riddle &puzzle
发布网友
发布时间：2024-05-15 10:21
共4个回答
热心网友
时间：2024-05-18 04:04
puzzle-1.难题；难解之事(It's a puzzle that...)2.谜；测验智力的玩具（或问题）(word puzzle填字谜)3.视觉误差4.使困惑，使为难，使伤脑筋(puzzle sb.)5.（常与over连用）苦想，使苦思6.（与out连用）想出；解开
热心网友
时间：2024-05-18 04:05
n.[C]
1. 谜,谜语
Can you guess the answer to this riddle?
你能猜出这个谜吗?
2. 谜一般的人;难题;莫名其妙的事情
All she said was a riddle to me.
她所说的一切对我来说是一个谜。
vt.
1. 解...的谜
2. 给...出谜;使困惑
vi.
1. 打谜似地说
2. 出谜,打谜
n.
1. 粗筛,格筛[C]
vt.
1. 筛(谷物,砂石等)
2. 把...打得满是窟窿[H][(+with)]
The car was riddled with bullets.
汽车给子弹打得车身上尽是窟窿。
3. 充斥,布满[H][(+with)]
The umbrella is riddled with holes and the rain's dripping down.
雨伞上满是洞,雨水正滴下来。
4. 连续质问;处处挑剔
puzzle
1. 使迷惑;使为难,使窘困[H]
What puzzles me is why they didn't show up.
令我百思不解的是他们为什么没有出现。
He looked a little puzzled.
他看上去有点困惑。
His recent behavior puzzles me.
他最近的行为使我迷惑不解。
2. 苦思而得出[(+out)]
I could not puzzle out her intentions.
我猜不出她的意图。
We finally puzzled out the meaning of the poem.
我们苦苦思索终于理解这首诗的意思。
vi.
1. 感到迷惑[(+at)]
I have been puzzling about this question for weeks now.
我对这个问题已冥思苦想了好几个星期。
2. 苦思,冥思苦想[(+about/over/as to)]
n.
1. (游戏的)猜谜,智力竞赛[C]
2. 难题,谜,难以理解之事[S1]
Her decision was a puzzle to him.
她的决定对他来说是个谜。
3. 困惑,迷惑[S]
I'm in a puzzle as to how to cope with the new situation.
我不知道该如何应付这新局面。
参考资料：Dr.eye
热心网友
时间：2024-05-18 04:05
riddle指供人猜的谜语 游戏``也指终究会有解答的难题``两者都有谜 难题的意思``
FOR
This letter puzzles me
这封信让我迷惑不解
I will put a riddle to you,boys
孩子们，我出个谜语给你们猜
热心网友
时间：2024-05-18 04:06
(a) The process cannot terminate, because before the last move a single lamp would have been on. But the last move could not have turned it off, because the adjacent lamp was off. There are only finitely many states (each lamp is on or off and the next move can be at one of finitely many lamps), hence the process must repeat. The outcome of each step is uniquely determined by the state, so either the process moves around a single large loop, or there is an initial sequence of steps as far as state k and then the process goes around a loop back to k. However, the latter is not possible because then state k would have had two different precursors. But a state has only one possible precursor which can be found by toggling the lamp at the current position if the previous lamp is on and then moving the position back one. Hence the process must move around a single large loop, and hence it must return to the initial state.
(b) Represent a lamp by X when on, by - when not. For 4 lamps the starting situation and the situation after 4, 8, 12, 16 steps is as follows:
X X X X
- X - X
X - - X
- - - X
X X X -
On its first move lamp n-2 is switched off and then remains off until each lamp has had n-1 moves. Hence for each of its first n-1 moves lamp n-1 is not toggled and it retains its initial state. After each lamp has had n-1 moves, all of lamps 1 to n-2 are off. Finally over the next n-1 moves, lamps 1 to n-2 are turned on, so that all the lamps are on. We show by inction on k that these statements are all true for n = 2k. By inspection, they are true for k = 2. So suppose they are true for k and consider 2n = 2k+1 lamps. For the first n-1 moves of each lamp the n left-hand and the n right-hand lamps are effectively insulated. Lamps n-1 and 2n-1 remain on. Lamp 2n-1 being on means that lamps 0 to n-2 are in just the same situation that they would be with a set of only n lamps. Similarly, lamp n-1 being on means that lamps n to 2n-2 are in the same situation that they would be with a set of only n lamps. Hence after each lamp has had n-1 moves, all the lamps are off except for n-1 and 2n-1. In the next n moves lamps 1 to n-2 are turned on, lamp n-1 is turned off, lamps n to 2n-2 remain off, and lamp 2n-1 remains on. For the next n-1 moves for each lamp, lamp n-1 is not toggled, so it remains off. Hence all of n to 2n-2 also remain off and 2n-1 remains on. Lamps 0 to n-2 go through the same sequence as for a set of n lamps. Hence after these n-1 moves for each lamp, all the lamps are off, except for 2n-1. Finally, over the next 2n-1 moves, lamps 0 to 2n-2 are turned on. This completes the inction. Counting moves, we see that there are n-1 sets of n moves, followed by n-1 moves, a total of n2 - 1.
(c) We show by inction on the number of moves that for n = 2k+ 1 lamps after each lamp has had m moves, for i = 0, 1, ... , 2k - 2, lamp i+2 is in the same state as lamp i is after each lamp has had m moves in a set of n - 1 = 2k lamps (we refer to this as lamp i in the reced case). Lamp 0 is off and lamp 1 is on. It is easy to see that this is true for m = 1 (in both cases odd numbered lamps are on and even numbered lamps are off). Suppose it is true for m. Lamp 2 has the same state as lamp 0 in the reced case and both toggle since their predecessor lamps are on. Hence lamps 3 to n - 1 behave the same as lamps 1 to n - 3 in the reced case. That means that lamp n - 1 remains off. Hence lamp 0 does not toggle on its m+1th move and remains off. Hence lamp 1 does not toggle on its m+1th move and remains on. The inction stops working when lamp n - 2 toggles on its nth move in the reced case, but it works up to and including m = n - 2. So after n - 2 moves for each lamp all lamps are off except lamp 1. In the next two moves nothing happens, then in the following n - 1 moves lamps 2 to n - 1 and lamp 0 are turned on. So all the lamps are on after a total of (n - 2)n + n + 1 = n2 + n + 1 moves.
