How hard is this MAZE?
958990
Geeze.....it scared the crap outta me! good thing i peed before i came back from lunch! lol
212031
Well, the question doesn't have all the required details, but why not:\n- make sure that you ar eusing the correct type of vaiable (string/number/etc)\n- make a simple if statement that says, IF the number of tries is less than 10 then continue (where number of tries is lets say "x")\nOnce an answer is chosen, if the answer is corret, then thats it, ELSE, if x < 10 then x = x + 1 \n(this makes sure that x is less than 10...and adds 1)\nIF x = 10 THEN MsgBox "HARD LUCK" (like that, if the previous number was 9, by adding 1 it became 10...this line finishes the game up)\n\nAt the end, make sur eto reset the value of "x"\nx = 0
807448
that might be a bit difficult.....
23859
I run through the numbers from 1-9, looking at each to see where the constraints ensure that there is only one spot left to place an additional number.  To do this, realize that each row, each column and each block need exactly one of each number, so starting with 1, I look for combos with limited possibilities.\n\nIf the blocks 1,2 have a 1 in them, then 3 should be easy to find, look at blocks 6,9 to see if they have a 1, and mentally cross out the rows and columns that those 1's eliminate and see whats left.  Similary, if 2 and 6 have 1's in them try looking at blocks 5 and 3 for enough eliminated squares to ensure there is only one remaining spot for 1 in those blocks. Once you've tried all the possibilities for 1, move on to 2, and keep going until you've been through the whole list of numbers once without finding any new entries.\n\nFinally, a trick I really like, is when a block has only one row or one column with space remaining. You know that the remaining numbers in this mostly complete block will be in that row or column, and so you can use that information and apply it to other blocks even though you could not exactly place the numbers in the mostly complete block.\n\nHope that helps!
582369
its only hard if you make it hard\n\nbe prepared to be very good in math and physics..
240341
Not that hard for most games.
579770
hi\nhere is a link to help u\ncheers\n\nhttp://www.harperchildrens.com/hch/picture/features/stuartlittle/maze.asp
1012939
the brain tree thing is too tough... how do ppl do it??
226806
Go to Gamewinners (link below), or a similar "game cheat" web site for hints on winning your cases. \nEach case will probably be just as, if not more hard for you to solve. They tried to make the game as realistic as possible, that's why it seems so hard.
779006
When you say "you may not retrace your steps" can you clarify that?\n\nDo you assume that the person is always trying to advance toward the destination?  So once I go one block east, I would never go back west, even if it was a different block?\n\nFor example, a possible route is to go north for 10 blocks, go east for 1 block, south for 10 blocks, east for 1 block, north for 10 blocks, etc.  It would seem this would sort of count as "retracing" so I want to make sure.\n\nIf it is finding the number of ways assuming you can only go north and east, then we can find an answer relatively easily.  If you are allowed to meander all over the map, but just not cross a prior path, it is much more difficult to answer...\n\nAssuming it is the former case, you can solve this using Pascal's triangle.\n\nPlace a 1 at your starting point (there is only one way to get there)\nNow put a 1 at a spot 1 block north, and a 1 at a spot 1 block east.  (Again, there is only one way to get to each of these points if we always want to advance).\nThe next spots are 2 blocks north (1 way), 1 block north-east (2 ways), 2 blocks east (1 way).\n\nNotice how the pattern corresponds to Pascal's triangle turned on its side with each intersection being the sum of the ways to get to the prior corners:\n1 \n1 ..7\n1 ..6 21\n1 ..5 15 35\n1 ..4 10 20 35\n1 ..3 ..6 10 15 21\n1 ..2 ..3 ..4 ..5 ..6 ..7\n1 ..1 ..1 ..1 ..1 ..1 ..1 ..1\n\nRather than filling out the whole grid, it helps to remember that each spot in Pascal's triangle is related to "n choose k" which can also be written C(n, k).\n\nC(0, 0) = 1\nC(1, 0) = 1, C(1, 1) = 1\nC(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1\nC(3, 0) = 1, C(3, 1) = 3, C(3, 2) = 3, C(3, 3) = 1\netc.\n\nSo the grid looks like:\n...\nC(5,5) ...\nC(4,4) C(5,4) ...\nC(3,3) C(4,3) C(5,3) ...\nC(2,2) C(3,2) C(4,2) C(5,2) ...\nC(1,1) C(2,1) C(3,1) C(4,1) C(5,1) ...\nC(0,0) C(1,0) C(2,0) C(3,0) C(4,0) C(5,0) ...\n\nLook along the diagonal:\nBlock 1: C(0,0)\nBlock 2: C(2,1)\nBlock 3: C(4,2)\nBlock 4: C(6,3)\nBlock 5: C(8,4)\n...\nBlock 9: C(16,8)\nBlock 10: C(18,9)\n\nSo the value corresponding to our destination corner will be C(18,9)\n\nThe formula for "n choose k" is:\nC(n, k) = n! / (n-k)! k!\n\nPlugging in our values of n = 18, k = 9:\nC(18, 9) = 18! / (18-9)! 9!\n\nIf you don't have a calculator or other way to figure this out, you can do this manually, just cancel out factors as you go:\nC(18, 9) = (18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10) / (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)\nC(18, 9) = (2 x 17 x 2 x 15 x 2 x 13 x 2 x 11 x 2) / (4 x 3 x 2 x 1)\nC(18, 9) = 17 x 5 x 13 x 2 x 11 x 2\nC(18, 9) = 48,620 ways\n\nYou could also use Google Calculator by typing '18 choose 9' into the search box.\n\nThe answer is there are 48,620 ways assuming you always make progress to the destination and never backtrack or sidetrack.
758440
Wow... this one is tough! LOL!   ;-) \n\nGWB
302564
I remember playing that game, it was on the Microsoft Encarta 98 Encyclopedia CD. Not sure if you can still get hold of a copy. I have a vague memory the game itself might have been called "Mind Maze". Hope that helps!
430201
Its tricky work, i need digrams to explain to you so your better just following this link, I think it will help you out.
822022
no its a screensaver meant to look like a maze \nu can buy maze games tho....
970581
Difficult ones...dont cheat!...;
104358
MAZE NAVIGATION ALGORITHM\n\nPart 1\n\n    A guaranteed way to to solve a maze is to stick to a wall and always try to turn one direction or the other.  It is not the most efficient way but it works all the time.  This is how the Boe-Bot solves mazes.  It hugs the left wall.  IR sensors are mounted on the left and front sides of the robot.  It begins the maze with a forward movement.  After that it checks the left IR.  If left is open, it turns left, moves forward, and repeats the process.  If left is closed, the robot checks the front.  If front is open, the robot moves forward and repeats the process.  If the front is also closed, the  robot turns right.  It then checks the front IR sensor to determine if the new front is open.  If its open, the robot moves forward and repeats the process.  If it is closed, the robot again turns right and moves forward (the way it just came) and repeats the process.  The flow chart summarizes all these instructions.  Each time the robot makes a movement (forward, left, or right) it records a F,L, or R to a memory block in the EEPROM by using the WRITE function in Stamp BASIC.  The variable 'total_moves' is incremented each time a movement is written.  \n\n    The forward, left, and right subroutines use the PULSOUT function to move the servos.  The pulsout variables for each servo (eg. 699, 879) are not set in stone.  They have been changed numerous times and will probably need to be played with more.   We determined them partly by experiment and largely by trial and error.  The results of those experiments are also published in this site.  \n\n    At the beginning of the forward subroutine is the BUTTON function.  Each time through this subroutine, the programs checks to see if the button has been depressed.  If it has not, the program continues.  If the button has been pressed, control is moved to the find_RR subroutine.\n\nPart 2\n\n    When the program enters part 2, the robot has already left the maze and the button has been pressed.  It enters the find_RR subroutine and first pauses for 10 seconds so the robot can be repositioned at the starting point to re-do the maze.  Find_RR is a superfluous piece of code that originated from poor planning.  It reads in the movement from each memory block and the following memory block (blocks n and n+1) and compares them to each other.  If it detects a R,R combo, it replaces the first R with an A.  It then copies each movement into the previous one.  For example, it would change a FFRRFLF to a FFAFLF.  Each time it finds an RR combo and replaces it with an A, all the code following the last R must be adjusted. Therefore, the 'total_moves' variable is reduced by 1 each time RR is detected.  The write_A subroutine handles that task.  If no RR combo is found (they've all been switched) control moves to the find_A subroutine.\n\n    Find_A searches each memory block for the 'A' movement.  If no 'A' is found, control moves to the done subroutine.  However, this will only occur after all the 'A' have been dealt with.  When an 'A' is found, the memory location is stored in variable 'counter' and control goes to subroutine surround_A.  The purpose of surround_A is to determine all of the movements before and after the 'A' that need to be ignored for successful completion of the maze.  \n\n    After the robot completes the maze for the first time, and changes the 'RR' to 'A' the code will something like:\n\n        F F F L F A F R F F L F  this is for the hypothetical situation shown here:\n\n\n\n To eliminate the unwanted code, the pairs of movements surrounding A are examined and compared to the green, blue and purple patterns.\n\nF                F                F            L                F            A                F                R                F                F            L            F\n\n0                1                 2            3                4
274789
It depends how good you're at foreign languages. Although, as for the average person--I'd say it's pretty hard.