Prompt:
# Prompt

## Database Storage

### Problem Context and Objective

Consider a B+ tree where each node can hold a maximum of 4 keys and 5 pointers. The tree is initially empty. The following keys are inserted into the tree one at a time, in the order shown:

era, ban, bat, kin, day, log, rye, max, won, ace, ado, bug, cop, gas, let, fax

Use lexicographic (dictionary) ordering to compare strings.

**B+ tree insertion rules:**
- Keys are always inserted into the appropriate leaf node.
- When a leaf node overflows (receives a 5th key), sort all 5 keys, place the first 2 in the left node, place the last 3 in a new right node, and copy the first key of the right node up into the parent. If no parent exists, create a new root containing that single key.
- When an internal node overflows (receives a 5th key), sort all 5 keys, place the first 2 in the left node, place the last 2 in a new right node, and push the 3rd (middle) key up into the parent. If no parent exists, create a new root.
- In a leaf split, the promoted key is copied (it remains in the leaf). In an internal split, the promoted key is pushed up (it is removed from the child).

Compute the tree utilization after every 4 insertions. Utilization is defined as:

$$\text{utilization} = \frac{\text{total number of keys stored across all nodes}}{4 \times \text{total number of nodes}}$$

Store the utilization values as decimal fractions (between 0 and 1) in variables $a$, $b$, $c$, $d$ corresponding to after 4, 8, 12, and 16 insertions respectively. Round each to exactly 3 decimal places.

---

Now consider a separate problem. Assume we have a relation $r(X, Y)$ where the domain of $X$ is $\{A, B, C, D\}$ and the domain of $Y$ is integers. The hash function on $X$ is defined as follows (4-bit binary values):

| Key | Hash |
|-----|------|
| A   | 0100 |
| B   | 0101 |
| C   | 0011 |
| D   | 1101 |

The hash structure uses **prefix-based matching starting from the most significant bit (bit 1 is leftmost)**.

Construct an extendible hash structure by inserting the following records in the order shown:

(B,8), (D,4), (C,1), (A,7), (D,3), (D,9), (B,5)

Each bucket can hold at most 2 records.

**Extendible hashing rules:**
- Each bucket has a Local Depth (LD). The directory has a Global Depth (GD). Initially GD = 0 and there is one bucket with LD = 0.
- When a bucket overflows: if LD < GD, increment LD and split the bucket by examining the next bit of the hash. If LD = GD, first double the directory (increment GD), then increment LD and split.
- After a split, if all overflowing records still map to the same new bucket because their hash values share the same prefix for all bits up to the new LD, repeat the split process (doubling the directory again if needed).
- If all records in an overflowing bucket have identical full 4-bit hash values (so no further bit can separate them), stop splitting and attach an overflow page to that bucket instead.

In the final hash structure, number the distinct buckets 1 through $N$ in the order they are first encountered when scanning the directory from binary index 0000...0 to 1111...1.

---

**Computation:**

$$e = \text{round}(10 \times a \times b \times c \times d)$$

where $\text{round}$ denotes rounding to the nearest integer.

Let $N$ be the total number of distinct buckets (not counting overflow pages). Compute:
- Bucket index $p = ((e - 1) \bmod N)$. If $p = 0$, set $p = N$.
- Bucket index $q = ((e + 1) \bmod N)$. If $q = 0$, set $q = N$.

Count the records in buckets $p$ and $q$ (only records in the main bucket, not in overflow pages). Compute the fraction of total records (out of 7) stored in these two buckets combined. Round to 3 decimal places.

**Report the following values in your answer: $a$, $b$, $c$, $d$, $e$, and the final fraction.**

Does the solution report the utilization after 4 insertions as a = 1.000 (or equivalent rounding)?

Does the solution report the utilization after 8 insertions as b = 0.625?

Does the solution report the utilization after 12 insertions as c = 0.667?

Does the solution report the utilization after 16 insertions as d = 0.833?

Does the solution correctly compute the value of e as 3?

Does the solution determine that the final extendible hash structure has a global depth of 4?

Does the solution determine that the final extendible hash structure contains exactly 6 distinct buckets (not counting overflow pages)?

Does the solution correctly apply the overflow page rule for key D, recognizing that all D records share the identical 4-bit hash 1101 and therefore cannot be separated by further splitting?

Does the solution correctly determine the bucket indices p = 2 and q = 4 using modulo arithmetic with 1-based adjustment?

Does the solution correctly identify the contents of bucket 2 ([(A,7)]) and bucket 4 (empty), counting only main bucket records?

Does the solution output the final fraction as 0.143?

# Golden Answer

## Part 1: B+ Tree Construction

The tree starts empty. Each node can hold at most 4 keys. The split rule for leaves: when a leaf has 5 keys, left gets keys at positions 1-2, right gets keys at positions 3-5, and the key at position 3 (first key of right node) is copied up to the parent.

**Insertions 1-4 (era, ban, bat, kin):**

Insert era: single leaf [era].
Insert ban: leaf [ban, era] (sorted).
Insert bat: leaf [ban, bat, era].
Insert kin: leaf [ban, bat, era, kin].

One node, 4 keys, 4 spots.
$a = 4/(4 \times 1) = 4/4 = 1.000$.

**Insertions 5-8 (day, log, rye, max):**

Insert day into leaf [ban, bat, era, kin]. Now has 5 keys: [ban, bat, day, era, kin].
Overflow: left = [ban, bat], right = [day, era, kin]. Copy "day" up to new root.
Tree: root=[day], leaves=[ban, bat] and [day, era, kin].

Insert log: goes to right leaf (log > day). Right leaf becomes [day, era, kin, log]. Fits (4 keys).

Insert rye: goes to right leaf. Now 5 keys: [day, era, kin, log, rye].
Overflow: left = [day, era], right = [kin, log, rye]. Copy "kin" up to root.
Root = [day, kin]. Leaves: [ban, bat], [day, era], [kin, log, rye].

Insert max: goes to rightmost leaf (max > kin). Leaf becomes [kin, log, max, rye]. Fits (4 keys).

Node count: 1 root + 3 leaves = 4.
Key count: root has 2, leaves have 2 + 2 + 4 = 8. Total = 10.
Spots: $4 \times 4 = 16$.
$b = 10/16 = 0.625$.

**Insertions 9-12 (won, ace, ado, bug):**

Insert won: goes to rightmost leaf [kin, log, max, rye]. Now 5 keys: [kin, log, max, rye, won].
Overflow: left = [kin, log], right = [max, rye, won]. Copy "max" up.
Root = [day, kin, max]. Leaves: [ban, bat], [day, era], [kin, log], [max, rye, won].

Insert ace: goes to leftmost leaf (ace < day, ace < ban). Leaf becomes [ace, ban, bat]. Fits.

Insert ado: goes to leftmost leaf. Leaf becomes [ace, ado, ban, bat]. Fits (4 keys, full).

Insert bug: goes to leftmost leaf (bug > ban, bug < day). Now 5 keys: [ace, ado, ban, bat, bug].
Overflow: left = [ace, ado], right = [ban, bat, bug]. Copy "ban" up.
Root = [ban, day, kin, max]. Leaves: [ace, ado], [ban, bat, bug], [day, era], [kin, log], [max, rye, won].

Node count: 1 root + 5 leaves = 6.
Key count: root has 4, leaves have 2 + 3 + 2 + 2 + 3 = 12. Total = 16.
Spots: $4 \times 6 = 24$.
$c = 16/24 = 0.66\overline{6} \approx 0.667$.

**Insertions 13-16 (cop, gas, let, fax):**

Insert cop: goes to leaf [ban, bat, bug] (cop > ban, cop < day). Leaf becomes [ban, bat, bug, cop]. Fits (full).

Insert gas: goes to leaf [day, era] (gas > day, gas < kin). Leaf becomes [day, era, gas]. Fits.

Insert let: goes to leaf [kin, log] (let > kin, let < log lexicographically). Leaf becomes [kin, let, log]. Fits.

Insert fax: goes to leaf [day, era, gas] (fax > day, fax < gas). Leaf becomes [day, era, fax, gas]. Fits (full).

Node count: 1 root + 5 leaves = 6. (No new splits occurred.)
Key count: root has 4, leaves have 2 + 4 + 4 + 3 + 3 = 16. Total = 20.
Spots: $4 \times 6 = 24$.
$d = 20/24 = 0.83\overline{3} \approx 0.833$.

**Summary:** $a = 1.000$, $b = 0.625$, $c = 0.667$, $d = 0.833$.

---

## Part 2: Extendible Hashing

Hash values (4-bit, MSB first): A = 0100, B = 0101, C = 0011, D = 1101.

**Step 1 - Initialization:** GD = 0. One bucket (LD = 0), empty.

**Step 2 - Insert (B,8):** Hash(B) = 0101. Bucket has room. Bucket: [(B,8)]. LD = 0.

**Step 3 - Insert (D,4):** Hash(D) = 1101. Same bucket (GD = 0, only one directory entry). Bucket: [(B,8), (D,4)]. Now full (capacity 2). LD = 0.

**Step 4 - Insert (C,1):** Hash(C) = 0011. Target bucket is full. Must split.
- LD = GD = 0, so double directory first. GD becomes 1. Directory now has 2 entries (0 and 1).
- Increment LD to 1. Split by bit 1:
  - B(0101) -> prefix 0 -> directory 0
  - C(0011) -> prefix 0 -> directory 0
  - D(1101) -> prefix 1 -> directory 1
- Directory 0 -> Bucket A (LD=1): [(B,8), (C,1)] - full
- Directory 1 -> Bucket B (LD=1): [(D,4)]

**Step 5 - Insert (A,7):** Hash(A) = 0100. Prefix "0" -> directory 0 -> Bucket A: [(B,8), (C,1)]. Full. Must split.
- LD = GD = 1, so double directory. GD becomes 2. Directory now has 4 entries (00, 01, 10, 11).
- Increment LD to 2. Split Bucket A by bit 2:
  - C(0011) -> prefix 00 -> directory 00
  - B(0101) -> prefix 01 -> directory 01
  - A(0100) -> prefix 01 -> directory 01
- Directory 00 -> Bucket C (LD=2): [(C,1)]
- Directory 01 -> Bucket D (LD=2): [(B,8), (A,7)] - full
- Directory 10 -> Bucket B (LD=1): [(D,4)] (unchanged, LD still 1)
- Directory 11 -> Bucket B (LD=1): [(D,4)] (same bucket as 10)

**Step 6 - Insert (D,3):** Hash(D) = 1101. Prefix "11" -> directory 11 -> Bucket B: [(D,4)]. Has room.
Bucket B: [(D,4), (D,3)]. Now full. LD = 1.

**Step 7 - Insert (D,9):** Hash(D) = 1101. Prefix "11" -> directory 11 -> Bucket B: [(D,4), (D,3)]. Full. Must split.
- LD(1) < GD(2), so no directory doubling needed. Increment LD to 2. Split by bit 2:
  - D(1101) -> prefix 11. All three D records (D,4), (D,3), (D,9) have hash 1101 -> prefix 11.
  - Directory 10 -> new Bucket E (LD=2): empty.
  - Directory 11 -> Bucket B (LD=2): All three D records still map here.
- Bucket B still has 3 records but capacity 2. Attempt further split:
  - LD(2) = GD(2), so must double directory to split further. But all three records have identical 4-bit hash 1101. Check: can bit 3 separate them? D's bit 3 is 0 for all. Can bit 4 separate them? D's bit 4 is 1 for all. Since all records have the identical full 4-bit hash value 1101, no further splitting can separate them.
  - Create overflow page.
- Directory 11 -> Bucket B (LD=2): [(D,4), (D,3)] + overflow page: [(D,9)]

State after step 7:
- 00 -> Bucket C (LD=2): [(C,1)]
- 01 -> Bucket D (LD=2): [(B,8), (A,7)]
- 10 -> Bucket E (LD=2): empty
- 11 -> Bucket B (LD=2): [(D,4), (D,3)] + overflow [(D,9)]

**Step 8 - Insert (B,5):** Hash(B) = 0101. Prefix "01" -> directory 01 -> Bucket D: [(B,8), (A,7)]. Full. Must split.
- LD(2) = GD(2), so double directory. GD becomes 3. Directory now has 8 entries (000-111).
- Increment LD to 3. Split Bucket D by bit 3:
  - A(0100) -> prefix 010
  - B(0101) -> prefix 010
  - B(0101) -> prefix 010 (the new record)
  - All three map to prefix 010. Prefix 011 is empty.
- Directory 010 -> Bucket D (LD=3): Still has A(0100), B(0101), B(0101) = 3 records. Still overflows.
- LD(3) = GD(3), so double directory again. GD becomes 4. Directory now has 16 entries (0000-1111).
- Increment LD to 4. Split Bucket D by bit 4:
  - A(0100) -> prefix 0100
  - B(0101) -> prefix 0101
  - B(0101) -> prefix 0101
  - Now separated: A goes to 0100, both B's go to 0101.
- Directory 0100 -> Bucket F (LD=4): [(A,7)]
- Directory 0101 -> Bucket D (LD=4): [(B,8), (B,5)]

**Final structure:** GD = 4, 16 directory entries.

Directory-to-bucket mapping (scanning 0000 to 1111, numbering new buckets in encounter order):

| Directory entries | Bucket | LD | Contents |
|---|---|---|---|
| 0000, 0001, 0010, 0011 | Bucket 1 | 2 | [(C,1)] |
| 0100 | Bucket 2 | 4 | [(A,7)] |
| 0101 | Bucket 3 | 4 | [(B,8), (B,5)] |
| 0110, 0111 | Bucket 4 | 3 | empty |
| 1000, 1001, 1010, 1011 | Bucket 5 | 2 | empty |
| 1100, 1101, 1110, 1111 | Bucket 6 | 2 | [(D,4), (D,3)] + overflow [(D,9)] |

Verification of directory entry counts per bucket:
- Bucket 1 (LD=2): $2^{4-2} = 4$ entries. Correct (0000-0011).
- Bucket 2 (LD=4): $2^{4-4} = 1$ entry. Correct (0100).
- Bucket 3 (LD=4): $2^{4-4} = 1$ entry. Correct (0101).
- Bucket 4 (LD=3): $2^{4-3} = 2$ entries. Correct (0110, 0111).
- Bucket 5 (LD=2): $2^{4-2} = 4$ entries. Correct (1000-1011).
- Bucket 6 (LD=2): $2^{4-2} = 4$ entries. Correct (1100-1111).
Total entries: 4+1+1+2+4+4 = 16. Correct.

$N = 6$ buckets. Total records = 7 (6 in main buckets + 1 in overflow).

---

## Final Computation

$e = \text{round}(10 \times 1.000 \times 0.625 \times 0.667 \times 0.833)$

$= \text{round}(10 \times 0.34722) = \text{round}(3.4722) = 3$

$p = ((e-1) \bmod N) = ((3-1) \bmod 6) = (2 \bmod 6) = 2$. Since $p \neq 0$, bucket $p = 2$.
$q = ((e+1) \bmod N) = ((3+1) \bmod 6) = (4 \bmod 6) = 4$. Since $q \neq 0$, bucket $q = 4$.

Bucket 2: [(A,7)] = 1 record.
Bucket 4: empty = 0 records.

Fraction $= (1 + 0)/7 = 1/7 = 0.14285... \approx 0.143$.

**Answer:** $a = 1.000$, $b = 0.625$, $c = 0.667$, $d = 0.833$, $e = 3$, final fraction = **0.143**.


Prompt:
You are given the following tree-path query problem.

A tree has $n$ nodes numbered from $1$ to $n$. Each node $i$ has an integer value $a_i$.

Each query gives two nodes $x$ and $y$. Let the simple path from $x$ to $y$ be

$$x = p_0,\; p_1,\; p_2,\; \ldots,\; p_r = y$$

Define

$$S = \sum_{k=0}^{r} \bigl(a_{p_k} \oplus k\bigr)$$

where $\oplus$ is bitwise XOR and $k$ is the 0-based position along the path.

The original input format is:

- First line: integer $t$, the number of test cases.
- For each test case:
  - First line: integer $n$ with $1 \le n \le 5 \times 10^5$
  - Next $n-1$ lines: edges $u, v$
  - Next line: $n$ integers $a_1, a_2, \ldots, a_n$ with $0 \le a_i \le 10^9$
  - Next line: integer $q$ with $1 \le q \le 10^5$
  - Next $q$ lines: queries $x, y$

Constraints:

- The edges form a valid tree.
- The sum of $n$ over all test cases is at most $5 \times 10^5$.
- The sum of $q$ over all test cases is at most $10^5$.

Use this sample instance:

**Sample Input**
```text
1
4
1 2
2 3
3 4
1 2 3 4
3
1 4
4 1
2 4
```

**Sample Output**
```text
12
8
10
```

Write a technical explanation of the solution. Your answer must contain all of the following:

1. A bitwise decomposition of $S$ using symbols $b_k(j)$ for the $j$-th bit of $a_{p_k}$ and $P_j(k)$ for the $j$-th bit of the index $k$, including the per-bit contribution formula.
2. Definitions of $S_0$, $S_1$, and $C_1$, and the identity
   $$\sum_{k=0}^{r} (b_k(j) \oplus P_j(k)) = S_0 + (C_1 - S_1).$$
3. A closed form for $C_1$ in terms of $r_{\text{len}} = r+1$, `full_blocks = floor(r_len / 2^{j+1})`, and `rem = r_len mod 2^{j+1}`.
4. A direct computation for the sample queries showing exactly that $(1,4) \mapsto 12$, $(4,1) \mapsto 8$, and $(2,4) \mapsto 10$.
5. A bitwise verification of the query $(1,4)$ showing per-bit contributions $4$, $4$, and $4$ for bits $j=0,1,2$.
6. A naive per-query algorithm that reconstructs the path and scans positions and bits, with time complexity $O(r \cdot B + \log n)$ per query.
7. An optimized method based on binary lifting with per-bit phase counts, with preprocessing complexity $O(n \cdot \log n \cdot B)$ and query complexity $O(B \cdot \log n)$.

Does the answer define b_k(j) as the j-th bit of a_{p_k}, define P_j(k) as the j-th bit of the path index k, and state that the contribution of bit j is 2^j times the sum of b_k(j) XOR P_j(k) over the path positions?

Does the answer define S0, S1, and C1 for a fixed bit j and state the identity sum_k (b_k(j) XOR P_j(k)) = S0 + (C1 - S1)?

Does the answer give the closed form C1 = full_blocks * 2^j + max(0, rem - 2^j), with full_blocks = floor(r_len / 2^{j+1}), rem = r_len mod 2^{j+1}, and r_len = r + 1?

Does the answer directly compute the sample query results (1,4) = 12, (4,1) = 8, and (2,4) = 10?

Does the answer verify the sample query (1,4) by bitwise decomposition with per-bit contributions 4, 4, and 4 for bits j = 0, 1, and 2?

Does the answer describe a naive per-query method that reconstructs the path and scans positions and bits, and state its per-query time complexity as O(r * B + log n)?

Does the answer describe an optimized binary-lifting approach with per-bit phase counts and state preprocessing complexity O(n * log n * B) and query complexity O(B * log n)?

## Bitwise Decomposition

Let the path for a query be $p_0, p_1, \ldots, p_r$. For each bit position $j \ge 0$, define

$$b_k(j) = ((a_{p_k} \gg j) \mathbin{\&} 1)$$

and

$$P_j(k) = ((k \gg j) \mathbin{\&} 1).$$

Then the $j$-th bit of $a_{p_k} \oplus k$ is

$$b_k(j) \oplus P_j(k).$$

So the contribution of bit $j$ to

$$S = \sum_{k=0}^{r} (a_{p_k} \oplus k)$$

is

$$2^j \cdot \sum_{k=0}^{r} (b_k(j) \oplus P_j(k)).$$

Therefore,

$$S = \sum_{j \ge 0} 2^j \cdot \sum_{k=0}^{r} (b_k(j) \oplus P_j(k)).$$

## Simplifying the Per-Bit Count

Fix a bit $j$. Let

- $S_0$ be the sum of $b_k(j)$ over positions with $P_j(k)=0$.
- $S_1$ be the sum of $b_k(j)$ over positions with $P_j(k)=1$.
- $C_1$ be the number of positions with $P_j(k)=1$.

Then

$$\sum_{k=0}^{r} (b_k(j) \oplus P_j(k)) = S_0 + (C_1 - S_1).$$

Reason:

- If $P_j(k)=0$, then $b_k(j) \oplus P_j(k)=b_k(j)$, so those positions contribute $S_0$.
- If $P_j(k)=1$, then $b_k(j) \oplus P_j(k)=1-b_k(j)$, so those positions contribute $C_1-S_1$.

Adding the two parts gives the identity.

## Closed Form for $C_1$

Let

$$r_{\text{len}} = r+1.$$

For a fixed bit $j$, the sequence $P_j(k)$ over consecutive indices has period $2^{j+1}$ and looks like

- $2^j$ zeros,
- then $2^j$ ones.

Define

$$\text{full\_blocks} = \left\lfloor \frac{r_{\text{len}}}{2^{j+1}} \right\rfloor,\qquad \text{rem} = r_{\text{len}} \bmod 2^{j+1}.$$

Each full block contributes exactly $2^j$ ones, and the remainder contributes $\max(0, \text{rem} - 2^j)$ ones. Hence

$$C_1 = \text{full\_blocks} \cdot 2^j + \max(0, \text{rem} - 2^j).$$

## Direct Computation on the Sample

The sample tree is the chain $1-2-3-4$ with values $a=[1,2,3,4]$.

For query $(1,4)$, the path is $[1,2,3,4]$:

$$S = (1 \oplus 0) + (2 \oplus 1) + (3 \oplus 2) + (4 \oplus 3) = 1 + 3 + 1 + 7 = 12.$$

For query $(4,1)$, the path is $[4,3,2,1]$:

$$S = (4 \oplus 0) + (3 \oplus 1) + (2 \oplus 2) + (1 \oplus 3) = 4 + 2 + 0 + 2 = 8.$$

For query $(2,4)$, the path is $[2,3,4]$:

$$S = (2 \oplus 0) + (3 \oplus 1) + (4 \oplus 2) = 2 + 2 + 6 = 10.$$

These are exactly the sample outputs: $12$, $8$, and $10$.

## Bitwise Verification for Query $(1,4)$

Here $r=3$, so the path has four positions $k=0,1,2,3$.

For $j=0$:

- $P_0(k) = [0,1,0,1]$
- $b_k(0)$ from values $[1,2,3,4]$ is $[1,0,1,0]$
- $S_0 = 1+1 = 2$
- $S_1 = 0+0 = 0$
- $C_1 = 2$

So the contribution is

$$2^0 \cdot (S_0 + C_1 - S_1) = 1 \cdot (2+2-0) = 4.$$

For $j=1$:

- $P_1(k) = [0,0,1,1]$
- $b_k(1) = [0,1,1,0]$
- $S_0 = 0+1 = 1$
- $S_1 = 1+0 = 1$
- $C_1 = 2$

So the contribution is

$$2^1 \cdot (1+2-1) = 4.$$

For $j=2$:

- $P_2(k) = [0,0,0,0]$
- $b_k(2) = [0,0,0,1]$
- $S_0 = 1$
- $S_1 = 0$
- $C_1 = 0$

So the contribution is

$$2^2 \cdot (1+0-0) = 4.$$

Thus the per-bit contributions for $j=0,1,2$ are $4$, $4$, and $4$, summing to $12$.

## Naive Per-Query Method

A direct method for one query is:

1. Reconstruct the path from $x$ to $y$ using an LCA-based path decomposition.
2. Enumerate the path positions $k=0$ to $r$.
3. For each relevant bit $j$, scan all positions, compute or accumulate the needed values, and add the contribution for that bit.

If $B$ is the number of relevant bit positions, this costs

$$O(r \cdot B + \log n)$$

per query: $O(\log n)$ to support path reconstruction with LCA machinery and $O(r \cdot B)$ to scan the path and bits. This is too slow in the worst case because $r$ can be $\Theta(n)$.

## Optimized Method

Use binary lifting with per-bit phase counts.

For each node $u$, jump size $2^k$, and bit $j$, store two counts over that upward segment:

- how many nodes with bit $j$ equal to $1$ lie at offsets whose $j$-th offset bit is $0$,
- how many lie at offsets whose $j$-th offset bit is $1$.

This lets two adjacent $2^k$ segments be merged in $O(1)$ per bit because the second half is shifted by exactly $2^k$, so for each bit $j$ its phase either stays the same or swaps depending on whether that shift flips the $j$-th offset bit.

During a query:

1. Split the path using the LCA.
2. Decompose each side into $O(\log n)$ lifted segments.
3. Maintain the global offset along the full path so each stored segment count is mapped to the correct phase.
4. Aggregate per-bit counts and then apply the formula above.

The complexities are:

- preprocessing: $O(n \cdot \log n \cdot B)$
- per query: $O(B \cdot \log n)$

This meets the given constraints.


Prompt:
A Count-Min Sketch (CMS) is used to estimate frequencies over a stream of integer keys.

-----------------------------------
SKETCH CONFIGURATION
-----------------------------------

- Depth: d = 4 rows
- Width: w = 11 columns
- Counters are indexed from 0 to 10
- All counters are initialized to 0
- No pre-warming is applied

-----------------------------------
HASH FUNCTIONS
-----------------------------------

For any non-negative integer x:

g₁(x) = (3x + 2) mod 11
g₂(x) = (5x + 7) mod 11
g₃(x) = (7x + 1) mod 11
g₄(x) = (9x + 4) mod 11

-----------------------------------
UPDATE RULE (STANDARD CMS)
-----------------------------------

For each update (x, Δ):

- If Δ ≤ 0, ignore the update
- Otherwise, for each row i = 1..4:
  - Increment counter[i][gᵢ(x)] by Δ

-----------------------------------
QUERY RULE
-----------------------------------

To query key x:

- Compute the 4 counters addressed by g₁(x), g₂(x), g₃(x), g₄(x)
- Return the minimum of these 4 values

-----------------------------------
UPDATES (IN ORDER)
-----------------------------------

(4, 6)
(9, 3)
(20, 5)
(1, 4)
(31, 2)
(9, 7)
(2, -3)
(13, 1)
(20, 4)
(0, 2)
(9, -2)
(22, 8)
(31, 5)
(20, 1)

-----------------------------------
TASK
-----------------------------------

After processing all updates, determine the final CMS state relevant to querying key x = 20.

Use row numbers 1 through 4.

-----------------------------------
OUTPUT FORMAT

Output exactly the following 7 lines, with no extra text:

ignored_updates: <integer>
hashes_20: <g1>,<g2>,<g3>,<g4>
counter_1_7: <integer>
counter_2_8: <integer>
counter_3_9: <integer>
counter_4_8: <integer>
query_20: <integer>

For `hashes_20`, list the four hash results for x = 20 in the order g1, g2, g3, g4, separated by commas and with no spaces.

The response states `ignored_updates: 2`.

The response states `hashes_20: 7,8,9,8`.

Reasoning

Only updates with positive delta are applied. Therefore `(2, -3)` and `(9, -2)` are ignored, so the number of ignored updates is 2.

For the query key `x = 20`, the four hash locations are:

- `g1(20) = (3*20 + 2) mod 11 = 62 mod 11 = 7`
- `g2(20) = (5*20 + 7) mod 11 = 107 mod 11 = 8`
- `g3(20) = (7*20 + 1) mod 11 = 141 mod 11 = 9`
- `g4(20) = (9*20 + 4) mod 11 = 184 mod 11 = 8`

So `hashes_20 = 7,8,9,8`.

To determine the final values of those four counters, sum the positive updates from every key that lands in the same bucket in each row.

For key `20`, the positive updates are `(20,5)`, `(20,4)`, and `(20,1)`, for a total contribution of `10`.

Key `9` hashes to the same four positions:

- `g1(9) = (27 + 2) mod 11 = 29 mod 11 = 7`
- `g2(9) = (45 + 7) mod 11 = 52 mod 11 = 8`
- `g3(9) = (63 + 1) mod 11 = 64 mod 11 = 9`
- `g4(9) = (81 + 4) mod 11 = 85 mod 11 = 8`

Its positive updates are `(9,3)` and `(9,7)`, for a total of `10`.

Key `31` also hashes to the same four positions:

- `g1(31) = (93 + 2) mod 11 = 95 mod 11 = 7`
- `g2(31) = (155 + 7) mod 11 = 162 mod 11 = 8`
- `g3(31) = (217 + 1) mod 11 = 218 mod 11 = 9`
- `g4(31) = (279 + 4) mod 11 = 283 mod 11 = 8`

Its positive updates are `(31,2)` and `(31,5)`, for a total of `7`.

The remaining keys do not change the queried counters for `x = 20`, because at least one of their row positions differs from the needed counter in each row. Therefore each queried counter receives exactly:

- `10` from key `20`
- `10` from key `9`
- `7` from key `31`

That gives:

- `counter_1_7 = 27`
- `counter_2_8 = 27`
- `counter_3_9 = 27`
- `counter_4_8 = 27`

The CMS query returns the minimum of these four addressed counters, so:

- `query_20 = min(27, 27, 27, 27) = 27`

Required final output

ignored_updates: 2
hashes_20: 7,8,9,8
counter_1_7: 27
counter_2_8: 27
counter_3_9: 27
counter_4_8: 27
query_20: 27


Prompt:
You are given a permutation `p` of size `n` (containing each integer from `1` to `n` exactly once).

For the current permutation, define for each current position `i` (1-indexed):

`c_i = |i - p_i|`

You may repeatedly remove elements from the current permutation. A removal is valid if the chosen element satisfies `c_i <= s`.

After each removal:
- The remaining elements shift left to fill the gap.
- The remaining values are relabeled to again form a permutation of `1..k`:
  subtract `1` from every remaining value that was greater than the removed value.
- Then all `c_i` values are recomputed on the new permutation.

To remove all ambiguity, if more than one position is currently removable, always remove the leftmost removable position.

Goal: find the minimum integer `s` such that this deterministic process removes the entire permutation.

Input:

`n = 5`
`p = [3,1,5,2,4]`

Required output:

Output exactly these 5 lines and nothing else:

`s=<minimum s>`
`positions=<comma-separated removed positions>`
`values=<comma-separated removed values>`
`states=[3,1,5,2,4]->...->[]`
`why_s_minus_1_fails=<one sentence>`

Use the deterministic leftmost-removable rule for every step.

States `s=1` as the minimum integer that allows the deterministic process to remove the entire permutation.

Explains that `s-1 = 0` fails because the initial permutation has no position with `i = p_i`, so no removal is valid at the start when `c_i` must equal 0.

Uses the exact 5-line output structure requested in the prompt with no extra lines or omitted lines.

s=1
positions=2,1,2,1,1
values=1,2,1,2,1
states=[3,1,5,2,4]->[2,4,1,3]->[3,1,2]->[2,1]->[1]->[]
why_s_minus_1_fails=s-1=0 fails because the initial permutation [3,1,5,2,4] has no position with i=p_i, so no element has c_i=0 and the process cannot start.

Reasoning

The key detail is that `c_i` must be recomputed after every removal because both positions and values change. The original version treated the initial `c_i` values as if they stayed fixed, which is not correct for this process.

Check `s = 0` first, because the minimum `s` must be the smallest nonnegative integer that works. With `s = 0`, a move is valid only when `|i - p_i| = 0`, which means `i = p_i`. In the initial permutation `[3,1,5,2,4]`, the position-value pairs are `(1,3)`, `(2,1)`, `(3,5)`, `(4,2)`, `(5,4)`. None of them satisfy `i = p_i`, so there is no valid first move. Therefore `s = 0` is impossible.

Now test `s = 1` using the required deterministic rule: always remove the leftmost currently valid position.

Start with `[3,1,5,2,4]`.
- Current `c` values are `[2,1,2,2,1]`.
- Valid positions are 2 and 5, so remove the leftmost one: position 2, value 1.
- After shifting and relabeling values greater than 1 down by 1, the new permutation is `[2,4,1,3]`.

For `[2,4,1,3]`:
- Current `c` values are `[1,2,2,1]`.
- Valid positions are 1 and 4, so remove the leftmost one: position 1, value 2.
- After shifting and relabeling values greater than 2 down by 1, the new permutation is `[3,1,2]`.

For `[3,1,2]`:
- Current `c` values are `[2,1,1]`.
- Valid positions are 2 and 3, so remove the leftmost one: position 2, value 1.
- After shifting and relabeling values greater than 1 down by 1, the new permutation is `[2,1]`.

For `[2,1]`:
- Current `c` values are `[1,1]`.
- Valid positions are 1 and 2, so remove the leftmost one: position 1, value 2.
- After shifting and relabeling, the new permutation is `[1]`.

For `[1]`:
- Current `c` value is `[0]`.
- Remove position 1, value 1, leaving `[]`.

So `s = 1` succeeds, and `s = 0` fails. Therefore the minimum valid `s` is `1`.


Programming Languages, Compilers & Formal Methods

Prompt:
In the secretary problem, an adversary chooses $n$ values $x_1 \ge x_2 \ge \dots \ge x_n$ in $[0,1]$ that arrive online at independent uniform random times in $[0,1]$. An online algorithm must irrevocably select exactly one value upon seeing it (selecting $0$ if none is chosen). The regret of an algorithm on instance $\vec{x}$ is $x_1 - \mathbb{E}[\text{selected value}]$, and the worst-case regret is its supremum over all $\vec{x}$.

A pricing-curve algorithm is defined by a measurable threshold function $f:[0,1] \to [0,1]$ and selects the first arriving value $v$ at time $t$ with $v \ge f(t)$, or $0$ if none arrives.

Prove rigorously that the pricing-curve algorithm with $f(t)=1-t$ has worst-case regret at most $1/4$, and that every pricing-curve algorithm (any $f$) has worst-case regret at least $1/4$. The proofs must be fully self-contained, use only elementary probability, Lebesgue measure, and calculus, and hold for all $n \ge 1$.

Does the proof explicitly consider the harder variant in which $t_1 \sim U[0,1]$ is random but the adversary chooses all arrival times $t_2, \dots, t_n$ in advance?

Does the proof remove every $x_i < f(t_i)$ for $i \ge 2$ and state that this does not change the algorithm’s behavior or regret?

Does the proof define $j = \arg\min_{i \ge 2}$?

Does the proof reduce the instance to keeping only $x_1$ and $x_j$ while preserving behavior and regret?

Does the proof state that the algorithm selects $x_1$ if and only if $t_1 \in [1-x_1, t_j]$?

Does the proof write the exact regret expression $(x_1 - x_j)(1 - (t_j - (1 - x_1)))$?

Does the proof invoke the inequality $x_j \ge 1 - t_j$ to bound the regret expression?

Does the proof claim that the quadratic $(a - b)(1 - a + b)$ for $a \ge b \in [0,1]$ is at most $1/4$?

Does the lower-bound proof define $\alpha = \mu(\{t \in [0,1] : f(t) < 0.5\})$?

Does the lower-bound proof separately handle the case $\alpha > 1/2$ by arguing that, as $n \to \infty$, the algorithm accepts some value by time $0.5$ with probability approaching $1$ when $t_1 \ge 0.5$?

Does the lower-bound proof correctly handle the $\alpha \le 1/2$ case by setting $\alpha = \mu(\{t : f(t) < 0.5\})$?

Upper bound ($f(t)=1-t$): Consider the harder variant where $t_1 \sim U[0,1]$ but $t_2, \dots, t_n$ are chosen adversarially in advance (any bound here implies the original). Remove every $x_i < f(t_i)$ for $i \ge 2$ (unchanged behavior). Let $j = \arg\min_{i \ge 2} t_i$ (add $x_2=0$ at $t_2=1$ if needed), then keep only $x_1, x_j$ (unchanged). The algorithm picks $x_1$ iff $t_1 \in [1-x_1, t_j]$, else $x_j$. Regret equals $(x_1 - x_j)(1 - (t_j - (1 - x_1)))$. Since $x_j \ge 1 - t_j$, this is at most $(x_1 - x_j)(1 - x_1 + x_j) \le 1/4$ (maximum of quadratic $(a-b)(1-a+b)$ on $a \ge b \in [0,1]$ is $1/4$).

Lower bound (any $f$): Let $\alpha = \mu(\{t : f(t) < 0.5\})$.
 * If $\alpha \le 1/2$, take $n=1$, $x_1=0.5$: expected value $\le \alpha \cdot 0.5 \le 0.25$, regret $\ge 0.25$.
 * If $\alpha > 1/2$, take $x_1=1$, $x_i=0.5$ for $i \ge 2$. With probability $1/2$, $t_1 \ge 0.5$. Also $\mu(\{t \in [0, 0.5] : f(t) \ge 0.5\}) \ge \alpha - 1/2 > 0$. As $n \to \infty$, probability of accepting some value by time $0.5$ approaches $1$. Thus regret approaches at least $0.5(1 - 0.5) = 0.25$.

In both cases worst-case regret $\ge 1/4$.


Prompt:
## Problem

Shivam is playing a game with his friends Varad (who controls the doll) and Tanmay.

There are `n` boxes in a line, numbered `1..n`. Varad hides a doll in one box. Shivam will make `m` guesses: on the `i`-th guess Shivam asks whether the doll is currently in box `a[i]`.

Varad can “cheat” by moving the doll **before the first guess** and also **after every guess**:
- she may move the doll to a neighboring box (`i -> i-1` or `i -> i+1`) or keep it in place,
- moves must stay within `1..n`.

So the order is:
- game starts (doll at start box `x`),
- Varad makes one move,
- Shivam guesses `a1`,
- Varad moves,
- Shivam guesses `a2`,
- ...
- Shivam guesses `am`,
- Varad moves,
- game ends (doll at end box `y`).

Shivam **loses** (i.e., Varad wins) if **none** of Shivam’s guesses ever equals the doll’s position at that moment.

You are given `n`, `m`, and the guess sequence `a1..am`.
Count the number of pairs `(x, y)` with `1 ≤ x, y ≤ n` such that Varad can choose moves so that:
- the doll starts at `x`,
- Shivam never guesses correctly,
- the doll ends at `y`.

### Input
- First line: `n m` (`1 ≤ n, m ≤ 1e5`)
- Second line: `m` integers `a1..am` (`1 ≤ ai ≤ n`)

### Output
Print the number of winning scenarios `(x,y)`.

The answer states that for each fixed start box x, the set of reachable end boxes y forms a consecutive segment [L[x], R[x]] (except trivial n=1 handling).

The answer states that the final count is computed as the sum over x of the segment lengths (e.g., sum max(0, R[x]-L[x]+1)).

The answer describes computing L[x] and R[x] as extreme reachable ends using a greedy strategy (always push left for L, always push right for R).

The answer states that the right-bound computation is obtained by a symmetric/mirrored procedure of the left-bound computation.

The answer uses a DSU (union-find) idea to merge starting positions whose greedy simulation leads to the same current box, so their future evolution is identical.

The answer specifies that at guess i, any simulated state that equals the forbidden box a[i] must be shifted to an alternative position to avoid losing immediately (e.g., forced move to a[i]+1 for the left-bound sweep).

The answer includes boundary handling for moves at boxes 1 and n (cannot move to 0 or n+1).

The answer explicitly describes that after processing guess i, the greedy-left sweep applies a move curPos := max(1, curPos−1) (or an equivalent uniform “try move left” update).

The answer states that when two DSU components move into the same current box during the sweep, they are unioned/merged.

The answer states an overall time complexity of O((n+m) α(n)) or O(n+m) (near-linear) for computing all bounds.

### 1) Model as reachable positions with “forbidden” points at guess times

Let the doll’s position **before** guess `i` be `pos[i]` (so `pos[1]` is after Varad’s initial move).
Constraints:
- `pos[i] != a[i]` for all `i = 1..m` (otherwise Shivam guesses correctly).
- `pos[i+1]` is obtained from `pos[i]` by a move in `{ -1, 0, +1 }` staying in `[1..n]`.
- The final position after the last extra move is `y = pos[m+1]`.

For a fixed start `x`, the set of reachable end positions `y` (over Varad’s choices) is always a **consecutive segment** `[L[x], R[x]]` (for `n=1` it is trivial). Intuition: the movement graph is 1D and “holes” created by forbidding a single point at a time get filled again by the ±1 closure; reachability remains interval-like.

Therefore the required answer is:
\[
\sum_{x=1}^{n} \max(0, R[x]-L[x]+1).
\]

So we only need to compute, for every start `x`, the **leftmost** possible end `L[x]` and the **rightmost** possible end `R[x]`.

---

### 2) Compute extreme ends via greedy, but for all starts simultaneously

To get `L[x]` (leftmost possible end), consider an optimal strategy that always tries to move the doll as far **left** as possible, only deviating when that would place it on the next forbidden guess box. Symmetrically, `R[x]` is obtained by always trying to move as far **right** as possible.

Doing this independently for each `x` would be `O(nm)`, too slow. The key is that many starting points quickly become indistinguishable under the greedy process: once two starts lead to the same current position under the greedy policy, their future behavior (and thus final bound) is identical. We can merge such starts.

---

### 3) DSU “merge-starts-by-current-cell” sweep for the left bound

We process guesses from `i=1` to `m` and maintain groups of starting positions whose greedy-simulated doll position (just before guess `i`) is the same cell.

Data structure:
- A disjoint set union over starting indices `x=1..n`.
- For each DSU component, we store the current simulated position `curPos(comp)` (a box number).
- We also maintain for each box `v` which DSU component currently sits at `v` (if any), so if two components move into the same `v`, we union them.

Update rule per step `i` for left-bound simulation:
1. **Avoid the forbidden box.**
   If some component currently has `curPos = a[i]`, then to keep losing, it must shift to the smallest possible alternative, i.e. to the right neighbor `a[i]+1` if it exists, otherwise the component becomes impossible (no valid path for those starts).
2. **Apply Varad’s move after guess `i` to prepare for the next step.**
   Greedy-left move: every component tries to move `curPos := max(1, curPos-1)`.

Whenever step (1) or (2) moves a component into a box already occupied by another component, we union them (because from this point on, their greedy evolution is identical).

At the end (after processing guess `m` and then the final move), the value `curPos` of the component containing start `x` equals `L[x]`.

Boundary handling:
- Moves that would go to `0` stay at `1`.
- If step (1) requires moving from `n` to `n+1`, that component is marked “dead” (contributes 0 end positions).

This sweep is near-linear because each union happens at most `n-1` times and each component changes position only when forced by a guess or by the uniform greedy shift.

Compute `R[x]` with the symmetric sweep (mirror the line):
- replace each guess `a[i]` by `a'[i] = n+1-a[i]`,
- compute left bounds `L'` on the mirrored instance,
- map back: `R[x] = n+1 - L'[n+1-x]`.

---

### 4) Final counting

After obtaining arrays `L[1..n]` and `R[1..n]` (with “dead” starts contributing 0),
sum:
\[
\text{ans} = \sum_{x=1}^n \max(0, R[x]-L[x]+1).
\]

---

### 5) Complexity

Both sweeps are `O((n+m) α(n))` with DSU (effectively linear for constraints), and memory is `O(n+m)`.


Prompt:
You are given an undirected tree with `n` vertices numbered `1` to `n`. Vertex `i` has an integer value `a_i`.

You may perform the following operation any number of times:

- Choose a vertex with current degree `≤ 1` and delete it (and its incident edge if it exists).

You may delete all vertices.

After deletions, perform **compression**:

- While there exists a vertex of degree exactly `2`, delete it and connect its two neighbors by an edge.

Repeat until no degree-2 vertices remain. The final compressed tree is unique (order-independent).

Compute the maximum possible **sum of values of vertices remaining after compression**.

### Input
- `n` (`2 ≤ n ≤ 5·10^5`)
- `a_1, ..., a_n` (`-10^9 ≤ a_i ≤ 10^9`)
- `n-1` undirected edges `u v` forming a tree

### Output
One integer: the maximum achievable sum.

Does the answer explicitly state: in the chosen remaining subtree S (before compression), a vertex contributes a_v to the final sum iff deg_S(v) ≠ 2 (and contributes 0 iff deg_S(v)=2 because it is removed during compression)?

Does the answer formally define dp1[v] as the maximum obtainable final sum using only vertices in subtree(v) under the condition that the edge (parent(v),v) is kept in the remaining subtree S?

Does the answer state the dp1[v] candidate for keeping 0 child edges as a_v (degree in S equals 1, so v is not compressed)?

Does the answer state the dp1[v] candidate for keeping exactly 1 child edge as max_u dp1[u] and explicitly exclude adding a_v (degree in S equals 2, so v is compressed)?

Does the answer state the dp1[v] candidate for keeping ≥2 child edges as a_v + bestSum_{≥2}?

Does the answer define the component (no-parent-edge) candidate for keeping 0 child edges as a_v (degree 0, v remains)?

Does the answer define the component candidate for keeping exactly 1 child edge as a_v + best1 (degree 1, v remains)?

Does the answer define the component candidate for keeping exactly 2 child edges as best2 and explicitly exclude adding a_v (degree 2, v is compressed)?

Does the answer state the final output is max(0, maximum over all vertices of the component candidates) to allow deleting all vertices?

## Expert Solution

Root the tree at node `1`. Any sequence of leaf deletions leaves a connected subtree `S` (possibly empty).

#### Key characterization (compression effect)
Compression repeatedly deletes every vertex of degree exactly `2`. It never deletes vertices of degree `0`, `1`, or `≥ 3`. Hence:

**A vertex `v` contributes `a_v` to the final sum iff `deg_S(v) ≠ 2`;**
**if `deg_S(v) = 2` then `v` is removed and contributes `0`.**

---

## DP definition
Let `parent(v)` and `children(v)` be w.r.t. the rooting.

`dp1[v] = maximum final (post-compression) sum using only subtree(v)`
**under the condition that** edge `(parent(v), v)` is kept in `S`.

Then `v` has a parent-edge, so:
`deg_S(v) = 1 + #{kept child edges (v, u)}`

For each child `u`, either keep `(v, u)` (gain `dp1[u]`) or delete that child side (gain `0`).
Let `b_u = dp1[u]` for children `u`.

---

## Deterministic helper values from `{b_u}`
- `best1 = max(b_u)` (requires `≥ 1` child)
- `best2 =` sum of the two largest `b_u` (requires `≥ 2` children)
- `bestSum_{≥k}` for `k ∈ {2, 3}`: maximum sum over any subset of children of size `≥ k`

Compute `bestSum_{≥k}` deterministically:
Let `P = ∑_{u: b_u > 0} b_u`, and `p = #{u: b_u > 0}`.
- If `p ≥ k`: `bestSum_{≥k} = P`
- Else: `bestSum_{≥k} = P +` (sum of the `k - p` largest values among the non-positive `b_u`).

This is optimal: take all positive contributions, then add the least harmful remaining values until reaching size `k`.

---

## Transition for `dp1[v]`
Let `k` be the number of kept child edges.

1) `k = 0`: `deg_S(v) = 1 ⇒ v` is **not compressed**, so include `a_v`:
   `cand0 = a_v`

2) `k = 1`: `deg_S(v) = 2 ⇒ v` **is compressed**, so **exclude `a_v`**:
   `cand1 = best1`

3) `k ≥ 2`: `deg_S(v) ≥ 3 ⇒ v` is **not compressed**, so include `a_v`:
   `candGe2 = a_v + bestSum_{≥2}`

Therefore:
`dp1[v] = max(cand0, cand1 (if exists), candGe2 (if exists))`

---

## Component candidates (no parent edge)
A remaining subtree `S` can be entirely inside `subtree(v)` without keeping the parent edge. Then `deg_S(v)` equals the number of kept child edges, giving four explicit cases:

- keep 0 children: `deg = 0 ⇒ v` remains
  `val0 = a_v`
- keep 1 child: `deg = 1 ⇒ v` remains
  `val1 = a_v + best1`
- keep 2 children: `deg = 2 ⇒ v` is compressed (**exclude `a_v`**)
  `val2 = best2`
- keep `≥ 3` children: `deg ≥ 3 ⇒ v` remains
  `valGe3 = a_v + bestSum_{≥3}`

Let `ans` be the maximum of these component candidates over all vertices `v`.

Finally, deleting all vertices is allowed, so the output is:
`max(0, ans)`

Compute `dp1[v]` in postorder and update `ans` at each vertex. Use 64-bit integers.


Prompt:
This problem tests knowledge of operating systems and CPU scheduling. A system implements Multilevel Feedback Queue (MLFQ) scheduling with I/O handling and priority boosting.

SYSTEM CONFIGURATION:
- 3 queues: Q0 (highest priority), Q1 (medium), Q2 (lowest)
- Time quantum: Q0=2ms, Q1=4ms, Q2=8ms
- Higher-priority queues preempt lower-priority queues (see Rule 8); within the same queue, round-robin applies (see Rule 6)
- 5 processes: P1, P2, P3, P4, P5

SCHEDULING RULES (apply exactly as stated):

RULE 1 (Priority): Always run the highest-priority (lowest queue number) ready process. Q0 > Q1 > Q2.

RULE 2 (New Process): New processes enter at Q0 with a full Q0 quantum.

RULE 3 (Quantum Exhaustion): If a process uses its entire time quantum without its CPU burst completing, demote it to the next lower queue (Q0->Q1->Q2). Q2 processes stay in Q2. Demotion applies even if the quantum boundary coincides with a preemption event. If a CPU burst completes at the exact same instant as the quantum expires, the burst completion takes precedence: the process is NOT demoted; it either blocks for I/O (Rule 4) or finishes, whichever applies next.

RULE 4 (I/O Block): If a process blocks for I/O before its quantum expires, it stays in its current queue. Its remaining quantum is discarded.

RULE 5 (I/O Completion): When I/O completes, the process returns to the BACK of its current queue with a fresh full quantum for that queue.

RULE 6 (Round Robin): Within the same queue, use round-robin. The process at the front runs first.

RULE 7 (Priority Boost): Every 20ms (at t=20, t=40, t=60, ...), ALL processes in the system are moved to Q0, including processes currently blocked on I/O. After the boost, the Q0 order is by ascending process ID (P1 before P2, etc.). Each process receives a fresh Q0 quantum.

RULE 8 (Preemption): A newly ready higher-priority (lower queue number) process immediately preempts the running process. The preempted process goes to the BACK of its current queue with its remaining quantum preserved.

RULE 9 (Tie-Breaking): If multiple processes enter the same queue at the exact same instant (e.g., after a priority boost or simultaneous I/O completions), they are ordered within that queue by ascending process ID (P1 before P2, etc.). Processes already in a queue before the simultaneous event retain their existing relative order; the newly arrived processes are appended after them in ascending PID order.

RULE 10 (I/O Duration): I/O takes exactly the specified duration. A process cannot use the CPU during I/O.

RULE 11 (Simultaneous Events): When multiple events occur at the same time, process them in this order: (1) I/O completion, (2) new process arrival, (3) quantum expiry or CPU burst completion, (4) priority boost.

RULE 12 (CPU Idle): The CPU idles only if all arrived and not-yet-finished processes are blocked on I/O.

INITIAL STATE (time 0):
All queues empty. CPU idle.

PROCESS ARRIVALS AND BEHAVIOR:
P1: Arrives t=0,  CPU burst pattern: 3ms, I/O 5ms, 4ms, I/O 3ms, 6ms, DONE
P2: Arrives t=1,  CPU burst pattern: 5ms, I/O 4ms, 2ms, I/O 6ms, 3ms, DONE
P3: Arrives t=3,  CPU burst pattern: 7ms, I/O 2ms, 5ms, DONE
P4: Arrives t=6,  CPU burst pattern: 2ms, I/O 8ms, 4ms, I/O 2ms, 1ms, DONE
P5: Arrives t=10, CPU burst pattern: 9ms, I/O 3ms, 2ms, DONE

(Pattern: CPU burst, then I/O, then CPU burst, etc., until DONE)

COMPUTE EXACTLY THESE TEN VALUES:
(1) V1: At what time does P1 complete (all bursts finished)?
(2) V2: The total time the CPU is idle (no process running).
(3) V3: How many queue demotions occur during the entire execution? (A demotion is a move from a higher queue to a lower queue caused by quantum exhaustion, not by a priority boost.)
(4) V4: At time t=25, which process is currently running? "At time t=25" means the state at exactly t=25ms, after all simultaneous events at t=25 have been resolved per Rule 11. Report the process ID (1-5), or 0 if the CPU is idle.
(5) V5: How many priority boosts (Rule 7) actually move at least one process from a non-Q0 queue (or from I/O-blocked state in a non-Q0 queue) to Q0?
(6) V6: The product of the completion times of P2 and P4.
(7) V7: How many context switches occur? Count each time the CPU switches from one process to another, or from idle to a process. Do not count a running process blocking for I/O or completing as a separate context switch; count only the subsequent transition when the CPU picks up the next process.
(8) V8: The total time P3 spends in the READY state (in a queue, not running and not blocked on I/O).
(9) V9: At time t=40, after the priority boost has been applied, how many processes are in Q2?
(10) V10: Verification checksum: U = (V1 + V3 * V5 + V7) mod 79.

FORMAT: V1=<int>;V2=<int>;V3=<int>;V4=<int>;V5=<int>;V6=<int>;V7=<int>;V8=<int>;V9=<int>;V10=<int>

Does the answer state that V1 (P1 completion time) equals exactly 49?

Does the answer state that V2 (total CPU idle time) equals exactly 3?

Does the answer state that V3 (total queue demotions) equals exactly 12?

Does the answer state that V4 (process running at t=25) equals exactly 5?

Does the answer state that V5 (priority boosts that moved at least one process) equals exactly 2?

Does the answer state that V6 (product of P2 and P4 completion times) equals exactly 1376?

Does the answer state that V7 (context switches) equals exactly 27?

Does the answer state that V8 (total time P3 spends in the ready state) equals exactly 23?

Does the answer state that V9 (number of processes in Q2 at t=40 after boost) equals exactly 0?

Does the answer state that V10 (verification checksum) equals exactly 21?

## Setup

Queues: Q0 (quantum=2ms, highest), Q1 (quantum=4ms), Q2 (quantum=8ms, lowest).
Priority boost fires at t=20, t=40, ...
All processes including I/O-blocked ones are moved to Q0 on a boost (Rule 7).
Demotion occurs on full quantum exhaustion, even when that coincides with a preemption (Rule 3).
On I/O completion, a process gets a fresh quantum for its current queue (Rule 5).

---

## Detailed Execution Timeline

### Phase 1: t=0 to t=20 (before first boost)

t=0:   P1 arrives -> Q0. CPU idle -> P1 starts. Q0=[P1 running].
t=0-2: P1 runs 2ms (full Q0 quantum). burst_rem=1ms. P1 demoted -> Q1.
       DEMOTION 1: P1 Q0->Q1.

t=1:   P2 arrives -> Q0 back. (P1 still running until t=2; no preemption, same queue.)

t=2:   P1's quantum expires -> P1->Q1 (burst_rem=1ms). Q0=[P2]. P2 starts.
       CONTEXT SWITCH 1: (idle->P1 at t=0)   [initial start]
       CONTEXT SWITCH 2: P1->P2 at t=2.

t=2-4: P2 runs 2ms (full Q0 quantum). burst_rem=3ms. P2 demoted -> Q1.
       DEMOTION 2: P2 Q0->Q1.

t=3:   P3 arrives -> Q0 back. (P2 running; no preemption.)

t=4:   P2's quantum expires -> P2->Q1 (burst_rem=3ms). Q0=[P3]. P3 starts.
       CONTEXT SWITCH 3: P2->P3 at t=4.

t=4-6: P3 runs 2ms (full Q0 quantum). burst_rem=5ms.

t=6:   P4 arrives (Rule 11 step 2: arrival before quantum expiry).
       P4 -> Q0. Then P3's quantum expires (Rule 11 step 3) -> P3 demoted Q0->Q1 (burst_rem=5ms).
       DEMOTION 3: P3 Q0->Q1.
       Q0=[P4]. P4 starts.
       CONTEXT SWITCH 4: P3->P4 at t=6.

t=6-8: P4 runs 2ms. P4's CPU burst is exactly 2ms = full Q0 quantum.
       P4's burst completes at t=8. P4 blocks for I/O (8ms, done at t=16). Rule 4: P4 stays Q0.

t=8:   P4 blocks -> I/O. Q0=[] (P4 blocked). Q1=[P1(burst_rem=1), P2(burst_rem=3), P3(burst_rem=5)].
       P1 (front of Q1) starts. Fresh Q1 quantum=4ms.
       CONTEXT SWITCH 5: P4->P1 at t=8.

t=8-9: P1 runs 1ms, completes burst 1. P1 blocks for I/O (5ms, done at t=14). Stays Q1.
       Q1=[P2, P3].
       CONTEXT SWITCH 6: P1->P2 at t=9.

t=9:   P2 starts in Q1. burst_rem=3ms, quantum=4ms.

t=10:  P5 arrives -> Q0. P5 (Q0) preempts P2 (Q1) per Rule 8.
       P2 back of Q1 with quantum_rem=3ms, burst_rem=2ms.
       Q0=[P5]. Q1=[P3, P2].
       CONTEXT SWITCH 7: P2->P5 at t=10.

t=10-12: P5 runs 2ms (full Q0 quantum). P5 demoted -> Q1 (burst_rem=7ms).
         DEMOTION 4: P5 Q0->Q1.

t=12:  P5->Q1 (back). Q1=[P3, P2, P5]. P3 starts. Fresh quantum=4ms.
       CONTEXT SWITCH 8: P5->P3 at t=12.

t=12-16: P3 runs 4ms (full Q1 quantum). burst_rem=1ms. Normally demoted Q1->Q2.
         At t=16: P4 I/O done (Rule 11: I/O before quantum expiry).
         P4 I/O completes -> P4 back of Q0. P4 (Q0) preempts P3 (Q1).
         P3 has used its full Q1 quantum -> per Rule 3, P3 is demoted Q1->Q2 (burst_rem=1ms).
         DEMOTION 5: P3 Q1->Q2.
         CONTEXT SWITCH 9: P3->P4 at t=16.

t=14:  P1 I/O done -> P1 back of Q1. Q1=[P2, P5, P1]. (P3 still running at that point.)

t=16-18: P4 runs 2ms (full Q0 quantum). burst_rem=2ms. P4 demoted -> Q1.
         DEMOTION 6: P4 Q0->Q1.

t=18:  P4->Q1 (back). Q1=[P2, P5, P1, P4]. P2 starts. quantum_rem=3ms (preserved), burst_rem=2ms.
       CONTEXT SWITCH 10: P4->P2 at t=18.

t=18-20: P2 runs 2ms, completing burst 1 at exactly t=20.

### Phase 2: Priority Boost at t=20

At t=20, per Rule 11:
  Step 1: P2 completes burst, blocks for I/O (4ms, done at t=24). P2 stays Q1 (Rule 4).
  Step 2: Priority boost fires.
    - P1 (Q1, ready) -> Q0.
    - P2 (Q1, I/O-blocked, non-Q0) -> Q0 (boost includes I/O-blocked, Rule 7).
    - P3 (Q2, ready) -> Q0.
    - P4 (Q1, ready) -> Q0.
    - P5 (Q1, ready) -> Q0.
    Q0=[P1, P2(blocked), P3, P4, P5] ordered by PID. Each gets fresh Q0 quantum=2ms.
BOOST 1 moves processes: YES. EFFECTIVE BOOST COUNT = 1.
  CPU: P1 (front of Q0, lowest PID among ready) starts.
  CONTEXT SWITCH 11: P2->P1 at t=20.

t=20-22: P1 runs 2ms (full Q0 quantum). burst_rem=2ms. P1 demoted -> Q1.
         DEMOTION 7: P1 Q0->Q1.

t=22:  Q0=[P3, P4, P5]. P3 starts.
       CONTEXT SWITCH 12: P1->P3 at t=22.

t=22-23: P3 runs 1ms, completing burst 1 at t=23. P3 blocks I/O (2ms, done at t=25). Stays Q0.
         CONTEXT SWITCH 13: P3->P4 at t=23.

t=23-25: P4 runs. burst_rem=2ms, quantum=2ms. P4's burst done at t=25 = quantum boundary.
         P4 blocks for I/O (2ms, done at t=27). Stays Q0.

t=24:  P2 I/O done -> P2 back of Q0. Q0=[P5, P2] (P4 still running).

t=25:  Per Rule 11: I/O before quantum expiry.
       P3 I/O done at t=25 -> P3 back of Q0. Q0=[P5, P2, P3].
       P4 burst done (at quantum boundary) -> P4 blocks I/O (done at t=27). Stays Q0.
       CPU: P5 starts.
       CONTEXT SWITCH 14: P4->P5 at t=25.

       V4 = 5 (P5 starts running at t=25).

t=25-27: P5 runs 2ms (full Q0 quantum). burst_rem=5ms. P5 demoted -> Q1.
         DEMOTION 8: P5 Q0->Q1.

t=27:  Per Rule 11: I/O before quantum expiry.
       P4 I/O done -> P4 back of Q0.
       P5 quantum expires -> P5 demoted Q0->Q1.
       Q0=[P2, P3, P4]. P2 starts.
       CONTEXT SWITCH 15: P5->P2 at t=27.

t=27-29: P2 runs 2ms (burst 2 = 2ms = full quantum). P2 burst done, blocks I/O (6ms, done at t=35). Stays Q0.

t=29:  Q0=[P3, P4]. P3 starts.
       CONTEXT SWITCH 16: P2->P3 at t=29.

t=29-31: P3 runs 2ms (full Q0 quantum). burst_rem=3ms. P3 demoted -> Q1.
         DEMOTION 9: P3 Q0->Q1.

t=31:  Q0=[P4]. P4 starts.
       CONTEXT SWITCH 17: P3->P4 at t=31.

t=31-32: P4 runs 1ms, completing burst 3 (1ms < quantum=2ms). P4 DONE at t=32.
         P4 completion = t=32.
         CONTEXT SWITCH 18: P4->P1 at t=32.

t=32:  Q0=[]. Q1=[P1(burst_rem=2ms), P5(burst_rem=5ms), P3(burst_rem=3ms)]. P1 starts.

t=32-34: P1 runs 2ms, completing burst 2 at t=34. P1 blocks I/O (3ms, done at t=37). Stays Q1.
         CONTEXT SWITCH 19: P1->P5 at t=34.

t=34-35: P5 runs 1ms in Q1.

t=35:  P2 I/O done -> P2 back of Q0. P2 (Q0) preempts P5 (Q1).
       P5 back of Q1: quantum_rem=3ms, burst_rem=4ms.
       Q0=[P2]. Q1=[P3, P5].
       CONTEXT SWITCH 20: P5->P2 at t=35.

t=35-37: P2 runs 2ms in Q0 (burst 3 = 3ms, quantum=2ms). Quantum expires at t=37. burst_rem=1ms.
         P2 demoted Q0->Q1.
         DEMOTION 10: P2 Q0->Q1.

t=37:  Per Rule 11: P1 I/O done first -> P1 back of Q1. Then P2 quantum expires -> P2 Q1 (back).
       Q1=[P3, P5, P1, P2]. P3 starts.
       CONTEXT SWITCH 21: P2->P3 at t=37.

t=37-40: P3 runs 3ms, completing burst 2 at t=40. P3 DONE at t=40.
         P3 completion = t=40.

### Phase 3: Priority Boost at t=40

At t=40, per Rule 11:
  Step 1: P3 burst complete -> P3 exits (done).
  Step 2: Priority boost fires.
    Remaining processes: P1 (Q1), P2 (Q1), P5 (Q1).
    All moved to Q0. Q0=[P1, P2, P5] (by PID). Fresh Q0 quantum=2ms.
BOOST 2 moves processes: YES. EFFECTIVE BOOST COUNT = 2.
  CONTEXT SWITCH 22: P3->P1 at t=40.

V5 = 2 (both boosts moved at least one process).

V9 = 0 (Q2 is empty after boost at t=40 -- all processes moved to Q0).

t=40-42: P1 runs 2ms (full Q0 quantum). burst_rem=4ms. P1 demoted -> Q1.
         DEMOTION 11: P1 Q0->Q1.

t=42:  Q0=[P2, P5]. P2 starts.
       CONTEXT SWITCH 23: P1->P2 at t=42.

t=42-43: P2 runs 1ms, completing burst 3. P2 DONE at t=43.
         P2 completion = t=43.
         CONTEXT SWITCH 24: P2->P5 at t=43.

t=43-45: P5 runs 2ms (full Q0 quantum). burst_rem=2ms. P5 demoted -> Q1.
         DEMOTION 12: P5 Q0->Q1.

t=45:  Q1=[P1(burst_rem=4ms), P5(burst_rem=2ms)]. P1 starts.
       CONTEXT SWITCH 25: P5->P1 at t=45.

t=45-49: P1 runs 4ms in Q1 (burst=4ms, quantum=4ms). P1 burst done at t=49. P1 DONE.
         P1 completion = t=49.
         CONTEXT SWITCH 26: P1->P5 at t=49.

t=49-51: P5 runs 2ms in Q1, completing burst 1 (total 9ms across all slices). P5 blocks I/O (3ms, done at t=54). Stays Q1.

t=51-54: CPU IDLE. Only P5 remains and is blocked on I/O.
         IDLE DURATION = 3ms.
         CONTEXT SWITCH 27: idle->P5 at t=54.

t=54-56: P5 runs 2ms in Q1, completing burst 2. P5 DONE at t=56.

---

## Summary of Values

V1 = 49     (P1 completion time)

V2 = 3      (CPU idle: t=51 to t=54, 3ms)

V3 = 12     (demotions)
     P1: Q0->Q1 (t=2)
     P2: Q0->Q1 (t=4)
     P3: Q0->Q1 (t=6)
     P5: Q0->Q1 (t=12)
     P3: Q1->Q2 (t=16, quantum exhausted even though preemption was simultaneous per Rule 3)
     P4: Q0->Q1 (t=18)
     P1: Q0->Q1 (t=22, after boost)
     P5: Q0->Q1 (t=27)
     P3: Q0->Q1 (t=31)
     P2: Q0->Q1 (t=37)
     P1: Q0->Q1 (t=42, after boost)
     P5: Q0->Q1 (t=45)

V4 = 5      (P5 starts running at t=25)

V5 = 2      (boost at t=20 and boost at t=40 each moved processes)

V6 = 43 * 32 = 1376  (P2 completes t=43, P4 completes t=32)

V7 = 27     (context switches: 25 process-to-process transitions
             + 2 idle-to-process transitions: idle->P1 at t=0, idle->P5 at t=54.
             Transitions from process to idle, e.g. at t=51, are not counted.)

V8 = 23     (P3 ready time)
     t=3-4:   1ms (Q0, P2 running)
     t=6-12:  6ms (Q1, P4 then P1 then P2 then P5 running)
     t=16-20: 4ms (Q2, P4 then P2 running)
     t=20-22: 2ms (Q0, P1 running after boost)
     t=25-29: 4ms (Q0, P5 then P2 running)
     t=31-37: 6ms (Q1, P4 completes then P1 then P5 preempted then P2)
     Total: 1+6+4+2+4+6 = 23ms

V9 = 0      (at t=40, after boost, Q0=[P1,P2,P5], Q1=[], Q2=[])

V10 = (V1 + V3*V5 + V7) mod 79
    = (49 + 12*2 + 27) mod 79
    = (49 + 24 + 27) mod 79
    = 100 mod 79
    = 21

---

## Final Answer

V1=49;V2=3;V3=12;V4=5;V5=2;V6=1376;V7=27;V8=23;V9=0;V10=21


Programming Languages, Compilers & Formal Methods

Prompt:
Simulate Tomasulo's Algorithm with speculative execution for the following instruction sequence.

CONFIGURATION:
Functional Units: 2 Add/Sub (2 cycles), 1 Mul (4 cycles), 1 Div (8 cycles), 1 Load (3 cycles).
Reservation Stations: 3 Add (Add1, Add2, Add3), 2 Mul (Mul1, Mul2), 2 Load (Load1, Load2).
Reorder Buffer: 8 entries (ROB1-ROB8).
Common Data Bus: 1 (only one result broadcast per cycle, priority: Add1 < Add2 < Add3 < Mul1 < Mul2 < Load1 < Load2).

Rules:
(1) Issue: One instruction issues per cycle in program order if RS and ROB available.
(2) Execute: Begins cycle after ALL operands ready (not same cycle operand broadcasts).
(3) Write Result: Broadcasts on CDB one cycle after execution completes.
(4) Commit: In-order from ROB after write completes.
(5) Structural Hazards: Stall issue if no RS or ROB entry available.
(6) RAW Hazards: Wait for operand via CDB broadcast.

REGISTER FILE (initial values):
F0=1.0, F2=2.0, F4=4.0, F6=6.0, F8=8.0, F10=10.0

INSTRUCTION SEQUENCE (I1-I12):
I1:  LOAD  F2, 0(R1)      ; F2 = Mem[R1+0]
I2:  LOAD  F4, 0(R2)      ; F4 = Mem[R2+0]
I3:  MUL   F6, F2, F4     ; F6 = F2 * F4
I4:  ADD   F8, F6, F2     ; F8 = F6 + F2
I5:  DIV   F10, F8, F4    ; F10 = F8 / F4
I6:  SUB   F2, F6, F4     ; F2 = F6 - F4
I7:  ADD   F4, F2, F8     ; F4 = F2 + F8
I8:  MUL   F0, F4, F6     ; F0 = F4 * F6
I9:  LOAD  F12, 0(R3)     ; F12 = Mem[R3+0]
I10: ADD   F14, F12, F0   ; F14 = F12 + F0
I11: SUB   F16, F14, F10  ; F16 = F14 - F10
I12: MUL   F18, F16, F6   ; F18 = F16 * F6

Track each instruction through: Issue (I), Execute Start (X), Execute End (E), Write Result (W), Commit (C).

Compute the following EXACT values:

(1) In which cycle does I3 begin execution?
(2) In which cycle does I5 write its result to CDB?
(3) What is the SUM of issue cycles for all 12 instructions?
(4) How many cycles have at least one instruction in execution phase?
(5) In which cycle does I8 begin execution?
(6) What is the PRODUCT of cycles where I1, I2, and I3 write results?
(7) How many total stall cycles occur due to structural hazards?
(8) What is the cycle number when I12 commits?
(9) At cycle 10, how many reservation stations are occupied?
(10) What is the ratio of ADD/SUB operations to MUL operations expressed as A:B in lowest terms?

Does the answer state that I3 begins execution in exactly cycle 7?

Does the answer state that I5 writes its result in exactly cycle 23?

Does the answer state that the sum of issue cycles for all 12 instructions is exactly 108?

Does the answer state that exactly 26 cycles have at least one instruction executing?

Does the answer state that I8 begins execution in exactly cycle 19?

Does the answer state that the product of write cycles for I1, I2, and I3 is exactly 330?

Does the answer state that exactly 7 stall cycles occur due to structural hazards?

Does the answer state that I12 commits in exactly cycle 36?

Does the answer state that exactly 5 reservation stations are occupied at cycle 10?

Does the answer state that the ratio of ADD/SUB to MUL operations is exactly 5:3?

TOMASULO'S ALGORITHM EXECUTION TRACE

=== CYCLE-BY-CYCLE EXECUTION ===

CYCLE 1:
- I1 (LOAD F2) ISSUES to Load1, ROB1
- Load1: Busy, Op=Load, Dest=ROB1
- ROB1: Busy, Instr=I1, Dest=F2, Ready=No

CYCLE 2:
- I1 begins EXECUTION (cycles 2-4, 3 cycles for Load)
- I2 (LOAD F4) ISSUES to Load2, ROB2
- Load2: Busy, Op=Load, Dest=ROB2

CYCLE 3:
- I1 executing (cycle 2 of 3)
- I2 begins EXECUTION (cycles 3-5)
- I3 (MUL F6, F2, F4) ISSUES to Mul1, ROB3
- Mul1: waiting for F2 (ROB1), F4 (ROB2)

CYCLE 4:
- I1 execution ENDS
- I2 executing (cycle 2 of 3)
- I3 waiting for operands
- I4 (ADD F8, F6, F2) ISSUES to Add1, ROB4
- Add1: waiting for F6 (ROB3), F2 (ROB1)

CYCLE 5:
- I1 WRITES to CDB (F2 ready) - Load1 freed
- I2 execution ENDS
- I3 captures F2, still waiting for F4
- I4 captures F2, still waiting for F6
- I5 (DIV F10, F8, F4) ISSUES to Mul2, ROB5
- Mul2: waiting for F8 (ROB4), F4 (ROB2)

CYCLE 6:
- I1 COMMITS (ROB1 freed)
- I2 WRITES to CDB (F4 ready) - Load2 freed
- I3 captures F4 (all operands ready, executes NEXT cycle)
- I5 captures F4, still waiting for F8
- I6 (SUB F2, F6, F4) ISSUES to Add2, ROB6
- Add2: waiting for F6 (ROB3), has F4

CYCLE 7:
- I2 COMMITS (ROB2 freed)
- I3 begins EXECUTION (cycles 7-10, 4 cycles for MUL)
- I6 waiting for F6
- I7 (ADD F4, F2, F8) ISSUES to Add3, ROB7
- Add3: waiting for F2 (ROB6), F8 (ROB4)

CYCLE 8:
- I3 executing (cycle 2 of 4)
- I8 (MUL F0, F4, F6) ISSUES to Mul1 (freed after I3 writes), ROB8
- Wait - Mul1 still busy with I3!
- I8 STALLS (no Mul RS available)

CYCLE 9:
- I3 executing (cycle 3 of 4)
- I8 STALLS (structural hazard - Mul1, Mul2 both busy)

CYCLE 10:
- I3 execution ENDS
- I8 STALLS

CYCLE 11:
- I3 WRITES to CDB (F6 ready) - Mul1 freed
- I4 captures F6 (all operands ready)
- I6 captures F6 (all operands ready)
- I8 (MUL F0, F4, F6) ISSUES to Mul1, ROB8
- Mul1: waiting for F4 (ROB7), has F6

CYCLE 12:
- I3 COMMITS (ROB3 freed)
- I4 begins EXECUTION (cycles 12-13, 2 cycles for ADD)
- I6 begins EXECUTION (cycles 12-13, 2 cycles for SUB)
- I8 waiting for F4
- I9 (LOAD F12) ISSUES to Load1, ROB1 (reused)

CYCLE 13:
- I4 execution ENDS
- I6 execution ENDS
- I9 begins EXECUTION (cycles 13-15)

CYCLE 14:
- I4 WRITES to CDB (F8 ready) - Add1 freed (priority)
- I5 captures F8 (all operands ready)
- I7 captures F8, still waiting for F2
- CDB conflict: I6 must wait

CYCLE 15:
- I4 COMMITS (ROB4 freed)
- I6 WRITES to CDB (F2 ready) - Add2 freed
- I7 captures F2 (all operands ready)
- I8 waiting for F4 (from I7)
- I5 begins EXECUTION (cycles 15-22, 8 cycles for DIV)
- I9 execution ENDS

CYCLE 16:
- I5 executing (cycle 2 of 8)
- I9 WRITES to CDB (F12 ready) - Load1 freed
- I7 begins EXECUTION (cycles 16-17)
- I10 (ADD F14, F12, F0) ISSUES to Add1, ROB2
- Add1: has F12, waiting for F0 (ROB8)

CYCLE 17:
- I5 executing
- I7 execution ENDS
- I6 COMMITS (ROB6 freed)

CYCLE 18:
- I5 executing
- I7 WRITES to CDB (F4 ready) - Add3 freed
- I8 captures F4 (all operands ready)
- I11 (SUB F16, F14, F10) ISSUES to Add2, ROB6
- Add2: waiting for F14 (ROB2), F10 (ROB5)

CYCLE 19:
- I5 executing
- I7 COMMITS (ROB7 freed)
- I8 begins EXECUTION (cycles 19-22, 4 cycles for MUL)
- I12 (MUL F18, F16, F6) ISSUES to Mul2 - Wait, Mul2 has I5!
- I12 STALLS (no Mul RS)

CYCLE 20:
- I5 executing
- I8 executing
- I12 STALLS

CYCLE 21:
- I5 executing
- I8 executing
- I12 STALLS

CYCLE 22:
- I5 execution ENDS
- I8 execution ENDS
- I12 STALLS

CYCLE 23:
- I5 WRITES to CDB (F10 ready) - Mul2 freed (priority over I8)
- I11 captures F10, waiting for F14
- I8 must wait for CDB
- I12 ISSUES to Mul2, ROB7
- Mul2: waiting for F16 (ROB6), has F6

CYCLE 24:
- I5 COMMITS (ROB5 freed)
- I8 WRITES to CDB (F0 ready) - Mul1 freed
- I10 captures F0 (all operands ready)
- I12 waiting for F16

CYCLE 25:
- I8 COMMITS (ROB8 freed)
- I10 begins EXECUTION (cycles 25-26)
- I9 COMMITS (ROB1 freed)

CYCLE 26:
- I10 execution ENDS

CYCLE 27:
- I10 WRITES to CDB (F14 ready) - Add1 freed
- I11 captures F14 (all operands ready)

CYCLE 28:
- I10 COMMITS (ROB2 freed)
- I11 begins EXECUTION (cycles 28-29)

CYCLE 29:
- I11 execution ENDS

CYCLE 30:
- I11 WRITES to CDB (F16 ready) - Add2 freed
- I12 captures F16 (all operands ready)

CYCLE 31:
- I11 COMMITS (ROB6 freed)
- I12 begins EXECUTION (cycles 31-34)

CYCLE 32-34:
- I12 executing

CYCLE 35:
- I12 WRITES to CDB (F18 ready)

CYCLE 36:
- I12 COMMITS (ROB7 freed)
- EXECUTION COMPLETE

=== INSTRUCTION TIMING SUMMARY ===

Inst | Issue | ExecStart | ExecEnd | Write | Commit
-----|-------|-----------|---------|-------|-------
I1   |   1   |     2     |    4    |   5   |   6
I2   |   2   |     3     |    5    |   6   |   7
I3   |   3   |     7     |   10    |  11   |  12
I4   |   4   |    12     |   13    |  14   |  15
I5   |   5   |    15     |   22    |  23   |  24
I6   |   6   |    12     |   13    |  15   |  17
I7   |   7   |    16     |   17    |  18   |  19
I8   |  11   |    19     |   22    |  24   |  25
I9   |  12   |    13     |   15    |  16   |  25
I10  |  16   |    25     |   26    |  27   |  28
I11  |  18   |    28     |   29    |  30   |  31
I12  |  23   |    31     |   34    |  35   |  36

=== ANSWER CALCULATIONS ===

(1) Cycle I3 begins execution: 7

(2) Cycle I5 writes result: 23

(3) Sum of issue cycles:
1+2+3+4+5+6+7+11+12+16+18+23 = 108

(4) Cycles with at least one instruction executing:
Cycles 2,3,4,5,7,8,9,10,12,13,14,15,16,17,19,20,21,22,25,26,28,29,31,32,33,34 = 26 cycles

(5) Cycle I8 begins execution: 19

(6) Product of write cycles for I1, I2, I3:
5 × 6 × 11 = 330

(7) Stall cycles due to structural hazards:
I8 stalls cycles 8,9,10 (3 cycles)
I12 stalls cycles 19,20,21,22 (4 cycles)
Total: 7 stall cycles

(8) Cycle I12 commits: 36

(9) Reservation stations occupied at cycle 10:
Mul1 (I3), Mul2 (I5), Add1 (I4), Add2 (I6), Add3 (I7) = 5 stations

(10) Ratio ADD/SUB to MUL:
ADD/SUB: I4, I6, I7, I10, I11 = 5 operations
MUL: I3, I8, I12 = 3 operations
Ratio = 5:3

=== FINAL ANSWERS ===

(1) 7
(2) 23
(3) 108
(4) 26
(5) 19
(6) 330
(7) 7
(8) 36
(9) 5
(10) 5:3


Prompt:
You are given a multiset `a` of size `n`: `a_1, a_2, ..., a_n`.

You generate a new multiset `s` as follows:

1. Partition `a` into any number of **non-empty** multisets `x_1, x_2, ..., x_k`, using every element exactly once.
2. For each `x_i`, choose **one of its modes** and insert it into `s`.

A **mode** of a multiset is any element with maximum frequency in that multiset (ties allowed).

Count how many **different multisets** `s` can be produced (multiplicities matter), modulo `998244353`.

### Input
- `n` (`1 ≤ n ≤ 5000`)
- `a_1, ..., a_n` (`1 ≤ a_i ≤ n`)

### Output
Output the number of obtainable multisets `s` modulo `998244353`.

Does the answer define cnt[v] as the frequency of value v in the input multiset a?

Does the answer state the feasibility condition for a chosen support T (values appearing at least once in s) as: for every x ∉ T, cnt[x] ≤ sum_{c∈T}?

Does the answer simplify feasibility to the single inequality sum_{c∈T}?

Does the answer state that for any feasible support T, the multiplicity of each c∈T in output multiset s can be any integer t_c with 1 ≤ t_c ≤ cnt[c], giving exactly cnt[c] choices per c?

Does the answer define a knapsack DP dp[sum] that equals the total number of outputs whose chosen support has total weight sum = sum_{c∈T}?

Does the answer give the per-color DP transition: dp_next[s] += dp[s] (exclude), and dp_next[s+cnt] += dp[s] * cnt (include), all modulo 998244353?

Does the answer compute the final result as ans = sum_{s = M..n}?

Does the answer explicitly state the modulus as 998244353 and apply it to DP and final answer?

Does the answer state the overall time complexity as O(n^2) and space complexity as O(n) (or O(n^2) if explicitly chosen), consistent with n ≤ 5000?

### Expert Solution

Let `cnt[v]` be the frequency of value `v` in the input multiset `a`, and let
`M = max_v cnt[v]`
Also let `MOD = 998244353`.

---

## 1) Feasibility condition

Let `T` be the **support** of the output multiset `s` (the set of values that appear at least once in `s`).

For any value `x ∉ T`, we must be able to place all `cnt[x]` copies of `x` into parts whose **chosen mode** is some `c ∈ T`, without making `x` the *unique* most frequent element in any part.

If a part’s chosen mode is `c`, then that part can contain at most as many `x`'s as it contains `c`'s (otherwise `x` would be strictly more frequent and thus the unique mode). Summing over all parts whose chosen mode lies in `T`, the total "hiding capacity" for `x` is at most `∑_{c ∈ T} cnt[c]`. Hence a necessary condition is:
`cnt[x] ≤ ∑_{c ∈ T} cnt[c]` for every `x ∉ T`.

This condition is also sufficient (we can distribute occurrences of each excluded `x` across the mode-parts so that in every part, `#x ≤ #(mode)`).

Since `cnt[x] ≤ M` for all `x`, this is equivalent to the single inequality:
`∑_{c ∈ T} cnt[c] ≥ M`

---

## 2) Counting multiplicities of `s` for a fixed support

If `c ∈ T`, then `c` must be chosen as a mode for some positive number `t_c` of parts.
Each such part must contain at least one `c`, so
`1 ≤ t_c ≤ cnt[c]`
Thus for each `c ∈ T` there are exactly `cnt[c]` choices for `t_c`, and for fixed support `T` the number of different multisets `s` is:
`∏_{c ∈ T} cnt[c]`

So the required answer is:
`∑_{T: ∑ cnt[c] ≥ M} ( ∏_{c ∈ T} cnt[c] ) (mod MOD)`

---

## 3) Knapsack DP

Let `dp[S]` be the total contribution (sum of products `∏ cnt[c]`) over supports `T` whose total weight
`S = ∑_{c ∈ T} cnt[c]`
equals `S`.

Initialize `dp[0] = 1`. For each value `v` with `cnt = cnt[v] > 0`:
- Exclude `v`: `dp_next[s] += dp[s]`
- Include `v`: `dp_next[s + cnt] += dp[s] · cnt`

All updates are modulo `998244353`.

Finally,
`ans = ∑_{s=M..n} dp[s] (mod 998244353)`

---

## 4) Complexity

The DP runs over sums `0...n` for at most `n` distinct values, so:
- Time: `O(n^2)`
- Memory: `O(n)`

This fits for `n ≤ 5000`.


Prompt:
Ashwini is trapped in a nightmare where Gauss appears and challenges her with a task.

Ram, her friend, explains the rule:

For a given integer **n**, Ashwini must produce a sequence **[a₁, a₂, ..., aₙ]** such that:

1. Each **aᵢ** is between **1** and **10¹⁸**.
2. For every adjacent pair **(aᵢ, aᵢ₊₁)**, the **GCD values must all be distinct**.
3. The sequence must use the **minimum possible number of distinct elements**.

To make the task deterministic, Ram adds a strict rule:

> “Ashwini must use the **fixed construction** described below, and output the **lexicographically smallest valid sequence** generated by that construction.”

### Fixed Construction (Deterministic)

1. Let **M = 10**.
2. Construct **100 distinct integers** using:

   \[
   x_{i,j} = 2^{i} \cdot 3^{j} \cdot 5^{9-i} \cdot 7^{9-j}
   \quad \text{for } 0 \le i, j < 10
   \]

3. This gives a fixed set **X** of size 100.
4. Build a complete graph on X, labeling each edge between **xₐ and x_b** with **gcd(xₐ, x_b)**.
5. Construct an **Eulerian trail** on this graph using **Hierholzer’s algorithm** with a fixed deterministic edge order (sorted by gcd value).
6. The resulting Eulerian trail produces a deterministic sequence of length **5050** (≥ 5000).
7. Output the **first n elements** of this Eulerian trail.

### Input format:
- First line contains **t**, the number of test cases.
- Each test case contains a single integer **n**.

### Output format:
- For each test case, output **n** space-separated integers on a new line.

Ashwini must solve the nightmare to wake up. She knows a solution always exists.

### Constraints:
- **1 ≤ t ≤ 200**
- **2 ≤ n ≤ 5000**

Does the answer state the final minimum number of distinct elements as k = 100?

Does the answer explicitly give the maximum sequence length formula for odd k as k(k+1)/2 via Euler tour?

Does the answer explicitly give the maximum sequence length formula for even k as k^2/2 + 1?

Does the answer explicitly show the inequality k^2/2 + 1 ≥ 5000 ⇒ k ≥ 100?

Does the answer explicitly define the construction formula x_{i,j}?

Does the answer explicitly set M = 10 for the construction?

Does the answer explicitly state the gcd formula using min-exponents for any two vertices?

Does the answer explicitly claim that the gcd mapping is unique for each unordered pair?

Does the answer specify deterministic Hierholzer traversal rule: choose unused edge with smallest gcd label, tie-break by smallest neighbor index?

Step 1: Minimum distinct elements (k)

For a set of k distinct numbers, define a complete graph with k vertices and edges labeled by gcd of endpoints.
Each adjacent pair in the sequence corresponds to traversing one edge, and all gcds must be distinct ⇒ no edge may be repeated.

- If k is odd, the complete graph has all even degrees ⇒ Euler tour exists covering all edges:
  length = k(k+1)/2
- If k is even, all degrees are odd. We must delete a matching of size (k−2)/2 to make degrees even, leaving:
  length = k(k+1)/2 − (k−2)/2 = k^2/2 + 1

To reach n up to 5000, we need:
k^2/2 + 1 ≥ 5000  ⇒  k ≥ 100

So **minimum k = 100**.

Step 2: Deterministic set construction

Let M = 10. For 0 ≤ i, j < 10:

x_{i,j} = 2^i * 3^j * 5^(9−i) * 7^(9−j)

This produces 100 distinct numbers, all ≤ 10^18 (max value = 5^9 * 7^9 = 7.88×10^13).

Example values:
- x_{0,0} = 2^0 * 3^0 * 5^9 * 7^9 = 78815638671875
- x_{9,9} = 2^9 * 3^9 * 5^0 * 7^0 = 10077696
- x_{1,0} = 2^1 * 3^0 * 5^8 * 7^9 = 31526255468750
- x_{8,9} = 2^8 * 3^9 * 5^1 * 7^0 = 62985600
- x_{9,8} = 2^9 * 3^8 * 5^0 * 7^1 = 23514624

Step 3: Unique gcd property

For any two vertices (i1,j1) and (i2,j2), gcd is:

2^{min(i1,i2)} * 3^{min(j1,j2)} * 5^{min(9−i1,9−i2)} * 7^{min(9−j1,9−j2)}

This mapping is unique for each unordered pair (i1,j1),(i2,j2), so all edge labels are distinct.

Step 4: Eulerian trail (deterministic)

The graph is Eulerian (all degrees even).
Use Hierholzer’s algorithm with deterministic edge choice:

- At each vertex, choose the unused edge with **smallest gcd label** (sorted order).
- If ties occur, choose the smallest neighbor index.

This produces a deterministic Euler trail of length 5050 edges (5051 vertices).

Step 5: First few elements of the Euler trail (actual computed values)

Using the deterministic Hierholzer’s traversal, the first vertices in the Euler trail are:

Index | Value
----- | -----
1 | 78815638671875
2 | 10077696
3 | 31526255468750
4 | 25194240
5 | 78815638671875
6 | 23514624
7 | 31526255468750
8 | 62985600
9 | 78815638671875
10 | 58786560

GCDs between consecutive values:

- gcd(78815638671875, 10077696) = 1
- gcd(10077696, 31526255468750) = 2
- gcd(31526255468750, 25194240) = 10
- gcd(25194240, 78815638671875) = 5
- gcd(78815638671875, 23514624) = 7
- gcd(23514624, 31526255468750) = 14
- gcd(31526255468750, 62985600) = 50
- gcd(62985600, 78815638671875) = 25
- gcd(78815638671875, 58786560) = 35

All GCDs are distinct, satisfying the condition.

Step 6: Output

For each test case n, output the first n vertices from this deterministic Euler trail.

Example outputs:

For n = 2:
78815638671875 10077696

For n = 3:
78815638671875 10077696 31526255468750

For n = 5:
78815638671875 10077696 31526255468750 25194240 78815638671875

Verification:
- All values ≤ 10^18.
- Adjacent gcds are distinct.
- Uses exactly 100 distinct elements (minimum).
- Deterministic due to fixed construction + fixed traversal.


Prompt:
You are given a distributed system with 4 processes: P0, P1, P2, P3.

Each process maintains a vector clock of size 4.
All clocks are initially: [0,0,0,0]

-----------------------------------
VECTOR CLOCK RULES
-----------------------------------

1. Internal event:
   Increment own component.

2. Send event:
   Increment own component and send the updated clock.

3. Receive event:
   Take component-wise max of local and received clock,
   then increment own component.

-----------------------------------
EVENT SEQUENCE
-----------------------------------

E1: P0 internal
E2: P1 internal
E3: P0 sends M1 to P2
E4: P2 receives M1 from P0
E5: P1 sends M2 to P3
E6: P3 receives M2 from P1
E7: P2 sends M3 to P1
E8: P3 internal
E9: P1 receives M3 from P2
E10: P0 sends M4 to P3
E11: P2 internal
E12: P3 receives M4 from P0
E13: P1 sends M5 to P0
E14: P0 receives M5 from P1
E15: P3 sends M6 to P2
E16: P2 receives M6 from P3

-----------------------------------
TASK
-----------------------------------

Compute vector clocks and answer the following:

1. Vector clock at E4
2. Vector clock at E9
3. Vector clock at E12
4. Vector clock at E14
5. Vector clock at E16
6. Total number of happened-before (→) pairs among all unordered event pairs {Ei, Ej} with i < j. Count a pair if the vector clock of one event is strictly less than the other component-wise.
7. Total number of concurrent (||) pairs among the same 120 unordered event pairs. Two events are concurrent if neither vector clock is strictly less than the other component-wise.
8. Event with maximum vector clock sum
9. Length of the longest causal chain, defined as the maximum number of distinct events in a sequence X1 → X2 → ... → Xk, where each adjacent pair satisfies happened-before.
10. Number of receive events where exactly 2 components increase when comparing the receiver's local clock immediately before the receive with its clock immediately after applying the receive rule

-----------------------------------
FINAL OUTPUT FORMAT

E4,E9,E12,E14,E16,HB,CONC,MAX_EVENT,CHAIN_LEN,RECV2

Where:
- Each vector clock is formatted as [a,b,c,d]
- MAX_EVENT is event label (e.g., E16)

Correctly computes the number of happened-before pairs among the 120 unordered event pairs as 67

Correctly computes the number of concurrent pairs among the 120 unordered event pairs as 53

Correctly identifies E16 as max sum event

Correctly computes the longest causal chain length as 7 events

Correctly computes that 3 receive events have exactly 2 component increases from the receiver's pre-receive clock to its post-receive clock

[2,0,1,0],[2,3,2,0],[3,2,0,3],[4,4,2,0],[3,2,4,4],67,53,E16,7,3

EXPLANATION

Apply the vector clock rules in order. The complete clock after each event is:

- E1 (P0 internal): [1,0,0,0]
- E2 (P1 internal): [0,1,0,0]
- E3 (P0 sends M1 to P2): [2,0,0,0]
- E4 (P2 receives M1): max([0,0,0,0],[2,0,0,0]) = [2,0,0,0], then increment P2 -> [2,0,1,0]
- E5 (P1 sends M2 to P3): [0,2,0,0]
- E6 (P3 receives M2): max([0,0,0,0],[0,2,0,0]) = [0,2,0,0], then increment P3 -> [0,2,0,1]
- E7 (P2 sends M3 to P1): [2,0,2,0]
- E8 (P3 internal): [0,2,0,2]
- E9 (P1 receives M3): max([0,2,0,0],[2,0,2,0]) = [2,2,2,0], then increment P1 -> [2,3,2,0]
- E10 (P0 sends M4 to P3): [3,0,0,0]
- E11 (P2 internal): [2,0,3,0]
- E12 (P3 receives M4): max([0,2,0,2],[3,0,0,0]) = [3,2,0,2], then increment P3 -> [3,2,0,3]
- E13 (P1 sends M5 to P0): [2,4,2,0]
- E14 (P0 receives M5): max([3,0,0,0],[2,4,2,0]) = [3,4,2,0], then increment P0 -> [4,4,2,0]
- E15 (P3 sends M6 to P2): [3,2,0,4]
- E16 (P2 receives M6): max([2,0,3,0],[3,2,0,4]) = [3,2,3,4], then increment P2 -> [3,2,4,4]

From this table:

1. E4 = [2,0,1,0]
2. E9 = [2,3,2,0]
3. E12 = [3,2,0,3]
4. E14 = [4,4,2,0]
5. E16 = [3,2,4,4]

For happened-before and concurrency, compare all unordered pairs of events using vector clock comparison. There are C(16,2) = 120 unordered pairs total.

- A pair is happened-before if one vector is strictly less than the other component-wise.
- A pair is concurrent if neither vector is strictly less than the other.

Counting all 120 pairs gives:

- Happened-before pairs = 67
- Concurrent pairs = 53

The event with maximum vector clock sum is E16 because:

- sum(E14) = 4 + 4 + 2 + 0 = 10
- sum(E15) = 3 + 2 + 0 + 4 = 9
- sum(E16) = 3 + 2 + 4 + 4 = 13

No earlier event has a sum greater than 13, so MAX_EVENT = E16.

For the longest causal chain, one maximum-length sequence is:

E1 -> E3 -> E4 -> E7 -> E9 -> E13 -> E14

Each event in this sequence happened-before the next one, so the chain length is 7 events. No longer chain exists.

For receive events with exactly 2 component increases, compare the receiver's local clock immediately before and immediately after each receive:

- E4: [0,0,0,0] -> [2,0,1,0], increases in components 0 and 2 only
- E6: [0,0,0,0] -> [0,2,0,1], increases in components 1 and 3 only
- E9: [0,2,0,0] -> [2,3,2,0], 3 components increase
- E12: [0,2,0,2] -> [3,2,0,3], increases in components 0 and 3 only
- E14: [3,0,0,0] -> [4,4,2,0], 3 components increase
- E16: [2,0,3,0] -> [3,2,4,4], 4 components increase

So exactly 3 receive events satisfy the condition: E4, E6, and E12.

FINAL ANSWER

[2,0,1,0],[2,3,2,0],[3,2,0,3],[4,4,2,0],[3,2,4,4],67,53,E16,7,3