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AlexRosenthal_ThinkLikeACoderS001E006_2019E-
Ethic , Hedge , and Octavia stand on the edge of a bottomless ravine . 艾斯克、海吉和奥克塔维亚 站在无底峡谷的边缘。
Ethic:n.伦理;道德规范;adj.伦理的;道德的(等于ethical); Hedge:n.树篱; v.避免正面回答; adj.树篱的; on the edge of:adv.几乎;濒于;在…边缘; bottomless:adj.无底的;深不可测的;难解的;取之不尽的; ravine:n.沟壑,山涧;峡谷;
It’s the only thing between them and the tower that houses the second of three powerful artifacts . 峡谷是他们与塔之间的唯一阻碍, 塔里藏有三个强大神器中的第二个。
artifacts:n.史前古器物;人工产品;
They’ve got a brief window of time to get across before the guards return. 他们在警卫回来前
With Hedge’s fuel gauge on empty he won’t be able to fly Ethic across, so the only option is to make a bridge. 海吉的燃料箱空了, 无法带艾斯克飞越过去, 所以,唯一选择是建一座桥。
gauge:n.计量器;标准尺寸;容量规格;vt.测量;估计;给…定规格; option:n.选择;可选择的东西;
Fortunately , the floating stacks of stones nearby are bridge components— invented by Octavia herself— called hover-blocks. 幸运的是,附近漂浮的石头堆 可以用来搭一座桥, 石头堆是奥克塔维亚 自己发明的——
Fortunately:adv.幸运地; stacks:n.堆叠;[图情]书库(stack的复数);进程堆栈应用; nearby:adj.附近的,邻近的;adv.在附近;prep.在…附近;
Activate a pile with a burst of energy, and they’ll self-assemble to span the ravine as Ethic walks across. 他们可以用一股能量 激活一堆飘浮块, 这样在这些飘浮块自动组装时, 艾斯克就能跨越峡谷。
Activate:vt.刺激;使活动;使活泼;使产生放射性;vi.激活;有活力; pile:n.桩;堆;摞;桩柱;v.堆放;摞起;叠放;放置; burst:v.破裂;爆炸;突破;冲进;n.爆发;点射;释放;(情感的)迸发; self-assemble:vi.自组装; span:n.持续时间; v.持续;
But there is, of course, a catch. 但现在存在一个问题。
The hover-blocks are only stable when they’re perfectly palindromic. 这些飘浮块只有在组合成 完美的回文结构时才能保持稳定。
stable:n.马厩;牛棚;adj.稳定的;牢固的;坚定的;vi.被关在马厩;赶入马房;
Meaning they have to form a sequence that’s the same when viewed forwards and backwards. 也就是说,它们得被组合成 正序看和倒序看都一致的结构才行。
sequence:n.顺序; v.按顺序排列;
The stacks start in random orders, but will always put themselves into a palindromic configuration if they can. 飘浮块最初是随机排列的, 但如果可行,
random:adj.[数]随机的;任意的;胡乱的;n.随意;adv.胡乱地; configuration:n.配置;结构;外形;
If they get to a point where a palindrome isn’t possible, the bridge will collapse , and whoever’s on it will fall into the ravine. 但如果它们无法被拼成回文结构, 桥就会坍塌掉, 走在上面的人就会坠入深谷。
palindrome:n.回文; collapse:vi.倒塌;瓦解;暴跌;vt.使倒塌,使崩溃;使萎陷;折叠;n.倒塌;失败;衰竭;
Let’s look at an example. 让我们来看一个例子。
This stack would make itself stable. 这堆飘浮块能被组成稳定的结构。
First the A blocks hold themselves in place. 首先,A 型方块 能被排列在左右两侧。
And finally the C would nestle right between the B’s. 最后,C 型方块正好 能被放在两块 B 之间。
finally:adv.终于;最终;(用于列举)最后;彻底地; nestle:vi.舒适地坐定;偎依;半隐半现地处于;vt.抱;安置;
However, suppose there was one more A. 但假设多一块 A 型方块的话。
suppose:v.推断:假定:假设:设想:
First two A blocks form up, then two B’s, but now the remaining C and A have nowhere to go, so the whole thing falls apart. 首先两块 A 排列好, 然后两块 B 也排列好, 但剩下的 A 块 和 C 块就没地方放了, 因此整座桥都会塌掉。
nowhere:v.无处; n.无处; adj.不存在的;
The Node of Power enables Hedge to energize a single stack of blocks. 力量节点晶石只够海吉 激活一堆飘浮块,
Node:n.节点;结点;茎节;(根或枝上的)节; enables:v.使得; (enable的第三人称单数) energize:v.使充满热情;增添能量(或精力、活力、干劲);为…提供电力(或能量);
What instructions can Ethic give Hedge to allow him to efficiently find and power a stable palindromic stack? 艾斯克要给海吉什么指令, 才能让它有效地找到并激活 一堆稳固且为回文结构的飘浮块呢?
efficiently:adv.有效地;效率高地(efficient的副词形式);
Pause now to figure it out for yourself. [ 可暂停播放自行解题 ]
Examples of palindromes include ANNA, RACECAR , and MADAM IM ADAM. 回文排列的例子如:ANNA、RACECAR、 以及 MADAM IM ADAM。
palindromes:n.回文;(细胞中遗传物质的)回文结构;(palindrome的复数) RACECAR:n.赛车;
Counting the number of times a given letter appears in a palindrome will reveal a helpful pattern. 通过数一个字母 在回文中出现的次数, 你会发现一个有用的模式。
reveal:v.显示;透露;揭露;泄露;n.揭露;暴露;门侧,窗侧;
Pause now to figure it out for yourself. [ 可暂停播放自行解题 ]
Let’s first look at a na?ve solution to this problem. 我们先来看一个简单的方案。
solution:n.解决方案;溶液;溶解;解答;
A na?ve solution is a simple, brute-force approach that isn’t optimized— but will get the job done. 简易方案是一种 但是它可以达成目标。
brute-force:n.强力攻击;蛮力; approach:n.方法;路径;v.接近;建议;着手处理;
Na?ve solutions are helpful ways to analyze problems, and work as stepping stones to better solutions. 简单方案是分析问题的有用方法, 是更优方案的铺路石。
analyze:v.对…进行分析,分解(等于analyse);
In this case, a na?ve solution is to approach a pile of blocks, try all the arrangements , and see if one is a palindrome by reading it forward and then backwards. 在这种情况下,一个简单方案是: 尝试这堆飘浮块的所有可能组合, 并通过正向,反向阅读 来确定其是否为回文结构。
arrangements:n.安排;筹备;布置;商定;(arrangement的复数)
The problem with this approach is that it would take a tremendous amount of time. 但这种方法有个缺陷, 它需要大量的时间。
tremendous:adj.极大的,巨大的;惊人的;极好的;
If Hedge tried one combination every second, 如果海吉每秒 能判断一种组合可不可行,
combination:n.结合;组合;联合;[化学]化合;
That’s because the total time is a function of the factorial of the number of blocks there are. 这是因为总时间 是总块数的阶乘。
factorial:adj.因子的,阶乘的;n.[数]阶乘;
10 blocks have over 3 million combinations . 10 个飘浮块就有 超过 300 万种组合方式。
combinations:n.[数]组合;制品(combination的复数);合谱;
What this na?ve solution shows is that we need a much faster way to tell whether a pile of blocks can form a palindrome. 这个简单的解决方案表明, 我们需要一种更快的方法 来判断一堆飘浮块 是否可以形成回文。
To start, it may be intuitively clear that a pile of all different blocks will never form one. 首先,很明显的是, 一堆各不相同的飘浮块 永远不可能形成回文结构。
intuitively:adv.直观地;直觉地;
Why? 为什么？
The first and last blocks can’t be the same if there are no repeats. 为什么？
first and last:总的说来,整体看来;始终;主要地;
So when can a given sequence become a palindrome? 那么什么时候一个给定的 序列可以形成回文呢?
One way to figure that out is to analyze a few existing palindromes. 一种方法是分析 一些现有的回文。
The pattern here is that most of the letters occur an even number of times, and there’s at most 1 that occurs just once. 模式是,大多数字母 会出现偶数次, 最多只有一个字母 可以只出现一次。
occurs:v.重现(occur的第三人称单数);
Is that it? 这就完了吗？
What if RACECAR had 3 E’s instead of 1? 如果 RACECAR 有3个 E 而不是1个呢?
What if:如果…怎么办?
We could tack the new E’s onto the ends and still get a palindrome, so 3 is ok. 我们可以把新的两个 E 钉在两端, 仍能得到一个回文, 所以字母出现 3 次是可以的。
But make that 3 E’s and 3 C’s, and there’s nowhere for the last C to go. 但如果有 3 个 E 和 3 个 C , 最后一个 C 就没有地方放了。
So the most generalized insight is that at most one letter can appear an odd number of times, but the rest have to be even. 让我们来概括一下思路, 最多只有一个字母能出现奇数次, 但其他字母必须出现偶数次。
generalized:adj.广义的,普遍的; v.推广(generalize的过去分词); insight:n.洞察力;洞悉; odd:adj.古怪的;奇数的;n.奇数;
Hedge can count the letters in each stack and organize them into a dictionary, which is a tidy way of storing information. 海吉可以数每个字母 在飘浮堆中出现的次数, 这是一种储存信息的简洁方式。
organize:v.组织;安排;处理;分配;管理; tidy:adj.整齐的:整洁的:v.使整洁:
A loop could then go through and count how many times odd numbers appear. 可以通过一个循环遍历这个字典, 数有几个字母出现了奇数次。
loop:n.循环;回路;环路;圈;v.使成环;使绕成圈;成环形移动;
If there are less than 2 odd characters, the stack can be made into a palindrome. 如果有少于两个字母出现了奇数次,
This approach is much, much faster than the na?ve solution. 这种方法比之前的 那个简单解决方案要快得多。
Instead of factorial time, it takes linear time. 它只需线性时间, 而不是阶乘时间。
linear:adj.线的,线型的;直线的,线状的;长度的;
That’s where the time increases in proportion to the number of blocks there are. 线性时间指的是 程序运行时间与方块数量成正比。
in proportion to:与…成比例,与…相称;
Now write a loop for Hedge to approach the piles individually , and stop when he finds a good one, and you’ll be ready to go. 现在,请你为海吉编写一个循环, 来分别处理这些飘浮块堆, 当它找到可行的方案时, 暂停程序,这样就成功了。
piles:n.痔疮;痔; individually:adv.个别地,单独地;
Here’s what happens: 接下来会发生的是:
Hedge is fast, but there are so many piles it takes a long time. 虽然海吉的速度很快,
Too long. 很长很长的时间。
Ethic and Hedge are safe. 艾斯克和海吉现在安全了,
But Octavia is not so lucky. 但奥克塔维亚就没那么幸运了。