43 lines
2.8 KiB
Markdown
43 lines
2.8 KiB
Markdown
# HW3
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## 10.1
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- 1.
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- a) a, c or b, d
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- b) f
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- 3.
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| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | ... |
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| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
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| 1 | 1 | 0 | 0 | 1 | 1 | 2 | 2 | 0 | 1 | 0 | 0 | 2 | 1 | 1 | 0 | 0 | 2 | 1 | 1 | 0 | 2 | 2 | 1 | 0 | 0 | 0 | 2 | 1 | 1 | 0 | 0 | 2 | 1 | 3 | 0 | 2 | 2 | 1 | 1 | 0 |
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3, 4, 9, 11, 12, 16, 17, 21, 25, 26, 27, 31, 32, 36, 40+
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- 8.
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- a)
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| a | b | c | d |
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| --- | --- | --- | --- |
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| 0 | 1 | 0 | 1 |
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- b) impossible. Since there is no successor for f, then f must be 0. Because the graph is symmetric, so assuming start from e and e is 1, then d is 2, c is 1, b is 2, a is 1, e is 2, thus contradiction.
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- 12.
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- a) Since $K_n = K_{n−1} ∪ \{x | l(x) = n\ and\ s(x) ∩ K_{n−1} = Ø\}$, $l(x) = n$ iff there exist a path P $X_n->X_{n-1}->X_{n-2}...->X_0$ where $X_n \in L(n)$. This is the longest because if we the previous vertex must have higher level than the successor, if we skip any of the vertex say from $X_n->X_{n-2}$, the total length of the path must be shorter than the longest path gave before.
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- 15.
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- Grundy function: value is the first non used non negative integer of value of successors, then it must be different then successor's, proved.
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- Level numbers: $l(x) = k \Leftrightarrow x \notin L_{k-1}\ and\ s(x) \subseteq L_{k-1}$, thus any vertex must have different level than its successor.
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## 10.2
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- 1.
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- a) 10, 11, 11, 101 -> 111, change 101 to 10 (move 3 from 4th)
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- b) 1, 11, 101, 111 -> 0, no winning condition
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- c) 10, 100, 100, 110 -> 100, change 100 to 0 or 101 to 1 (move 4 from 2nd, 3rd or 4th)
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- 2.
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- a) 10, 0, 0, 10 -> 0, no winning condition
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- b) 1, 0, 10, 1 -> 10, change 10 to 0 or 0 to 10 (move 2 from 3rd or 1 from 2nd)
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- c) 10, 1, 1, 0 -> 10, change 10 to 0 or 0 to 10 (move 2 from 1st or 1 from 4th)
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- 3.
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- a) 0, 0, 11, 0 -> 11, change 11 to 0 or 0 to 11 (move 3 from 3rd or 2 from 4th)
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- b) 1, 0, 1, 10 -> 10, change 10 to 0 or 1 to 11 (move 2 from 3rd or 4th)
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- c) 0, 1, 0, 1 -> 0, no winning condition
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- 4. Grundy value for number of sticks less than or equal to 7 is the same as exercise 2. Thus the answers are the same except for b), add a winning strategy by moving 5 from 3rd.
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- 6.
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- a) 100 to 1 or 110 to 11, move 3 from 2nd, 3rd or 4th
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- b) 100 to 10 or 110 to 0, move 2 from 2nd or 3rd or all from 4th |