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 −96 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −968 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

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

1000 1000OR1001 1000
R =
3.Propositional logic & proofsMedium
¬p ∧ r¬p ∨ ¬r¬p ∧ ¬rp ∨ ¬r

Which formula is logically equivalent to ¬(p ∨ r)? Click it.

¬(p ∨ r)
  1. 1¬p ∧ r
  2. 2¬p ∨ ¬r
  3. 3¬p ∧ ¬r
  4. 4p ∨ ¬r
4.Propositional logic & proofsHard
∃x (A(x) → ∀y ¬R(x, y))∀x (A(x) ∧ ∀y ¬R(x, y))∃x (A(x) ∧ ∀y ¬R(x, y))∃x (¬A(x) ∧ ∃y R(x, y))

Which statement is equivalent to ¬∀x (A(x) → ∃y R(x, y)), with the negation moved all the way inside? Click it.

¬∀x (A(x) → ∃y R(x, y))
  1. 1∃x (A(x) → ∀y ¬R(x, y))
  2. 2∀x (A(x) ∧ ∀y ¬R(x, y))
  3. 3∃x (A(x) ∧ ∀y ¬R(x, y))
  4. 4∃x (¬A(x) ∧ ∃y R(x, y))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = AB′C′ + AB′C + ABC′ + BC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110101103121405170601201311501408X9011X10

For F(A, B, C, D) = Σm(0, 1, 2, 4, 7, 15) + d(9, 10), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 2, 4, 7, 15) + d(9, 10)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 5) and x ≡ 4 (mod 9).

x ≡ 2 (mod 5)x ≡ 4 (mod 9)
x =
8.Number theory & modular arithmeticHard
372c = me​ mod n?m = cd​ mod nm(n, e) = (629, 7)d

Encrypt the message m = 372 with the RSA public key (n, e) = (629, 7). Find the ciphertext c.

n = 629e = 7m = 372
c =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(n) solve to?

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

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AIt counts the cycles with depth-first search
  2. BIt checks whether any edge has a negative weight
  3. CAfter V − 1 rounds, it checks whether any edge can still be relaxed
  4. DIt runs Dijkstra's algorithm twice and compares the two distance results
11.Data structuresMedium

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

  1. Ai + 1
  2. B(i + 1) mod n
  3. C(i − 1) mod n
  4. D2i + 1
12.Data structuresHard

You must keep the k smallest values from a stream of n numbers using O(k) memory. Which approach works in O(n log k) time?

  1. AA FIFO queue of the last k values
  2. BA max-heap of size k, replacing its root when a smaller value arrives
  3. CA sorted array of all n values
  4. DA min-heap holding all n values, from which the k smallest are popped at the end
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
14.Graph theoryHard
101333463510127ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

Which ACID property means concurrent transactions do not see each other's unfinished changes?

  1. AAtomicity
  2. BDurability
  3. CAvailability
  4. DIsolation
16.Databases & SQLHard

Two transactions both read stock = 10, each subtracts 1 and writes the result, giving 9 instead of 8. Which anomaly is this?

  1. APhantom read
  2. BDeletion anomaly
  3. CDirty read
  4. DLost update
17.Computer networksMedium

What is the main difference between the HTTP methods GET and POST?

  1. AGET uses UDP, POST uses TCP
  2. BGET deletes a resource, POST reads it from the server's database
  3. CGET retrieves a resource, POST submits data in the request body
  4. DGET is encrypted, POST is not
18.Computer networksHard

A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?

  1. A50 kB
  2. B500 kB
  3. C4 kB
  4. D2.5 MB
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.137.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?

192.168.137.0/24/26subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 162112806413201608041201networkhost

A host has the address 10.20.251.162/29. What is the fourth octet x of its network address?

10.20.251.162/29
x =

3Solutions

1.Number systems

Answer: U = 160

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

96 = 0110 0000
inverted: 1001 1111
+1 → 1010 0000

Read it as unsigned

1010 0000₂ = 256 − 96 = 160

Answer

U = 160
2.Number systems

Answer: R = 152

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.

1000 1000
OR 1001 1000
= 1001 1000

Value of the result

1001 1000₂ = 152

Answer

R = 152
3.Propositional logic & proofs

Answer: ¬p ∧ ¬r

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

De Morgan: ¬(a ∨ b) ≡ ¬a ∧ ¬b. The ∨ turns into ∧ and both parts are negated.
¬(p ∨ r) ≡ ¬p ∧ ¬r

Check with the truth table

p = 1, r = 1: 0 | 0 0 0 1
p = 1, r = 0: 0 | 0 1 0 1
p = 0, r = 1: 0 | 1 1 0 0
p = 0, r = 0: 1 | 0 1 1 1

Answer

¬p ∧ ¬r

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

4.Propositional logic & proofs

Answer: ∃x (A(x) ∧ ∀y ¬R(x, y))

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 (A(x) → ∃y R(x, y)) ≡ ∃x (A(x) ∧ ∀y ¬R(x, y))

Answer

∃x (A(x) ∧ ∀y ¬R(x, y))

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: L = 3

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

AB′C′ → m(4)
AB′C → m(5)
ABC′ → m(6)
BC → m(3, 7)
ABC00011110010001130214151716

Group and read off

F = BC + A

Count the literals

L = 3

Answer

L = 3
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 2

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.

B′C′D, B′CD′, BCD, A′C′D′, A′B′D′, A′B′C′
ABCD0001111000011110101103121405170601201311501408X9011X10

Essential prime implicants

BCD, A′C′D′
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: x = 22

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 = 2 + 5k

Second congruence

2 + 5k ≡ 4 (mod 9) → 5k ≡ 2 (mod 9)
k ≡ 2·2 ≡ 4 (mod 9)

Result

x = 2 + 5·4 = 22
x = 22
check: 22 mod 5 = 2, 22 mod 9 = 4

Answer

x = 22
8.Number theory & modular arithmetic

Answer: c = 535

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

7 = 4 + 2 + 1
3721 ≡ 372 (mod 629)
3722 ≡ 4 (mod 629)
3724 ≡ 16 (mod 629)
3727 ≡ 372·4·16 ≡ 535 (mod 629)

Ciphertext

c = 535

Answer

c = 535
9.Algorithms & complexity

Answer: D · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

D · O(n²)
10.Algorithms & complexity

Answer: C · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

C · After V − 1 rounds, it checks whether any edge can still be relaxed
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 · A max-heap of size k, replacing its root when a smaller value arrives

Why this is the answer

The max-heap's root is the largest kept value; any smaller new value replaces it in O(log k), so the heap always holds the k smallest so far.

Answer

B · A max-heap of size k, replacing its root when a smaller value arrives
13.Graph theory

Answer: H = 60

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.

Count

H = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
14.Graph theory

Answer: d(F) = 16

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 10, C: 13
B (10): D: 13, G: 20
C (13): E: 17
D (13): F: 16
F (16)
E (17)
G (20)

Shortest path

A → B → D → F
d(F) = 16

Answer

d(F) = 16
15.Databases & SQL

Answer: D · Isolation

Why this is the answer

Isolation keeps concurrent transactions apart; isolation levels decide how strictly this is enforced.

Answer

D · Isolation
16.Databases & SQL

Answer: D · Lost update

Why this is the answer

In a lost update, both transactions work from the same old value and the second write erases the first; locking (SELECT ... FOR UPDATE) or atomic updates prevent it.

Answer

D · Lost update
17.Computer networks

Answer: C · GET retrieves a resource, POST submits data in the request body

Why this is the answer

GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.

Answer

C · GET retrieves a resource, POST submits data in the request body
18.Computer networks

Answer: B · 500 kB

Why this is the answer

The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.

Answer

B · 500 kB
19.IPv4 subnetting

Answer: x = 127

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

26 = 64

Subnet number 2

network: (2 − 1)·64 = 64
broadcast: 64 + 64 − 1 = 127

Answer

x = 127
20.IPv4 subnetting

Answer: x = 160

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

23 = 8

Subnet of this host

network: ⌊162/8⌋·8 = 160
broadcast: 160 + 8 − 1 = 167
usable hosts: 161 … 166

Answer

x = 160

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