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| User | Post | ||||||||||||||||||||||
| Squash Monster Posts: 18/296 |
||bass wanted a strategy that would be gaurinteed. He didn't care which player would be the one able to use it.
But it was easilly possible to misread as wanting a strategy that would work for both players. Just wanted to save you two pages of back and forth confusion. | ||||||||||||||||||||||
| Danielle Posts: 316/6737 |
Well that's what I said jesus christ >8( | ||||||||||||||||||||||
| Alastor Posts: 287/8204 |
Which is why YOU GO FIRST. If you read it, it doesn't HAVE to work for both players. "Find a strategy guarinteed to win. (It may be for either player.)" | ||||||||||||||||||||||
| Danielle Posts: 312/6737 |
Well if they go first.. isn't there always one of each letter available to them no mattter what you do? | ||||||||||||||||||||||
| Xkeeper Posts: 254/5653 |
Originally posted by Danielle It just says to pick one and run with it; you aren't forced to use a player that doesn't go first. | ||||||||||||||||||||||
| Danielle Posts: 308/6737 |
How can either player win? I don't see how you can use the same strategy if you don't go first. | ||||||||||||||||||||||
| ||bass Posts: 10/594 |
You people need to learn how to read. The question explicitly states that the strategy may be for either player. In this case, it was the first player. | ||||||||||||||||||||||
| NSNick Posts: 108/2228 |
A ha. Nice. | ||||||||||||||||||||||
| Danielle Posts: 285/6737 |
Originally posted by mcw That's the catch. You have to go first.  | ||||||||||||||||||||||
| emcee Posts: 33/867 |
EDIT: Ah, crap, I spent so long trying fix the color you posted the answer while I was typing. I thought I needed closing tags.
Ok, I get it. Taking your example: 1 8 5 4 7 9 6 3 2 0 Add up ever other number starting with the first one. 1+5+7+6+2=21 Now starting with the second: 8+4+9+3+0=24 The total of these is more, so those are the ones you want. Here they are color coded: 1854796320 You want the blue ones, so you take the zero: 185479632 They now have to take a red one. It doesn't matter which they take, but lets say they take the 2: 18547963 You of course take the blue: 1854796 They must take a red: 185479 You take a blue: 18547 They take a red: 1854 You a blue: 185 A red for them: 18 You take the last blue and they're left with the red, as long as you get all blues you win because they add up to more. For clarification, the color were for demostration, the "blues" are simply in the group that adds up to the most, and the "reds" are the others. jeff |
Posts: 11/213 what if he picks a in there and you have to be b? =p | ||bass |
Posts: 9/594 
Begin by seperating the series into sets A and B, alternating as shown. Sum the values of all the integers in each set. In this example, A is 25 and B is 20. Clearly you want to have the elements in set A. Draw from set A (the left). Your opponant is now forced to draw from set B. When your opponant draws, he opens up another element of set A for you to draw. Draw the open element from set A. Repeat this process until all elements have been drawn. This method is guarinteed to win 100% of the time for all arrangements of pegs. Deleted User |
Posts: 138/-7750 I can't wait to hear the answer on this, ||bass has me all excited. D: | ||bass |
Posts: 8/594 Hint 2: The "correct" soloution looks at all the pegs before the game starts, and then ignores them for the rest of the game. | Kattwah |
Posts: 287/3349 I guess its time for me to throw out a guess here...
| Take only the pegs with odd values? Thats a simple solution that works out either way (However, it isnt 100% if they take an odd number.) Heres a question, does it matter who goes first or not? Anyways, explaining my theory with a random example 1 8 5 4 7 9 6 3 2 0 You take 1, he takes 8 (1; 8) You take 5, he takes 4 (6; 12) You take 7, he takes 2 (13; 14) You take 9, he takes 6 (22; 20) You take 3, he takes 0 (25; 20) You win. However, here is a problem that presents itself with a "Not odd every turn" situation... 1 8 6 4 7 9 5 3 2 0 So... I just want to know if Im close or not right now  Light PKMN |
Posts: 14/51 left peg right peg left peg.....ect
| But im wrong ||bass |
Posts: 7/594 Hint time: The correct answer will work for any even number of pegs. | ||bass |
Posts: 6/594 Your algo plays a SMART strategy but not an unbeatable one. One could construct very specific arrangements of pegs that would defeat this algorithim. There is an answer that works for 100% of all possible arrangements of pegs and it's a MUCH simpler algo, you could do it in under 10 lines easily. | NSNick |
Posts: 97/2228 For some reason, I couldn't think in straight algebra, so here's a C code thing I think should work:
| function char Which_To_Choose ( int a, int b, int c, int d, ) /*! requires [a, b, c, and d are the numbers of the pegs, where a is the leftmost, b is the second to left, c is the second to right, and d is the rightmost] !*/ { object int ares, dres, hidb, hiac; if (d > b){ hidb = d; } else{ hidb = b; } if (a > c){ hiac = a; } else{ hiac = c; } takea = a - hidb; taked = d - hiac; if (takea > taked){ return 'a'; } return 'd'; ||bass |
Posts: 5/594 People keep asking me how this is possible if you can't see the values of the pegs. You CAN see them. Read the damn question people. I NEVER said you can't see the values of the pegs. You can. | This is a long thread. Click here to view it. | | ||
