Medium riddles

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Unusual connections

What is unusual about the following words: revive, banana, grammar, voodoo, assess, potato, dresser, uneven?
Take the first letter of each word and place it at the end. It will spell the same word backwards.
87.54 %
33 votes

cleanpoemswho am I

Two Sisters

Two sisters we are, one is dark and one is fair, In twin towers dwelling we're quite the pair, One from land and one from sea, Tell us truly, who are we?
The answer is salt and pepper.
87.14 %
55 votes

logicmath

Flipping Lockers

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
86.92 %
54 votes

logicmath

Anagram Checker

Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not). Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.
First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams. Here is pseudocode for the procedure that was described: def areAnagrams(S1, S2) A = new Array(26) A.initializeValues(0) for each character in S1 arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc... A[arrayIndex] += 1 end for each character in S2 arrayIndex = mapCharacterToNumber(character) A[arrayIndex] -= 1 end for (i = 0; i < 26; i++) if A[i] != 0 return false end end return true end
86.91 %
43 votes

logicshort

The mystery of Cindy

The day before yesterday Cindy was 17. Next year she will be 20. How can this be?
The statement was made on January 1. Cindy's birthday is on December 31. She was 17 the day before yesterday (Dec 30). She was 18 yesterday. She will be 19 this year (Dec 31) and 20 next year.
86.69 %
53 votes

cleanfunnylogic

A woman with no driver license

A woman with no driver license goes the wrong way on a one-way street and turns left at a corner with a no left turn sign. A policeman sees her but does nothing... Why?
She is walking.
86.69 %
53 votes