Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 −99 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −998 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0101 1100 << 1. What is the result's value R in decimal?

0101 1100<< 1
R =
3.Propositional logic & proofsMedium
pqr¬p¬p → q(¬r → p) ∧ (¬p → q)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (¬r → p) ∧ (¬p → q) true? Call this number N.

(¬r → p) ∧ (¬p → q)
N =
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((r ∨ s) ∧ (s ∧ q)) ∨ ¬((q → ¬p) ∧ (p → r)) true? Call this number N.

((r ∨ s) ∧ (s ∧ q)) ∨ ¬((q → ¬p) ∧ (p → r))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111111

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, 5, 6, 8, 11, 14, 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
ABCD000111100001111010X1131214X507060121130151140819111010

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 2, 3, 4, 9, 11, 13, 14) + d(1, 5), 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, 9, 11, 13, 14) + d(1, 5)
N =
7.Number theory & modular arithmeticMedium

Solve 71x ≡ 40 (mod 81). Give the solution x with 0 < x < 81.

71x ≡ 40 (mod 81)
x =
8.Number theory & modular arithmeticHard

97 is prime. Use Fermat’s little theorem to find 724105 mod 97.

p = 97k = 24105
724105 mod 97 =
9.Algorithms & complexityMedium

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

  1. AO(n²)
  2. BO(n log n)
  3. CO(n³)
  4. DO(n)
10.Algorithms & complexityHard

Which definition of the class NP is correct?

  1. AProblems that are not polynomial
  2. BProblems that no algorithm can solve
  3. CDecision problems whose yes-answers can be verified in polynomial time given a certificate
  4. DProblems that cannot be solved in polynomial time
11.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i − 1) mod n
  2. B(i + 1) mod n
  3. C2i + 1
  4. Di + 1
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery node must be red
  2. BEvery path from a node to its descendant leaves has the same number of black nodes
  3. CThe root must be red
  4. DEvery red node must have red children, so red nodes always form unbroken chains to the leaves
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1378157635ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

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

Why are indexes created on columns?

  1. ATo prevent NULL values
  2. BTo save disk space
  3. CTo make inserts and updates faster on those columns
  4. DTo speed up searches and joins on those columns
16.Databases & SQLHard

Table employee(emp_id, dept_id, dept_name) has the key emp_id, and dept_id → dept_name. Which statement is correct?

  1. AIt is not in 1NF
  2. BIt is not in 2NF
  3. CIt is in 2NF but not in 3NF
  4. DIt is in BCNF and therefore in 3NF
17.Computer networksMedium

Which protocol does the "ping" command use?

  1. AHTTP
  2. BUDP
  3. CTCP
  4. DICMP
18.Computer networksHard

You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?

  1. A/25
  2. B/26
  3. C/27
  4. D/28
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1554 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1554 hosts
p =
20.IPv4 subnettingHard
last octet = 160112806413201608040201networkhost

A host has the address 10.20.23.160/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.23.160/27
x =

3Solutions

1.Number systems

Answer: U = 157

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

99 = 0110 0011
inverted: 1001 1100
+1 → 1001 1101

Read it as unsigned

1001 1101₂ = 256 − 99 = 157

Answer

U = 157
2.Number systems

Answer: R = 184

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 left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0101 1100 << 1
= 1011 1000

Value of the result

1011 1000₂ = 184

Answer

R = 184
3.Propositional logic & proofs

Answer: N = 5

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.

Truth table row by row

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

Count the rows with value 1

(¬r → p) ∧ (¬p → q): 1, 1, 1, 1, 1, 0, 0, 0
N = 5

Answer

N = 5
4.Propositional logic & proofs

Answer: N = 8

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

r ∨ s; s ∧ q; (r ∨ s) ∧ (s ∧ q); ¬p; q → ¬p; p → r; (q → ¬p) ∧ (p → r); ¬((q → ¬p) ∧ (p → r)); ((r ∨ s) ∧ (s ∧ q)) ∨ ¬((q → ¬p) ∧ (p → r))

Split on p

p = 1: 6
p = 0: 2
N = 6 + 2 = 8

Answer

N = 8
5.Digital logic: gates, K-maps & flip-flops

Answer: AB = 01, CD = 01

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.

m5: m4, m7, m1, m13 → 0
ABCD000111100001111011111111

The isolated 1

m5 = A′BC′D → AB = 01, CD = 01

Answer

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

Answer: N = 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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

C′D, B′D, A′C′, A′B′, ABCD′
ABCD000111100001111010X1131214X507060121130151140819111010

Minimal cover

F = C′D + B′D + A′C′ + A′B′ + ABCD′
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: x = 77

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 71

r = 81, t = 0
r = 71, q = 1, t = 1
r = 10, q = 7, t = −1
r = 1, q = 10, t = 8
71−1 ≡ 8 (mod 81)

Solve

x ≡ 8·40 = 320 ≡ 77 (mod 81)
check: 71·77 = 5467 ≡ 40 (mod 81)

Answer

x = 77
8.Number theory & modular arithmetic

Answer: 724105 mod 97 = 55

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

24105 mod 96 = 9
724105 ≡ 79 (mod 97)

Small power

79 mod 97 = 55

Answer

724105 mod 97 = 55
9.Algorithms & complexity

Answer: A · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

A · O(n²)
10.Algorithms & complexity

Answer: C · Decision problems whose yes-answers can be verified in polynomial time given a certificate

Why this is the answer

For an NP problem, a proposed solution (certificate) can be checked quickly, for example verifying a given assignment satisfies a Boolean formula.

Answer

C · Decision problems whose yes-answers can be verified in polynomial time given a certificate
11.Data structures

Answer: B · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

B · (i + 1) mod n
12.Data structures

Answer: B · 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

B · Every path from a node to its descendant leaves has the same number of black nodes
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) = 4, deg(F) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: E

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): B: 13, C: 7
2. C (7): E: 14
3. B (13): D: 28
4. E (14): D: 20, F: 19
5. F (19)
6. D (20)

round 4

E

Answer

E
15.Databases & SQL

Answer: D · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

D · To speed up searches and joins on those columns
16.Databases & SQL

Answer: C · It is in 2NF but not in 3NF

Why this is the answer

The key has one column, so 2NF holds, but emp_id → dept_id → dept_name is a transitive dependency, which violates 3NF.

Answer

C · It is in 2NF but not in 3NF
17.Computer networks

Answer: D · ICMP

Why this is the answer

Ping sends ICMP echo requests and measures the time until echo replies return.

Answer

D · ICMP
18.Computer networks

Answer: C · /27

Why this is the answer

/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.

Answer

C · /27
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 < 1554
211 − 2 = 2046 ≥ 1554 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 191

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 of this host

network: ⌊160/32⌋·32 = 160
broadcast: 160 + 32 − 1 = 191
usable hosts: 161 … 190

Answer

x = 191

ExamProblems · examproblems.com/courses/computer-science-fundamentals