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
Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
Set the bits so they store −6 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
Set the bits to the 8-bit result of 1100 0110 XOR 1000 1011. What is the result's value R in decimal?
Is ¬(b ∨ ¬b) a tautology, a contradiction or a contingency? Click your answer.
How many of the 16 rows of the truth table make ¬((¬p → s) ↔ ¬p) ∧ ((r ∧ q) ∨ ¬r) true? Call this number N.
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.
A 10 kHz clock drives a mod-10 counter. Find the frequency fout of its most significant output, which completes one cycle per counting cycle.
Find the inverse of 8 modulo 31: the number x with 0 < x < 31 and 8·x ≡ 1 (mod 31).
An RSA key uses the primes p = 5 and q = 47 and the public exponent e = 3. Find the private exponent d (0 < d < φ(n)).
What does it mean that a problem has optimal substructure?
What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?
Why is traversing an array often much faster in practice than traversing a linked list of the same length?
Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.
Why can a table have only one clustered index?
Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?
What is the IPv6 loopback address?
In a TCP connection, what does the receive window (rwnd) control?
A department needs 1011 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
The network 192.168.8.0/24 is split with VLSM for these LANs: A: 29 hosts, B: 118 hosts, C: 21 hosts, D: 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?
Answer: U = 250
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.
Answer: R = 77
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.
XOR gives 1 where the two bits differ.
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).
Answer: N = 9
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.
Answer: AB = 00, CD = 01
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.
A 1 with a neighbouring 1 can join a group of two or more.
Answer: fout = 1 kHz
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.
Answer: 8−1 mod 31 = 4
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).
Each remainder r is written as r ≡ t·8 (mod 31).
Answer: d = 123
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.
Answer: A · An optimal solution can be built from optimal solutions of its subproblems
Optimal substructure is required for both greedy algorithms and dynamic programming to produce correct answers.
Answer: D · O(n) time and O(1) space
Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.
Answer: A · Array elements are contiguous, so they make good use of the CPU cache
Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.
Answer: C · Those bits may also belong to other elements, which would cause false negatives
Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.
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.
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.
Answer: D · The rows can be stored physically in only one order
A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.
Answer: A · 1, 1, 3
RANK gives tied rows the same rank and skips the following numbers (1, 1, 3); DENSE_RANK would give 1, 1, 2.
Answer: B · ::1
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
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).
Answer: x = 192
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.
LAN D: 192.168.8.192/29.
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