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 −16 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 1011 1101 << 2. What is the result's value R in decimal?
Which formula is logically equivalent to (c ∧ b) → a? Click it.
How many of the 16 rows of the truth table make (q ∨ (r ∨ ¬p)) ∧ (s ∨ ¬r) true? Call this number N.
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?
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?
Find the smallest positive integer x with x ≡ 2 (mod 6) and x ≡ 5 (mod 11).
An RSA key uses p = 5, q = 29 and e = 13. Decrypt the ciphertext c = 91: find the message m.
What is the time complexity of the Bellman–Ford algorithm on a graph with V vertices and E edges?
What would follow if someone found a polynomial-time algorithm for one NP-complete problem?
What is the cost of inserting an element in the middle of an array of n elements?
For each element of an array you want the next greater element to its right, in O(n) total time. Which structure helps?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.
In relational algebra, the selection operator (σ) corresponds to which SQL clause?
When is decomposing R into R1 and R2 guaranteed to be lossless?
NAT (network address translation) is mainly used to:
The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?
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?
A host has the address 10.20.69.152/26. What is the last number (fourth octet) x of its subnet's broadcast address?
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.
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.
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.
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 columns after the bar are the options in the order shown.
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.
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.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Another grouping of the same size exists; it has the same number of terms and literals.
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.
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).
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.
Answer: A · O(V × E)
Each of the V − 1 rounds relaxes all E edges, giving O(VE); in exchange it handles negative edge weights.
Answer: A · Every problem in NP could be solved in polynomial time, so P = NP
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: D · O(n), because later elements must shift
All elements after the insertion point move one position to the right, which takes time proportional to n.
Answer: C · A stack that keeps elements waiting for their 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: 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: 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.
Answer: D · WHERE
σcondition(R) returns the tuples of R that satisfy the condition, like a WHERE clause.
Answer: B · When the common attributes of R1 and R2 form a superkey of R1 or of R2
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: A · Let many devices with private addresses share one public IP address
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer: B · Distance-vector; split horizon and poison reverse
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: 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).
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).
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