, since the attacker does not know which function was chosen.

Answer

C · To make it very unlikely that an attacker can pick many keys that all collide
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: N = 2

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

One more step

(M³)BB = 0·1 + 3·0 + 1·1 + 1·0 + 1·1 + 1·0 = 2

Answer

N = 2
15.Databases & SQL

Answer: D · Undoes the changes made after savepoint s1, keeping the transaction open

Why this is the answer

A savepoint marks a point inside a transaction; rolling back to it cancels later work while earlier changes remain pending until COMMIT.

Answer

D · Undoes the changes made after savepoint s1, keeping the transaction open
16.Databases & SQL

Answer: C · Detects the deadlock and aborts one transaction as the victim

Why this is the answer

A cycle in the wait-for graph is a deadlock; the DBMS rolls back one transaction so the other can continue, and the victim may be retried.

Answer

C · Detects the deadlock and aborts one transaction as the victim
17.Computer networks

Answer: C · Many compromised machines flood a target with traffic

Why this is the answer

In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.

Answer

C · Many compromised machines flood a target with traffic
18.Computer networks

Answer: C · OSPF

Why this is the answer

OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.

Answer

C · OSPF
19.IPv4 subnetting

Answer: H = 62

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

255.255.255.192 = /26

Usable hosts

26 − 2 = 62

Answer

H = 62
20.IPv4 subnetting

Answer: p = 20

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

3 = 00000011
8 = 00001000
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.119.0.0/20 (it also covers 10 network(s) that are not in the list).

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