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 −23 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −238 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1110 0001 OR 0110 0010. What is the result's value R in decimal?

1110 0001OR0110 0010
R =
3.Propositional logic & proofsMedium
b → a¬b ∧ ¬a¬a → ¬b¬b → ¬a

Which formula is logically equivalent to a → b? Click it.

a → b
  1. 1b → a
  2. 2¬b ∧ ¬a
  3. 3¬a → ¬b
  4. 4¬b → ¬a
4.Propositional logic & proofsHard
qq ∧ pr¬r

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

q ∨ pq → rp → r
  1. 1q
  2. 2q ∧ p
  3. 3r
  4. 4¬r
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010312140507160121131150140809011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 2, 4, 6, 10, 13, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 2, 4, 6, 10, 13, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111100001031214051706012113X1511408X9111010

For F(A, B, C, D) = Σm(2, 4, 7, 11, 13, 14) + d(9, 15), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(2, 4, 7, 11, 13, 14) + d(9, 15)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 1 (mod 5) and x ≡ 4 (mod 11).

x ≡ 1 (mod 5)x ≡ 4 (mod 11)
x =
8.Number theory & modular arithmeticHard
1542c = me​ mod n?m = cd​ mod nm(n, e) = (1739, 17)d

Encrypt the message m = 1542 with the RSA public key (n, e) = (1739, 17). Find the ciphertext c.

n = 1739e = 17m = 1542
c =
9.Algorithms & complexityMedium

What is the time complexity of the Bellman–Ford algorithm on a graph with V vertices and E edges?

  1. AO(E log V)
  2. BO(V + E)
  3. CO(V³ × E)
  4. DO(V × E)
10.Algorithms & complexityHard

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. ACircuit satisfiability
  2. B2-SAT, where each clause has at most two literals
  3. CGeneral SAT on formulas in conjunctive normal form
  4. D3-SAT, where each clause has exactly three literals
11.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AThat keys are stored in insertion order
  2. BO(log n) height through rebalancing
  3. CConstant-time search in every case
  4. DThat no node has two children
12.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. AEmptying the slot could break the probe sequence for other keys
  2. BTombstones speed up hashing
  3. CDeleted keys must be kept forever for security, so slots can never be reused
  4. DIt makes the table sorted
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 12
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
1012615127615ABCDEF

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

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

In relational algebra, the selection operator (σ) corresponds to which SQL clause?

  1. AORDER BY
  2. BSELECT list
  3. CGROUP BY
  4. DWHERE
16.Databases & SQLHard

What does a FULL OUTER JOIN return?

  1. AThe Cartesian product, pairing each row of one table with every row of the other
  2. BOnly unmatched rows
  3. CAll rows from both tables, matched where possible and NULL-filled otherwise
  4. DOnly matching rows
17.Computer networksMedium

How does a stateful firewall differ from a simple packet filter?

  1. AIt only checks the MAC address of each frame
  2. BIt blocks all incoming traffic without exceptions, including replies to outgoing requests
  3. CIt encrypts every packet it forwards
  4. DIt tracks connections and allows replies that belong to established sessions
18.Computer networksHard

In a SYN flood attack, what resource is the attacker trying to exhaust?

  1. AThe switch's MAC address table
  2. BThe DNS cache of the client
  3. CThe server's table of half-open connections
  4. DThe server's disk space
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 10.226.61.0/24 is divided into equal subnets with prefix /26. How many subnets N are there?

10.226.61.0/24/26
N =
20.IPv4 subnettingHard
third octet = 53012806413211608140211networkhost

A host has the address 172.20.53.199/20. What is the third octet x of its network address?

172.20.53.199/20
x =

3Solutions

1.Number systems

Answer: U = 233

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

23 = 0001 0111
inverted: 1110 1000
+1 → 1110 1001

Read it as unsigned

1110 1001₂ = 256 − 23 = 233

Answer

U = 233
2.Number systems

Answer: R = 227

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.

1110 0001
OR 0110 0010
= 1110 0011

Value of the result

1110 0011₂ = 227

Answer

R = 227
3.Propositional logic & proofs

Answer: ¬b → ¬a

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

An implication is equivalent to its contrapositive ¬b → ¬a, not to its converse b → a or its inverse ¬a → ¬b.
a → b ≡ ¬b → ¬a

Check with the truth table

a = 1, b = 1: 1 | 1 0 1 1
a = 1, b = 0: 0 | 1 0 1 0
a = 0, b = 1: 1 | 0 0 0 1
a = 0, b = 0: 1 | 1 1 1 1

Answer

¬b → ¬a

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

4.Propositional logic & proofs

Answer: r

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

Conclusion

r (proof by cases)

Answer

r
5.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.

ABCD000111100001111010010312140507160121131150140809011110

Group the 0s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

(B + D′)
(A + D′)
(A′ + C + D)
(A′ + B′ + D)

Minimal POS form

F = (B + D′)(A + D′)(A′ + C + D)(A′ + B′ + D)
N = 4

Answer

N = 4

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

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.

BCD, AD, ABC, A′B′CD′, A′BC′D′
ABCD00011110000111100001031214051706012113X1511408X9111010

Essential prime implicants

BCD, AD, ABC, A′B′CD′, A′BC′D′
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: x = 26

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 = 1 + 5k

Second congruence

1 + 5k ≡ 4 (mod 11) → 5k ≡ 3 (mod 11)
k ≡ 9·3 ≡ 5 (mod 11)

Result

x = 1 + 5·5 = 26
x = 26
check: 26 mod 5 = 1, 26 mod 11 = 4

Answer

x = 26
8.Number theory & modular arithmetic

Answer: c = 1594

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

17 = 16 + 1
15421 ≡ 1542 (mod 1739)
15422 ≡ 551 (mod 1739)
15424 ≡ 1015 (mod 1739)
15428 ≡ 737 (mod 1739)
154216 ≡ 601 (mod 1739)
154217 ≡ 1542·601 ≡ 1594 (mod 1739)

Ciphertext

c = 1594

Answer

c = 1594
9.Algorithms & complexity

Answer: D · O(V × E)

Why this is the answer

Each of the V − 1 rounds relaxes all E edges, giving O(VE); in exchange it handles negative edge weights.

Answer

D · O(V × E)
10.Algorithms & complexity

Answer: B · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

B · 2-SAT, where each clause has at most two literals
11.Data structures

Answer: B · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

B · O(log n) height through rebalancing
12.Data structures

Answer: A · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

A · Emptying the slot could break the probe sequence for other keys
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) = 4, deg(C) = 4, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 4

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: d(D) = 25

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: 12
B (10): D: 25
C (12): E: 24
E (24): F: 39
D (25): F: 31
F (31)

Shortest path

A → B → D
d(D) = 25

Answer

d(D) = 25
15.Databases & SQL

Answer: D · WHERE

Why this is the answer

σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.

Answer

D · WHERE
16.Databases & SQL

Answer: C · All rows from both tables, matched where possible and NULL-filled otherwise

Why this is the answer

A full outer join keeps every row of both tables, pairing matches and filling the missing side with NULLs.

Answer

C · All rows from both tables, matched where possible and NULL-filled otherwise
17.Computer networks

Answer: D · It tracks connections and allows replies that belong to established sessions

Why this is the answer

A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.

Answer

D · It tracks connections and allows replies that belong to established sessions
18.Computer networks

Answer: C · The server's table of half-open connections

Why this is the answer

Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.

Answer

C · The server's table of half-open connections
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

26 − 24 = 2

Number of subnets

N = 22 = 4

Answer

N = 4

Each /26 subnet has 26 − 2 = 62 usable hosts.

20.IPv4 subnetting

Answer: x = 48

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

24 = 16

Subnet of this host

network: 172.20.48.0
broadcast: 172.20.63.255

Answer

x = 48

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