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 complement1−12806413211608141201

The 8-bit pattern 1011 0110 stores a signed integer in two's complement. What is its value N in decimal?

1011 01108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0001 1010 << 2. What is the result's value R in decimal?

0001 1010<< 2
R =
3.Propositional logic & proofsMedium
r ∧ q(r → q) ∧ (q → r)(r → q) ∨ (q → r)(r → q) ∧ (¬q → ¬r)

Which formula is logically equivalent to r ↔ q? Click it.

r ↔ q
  1. 1r ∧ q
  2. 2(r → q) ∧ (q → r)
  3. 3(r → q) ∨ (q → r)
  4. 4(r → q) ∧ (¬q → ¬r)
4.Propositional logic & proofsHard
prr → p¬r¬p11100100

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

r → p¬r∴ ¬p
  1. 1p = 1, r = 1
  2. 2p = 1, r = 0
  3. 3p = 0, r = 1
  4. 4p = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′C′ + AB′C + B′C′ + A′BC
L =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​CLKMOD-7

A mod-7 counter is built from the minimum number of flip-flops. How many of the possible states U are never used?

N = 7
U =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(99), the number of integers from 1 to 99 that are coprime to 99.

n = 99 = 32·11
φ(99) =
8.Number theory & modular arithmeticHard
mc = me​ mod n485m = cd​ mod n?(n, e) = (1189, 3)d

An RSA key uses p = 29, q = 41 and e = 3. Decrypt the ciphertext c = 485: find the message m.

p = 29q = 41e = 3c = 485
m =
9.Algorithms & complexityMedium

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

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

Topological sorting is possible only for:

  1. AGraphs with negative weights
  2. BComplete graphs
  3. CUndirected graphs
  4. DDirected acyclic graphs
11.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. AInsertion order
  2. BLevel order
  3. CAscending sorted order
  4. DDescending order always
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 binary search tree rebuilt for each element
  2. BA hash table of all elements
  3. CA stack that keeps elements waiting for their answer
  4. DA queue that holds every pair of elements
13.Graph theoryMedium

A tree has 3 vertices of degree 4, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 43 × deg 2
L =
14.Graph theoryHard
1421510712812311ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

start: Around 5
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G
15.Databases & SQLMedium

Which query returns the number of different cities in the customers table?

  1. ASELECT COUNT(DISTINCT city) FROM customers
  2. BSELECT COUNT(city) FROM customers GROUP BY city
  3. CSELECT SUM(DISTINCT city) FROM customers
  4. DSELECT DISTINCT COUNT(city) FROM customers
16.Databases & SQLHard

What does SELECT day, SUM(amount) OVER (ORDER BY day) FROM sales compute?

  1. AThe total of all amounts on every row
  2. BOne row per day with that day's total only
  3. CA running total of amount ordered by day
  4. DThe average amount per day
17.Computer networksMedium

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

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

Why was the TCP window scale option introduced?

  1. AUDP needed a way to control its sending rate on congested paths
  2. BWithout it, TCP could not detect lost segments on long-distance links
  3. CThe 32-bit window field was too large and wasted header space in every TCP segment sent
  4. DThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/245 subnets
H =
20.IPv4 subnettingHard
third octet = 8012806403201618040201networkhost

A host has the address 172.20.8.246/23. What is the third octet x of its network address?

172.20.8.246/23
x =

3Solutions

1.Number systems

Answer: N = −74

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 + 32 + 16 + 4 + 2 = −74

Answer

N = −74
2.Number systems

Answer: R = 104

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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.

0001 1010 << 2
= 0110 1000

Value of the result

0110 1000₂ = 104

Answer

R = 104
3.Propositional logic & proofs

Answer: (r → q) ∧ (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

a ↔ b means the implication holds in both directions: (a → b) ∧ (b → a).
r ↔ q ≡ (r → q) ∧ (q → r)

Check with the truth table

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

Answer

(r → q) ∧ (q → r)

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

4.Propositional logic & proofs

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

Counterexample

p = 1, r = 0

Answer

p = 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 = 6

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

A′C′ → m(0, 2)
AB′C → m(5)
B′C′ → m(0, 4)
A′BC → m(3)
ABC00011110011001131214150706

Group and read off

F = B′C′ + A′B + AB′

Count the literals

L = 6

Answer

L = 6

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

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

Answer: U = 1

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.

Number of flip-flops

2ⁿ ≥ 7 → n = 3 (2³ = 8)

Unused states

U = 8 − 7 = 1

Answer

U = 1
7.Number theory & modular arithmetic

Answer: φ(99) = 60

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

99 = 32·11

Totient

φ(99) = φ(9)·φ(11) = 6·10 = 60

Answer

φ(99) = 60
8.Number theory & modular arithmetic

Answer: m = 1042

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.

Private exponent

n = 1189, φ(n) = 1120
d = 3−1 mod 1120 = 747

Decrypt by repeated squaring

747 = 512 + 128 + 64 + 32 + 8 + 2 + 1
4851 ≡ 485 (mod 1189)
4852 ≡ 992 (mod 1189)
4854 ≡ 761 (mod 1189)
4858 ≡ 78 (mod 1189)
48516 ≡ 139 (mod 1189)
48532 ≡ 297 (mod 1189)
48564 ≡ 223 (mod 1189)
485128 ≡ 980 (mod 1189)
485256 ≡ 877 (mod 1189)
485512 ≡ 1035 (mod 1189)
485747 ≡ 485·992·78·297·223·980·1035 ≡ 1042 (mod 1189)

Message

m = 1042
check: 10423 mod 1189 = 485

Answer

m = 1042
9.Algorithms & complexity

Answer: C · 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

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

Answer: D · Directed acyclic graphs

Why this is the answer

A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.

Answer

D · Directed acyclic graphs
11.Data structures

Answer: C · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

C · Ascending sorted order
12.Data structures

Answer: C · 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

C · A stack that keeps elements waiting for their answer
13.Graph theory

Answer: L = 8

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.

Handshake lemma for a tree

L + 18 = 2(L + 6) − 2
L = 8

Answer

L = 8

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: D

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

1. A (0): B: 1, C: 4
2. B (1): C: 3, G: 9
3. C (3): E: 18
4. G (9): D: 12
5. D (12): F: 19
6. E (18)
7. F (19)

round 5

D

Answer

D
15.Databases & SQL

Answer: A · SELECT COUNT(DISTINCT city) FROM customers

Why this is the answer

COUNT(DISTINCT city) counts unique non-NULL cities; DISTINCT outside COUNT only removes duplicate result rows.

Answer

A · SELECT COUNT(DISTINCT city) FROM customers
16.Databases & SQL

Answer: C · A running total of amount ordered by day

Why this is the answer

With ORDER BY and no explicit frame, the window runs from the first row up to the current row (and its ties), which gives a cumulative sum.

Answer

C · A running total of amount ordered by day
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: D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths

Why this is the answer

The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.

Answer

D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
19.IPv4 subnetting

Answer: H = 30

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).

Bits to borrow

23 = 8 ≥ 5 → b = 3, /27

Hosts per subnet

25 − 2 = 30

Answer

H = 30
20.IPv4 subnetting

Answer: x = 8

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

21 = 2

Subnet of this host

network: 172.20.8.0
broadcast: 172.20.9.255

Answer

x = 8

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