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.
The 8-bit pattern 1011 0110 stores a signed integer in two's complement. What is its value N in decimal?
Set the bits to the 8-bit result of 0101 1010 >> 1. What is the result's value R in decimal?
Which formula is logically equivalent to (q ∧ r) → p? Click it.
The premises a ∨ c; ¬a are given. Which conclusion follows from them? Click it.
A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = 1, I₂ = 1, I₃ = C. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.
Find the minimal SOP form of F(A, B, C, D) = Σm(0, 2, 3, 6, 11, 12) + d(9, 10), where d marks the don't-care conditions. How many product terms N does it have?
Find Euler’s totient φ(169), the number of integers from 1 to 169 that are coprime to 169.
Encrypt the message m = 167 with the RSA public key (n, e) = (247, 5). Find the ciphertext c.
What does the recurrence T(n) = T(n − 1) + O(n) solve to?
Which statement about little-o notation is correct?
In a hash table using open addressing, what is the maximum possible load factor?
A Fenwick tree (binary indexed tree) over an array of n numbers supports which operations in O(log n)?
How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.
What does EXCEPT (called MINUS in Oracle) return?
Scores are 90, 90 and 80. Which values does RANK() OVER (ORDER BY score DESC) give?
At which OSI layer does a switch normally operate?
How does fragmentation differ in IPv6 compared with IPv4?
The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A host has the address 10.20.56.102/25. What is the last number (fourth octet) x of its subnet's broadcast address?
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.
The other bits keep their normal positive values.
Answer: R = 45
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.
Shifting right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
Answer: 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 columns after the bar are the options in the order shown.
Answer: c
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.
Answer: N = 7
Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.
Answer: N = 4
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.
The largest groups that can be formed from 1s and Xs.
Answer: φ(169) = 156
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).
Answer: c = 59
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: B · O(n²)
The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.
Answer: A · n = o(n²), but n is not o(n)
f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.
Answer: D · 1
With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.
Answer: C · Prefix-sum queries and single-element updates
A Fenwick tree stores partial sums over ranges determined by the low bits of each index, so both prefix sums and point updates take O(log n).
Answer: H = 3
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: B
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 · Rows of the first query that do not appear in the second
A EXCEPT B gives the rows in A that are not in B, so swapping the queries gives a different result.
Answer: B · 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: D · Data link (layer 2)
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer: C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer: H = 6
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 = 127
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).
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