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 complement0−12806403201608040201

Set the bits so they store −30 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −308 bit
U =
2.Number systemsHard
24-bit two's complement, N = ??sign???????????????????????

What is the most negative integer N that can be stored in 24-bit two's complement?

24 bit
N =
3.Propositional logic & proofsMedium
¬q ∨ ¬r¬q ∧ r¬q ∧ ¬r¬q ∨ r

Which formula is logically equivalent to ¬(q ∧ r)? Click it.

¬(q ∧ r)
  1. 1¬q ∨ ¬r
  2. 2¬q ∧ r
  3. 3¬q ∧ ¬r
  4. 4¬q ∨ r
4.Propositional logic & proofsHard
pqrq → (r ∨ p)qr111110101100011010001000

The argument with premises q → (r ∨ p); q and conclusion r is invalid. Click the row of the truth table that is a counterexample.

q → (r ∨ p)q∴ r
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = ABC′ + AB + B′C′ + A′B′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKJOHNSON

A 4-bit Johnson counter Q₃Q₂Q₁Q₀ starts at 0000. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives the inverse of Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 9 clock edges.

Q₃Q₂Q₁Q₀ = 00009
Q =
7.Number theory & modular arithmeticMedium

Solve 11x ≡ 10 (mod 14). Give the solution x with 0 < x < 14.

11x ≡ 10 (mod 14)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 755 mod 93.

a = 7k = 55n = 93
755 mod 93 =
9.Algorithms & complexityMedium

What does the Floyd–Warshall algorithm compute, and how fast?

  1. AShortest paths between all pairs of vertices in O(V³)
  2. BShortest paths from one source vertex to all others in O(V + E)
  3. CA minimum spanning tree in O(E log V)
  4. DA topological order in O(V²)
10.Algorithms & complexityHard

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

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

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AIts right child, always
  2. BIts parent, always
  3. CThe leftmost node of its right subtree
  4. DThe rightmost node of its left subtree
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA deque holding indices in decreasing order of value
  2. BA sorted array updated by shifting
  3. CA min-heap rebuilt for every window
  4. DA singly linked list of the window's elements
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| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
16134191281159ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 6; skipped edges do not count.

k = 6
  1. 1edge AB
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  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. ACombine the table with itself using UNION ALL and a WHERE filter
  2. BUse GROUP BY manager_id only
  3. CJoin the employees table with itself using two aliases
  4. DCreate a second copy of the table on disk
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. BWriters are not allowed while any reader is active
  3. CReaders read uncommitted data directly
  4. DReaders always wait until all writers have committed
17.Computer networksMedium

What is the IPv6 loopback address?

  1. A::
  2. B::1
  3. Cfe80::1
  4. Dff02::1
18.Computer networksHard

The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?

  1. AStatic routing; adding default routes
  2. BLink-state; flooding link updates
  3. CDistance-vector; split horizon and poison reverse
  4. DPath-vector; longer AS paths
19.IPv4 subnettingMedium
/?11111111111111111111111100000000network bitshost bits

A department needs 184 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

184 hosts
p =
20.IPv4 subnettingHard

The network 192.168.167.0/24 is split with VLSM for these LANs: A: 6 hosts, B: 28 hosts, C: 14 hosts, D: 16 hosts, E: 4 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?

A: 6B: 28C: 14D: 16E: 4
x =

3Solutions

1.Number systems

Answer: U = 226

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

30 = 0001 1110
inverted: 1110 0001
+1 → 1110 0010

Read it as unsigned

1110 0010₂ = 256 − 30 = 226

Answer

U = 226
2.Number systems

Answer: N = −8.389 × 10⁶

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.

Most negative value

−223 = −8388608

Answer

N = −8388608
3.Propositional logic & proofs

Answer: ¬q ∨ ¬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.
¬(q ∧ r) ≡ ¬q ∨ ¬r

Check with the truth table

q = 1, r = 1: 0 | 0 0 0 1
q = 1, r = 0: 1 | 1 0 0 0
q = 0, r = 1: 1 | 1 1 0 1
q = 0, r = 0: 1 | 1 0 1 1

Answer

¬q ∨ ¬r

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

4.Propositional logic & proofs

Answer: p = 1, q = 1, r = 0

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.

Premises and conclusion row by row

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

Counterexample

p = 1, q = 1, r = 0

Answer

p = 1, q = 1, r = 0

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: L = 4

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

ABC′ → m(6)
AB → m(6, 7)
B′C′ → m(0, 4)
A′B′C′ → m(0)
ABC00011110011001030214051716

Group and read off

F = B′C′ + AB

Count the literals

L = 4

Answer

L = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 8

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.

State after every edge

0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 → 1000

Decimal value

1000₂ = 8

Answer

Q = 8
7.Number theory & modular arithmetic

Answer: x = 6

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 11

r = 14, t = 0
r = 11, q = 1, t = 1
r = 3, q = 3, t = −1
r = 2, q = 1, t = 4
r = 1, q = 2, t = −5
11−1 ≡ 9 (mod 14)

Solve

x ≡ 9·10 = 90 ≡ 6 (mod 14)
check: 11·6 = 66 ≡ 10 (mod 14)

Answer

x = 6
8.Number theory & modular arithmetic

Answer: 755 mod 93 = 25

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.

Binary exponent

55 = 1101112 = 32 + 16 + 4 + 2 + 1

Repeated squaring

71 ≡ 7 (mod 93)
72 ≡ 49 (mod 93)
74 ≡ 76 (mod 93)
78 ≡ 10 (mod 93)
716 ≡ 7 (mod 93)
732 ≡ 49 (mod 93)

Multiply the needed powers

755 ≡ 7·49·76·7·49 ≡ 25 (mod 93)

Answer

755 mod 93 = 25
9.Algorithms & complexity

Answer: A · Shortest paths between all pairs of vertices in O(V³)

Why this is the answer

Floyd–Warshall is a dynamic programming algorithm that allows each vertex in turn as an intermediate point, using O(V³) time.

Answer

A · Shortest paths between all pairs of vertices in O(V³)
10.Algorithms & complexity

Answer: B · 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

B · After V − 1 rounds, it checks whether any edge can still be relaxed
11.Data structures

Answer: C · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

C · The leftmost node of its right subtree
12.Data structures

Answer: A · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

A · A deque holding indices in decreasing order of value
13.Graph theory

Answer: neither

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) = 3, deg(D) = 3, deg(E) = 2, deg(F) = 1, deg(G) = 3

Odd vertices

C, D, F, G → 4
neither

Answer

neither
14.Graph theory

Answer: edge AB

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. GC (1)
2. BD (4)
3. GB (8)
4. GE (9)
5. DF (12)
6. AB (16)

edge number 6

AB

Answer

edge AB
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: B · ::1

Why this is the answer

::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.

Answer

B · ::1
18.Computer networks

Answer: C · Distance-vector; split horizon and poison reverse

Why this is the answer

After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.

Answer

C · Distance-vector; split horizon and poison reverse
19.IPv4 subnetting

Answer: p = 24

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

27 − 2 = 126 < 184
28 − 2 = 254 ≥ 184 → h = 8

Prefix length

p = 32 − 8 = 24

Answer

p = /24
20.IPv4 subnetting

Answer: x = 32

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

B: 28 hosts → 25 = 32 (/27)
D: 16 hosts → 25 = 32 (/27)
C: 14 hosts → 24 = 16 (/28)
A: 6 hosts → 23 = 8 (/29)
E: 4 hosts → 23 = 8 (/29)

Allocation, largest first

B: .0 – .31
D: .32 – .63
C: .64 – .79
A: .80 – .87
E: .88 – .95

Answer

x = 32

LAN D: 192.168.167.32/27.

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