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 −81 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −818 bit
U =
2.Number systemsHard
4060 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4060 items its own binary code of the same length. What is the smallest number of bits b needed?

4060
b = bit
3.Propositional logic & proofsMedium
(c → b) ∧ (b → c)c ∧ b(c → b) ∨ (b → c)(c → b) ∧ (¬b → ¬c)

Which formula is logically equivalent to c ↔ b? Click it.

c ↔ b
  1. 1(c → b) ∧ (b → c)
  2. 2c ∧ b
  3. 3(c → b) ∨ (b → c)
  4. 4(c → b) ∧ (¬b → ¬c)
4.Propositional logic & proofsHard
¬pqq ∧ rp

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

q ∨ rq → pr → p
  1. 1¬p
  2. 2q
  3. 3q ∧ r
  4. 4p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000010302041507060121131151140809111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(5, 10, 11, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(5, 10, 11, 13, 14, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 34 modulo 37: the number x with 0 < x < 37 and 34·x ≡ 1 (mod 37).

a = 34n = 37
34−1 mod 37 =
8.Number theory & modular arithmeticHard
96c = me​ mod n?m = cd​ mod nm(n, e) = (161, 7)d

Encrypt the message m = 96 with the RSA public key (n, e) = (161, 7). Find the ciphertext c.

n = 161e = 7m = 96
c =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n/2) + O(1) solve to?

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

A problem is NP-complete if it is:

  1. AIn NP and every NP problem reduces to it in polynomial time
  2. BSolvable in polynomial time
  3. COutside NP but harder than all NP problems, such as the halting problem
  4. DImpossible to solve
11.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 0
  2. BIndex 4
  3. CIndex 3
  4. DIndex 6
12.Data structuresHard

Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?

  1. AO(1)
  2. BO(n)
  3. CO(α(n))
  4. DO(log n)
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1115915101313ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

Which statement about HAVING is correct?

  1. AIt must come before WHERE in the SELECT statement
  2. BIt cannot be used with GROUP BY
  3. CIt sorts the results
  4. DIt can use aggregate functions to filter groups
16.Databases & SQLHard

A reporting database copies customer names into the orders table to avoid joins. What is the main cost of this denormalisation?

  1. AThe database cannot use indexes any more
  2. BPrimary keys can no longer be used
  3. CQueries on the orders table become slower
  4. DRedundant copies must be kept consistent whenever customer data changes
17.Computer networksMedium

Using even parity, which parity bit is added to the data bits 1011001?

  1. A1
  2. B4
  3. C0
  4. D2
18.Computer networksHard

What is the maximum transmission unit (MTU) of standard Ethernet payload?

  1. A1500 bytes
  2. B65,535 bytes
  3. C9000 bytes in every network
  4. D64 bytes
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1151 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1151 hosts
p =
20.IPv4 subnettingHard
third octet = 56012806413211618040201networkhost

A host has the address 172.20.56.235/23. What is the third octet x of its network address?

172.20.56.235/23
x =

3Solutions

1.Number systems

Answer: U = 175

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

81 = 0101 0001
inverted: 1010 1110
+1 → 1010 1111

Read it as unsigned

1010 1111₂ = 256 − 81 = 175

Answer

U = 175
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 < 4060
212 = 4096 ≥ 4060

Answer

b = 12
3.Propositional logic & proofs

Answer: (c → b) ∧ (b → c)

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 means the implication holds in both directions: (a → b) ∧ (b → a).
c ↔ b ≡ (c → b) ∧ (b → c)

Check with the truth table

b = 1, c = 1: 1 | 1 1 1 1
b = 1, c = 0: 0 | 0 0 1 1
b = 0, c = 1: 0 | 0 0 1 0
b = 0, c = 0: 1 | 1 0 1 1

Answer

(c → b) ∧ (b → c)

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

4.Propositional logic & proofs

Answer: p

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

Conclusion

p (proof by cases)

Answer

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

Answer: L = 5

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.

ABCD000111100001111000010302041507060121131151140809111110

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

BC′D
AC

Minimal SOP form

F = BC′D + AC
L = 5

Answer

L = 5
6.Digital logic: gates, K-maps & flip-flops

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

Count the changes

0 → 0 → 1 → 1 → 0 → 0 → 0 → 1 → 1
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 34−1 mod 37 = 12

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·34 (mod 37).

r = 37, t = 0
r = 34, q = 1, t = 1
r = 3, q = 11, t = −1
r = 1, q = 3, t = 12

Inverse

1 ≡ 12·34 (mod 37)
x = 12
check: 34·12 = 408 = 11·37 + 1

Answer

34−1 mod 37 = 12
8.Number theory & modular arithmetic

Answer: c = 54

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.

Repeated squaring

7 = 4 + 2 + 1
961 ≡ 96 (mod 161)
962 ≡ 39 (mod 161)
964 ≡ 72 (mod 161)
967 ≡ 96·39·72 ≡ 54 (mod 161)

Ciphertext

c = 54

Answer

c = 54
9.Algorithms & complexity

Answer: D · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

D · O(log n)
10.Algorithms & complexity

Answer: A · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

A · In NP and every NP problem reduces to it in polynomial time
11.Data structures

Answer: B · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

B · Index 4
12.Data structures

Answer: B · O(n)

Why this is the answer

Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.

Answer

B · O(n)
13.Graph theory

Answer: Euler circuit

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) = 2, deg(B) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(F) = 25

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): C: 11
C (11): B: 26, E: 12
E (12): D: 17, F: 25, B: 25
D (17)
B (25)
F (25)

Shortest path

A → C → E → F
d(F) = 25

Answer

d(F) = 25
15.Databases & SQL

Answer: D · It can use aggregate functions to filter groups

Why this is the answer

HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.

Answer

D · It can use aggregate functions to filter groups
16.Databases & SQL

Answer: D · Redundant copies must be kept consistent whenever customer data changes

Why this is the answer

Denormalisation speeds up reads but brings back redundancy, so updates must change every copy or anomalies appear.

Answer

D · Redundant copies must be kept consistent whenever customer data changes
17.Computer networks

Answer: C · 0

Why this is the answer

1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.

Answer

C · 0
18.Computer networks

Answer: A · 1500 bytes

Why this is the answer

Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.

Answer

A · 1500 bytes
19.IPv4 subnetting

Answer: p = 21

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

210 − 2 = 1022 < 1151
211 − 2 = 2046 ≥ 1151 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 56

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

21 = 2

Subnet of this host

network: 172.20.56.0
broadcast: 172.20.57.255

Answer

x = 56

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