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 −14 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −148 bit
U =
2.Number systemsHard
3306 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 3306 items its own binary code of the same length. What is the smallest number of bits b needed?

3306
b = bit
3.Propositional logic & proofsMedium
¬c ∧ ¬a¬c ∨ ¬a¬c ∨ a¬c ∧ a

Which formula is logically equivalent to ¬(c ∧ a)? Click it.

¬(c ∧ a)
  1. 1¬c ∧ ¬a
  2. 2¬c ∨ ¬a
  3. 3¬c ∨ a
  4. 4¬c ∧ a
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ¬((¬b → ¬a) ∨ c) ∨ (¬d → (c ↔ a)) true? Call this number N.

¬((¬b → ¬a) ∨ c) ∨ (¬d → (c ↔ a))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = AB + ABC + A′BC + A′B′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
7.Number theory & modular arithmeticMedium

Solve 6x ≡ 7 (mod 17). Give the solution x with 0 < x < 17.

6x ≡ 7 (mod 17)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 61155 mod 74.

a = 61k = 155n = 74
61155 mod 74 =
9.Algorithms & complexityMedium

In a sorted array, you look for two elements that sum to a target by moving one pointer from each end inwards. What is the time complexity?

  1. AO(n)
  2. BO(n²)
  3. CO(n log n)
  4. DO(log n)
10.Algorithms & complexityHard

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

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

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA binary search tree
  2. BA hash table
  3. CA queue
  4. DA stack
12.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

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

A tree has 4 vertices of degree 4, 2 vertices of degree 5, 2 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 42 × deg 52 × deg 2
L =
14.Graph theoryHard
5132102041518ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 5.

start: Ck = 5
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF
15.Databases & SQLMedium

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

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

Which isolation level prevents dirty reads but still allows non-repeatable reads?

  1. ARead uncommitted
  2. BRepeatable read
  3. CRead committed
  4. DSerializable
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ABusiness policy
  2. BLink bandwidth
  3. CMeasured delay
  4. DHop count
18.Computer networksHard

In a SYN flood attack, what resource is the attacker trying to exhaust?

  1. AThe server's table of half-open connections
  2. BThe server's disk space
  3. CThe switch's MAC address table
  4. DThe DNS cache of the client
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.59.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 5?

192.168.59.0/24/27subnet 5 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 129112806403201608040211networkhost

A host has the address 10.20.234.129/30. What is the fourth octet x of its network address?

10.20.234.129/30
x =

3Solutions

1.Number systems

Answer: U = 242

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

14 = 0000 1110
inverted: 1111 0001
+1 → 1111 0010

Read it as unsigned

1111 0010₂ = 256 − 14 = 242

Answer

U = 242
2.Number systems

Answer: b = 12 bit

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.

Powers of two

211 = 2048 < 3306
212 = 4096 ≥ 3306

Answer

b = 12
3.Propositional logic & proofs

Answer: ¬c ∨ ¬a

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

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(c ∧ a) ≡ ¬c ∨ ¬a

Check with the truth table

a = 1, c = 1: 0 | 0 0 1 0
a = 1, c = 0: 1 | 0 1 1 1
a = 0, c = 1: 1 | 0 1 0 0
a = 0, c = 0: 1 | 1 1 1 0

Answer

¬c ∨ ¬a

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

4.Propositional logic & proofs

Answer: N = 13

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

¬b; ¬a; ¬b → ¬a; (¬b → ¬a) ∨ c; ¬((¬b → ¬a) ∨ c); ¬d; c ↔ a; ¬d → (c ↔ a); ¬((¬b → ¬a) ∨ c) ∨ (¬d → (c ↔ a))

Split on a

a = 1: 7
a = 0: 6
N = 7 + 6 = 13

Answer

N = 13
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 7

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

AB → m(6, 7)
ABC → m(7)
A′BC → m(3)
A′B′C′ → m(0)
ABC00011110011001130204051716

Group and read off

F = BC + AB + A′B′C′

Count the literals

L = 7

Answer

L = 7
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 3

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.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 1 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 1 → Q = 0
4: J = 1, K = 0 → Q = 1
5: J = 1, K = 0 → Q = 1
6: J = 1, K = 1 → Q = 0
7: J = 1, K = 1 → Q = 1

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: x = 4

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 6

r = 17, t = 0
r = 6, q = 2, t = 1
r = 5, q = 1, t = −2
r = 1, q = 5, t = 3
6−1 ≡ 3 (mod 17)

Solve

x ≡ 3·7 = 21 ≡ 4 (mod 17)
check: 6·4 = 24 ≡ 7 (mod 17)

Answer

x = 4
8.Number theory & modular arithmetic

Answer: 61155 mod 74 = 59

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.

Binary exponent

155 = 100110112 = 128 + 16 + 8 + 2 + 1

Repeated squaring

611 ≡ 61 (mod 74)
612 ≡ 21 (mod 74)
614 ≡ 71 (mod 74)
618 ≡ 9 (mod 74)
6116 ≡ 7 (mod 74)
6132 ≡ 49 (mod 74)
6164 ≡ 33 (mod 74)
61128 ≡ 53 (mod 74)

Multiply the needed powers

61155 ≡ 61·21·9·7·53 ≡ 59 (mod 74)

Answer

61155 mod 74 = 59
9.Algorithms & complexity

Answer: A · O(n)

Why this is the answer

If the sum is too small the left pointer moves right, if too large the right pointer moves left, so there are at most n steps.

Answer

A · O(n)
10.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

C · O(n log n)
11.Data structures

Answer: D · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

D · A stack
12.Data structures

Answer: B · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

B · O(log n)
13.Graph theory

Answer: L = 16

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.

Handshake lemma for a tree

L + 30 = 2(L + 8) − 2
L = 16

Answer

L = 16

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: edge DF

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.

Prim’s algorithm

1. BC (2)
2. AB (5)
3. BD (10)
4. DE (4)
5. DF (15)

edge number 5

DF

Answer

edge DF
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: C · Read committed

Why this is the answer

Read committed hides uncommitted changes; repeatable read also prevents non-repeatable reads; serializable prevents phantoms too.

Answer

C · Read committed
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 · The server's table of half-open connections

Why this is the answer

Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.

Answer

A · The server's table of half-open connections
19.IPv4 subnetting

Answer: x = 128

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 number 5

network: (5 − 1)·32 = 128
broadcast: 128 + 32 − 1 = 159

Answer

x = 128
20.IPv4 subnetting

Answer: x = 128

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

22 = 4

Subnet of this host

network: ⌊129/4⌋·4 = 128
broadcast: 128 + 4 − 1 = 131
usable hosts: 129 … 130

Answer

x = 128

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