superprismatic

Why here? This REALLY isn't a Logic/Math Puzzle. Why don't you put it in the appropriate Place? 1. Season: winter, fall, summer 2. Fish: trout, grouper, haddock 3. Mountains: andies, alps, rockies 4. Crime: arson, robbery, murder

[spoiler=I see what you're driving at. ]But, the math would not work out to having 4,625 students saying yes.

Neither, it is 4,625 students out of 10,000 students. That would be 46.25% of the student body.

That sounds like a great idea! Please feel free to post some of these. You could even alter the topology of the grid. Besides a torus, one could use a projective plane, or a Klein bottle.

The University of Southern North Carolina at Hoople wants to estimate the number of its students who have experimented with illegal drugs at some point in their lives. But, such drug experimentation is illegal and subject to severe punishment. So a polling procedure was designed such that an admission of guilt during the poll would not be sufficiently convincing for a person to be found guilty of this offence. Here are the details of the procedure which USNCH has instituted and which each of USNCH's 10,000 students is required to take: 1. The student is placed in a secluded room where there are 100 marbles in 4 lines of 25 each arrayed on a board which has been dimpled to hold the marbles in this configuration. Three lines contain green marbles and one line contains red marbles. The student is supplied with an empty black bag into which he places all the marbles. 2. The student has been told to randomize the bag (by shaking it), reach in and randomly pick one marble, and surreptitiously peek at it to determine its color. 3. The student is told to tell the truth if the marble he picked is green, but to lie if the picked marble is red. 4. The student is asked the question, "Have you ever experimented with illegal drugs?" He is to answer according to step 3. 5. The answer is tallied with those of all the other students. It is known (I don't know how) that 75% of the students will follow the instructions correctly. It is also known that 10% of the student body (but their identities are unknown) are members of the Verititian religion who always tell the truth when asked a question, regardless of any other circumstances. A further 10% are anarchists who, out of sheer contempt for authority, will always do the opposite of what they are told -- in this case they would tell the truth if they saw a red marble and lie if they see a green one. The remaining 5% belong to the Nastyranian cult and will always lie regardless of the marble color picked. Which students are in which of these four groups is unknown. 4,625 of the students answered the question in the affirmative. What is the most likely number of students who have actually experimented with illegal drugs?

a most enjoyable puzzle! might not be able to stop working at it. thanks superprismatic Thanks for the kind words. Wow, your score of 62 is way more than I thought was possible! Perhaps someone can do even better?

Wow! I count your score to be 54. Still, it's way better than my best answer of 45. I wonder how high this can go?

You are correct, names with double letters, like HOOVER are not allowed because the OP states that one must move to an adjacent letter. This bring up another point, though: unlike the rules of the game Boggle, a letter may be used twice for the same word, e.g. the A in OBAMA. So, HOOVER is a no-no, but OBAMA is OK (no political pun intended).

Have you ever played Boggle on a torus? I'd wager you hadn't. That is what this puzzle is all about. In this case, however, we will use only a limited number of words -- the 38 distinct surnames of all of the U.S. presidents: EISENHOWER WASHINGTON JEFFERSON CLEVELAND ROOSEVELT VANBUREN HARRISON FILLMORE BUCHANAN GARFIELD MCKINLEY COOLIDGE MADISON JACKSON LINCOLN JOHNSON HARDING KENNEDY CLINTON MONROE TAYLOR PIERCE ARTHUR WILSON HOOVER TRUMAN CARTER REAGAN TYLER ADAMS GRANT HAYES NIXON OBAMA POLK TAFT FORD BUSH [/code] Of all the letters in all these names, only two letters are not used, Q and Z. So, the grid we will use is a 6 by 4 grid filled with the other 24 letters, a different one for each cell of the grid, like so: [code] ------------------------- | A | B | C | D | E | F | ------------------------- | G | H | I | J | K | L | ------------------------- | M | N | O | P | R | S | ------------------------- | T | U | V | W | X | Y | ------------------------- What makes this a torus is that the upper edge is considered to be the same as the lower edge, and the left edge is identified with the right edge. In the above example, the adjacent letters to A are T, U, B, H, G, L, F, and Y (clockwise from the letter directly above A), just as the adjacent letters to H are B, C, I, O, N, M, G, and A. A president's name can be considered to be in the grid if the letters in his name, in order, form a chain of adjacent letters in the torus grid. So, for example, if my grid were chosen to be ------------------------- | X | Y | L | E | V | S | ------------------------- | T | A | N | C | I | H | ------------------------- | J | D | F | M | G | B | ------------------------- | U | K | R | O | P | W | ------------------------- [/code] the following presidential names (and no others) can be placed in this grid: [code] CLEVELAND TAYLOR TYLER POLK FORD BUSH Your task is to place these 24 letters (A through Y, without Q) in the torus grid so that the sum of the letters in all the presidential surnames which could be placed in the grid is as large as possible. In the above example, this number would be 32. Ties will be broken in favor of the largest number of presidential names (6 in this case).

I'm afraid you misread the OP. I was asking for the largest number of words.

You could possibly make a 15 by 15 with no black squares, similar to what I did

Here's a typed version. I can't vouch for its accuracy. P1 (MHD MKNE A RSE-S-M-KNARE) (ACSM) OTFRNE NPtNSE NPBSE RCBBNSE NPRSE INC PRSE NMRSE OPRE HLDWLDNCBE (TFXLF TCXL NCBE) AL-PRPPIT XLYPPIY NCBE MGKSE INCDRCBRNSE PRSE WLDRCBRNSE NTSBNEN TXSE-CRSLE-CITRSE WLD NCBE ALWLP NCBE TSME LRSE RLSE URGLSNEASN WLD NCBE (NO PFSE NLSRE NCBE) NTEGDDMNSENCURERCBRNE (TENE TFRNE NCBRTSE NCBE INC) (FLRSE PRSE ONDE 71 NCBE) (CDNSE PRSE ONSDE 74 NCBE) (PRTSE PRSE ONREDE 75 NCBE) (TFNR CMSP SOLE MRDE LUSE TOTE WLD NWLD NCBE) (194 WLD's NCBE)(TRFXL) ---------------------------------------… NOTES ALPNTE GLSE - SE ERTE VLSE MTSE-CTSE-WSE-FRTSE PNRTRSE ON DRSE WLD NCBE N WLD XLR CMSPNE WLDSTS 'ME XL DULMT 6 TUNSE NCBE XL -------------------------------- (MUNSARSTEN M4 N A RSE) KLSE-LRSTE-TRSE-TRSE-MRSEN-MRSE (SAE6NSE SE N MBSE) -------------------------------- NMNRCBRNSEPTE2PTEWSREBRNSE 36 MLSE 74 SPRKSE 29 KENOS OLE + 73 RTRSE 35 SLE CLGSE OUNUTRE DKRSE PSESHLE 651 MTCSE HTLSE N CU TC TRS NMRE 99.84.8 2 UNE PLSE VCRSE AOLTSE NSKSE NBSE NSRE ONSE PUT SE WLD NCBE (3 X ORL) -------------------------------- ?NMSE NRSE I N 2 N TRLERCB ANSE NTSRCR O NE LSPNSE N G-SPSE MSKE R 8 SE NEBE AU XL R HM CRENMRE NCBE 1/2 MUNDDLSE -------------------------------- D-W-M-4 HPL XDRLX

Consider an American N by M crossword grid containing precisely B black squares. What is the largest number of words that this crossword can contain? (For those who don't know: Every letter in an American crossword is in an across word as well as in a down word. The number of words is equal to the number of clues in the crossword puzzle, so words that just happen to be imbedded in clued words do not count.)

Well, I tried Plasmid's experiment. I placed a white piece of cardboard in such a way that its perpendicular pointed roughly at the sun. Then I took a foot-long stick, placed it on the cardboard and marked the cardboard at the end of the stick. Then I moved it about 10 feet away and its shadow still hit those marks, indicating that the shadow's length didn't change. I wasn't willing to get up a tree or on the roof to get a larger distance. So, I think d3k3 is right, and this is born out by my little experiment.

I just want to mention that some of your picks have scores which are only slightly less than the maximum. I was a bit nasty in picking the hands that I did. Most have at least one which is reasonably close to the maximum. I wouldn't expect intuition alone to do very well in this test. Some calculation would be necessary to tease out the close ones.

Indeed, my optimal draw would not be good with a poker machine because hands of the same type are considered equal. I did not design the problem for that situation. I was trying to model how you should draw in an actual poker game. Ignoring bluffing, my optimal draw would be what you would use in a real game of draw poker with several other opponents. This is true because of three things: 1. Hands of the same type have different rankings (which they don't have in a poker machine). 2. Before the draw, you only get to see your cards. So, as far as you can tell, the draw is random from the 47 other cards you don't see. 3. The payoff is not decided until after the draw, so it is a variable which must be neglected. The only reason I made it a one player thing is to remove any effects of human interactions, like bluffing. Other than that, my drawing function is exactly what you would want to use in a real game with other people.