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 −16 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −168 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 1101 << 2. What is the result's value R in decimal?

1011 1101<< 2
R =
3.Propositional logic & proofsMedium
(c → a) ∧ (b → a)(c → b) → ac ∧ (b → a)c → (b → a)

Which formula is logically equivalent to (c ∧ b) → a? Click it.

(c ∧ b) → a
  1. 1(c → a) ∧ (b → a)
  2. 2(c → b) → a
  3. 3c ∧ (b → a)
  4. 4c → (b → a)
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make (q ∨ (r ∨ ¬p)) ∧ (s ∨ ¬r) true? Call this number N.

(q ∨ (r ∨ ¬p)) ∧ (s ∨ ¬r)
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011031204050716

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

F(A, B, C) = Σm(1, 2, 6)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​CLKRING

A ring counter is made of 4 D flip-flops with the last output fed back to the first input, and starts with a single 1. How many different states N does it go through?

n = 4
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 6) and x ≡ 5 (mod 11).

x ≡ 2 (mod 6)x ≡ 5 (mod 11)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n91m = cd​ mod n?(n, e) = (145, 13)d

An RSA key uses p = 5, q = 29 and e = 13. Decrypt the ciphertext c = 91: find the message m.

p = 5q = 29e = 13c = 91
m =
9.Algorithms & complexityMedium

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

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

What would follow if someone found a polynomial-time algorithm for one NP-complete problem?

  1. AEvery problem in NP could be solved in polynomial time, so P = NP
  2. BThe halting problem would become decidable
  3. CAll exponential-time algorithms would immediately run in linear time
  4. DOnly that one problem would move into P
11.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(1), because the index is known
  2. BO(n²), because each element moves n times
  3. CO(log n), because of binary search
  4. DO(n), because later elements must shift
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 hash table of all elements
  2. BA binary search tree rebuilt for each element
  3. CA stack that keeps elements waiting for their answer
  4. DA queue that holds every pair of 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| = 12
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
8147118172016592ABCDEFG

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 BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF
  9. 9edge GC
  10. 10edge GD
  11. 11edge GE
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

When is decomposing R into R1 and R2 guaranteed to be lossless?

  1. AWhen every attribute of R appears in both R1 and R2
  2. BWhen the common attributes of R1 and R2 form a superkey of R1 or of R2
  3. CWhen R1 and R2 have no attributes in common
  4. DWhen R1 and R2 have the same number of columns
17.Computer networksMedium

NAT (network address translation) is mainly used to:

  1. ALet many devices with private addresses share one public IP address
  2. BTranslate domain names into IP addresses for devices on a local network
  3. CIncrease the speed of a network cable
  4. DEncrypt traffic between two sites
18.Computer networksHard

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

  1. ALink-state; flooding link updates
  2. BDistance-vector; split horizon and poison reverse
  3. CStatic routing; adding default routes
  4. DPath-vector; longer AS paths
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.24.0/24 is divided into /29 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 12?

192.168.24.0/24/29subnet 12 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 152112806403211618040201networkhost

A host has the address 10.20.69.152/26. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.69.152/26
x =

3Solutions

1.Number systems

Answer: U = 240

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

16 = 0001 0000
inverted: 1110 1111
+1 → 1111 0000

Read it as unsigned

1111 0000₂ = 256 − 16 = 240

Answer

U = 240
2.Number systems

Answer: R = 244

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.

1011 1101 << 2
= 1111 0100

Value of the result

1111 0100₂ = 244

Answer

R = 244
3.Propositional logic & proofs

Answer: c → (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

Exportation: (a ∧ b) → c ≡ a → (b → c). Both are 0 only when a = 1, b = 1 and c = 0.
(c ∧ b) → a ≡ c → (b → a)

Check with the truth table

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

Answer

c → (b → a)

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

4.Propositional logic & proofs

Answer: N = 10

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

¬p; r ∨ ¬p; q ∨ (r ∨ ¬p); ¬r; s ∨ ¬r; (q ∨ (r ∨ ¬p)) ∧ (s ∨ ¬r)

Split on p

p = 1: 4
p = 0: 6
N = 4 + 6 = 10

Answer

N = 10
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

ABC00011110010011031204050716

Group the 0s

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

(B + C)
(B′ + C′)
(A′ + C′)

Minimal POS form

F = (B + C)(B′ + C′)(A′ + C′)
N = 3

Answer

N = 3

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 = 4

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.

Sequence for 4 flip-flops

1000 → 0100 → 0010 → 0001

Number of states

N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 38

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 = 2 + 6k

Second congruence

2 + 6k ≡ 5 (mod 11) → 6k ≡ 3 (mod 11)
k ≡ 2·3 ≡ 6 (mod 11)

Result

x = 2 + 6·6 = 38
x = 38
check: 38 mod 6 = 2, 38 mod 11 = 5

Answer

x = 38
8.Number theory & modular arithmetic

Answer: m = 51

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 = 145, φ(n) = 112
d = 13−1 mod 112 = 69

Decrypt by repeated squaring

69 = 64 + 4 + 1
911 ≡ 91 (mod 145)
912 ≡ 16 (mod 145)
914 ≡ 111 (mod 145)
918 ≡ 141 (mod 145)
9116 ≡ 16 (mod 145)
9132 ≡ 111 (mod 145)
9164 ≡ 141 (mod 145)
9169 ≡ 91·111·141 ≡ 51 (mod 145)

Message

m = 51
check: 5113 mod 145 = 91

Answer

m = 51
9.Algorithms & complexity

Answer: A · 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

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

Answer: A · Every problem in NP could be solved in polynomial time, so P = NP

Why this is the answer

Since all of NP reduces to any NP-complete problem in polynomial time, a fast algorithm for one gives a fast algorithm for all.

Answer

A · Every problem in NP could be solved in polynomial time, so P = NP
11.Data structures

Answer: D · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

D · O(n), because later elements must shift
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: 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: edge BC

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. BD (1)
2. GE (2)
3. GC (5)
4. BC (7)
5. AB (8)
6. EF (16)

edge number 4

BC

Answer

edge BC
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: B · When the common attributes of R1 and R2 form a superkey of R1 or of R2

Why this is the answer

If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.

Answer

B · When the common attributes of R1 and R2 form a superkey of R1 or of R2
17.Computer networks

Answer: A · Let many devices with private addresses share one public IP address

Why this is the answer

NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.

Answer

A · Let many devices with private addresses share one public IP address
18.Computer networks

Answer: B · 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

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

Answer: x = 95

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

23 = 8

Subnet number 12

network: (12 − 1)·8 = 88
broadcast: 88 + 8 − 1 = 95

Answer

x = 95
20.IPv4 subnetting

Answer: x = 191

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

26 = 64

Subnet of this host

network: ⌊152/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190

Answer

x = 191

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