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 complement1−12806413211618141201

The 8-bit pattern 1011 1110 stores a signed integer in two's complement. What is its value N in decimal?

1011 11108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 1001 AND 0010 1000. What is the result's value R in decimal?

1001 1001AND0010 1000
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

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

r → (r ∧ q)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
q¬qpp ∧ r

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

p ∨ rp → qr → q
  1. 1q
  2. 2¬q
  3. 3p
  4. 4p ∧ r
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = BC + ABC′ + A′C + A′B′C′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = BC + ABC′ + A′C + A′B′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101001130204150716112013X15X140809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(0, 3, 5, 6, 10, 12) + d(14, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(0, 3, 5, 6, 10, 12) + d(14, 15)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 4 (mod 5) and x ≡ 11 (mod 12).

x ≡ 4 (mod 5)x ≡ 11 (mod 12)
x =
8.Number theory & modular arithmeticHard

Solve 15x ≡ 9 (mod 21). Give the smallest positive solution x.

15x ≡ 9 (mod 21)
x =
9.Algorithms & complexityMedium

How much stack space does a simple recursive factorial function fact(n) = n × fact(n − 1) use?

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

Karatsuba multiplication satisfies T(n) = 3T(n/2) + O(n). What does this solve to?

  1. AΘ(n²)
  2. BΘ(n3)
  3. CΘ(n log n)
  4. DΘ(nlog₂ 3), about n1.585
11.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(log n), because of binary search
  2. BO(n), because later elements must shift
  3. CO(1), because the index is known
  4. DO(n²), because each element moves n times
12.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. APrimary clustering, where long runs of occupied slots form
  2. BThe need to store keys in the table
  3. CThe need for a hash function
  4. DThe O(n) cost of reading a key's value
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| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1739207118ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 2.

start: Ck = 2
  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 without a WHERE clause
  3. CReading data from a corrupted disk
  4. DReading the same row twice and getting the same value
16.Databases & SQLHard

Two doctors on call each check that the other is still on call, then both sign off at the same time. Which anomaly can snapshot isolation still allow here?

  1. ANon-repeatable read within one transaction
  2. BDirty read
  3. CWrite skew
  4. DDirty write
17.Computer networksMedium

A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?

  1. ABoth routes, splitting the traffic equally
  2. B10.1.0.0/16, because it is the longest matching prefix
  3. C10.0.0.0/8, because it covers more addresses than the /16
  4. D10.0.0.0/8, because it was added first
18.Computer networksHard

Why does a browser warn when a site presents a self-signed certificate?

  1. ANo trusted certificate authority vouches that the key belongs to that site
  2. BSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
  3. CThe certificate is always expired
  4. DThe site must be using HTTP instead of HTTPS
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 8 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/248 subnets
H =
20.IPv4 subnettingHard
third octet = 234112816413201618041201networkhost

A host has the address 172.20.234.120/18. What is the third octet x of its subnet's broadcast address?

172.20.234.120/18
x =

3Solutions

1.Number systems

Answer: N = −66

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 + 32 + 16 + 8 + 4 + 2 = −66

Answer

N = −66
2.Number systems

Answer: R = 8

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

AND gives 1 only where both bits are 1.

1001 1001
AND 0010 1000
= 0000 1000

Value of the result

0000 1000₂ = 8

Answer

R = 8
3.Propositional logic & proofs

Answer: contingency

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

Classify

r → (r ∧ q): 1, 1, 0, 1
1 in 3 of the 4 rows, 0 in the others: a contingency.

Answer

contingency
4.Propositional logic & proofs

Answer: q

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

Conclusion

q (proof by cases)

Answer

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

Answer: L = 6

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

BC → m(3, 7)
ABC′ → m(6)
A′C → m(1, 3)
A′B′C′ → m(0)
ABC00011110011011130204051716

Group and read off

F = BC + A′B′ + AB

Count the literals

L = 6

Answer

L = 6

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

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

Answer: N = 6

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.

BCD′, ACD′, ABD′, A′B′C′D′, A′B′CD, A′BC′D
ABCD00011110000111101001130204150716112013X15X140809011110

Minimal cover

F = BCD′ + ACD′ + ABD′ + A′B′C′D′ + A′B′CD + A′BC′D
N = 6

Answer

N = 6
7.Number theory & modular arithmetic

Answer: x = 59

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 = 4 + 5k

Second congruence

4 + 5k ≡ 11 (mod 12) → 5k ≡ 7 (mod 12)
k ≡ 5·7 ≡ 11 (mod 12)

Result

x = 4 + 5·11 = 59
x = 59
check: 59 mod 5 = 4, 59 mod 12 = 11

Answer

x = 59
8.Number theory & modular arithmetic

Answer: x = 2

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(15, 21) = 3
5x ≡ 3 (mod 7)

Solve the reduced congruence

5−1 ≡ 3 (mod 7) → x ≡ 2 (mod 7)

Solutions modulo 21

x ∈ {2, 9, 16}
x = 2

Answer

x = 2
9.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

The recursion is n calls deep before any returns, and each call needs a stack frame, so it uses O(n) space.

Answer

B · O(n)
10.Algorithms & complexity

Answer: D · Θ(nlog₂ 3), about n1.585

Why this is the answer

Since log₂ 3 ≈ 1.585 > 1, the leaves dominate by the master theorem, beating the O(n²) schoolbook method.

Answer

D · Θ(nlog₂ 3), about n1.585
11.Data structures

Answer: B · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

B · O(n), because later elements must shift
12.Data structures

Answer: A · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

A · Primary clustering, where long runs of occupied slots form
13.Graph theory

Answer: Euler path, no 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) = 1, deg(B) = 4, deg(C) = 2, deg(D) = 3, deg(E) = 4, deg(F) = 2

Odd vertices

A, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

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. AC (3)
2. CE (7)
3. BE (8)
4. DE (11)

edge number 2

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: C · Write skew

Why this is the answer

Under snapshot isolation both transactions see the old state and change different rows, so no write conflict is detected; serializable isolation prevents this.

Answer

C · Write skew
17.Computer networks

Answer: B · 10.1.0.0/16, because it is the longest matching prefix

Why this is the answer

IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.

Answer

B · 10.1.0.0/16, because it is the longest matching prefix
18.Computer networks

Answer: A · No trusted certificate authority vouches that the key belongs to that site

Why this is the answer

Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.

Answer

A · No trusted certificate authority vouches that the key belongs to that site
19.IPv4 subnetting

Answer: H = 30

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).

Bits to borrow

23 = 8 ≥ 8 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
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 in the third octet

26 = 64

Subnet of this host

network: 172.20.192.0
broadcast: 172.20.255.255

Answer

x = 255

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