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 complement1−12816413211608141211

The 8-bit pattern 1111 0111 stores a signed integer in two's complement. What is its value N in decimal?

1111 01118 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 1100 << 1. What is the result's value R in decimal?

1001 1100<< 1
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is q ↔ ¬q a tautology, a contradiction or a contingency? Click your answer.

q ↔ ¬q
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
¬pqpp ∧ ¬q

The premises (r ∧ p) → q; r; ¬q are given. Which conclusion follows from them? Click it.

(r ∧ p) → qr¬q
  1. 1¬p
  2. 2q
  3. 3p
  4. 4p ∧ ¬q
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(2, 3, 6, 7, 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
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
  8. 8clock edge 8
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(1183), the number of integers from 1 to 1183 that are coprime to 1183.

n = 1183 = 7·132
φ(1183) =
8.Number theory & modular arithmeticHard

97 is prime. Use Fermat’s little theorem to find 244791 mod 97.

p = 97k = 44791
244791 mod 97 =
9.Algorithms & complexityMedium

What is the running time of LSD radix sort on n integers with d digits in base k?

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

Which two things must be shown to prove that a decision problem X is NP-complete?

  1. AX is in P, and X reduces to SAT
  2. BX reduces to a problem in P, and X is in NP
  3. CX is in NP, and a known NP-complete problem reduces to X in polynomial time
  4. DX has no polynomial algorithm, and X is decidable
11.Data structuresMedium

Which abstract data type does Dijkstra's algorithm use to choose the next vertex to process?

  1. AA set with no ordering
  2. BA FIFO queue
  3. CA min-priority queue
  4. DA stack
12.Data structuresHard

For each element of an array you want the next greater element to its right, in O(n) total time. Which structure helps?

  1. AA stack that keeps elements waiting for their answer
  2. BA binary search tree rebuilt for each element
  3. CA hash table of all elements
  4. DA queue that holds every pair of elements
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| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
201312926101118ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.

k = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BD
  4. 4edge DE
  5. 5edge DF
  6. 6edge GB
  7. 7edge GC
  8. 8edge GD
  9. 9edge GE
15.Databases & SQLMedium

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. ACreate a second copy of the table on disk
  2. BUse GROUP BY manager_id only
  3. CJoin the employees table with itself using two aliases
  4. DCombine the table with itself using UNION ALL and a WHERE filter
16.Databases & SQLHard

With multiversion concurrency control (MVCC), how can readers avoid blocking writers?

  1. AReaders see a consistent snapshot of committed row versions while writers create new versions
  2. BReaders always wait until all writers have committed
  3. CReaders read uncommitted data directly
  4. DWriters are not allowed while any reader is active
17.Computer networksMedium

TLS provides which security properties for HTTPS?

  1. AAnonymity of the user
  2. BProtection against all malware and phishing on the visited site
  3. COnly faster loading
  4. DConfidentiality, integrity and server authentication
18.Computer networksHard

A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?

  1. AThe network card driver
  2. BThe physical cable
  3. CThe default gateway
  4. DName resolution (DNS)
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 878 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

878 hosts
p =
20.IPv4 subnettingHard
third octet = 92012816403211618140201networkhost

A host has the address 172.20.92.240/17. What is the third octet x of its subnet's broadcast address?

172.20.92.240/17
x =

3Solutions

1.Number systems

Answer: N = −9

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 + 64 + 32 + 16 + 4 + 2 + 1 = −9

Answer

N = −9
2.Number systems

Answer: R = 56

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.

1001 1100 << 1
= 0011 1000

Value of the result

0011 1000₂ = 56

Answer

R = 56
3.Propositional logic & proofs

Answer: contradiction

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: ¬q = 0, q ↔ ¬q = 0
q = 0: ¬q = 1, q ↔ ¬q = 0

Classify

q ↔ ¬q: 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ¬p

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

Conclusion

¬p (modus tollens with a conjunction)

Answer

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

Answer: AB = 10, 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.

m8: m9, m10, m12, m0 → 0
ABCD000111100001111011111111

The isolated 1

m8 = AB′C′D′ → AB = 10, CD = 00

Answer

AB = 10, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 3

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 0 → Q = 1
2: J = 1, K = 0 → Q = 1
3: J = 0, K = 1 → Q = 0
4: J = 1, K = 0 → Q = 1
5: J = 1, K = 0 → Q = 1
6: J = 0, K = 0 → Q = 1
7: J = 0, K = 0 → Q = 1
8: J = 0, K = 1 → Q = 0

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: φ(1183) = 936

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).

Prime factors

1183 = 7·132

Totient

φ(1183) = φ(7)·φ(169) = 6·156 = 936

Answer

φ(1183) = 936
8.Number theory & modular arithmetic

Answer: 244791 mod 97 = 31

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

44791 mod 96 = 55
244791 ≡ 255 (mod 97)

Small power

255 mod 97 = 31

Answer

244791 mod 97 = 31
9.Algorithms & complexity

Answer: B · O(d(n + k))

Why this is the answer

Each of the d passes is a counting sort costing O(n + k), so for fixed d and k the total is linear in n.

Answer

B · O(d(n + k))
10.Algorithms & complexity

Answer: C · X is in NP, and a known NP-complete problem reduces to X in polynomial time

Why this is the answer

Membership in NP shows X is not harder than NP; the reduction from a known NP-complete problem shows X is at least as hard as all of NP.

Answer

C · X is in NP, and a known NP-complete problem reduces to X in polynomial time
11.Data structures

Answer: C · A min-priority queue

Why this is the answer

A min-priority queue, usually a binary heap, returns the unvisited vertex with the smallest distance efficiently.

Answer

C · A min-priority queue
12.Data structures

Answer: A · A stack that keeps elements waiting for their answer

Why this is the answer

Scanning left to right, pop every stacked element smaller than the current one (the current element is their answer), then push it; each element moves at most twice.

Answer

A · A stack that keeps elements waiting for their answer
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) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 2, deg(E) = 3

Odd vertices

C, E → 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 GC

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.

Kruskal’s algorithm

1. DF (2)
2. GB (6)
3. DE (9)
4. GC (10)
5. GD (11)
6. AC (13)

edge number 4

GC

Answer

edge GC
15.Databases & SQL

Answer: C · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

C · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: A · Readers see a consistent snapshot of committed row versions while writers create new versions

Why this is the answer

MVCC keeps old row versions so a reader can see the data as of its snapshot, while a writer creates a new version instead of overwriting in place.

Answer

A · Readers see a consistent snapshot of committed row versions while writers create new versions
17.Computer networks

Answer: D · Confidentiality, integrity and server authentication

Why this is the answer

TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).

Answer

D · Confidentiality, integrity and server authentication
18.Computer networks

Answer: D · Name resolution (DNS)

Why this is the answer

A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.

Answer

D · Name resolution (DNS)
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 878
210 − 2 = 1022 ≥ 878 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.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 in the third octet

27 = 128

Subnet of this host

network: 172.20.0.0
broadcast: 172.20.127.255

Answer

x = 127

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