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 complement1−12806403201618040211

The 8-bit pattern 1000 1001 stores a signed integer in two's complement. What is its value N in decimal?

1000 10018 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 0011 OR 1101 1111. What is the result's value R in decimal?

0101 0011OR1101 1111
R =
3.Propositional logic & proofsMedium
p ↔ ¬q¬p ∨ ¬q¬p ↔ ¬qp ∧ ¬q

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

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

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

p ∨ q¬p ∨ r
  1. 1¬q
  2. 2p ∨ r
  3. 3q ∨ r
  4. 4q ∧ r
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD0001111000011110111111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(1, 3, 6, 9, 10, 11)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​DQ4​Q4​DQ5​Q5​CLKJOHNSON

A Johnson (twisted-ring) counter is made of 6 D flip-flops, with the inverted last output fed back to the first input, and starts at all zeros. How many different states N does it go through?

n = 6
N =
7.Number theory & modular arithmeticMedium

Solve 24x ≡ 7 (mod 31). Give the solution x with 0 < x < 31.

24x ≡ 7 (mod 31)
x =
8.Number theory & modular arithmeticHard
589c = me​ mod n?m = cd​ mod nm(n, e) = (943, 13)d

Encrypt the message m = 589 with the RSA public key (n, e) = (943, 13). Find the ciphertext c.

n = 943e = 13m = 589
c =
9.Algorithms & complexityMedium

Why do many library sorting routines switch to insertion sort for very small subarrays?

  1. ARecursive algorithms cannot handle arrays below 10 elements
  2. BInsertion sort has little overhead and is fast on small or nearly sorted inputs
  3. CInsertion sort is O(n log n) in the worst case
  4. DInsertion sort uses less memory than any other sorting algorithm on large inputs
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. ABinary search is the only possible algorithm, and it needs log₂ n comparisons
  2. BComparisons take logarithmic time each
  3. CA decision tree with n possible outcomes must have depth at least log₂ n
  4. DEvery element must be compared at least once
11.Data structuresMedium

A trie (prefix tree) is especially efficient for:

  1. AStoring matrices
  2. BFinding shortest paths
  3. CSorting numbers
  4. DPrefix searches such as autocomplete
12.Data structuresHard

A segment tree is typically used to:

  1. AImplement a stack
  2. BAnswer range queries (such as range sums) and updates in O(log n)
  3. CSort strings alphabetically by storing each prefix in a separate tree node
  4. DStore a graph's edges
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
24157101113ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.

k = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

In relational algebra, the projection operator (π) corresponds to:

  1. ACombining two tables on a key
  2. BSorting the result
  3. CChoosing which rows satisfy a given condition
  4. DChoosing which columns appear in the result
16.Databases & SQLHard

What does the following return: SELECT name FROM students WHERE grade > (SELECT AVG(grade) FROM students)?

  1. AStudents with the highest grade only
  2. BAn error because subqueries are not allowed in WHERE
  3. CStudents with a grade above the average grade
  4. DThe average grade of all students
17.Computer networksMedium

What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?

  1. A/24
  2. B/8
  3. C/32
  4. D/16
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIts 128-bit address space is large enough to give every device a global address
  2. BIPv6 encrypts all addresses
  3. CIPv6 does not use addresses
  4. DIPv6 only works on local networks, so addresses never need to be shared globally
19.IPv4 subnettingMedium
255.255.248.011111111111111111111100000000000network bitshost bits

A subnet uses the mask 255.255.248.0. How many usable host addresses H does it have?

255.255.248.0
H =
20.IPv4 subnettingHard
third octet 19 … 23012806403211608041211commonvaries

A router has routes to these networks: 10.55.19.0/24, 10.55.20.0/24, 10.55.21.0/24, 10.55.22.0/24, 10.55.23.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.55.19.0/24 … 10.55.23.0/24
p =

3Solutions

1.Number systems

Answer: N = −119

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 + 8 + 1 = −119

Answer

N = −119
2.Number systems

Answer: R = 223

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

OR gives 1 where at least one bit is 1.

0101 0011
OR 1101 1111
= 1101 1111

Value of the result

1101 1111₂ = 223

Answer

R = 223
3.Propositional logic & proofs

Answer: p ↔ ¬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.
¬(p ↔ q) ≡ p ↔ ¬q

Check with the truth table

p = 1, q = 1: 0 | 0 0 1 0
p = 1, q = 0: 1 | 1 1 0 1
p = 0, q = 1: 1 | 1 1 0 0
p = 0, q = 0: 0 | 0 1 1 0

Answer

p ↔ ¬q

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

4.Propositional logic & proofs

Answer: q ∨ 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

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

Conclusion

q ∨ r (resolution)

Answer

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

Answer: AB = 01, CD = 10

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m6: m7, m4, m2, m14 → 0
ABCD0001111000011110111111

The isolated 1

m6 = A′BCD′ → AB = 01, CD = 10

Answer

AB = 01, CD = 10
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 12

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.

Sequence for 6 flip-flops

000000 → 100000 → 110000 → 111000 → 111100 → 111110 → 111111 → 011111 → 001111 → 000111 → 000011 → 000001

Number of states

N = 12

Answer

N = 12
7.Number theory & modular arithmetic

Answer: x = 30

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 24

r = 31, t = 0
r = 24, q = 1, t = 1
r = 7, q = 3, t = −1
r = 3, q = 2, t = 4
r = 1, q = 3, t = −9
24−1 ≡ 22 (mod 31)

Solve

x ≡ 22·7 = 154 ≡ 30 (mod 31)
check: 24·30 = 720 ≡ 7 (mod 31)

Answer

x = 30
8.Number theory & modular arithmetic

Answer: c = 908

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

13 = 8 + 4 + 1
5891 ≡ 589 (mod 943)
5892 ≡ 840 (mod 943)
5894 ≡ 236 (mod 943)
5898 ≡ 59 (mod 943)
58913 ≡ 589·236·59 ≡ 908 (mod 943)

Ciphertext

c = 908

Answer

c = 908
9.Algorithms & complexity

Answer: B · Insertion sort has little overhead and is fast on small or nearly sorted inputs

Why this is the answer

For small n the simple loop of insertion sort beats the recursion overhead of quicksort or merge sort, even though it is O(n²).

Answer

B · Insertion sort has little overhead and is fast on small or nearly sorted inputs
10.Algorithms & complexity

Answer: C · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

C · A decision tree with n possible outcomes must have depth at least log₂ n
11.Data structures

Answer: D · Prefix searches such as autocomplete

Why this is the answer

Tries store strings by characters along paths, so lookups and prefix queries take time proportional to the key length.

Answer

D · Prefix searches such as autocomplete
12.Data structures

Answer: B · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

B · Answer range queries (such as range sums) and updates in O(log 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) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: edge CE

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.

Kruskal’s algorithm

1. AB (2)
2. AC (4)
3. BD (7)
4. CE (10)

edge number 4

CE

Answer

edge CE
15.Databases & SQL

Answer: D · Choosing which columns appear in the result

Why this is the answer

Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).

Answer

D · Choosing which columns appear in the result
16.Databases & SQL

Answer: C · Students with a grade above the average grade

Why this is the answer

The scalar subquery computes the average, and the outer query keeps students above it.

Answer

C · Students with a grade above the average grade
17.Computer networks

Answer: A · /24

Why this is the answer

Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.

Answer

A · /24
18.Computer networks

Answer: A · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

A · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: H = 2046

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

255.255.248.0 = /21

Usable hosts

211 − 2 = 2046

Answer

H = 2046
20.IPv4 subnetting

Answer: p = 21

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

19 = 00010011
23 = 00010111
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.55.16.0/21 (it also covers 3 network(s) that are not in the list).

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