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 −61 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −618 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 0100 XOR 0100 0100. What is the result's value R in decimal?

0111 0100XOR0100 0100
R =
3.Propositional logic & proofsMedium
(r ∧ q) ∨ (r ∧ p)(r ∨ q) ∧ p(r ∨ q) ∧ (r ∨ p)r ∧ (q ∨ p)

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

r ∨ (q ∧ p)
  1. 1(r ∧ q) ∨ (r ∧ p)
  2. 2(r ∨ q) ∧ p
  3. 3(r ∨ q) ∧ (r ∨ p)
  4. 4r ∧ (q ∨ p)
4.Propositional logic & proofsHard
qq ∧ p¬pp

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

q ∨ p¬q
  1. 1q
  2. 2q ∧ p
  3. 3¬p
  4. 4p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000111312140507160121130151140819111010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 2, 3, 4, 6, 9, 11, 13, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 2, 3, 4, 6, 9, 11, 13, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

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
7.Number theory & modular arithmeticMedium

Find the inverse of 40 modulo 73: the number x with 0 < x < 73 and 40·x ≡ 1 (mod 73).

a = 40n = 73
40−1 mod 73 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (85, 19)d = ?

An RSA key uses the primes p = 5 and q = 17 and the public exponent e = 19. Find the private exponent d (0 < d < φ(n)).

p = 5q = 17e = 19
d =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the activity that finishes earliest, then repeat with the compatible ones
  2. BPick the shortest activity first
  3. CPick the activity that starts earliest
  4. DPick the activity with the fewest overlaps first, then repeat with the compatible ones
10.Algorithms & complexityHard

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

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

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA queue
  2. BA hash table
  3. CA binary search tree
  4. DA stack
12.Data structuresHard

Which pair of traversals always determines a unique binary tree with distinct keys?

  1. ALevel-order alone
  2. BPreorder alone
  3. CPreorder and postorder
  4. DPreorder and inorder
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
21914978153ABCDEF

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 2; skipped edges do not count.

k = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE
15.Databases & SQLMedium

Why are indexes created on columns?

  1. ATo save disk space
  2. BTo make inserts and updates faster on those columns
  3. CTo prevent NULL values
  4. DTo speed up searches and joins on those columns
16.Databases & SQLHard

What does the write-ahead logging (WAL) rule require?

  1. AA change's log record must reach stable storage before the changed data page does
  2. BThe log is written only after a crash
  3. CEvery query must be logged before it is parsed
  4. DData pages must be written before the log
17.Computer networksMedium

How many usable host addresses does a /26 IPv4 subnet have?

  1. A62
  2. B30
  3. C64
  4. D126
18.Computer networksHard

What is the maximum transmission unit (MTU) of standard Ethernet payload?

  1. A65,535 bytes
  2. B64 bytes
  3. C1500 bytes
  4. D9000 bytes in every network
19.IPv4 subnettingMedium
/16 → /2311111111111111111111111000000000network bitshost bits

The network 172.27.0.0/16 is divided into equal subnets with prefix /23. How many subnets N are there?

172.27.0.0/16/23
N =
20.IPv4 subnettingHard
last octet = 157112806403211618140211networkhost

A host has the address 10.20.240.157/27. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.240.157/27
x =

3Solutions

1.Number systems

Answer: U = 195

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

61 = 0011 1101
inverted: 1100 0010
+1 → 1100 0011

Read it as unsigned

1100 0011₂ = 256 − 61 = 195

Answer

U = 195
2.Number systems

Answer: R = 48

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

XOR gives 1 where the two bits differ.

0111 0100
XOR 0100 0100
= 0011 0000

Value of the result

0011 0000₂ = 48

Answer

R = 48
3.Propositional logic & proofs

Answer: (r ∨ q) ∧ (r ∨ p)

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

∨ distributes over ∧: a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
r ∨ (q ∧ p) ≡ (r ∨ q) ∧ (r ∨ p)

Check with the truth table

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

Answer

(r ∨ q) ∧ (r ∨ p)

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

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

Conclusion

p (disjunctive syllogism)

Answer

p
5.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.

ABCD000111100001111000111312140507160121130151140819111010

Group the 0s

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

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

Minimal POS form

F = (B + C + D)(B′ + C′ + D′)(A′ + C + D)(A + B′ + D′)(A′ + B + D)
N = 5

Answer

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

Answer: clock edge 5

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

First change from 1 to 0

clock edge 5

Answer

clock edge 5
7.Number theory & modular arithmetic

Answer: 40−1 mod 73 = 42

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).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·40 (mod 73).

r = 73, t = 0
r = 40, q = 1, t = 1
r = 33, q = 1, t = −1
r = 7, q = 4, t = 2
r = 5, q = 1, t = −9
r = 2, q = 2, t = 11
r = 1, q = 2, t = −31

Inverse

1 ≡ −31·40 (mod 73)
x = 42
check: 40·42 = 1680 = 23·73 + 1

Answer

40−1 mod 73 = 42
8.Number theory & modular arithmetic

Answer: d = 27

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.

Modulus and totient

n = 5·17 = 85
φ(n) = 4·16 = 64

Extended Euclidean algorithm

r = 64, t = 0
r = 19, q = 3, t = 1
r = 7, q = 2, t = −3
r = 5, q = 1, t = 7
r = 2, q = 2, t = −10
r = 1, q = 2, t = 27

Private exponent

d = 27
check: 19·27 = 513 = 8·64 + 1

Answer

d = 27
9.Algorithms & complexity

Answer: A · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

A · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: A · 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

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

Answer: D · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

D · A stack
12.Data structures

Answer: D · Preorder and inorder

Why this is the answer

Preorder gives the root, and its position in the inorder sequence splits the left and right subtrees; preorder with postorder is ambiguous when a node has one child.

Answer

D · Preorder and inorder
13.Graph theory

Answer: H = 2520

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 = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: edge BE

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. AB (2)
2. BE (3)
3. DE (7)
4. DF (8)
5. CE (9)

edge number 2

BE

Answer

edge BE
15.Databases & SQL

Answer: D · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

D · To speed up searches and joins on those columns
16.Databases & SQL

Answer: A · A change's log record must reach stable storage before the changed data page does

Why this is the answer

Writing the log first lets the DBMS redo committed work and undo uncommitted work after a crash, supporting atomicity and durability.

Answer

A · A change's log record must reach stable storage before the changed data page does
17.Computer networks

Answer: A · 62

Why this is the answer

A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.

Answer

A · 62
18.Computer networks

Answer: C · 1500 bytes

Why this is the answer

Standard Ethernet carries up to 1500 bytes of payload; larger IP packets must be fragmented or use path MTU discovery.

Answer

C · 1500 bytes
19.IPv4 subnetting

Answer: N = 128

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

23 − 16 = 7

Number of subnets

N = 27 = 128

Answer

N = 128

Each /23 subnet has 29 − 2 = 510 usable hosts.

20.IPv4 subnetting

Answer: x = 159

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

25 = 32

Subnet of this host

network: ⌊157/32⌋·32 = 128
broadcast: 128 + 32 − 1 = 159
usable hosts: 129 … 158

Answer

x = 159

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