Solution To A Word Search

Apply a=1, b=2, ...y=25, z=26 and find the solution for the word discipline?

(4)(9)(19)(3)(9)(16)(12)(9)(14)(5) = 2233889280 Vola!Interesting indeed! Happy to help and discovered that my PC had a calculator app. I am not sure what your motivation in asking this question was but I can tell you with 100% surety that you will be made fun of.I am 100% sure that you know the answer to this problem, so you are either a 100% fool, just showing off or someone  conducting a kind of lame psychological experiment.

Can words overlap each other in word searches?

Good word searches do not have words that overlap completely. In a properly constructed word search puzzle:each word should occur exactly once, andeach word should contain at least one letter not contained in any other word.(In some word search puzzles the unused letters form the solution -- a secret message -- and the second "rule" makes sure that you need to find all words in order to reveal the entire secret message perfectly.)As a consequence of the second rule, a proper word search puzzle cannot have two words such that one is a substring of the other.That being said, I have seen puzzles that did contain such words, and then it was usually the case that if you had to find both "car" and "care" in the same puzzle, there was one occurrence of "care" and one different occurrence of "car". This meant that "car" appeared twice in the puzzle, and you had to reason that the 'correct' "car" is the one that is not a part of "care". Personally, I consider such puzzles ugly.

What is the solution for the word problem: divide 29 into two parts such that the sum of the squares of the parts is 425?

Ans:13 and 16..How???Given let..x+y=29,x^2+y^2=425now,(x+y)^2=x^2+y^2+2xy841=425+2xy.416=2xyxy=208now,(x-y)^2=(x+y)^2–4xy(x-y)^2=841–4(208).(x-y)^2=9.x-y=3.now we have x+y=29..and x-y=3..on solving above equations…we get x=16 and y=13…

Is there a hosted search solution like Swiftype Search or Relevanssi that can search a database and not just meta-data?

Here at Swiftype , if you use a crawler-based search engine, meta-tags would be the best way to associate data that wouldn't be immediately visible to users. However, this would still show in the page source, so it is just obscured, not completely hidden. Are you using WordPress? Our   Swiftype for WordPress plugin was recently named the #1 WordPress Plugins Of 2014, and would give you the functionality you are looking for by just copying and pasting a snippet of code to add custom metadata.  If you're not on a platform like WordPress and meta-tags are not an option, you could use our API to associate data that is invisible to the user. If you need any more help, please feel free to reach out, either on Quora or via email.

What word's meaning is "a solution that solves multiple problems?"

Multifaceted solution is the simplest way of saying this

Why is the runtime of the naive recursive solution to the word-break problem O(2^N)?

Consider the sample recursive solution given in the link you mentioned:String SegmentString(String input, Set dict) {
if (dict.contains(input)) return input;
int len = input.length();
for (int i = 1; i < len; i++) {
String prefix = input.substring(0, i);
if (dict.contains(prefix)) {
String suffix = input.substring(i, len);
String segSuffix = SegmentString(suffix, dict);
if (segSuffix != null) {
return prefix + " " + segSuffix;
}
}
}
return null;
}The worst case occurs when we are forced to analyze all the input suffixes. As the link you gave mentions, one such example is when the input is a string like "aaaa...aaab" and the dictionary contains all words "a..a" but not a "b".Consider an input formed by [math]n[/math] a's followed by one "b". Line 2 condition will never be true as we don't have a "b" in the dictionary.The for loop will always match [math]n-2[/math] times as we have all words formed by a's. So there are [math]n-2[/math] recursive calls. What we are effectively doing is checking all subsets of the given input. E.g. we are splitting "aaab" into"a" + "a" + "a" + "b"
"a" + "a" + "ab"
"a" + "aab"
"a" + "aa" + "b"
"aa" + "a" + "b"
"aa" + "ab"
"aaa" + "b"We will split the [math]n[/math] a's into [math]2^n - 1[/math] subsets. Hence the [math]O(2^n)[/math] time bound.