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 complement?−128?64?32?16?8?4?2?1

Write −87 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −878 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1110 XOR 0100 1011. What is the result's value R in decimal?

1101 1110XOR0100 1011
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (r ∧ q) → r a tautology, a contradiction or a contingency? Click your answer.

(r ∧ q) → r
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
∀x (¬P(x) → ¬Q(x))∀x (P(x) → ¬Q(x))∀x (P(x) ∧ ¬Q(x))∃x (P(x) ∧ ¬Q(x))

Which statement is equivalent to ¬∃x (P(x) → Q(x)), with the negation moved all the way inside? Click it.

¬∃x (P(x) → Q(x))
  1. 1∀x (¬P(x) → ¬Q(x))
  2. 2∀x (P(x) → ¬Q(x))
  3. 3∀x (P(x) ∧ ¬Q(x))
  4. 4∃x (P(x) ∧ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

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(0, 3, 7, 11, 12, 13, 15)
  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
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKMOD-12

A synchronous mod-12 counter counts 0, 1, …, 11 and then returns to 0. It starts at 6. Which count Q (in decimal) does it show after 41 clock pulses?

Q0 = 641N = 12
Q =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 3 (mod 4) and x ≡ 2 (mod 7).

x ≡ 3 (mod 4)x ≡ 2 (mod 7)
x =
8.Number theory & modular arithmeticHard

61 is prime. Use Fermat’s little theorem to find 927884 mod 61.

p = 61k = 27884
927884 mod 61 =
9.Algorithms & complexityMedium

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

  1. AMerge sort is already O(n)
  2. BIts subproblems do not overlap, so there is nothing to reuse
  3. CMerge sort has no recursive structure
  4. DDynamic programming only works on graph problems, not on arrays
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. BEvery element must be compared at least once
  3. CComparisons take logarithmic time each
  4. DA decision tree with n possible outcomes must have depth at least log₂ n
11.Data structuresMedium

Which data structure does a running program use to keep track of active function calls and their local variables?

  1. AA hash table of functions
  2. BThe heap memory queue
  3. CThe call stack
  4. DA binary search tree of frames
12.Data structuresHard

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. AA single left rotation at the unbalanced node
  2. BA left rotation at the child, then a right rotation at the node
  3. CRecolouring the node and its child
  4. DA single right rotation at the unbalanced node
13.Graph theoryMedium

A tree has 2 vertices of degree 3, 3 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?

2 × deg 33 × deg 52 × deg 2
L =
14.Graph theoryHard
4719931118ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 4.

start: Dk = 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

What is a dirty read?

  1. AReading data written by another transaction that has not yet committed
  2. BReading data from a corrupted disk
  3. CReading the same row twice and getting the same value
  4. DReading data without a WHERE clause
16.Databases & SQLHard

What is the result of SELECT COUNT(col) FROM t if all values in col are NULL and the table has 10 rows?

  1. A0
  2. BNULL
  3. CAn error
  4. D10
17.Computer networksMedium

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

  1. A0
  2. B2
  3. C4
  4. D1
18.Computer networksHard

Why does TCP choose an unpredictable initial sequence number for each connection?

  1. ATo avoid confusion with old segments and make spoofed segments harder to inject
  2. BTo encrypt the payload of each segment
  3. CTo select the best route to the destination for the segments of that connection
  4. DTo compress the data more efficiently
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1320 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1320 hosts
p =
20.IPv4 subnettingHard

The network 192.168.206.0/24 is split with VLSM for these LANs: A: 23 hosts, B: 3 hosts, C: 19 hosts, D: 90 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 B?

A: 23B: 3C: 19D: 90
x =

3Solutions

1.Number systems

Answer: U = 169

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.

Invert and add one

87 = 0101 0111
inverted: 1010 1000
+1 → 1010 1001

Read it as unsigned

1010 1001₂ = 256 − 87 = 169

Answer

U = 169
2.Number systems

Answer: R = 149

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.

1101 1110
XOR 0100 1011
= 1001 0101

Value of the result

1001 0101₂ = 149

Answer

R = 149
3.Propositional logic & proofs

Answer: tautology

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).

Truth table row by row

q = 1, r = 1: r ∧ q = 1, (r ∧ q) → r = 1
q = 1, r = 0: r ∧ q = 0, (r ∧ q) → r = 1
q = 0, r = 1: r ∧ q = 0, (r ∧ q) → r = 1
q = 0, r = 0: r ∧ q = 0, (r ∧ q) → r = 1

Classify

(r ∧ q) → r: 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: ∀x (P(x) ∧ ¬Q(x))

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.

Move the negation inwards

¬∃x (P(x) → Q(x)) ≡ ∀x (P(x) ∧ ¬Q(x))

Answer

∀x (P(x) ∧ ¬Q(x))

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: AB = 00, CD = 00

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.

m0: m1, m2, m4, m8 → 0
ABCD00011110000111101111111

The isolated 1

m0 = A′B′C′D′ → AB = 00, CD = 00

Answer

AB = 00, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 11

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.

Full cycles

Every 12 pulses bring the counter back to the same state.

41 = 3·12 + 5

Count after the remaining pulses

Q = (6 + 5) mod 12 = 11
Q = 1011₂

Answer

Q = 11
7.Number theory & modular arithmetic

Answer: x = 23

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 = 3 + 4k

Second congruence

3 + 4k ≡ 2 (mod 7) → 4k ≡ 6 (mod 7)
k ≡ 2·6 ≡ 5 (mod 7)

Result

x = 3 + 4·5 = 23
x = 23
check: 23 mod 4 = 3, 23 mod 7 = 2

Answer

x = 23
8.Number theory & modular arithmetic

Answer: 927884 mod 61 = 34

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

27884 mod 60 = 44
927884 ≡ 944 (mod 61)

Small power

944 mod 61 = 34

Answer

927884 mod 61 = 34
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: D · 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

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

Answer: C · The call stack

Why this is the answer

Each call pushes a stack frame and each return pops it, which is why the most recent call finishes first.

Answer

C · The call stack
12.Data structures

Answer: D · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

D · A single right rotation at the unbalanced node
13.Graph theory

Answer: L = 13

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 + 25 = 2(L + 7) − 2
L = 13

Answer

L = 13

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

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.

Prim’s algorithm

1. BD (9)
2. AB (4)
3. AC (7)
4. CE (3)

edge number 4

CE

Answer

edge CE
15.Databases & SQL

Answer: A · Reading data written by another transaction that has not yet committed

Why this is the answer

A dirty read sees uncommitted changes that might be undone; it is possible only at the read uncommitted level.

Answer

A · Reading data written by another transaction that has not yet committed
16.Databases & SQL

Answer: A · 0

Why this is the answer

COUNT(col) counts non-NULL values, so it returns 0; COUNT(*) would return 10.

Answer

A · 0
17.Computer networks

Answer: A · 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

A · 0
18.Computer networks

Answer: A · To avoid confusion with old segments and make spoofed segments harder to inject

Why this is the answer

Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.

Answer

A · To avoid confusion with old segments and make spoofed segments harder to inject
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 < 1320
211 − 2 = 2046 ≥ 1320 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 192

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

D: 90 hosts → 27 = 128 (/25)
A: 23 hosts → 25 = 32 (/27)
C: 19 hosts → 25 = 32 (/27)
B: 3 hosts → 23 = 8 (/29)

Allocation, largest first

D: .0 – .127
A: .128 – .159
C: .160 – .191
B: .192 – .199

Answer

x = 192

LAN B: 192.168.206.192/29.

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