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 −74 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −748 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0011 1100 >> 2. What is the result's value R in decimal?

0011 1100>> 2
R =
3.Propositional logic & proofsMedium
¬p ∨ q¬p ∧ qp ∨ ¬q¬p → ¬q

Which formula is logically equivalent to p → q? Click it.

p → q
  1. 1¬p ∨ q
  2. 2¬p ∧ q
  3. 3p ∨ ¬q
  4. 4¬p → ¬q
4.Propositional logic & proofsHard
∀x P(x)∃x P(x)∀x ¬P(x)∃x ¬P(x)

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

¬∀x P(x)
  1. 1∀x P(x)
  2. 2∃x P(x)
  3. 3∀x ¬P(x)
  4. 4∃x ¬P(x)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

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(4, 8, 9, 11, 13)
  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
ABCD00011110000111101001131214150706X12113015X141809111010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 2, 3, 4, 5, 8, 11, 13) + d(12, 14), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 2, 3, 4, 5, 8, 11, 13) + d(12, 14)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 31 modulo 37: the number x with 0 < x < 37 and 31·x ≡ 1 (mod 37).

a = 31n = 37
31−1 mod 37 =
8.Number theory & modular arithmeticHard

Solve 70x ≡ 58 (mod 72). Give the smallest positive solution x.

70x ≡ 58 (mod 72)
x =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?

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

Why can Dijkstra's algorithm give wrong results when some edge weights are negative?

  1. ANegative numbers cannot be stored in a priority queue
  2. BA vertex is finalised when extracted, but a later negative edge could still give it a shorter path
  3. CIt goes into an infinite loop on every negative edge
  4. DIt only works on undirected graphs
11.Data structuresMedium

How is a new element inserted into a binary heap?

  1. APlace it at the end of the array and sift it up while it violates the heap property
  2. BRebuild the whole heap from scratch
  3. CPlace it at the root and sift it down
  4. DInsert it at its sorted position by shifting all of the larger elements one place right
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  2. BEvery node must be red
  3. CEvery path from a node to its descendant leaves has the same number of black nodes
  4. DThe root must be red
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
124713153134ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
15.Databases & SQLMedium

Which statement removes all rows of a table, keeps its structure and cannot take a WHERE clause?

  1. AALTER TABLE
  2. BTRUNCATE TABLE
  3. CDROP TABLE
  4. DDELETE FROM with a WHERE clause
16.Databases & SQLHard

What does the write-ahead logging (WAL) rule require?

  1. AA change's log record must reach stable storage before the changed data page does
  2. BData pages must be written before the log
  3. CThe log is written only after a crash
  4. DEvery query must be logged before it is parsed
17.Computer networksMedium

Which subnet mask corresponds to /20?

  1. A255.255.248.0
  2. B255.255.224.0
  3. C255.255.240.0
  4. D255.255.255.240
18.Computer networksHard

How does fragmentation differ in IPv6 compared with IPv4?

  1. AIPv6 fragments at the transport layer inside TCP
  2. BIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
  3. CRouters never fragment IPv6 packets; only the source can, using path MTU discovery
  4. DIPv6 does not allow packets larger than 576 bytes
19.IPv4 subnettingMedium
/22 → /2411111111111111111111111100000000network bitshost bits

The network 10.211.204.0/22 is divided into equal subnets with prefix /24. How many subnets N are there?

10.211.204.0/22/24
N =
20.IPv4 subnettingHard
last octet = 197112816403201608140211networkhost

A host has the address 10.20.118.197/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.118.197/25
x =

3Solutions

1.Number systems

Answer: U = 182

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

74 = 0100 1010
inverted: 1011 0101
+1 → 1011 0110

Read it as unsigned

1011 0110₂ = 256 − 74 = 182

Answer

U = 182
2.Number systems

Answer: R = 15

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

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0011 1100 >> 2
= 0000 1111

Value of the result

0000 1111₂ = 15

Answer

R = 15
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 0 only when a = 1 and b = 0, so it is the same as ¬a ∨ b.
p → q ≡ ¬p ∨ q

Check with the truth table

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

Answer

¬p ∨ q

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

4.Propositional logic & proofs

Answer: ∃x ¬P(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) ≡ ∃x ¬P(x)

Answer

∃x ¬P(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 = 01, 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.

m4: m5, m6, m0, m12 → 0
ABCD000111100001111011111

The isolated 1

m4 = A′BC′D′ → AB = 01, CD = 00

Answer

AB = 01, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 4

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.

C′D′, BC′, B′CD, A′B′D′, A′B′C
ABCD00011110000111101001131214150706X12113015X141809111010

Minimal cover

F = C′D′ + BC′ + B′CD + A′B′D′
N = 4

Answer

N = 4

Another grouping of the same size exists; it has the same number of terms and literals.

7.Number theory & modular arithmetic

Answer: 31−1 mod 37 = 6

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·31 (mod 37).

r = 37, t = 0
r = 31, q = 1, t = 1
r = 6, q = 5, t = −1
r = 1, q = 6, t = 6

Inverse

1 ≡ 6·31 (mod 37)
x = 6
check: 31·6 = 186 = 5·37 + 1

Answer

31−1 mod 37 = 6
8.Number theory & modular arithmetic

Answer: x = 7

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).

Divide by the gcd

gcd(70, 72) = 2
35x ≡ 29 (mod 36)

Solve the reduced congruence

35−1 ≡ 35 (mod 36) → x ≡ 7 (mod 36)

Solutions modulo 72

x ∈ {7, 43}
x = 7

Answer

x = 7
9.Algorithms & complexity

Answer: A · O(n log n)

Why this is the answer

The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).

Answer

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

Answer: B · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path

Why this is the answer

The greedy step relies on non-negative weights so that extracted distances are final; Bellman–Ford is used when weights may be negative.

Answer

B · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path
11.Data structures

Answer: A · Place it at the end of the array and sift it up while it violates the heap property

Why this is the answer

Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.

Answer

A · Place it at the end of the array and sift it up while it violates the heap property
12.Data structures

Answer: C · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

C · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: neither

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) = 1, deg(B) = 3, deg(C) = 1, deg(D) = 2, deg(E) = 3

Odd vertices

A, B, C, E → 4
neither

Answer

neither
14.Graph theory

Answer: B

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

1. A (0): C: 12
2. C (12): B: 16, E: 25
3. B (16): D: 23, E: 20
4. E (20): F: 33
5. D (23): F: 26
6. F (26)

round 3

B

Answer

B
15.Databases & SQL

Answer: B · TRUNCATE TABLE

Why this is the answer

TRUNCATE quickly empties a table and keeps its definition; DELETE can remove selected rows, and DROP removes the table itself.

Answer

B · TRUNCATE TABLE
16.Databases & SQL

Answer: A · A change's log record must reach stable storage before the changed data page does

Why this is the answer

Writing the log first lets the DBMS redo committed work and undo uncommitted work after a crash, supporting atomicity and durability.

Answer

A · A change's log record must reach stable storage before the changed data page does
17.Computer networks

Answer: C · 255.255.240.0

Why this is the answer

/20 means 8 + 8 + 4 one-bits. Four one-bits in the third octet are 11110000 = 240, so the mask is 255.255.240.0.

Answer

C · 255.255.240.0
18.Computer networks

Answer: C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery

Why this is the answer

An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.

Answer

C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
19.IPv4 subnetting

Answer: N = 4

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).

Borrowed bits

24 − 22 = 2

Number of subnets

N = 22 = 4

Answer

N = 4

Each /24 subnet has 28 − 2 = 254 usable hosts.

20.IPv4 subnetting

Answer: x = 255

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

27 = 128

Subnet of this host

network: ⌊197/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 255

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