Author Topic: Camp Ivanhoe  (Read 5878 times)

Anthony Adams

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Re: Camp Ivanhoe
« Reply #375 on: September 09, 2021, 11:42:42 am »
The next highest is 99985, then 99983, 99976, 99972


99976, might be promising for disguising?

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #376 on: September 09, 2021, 11:43:38 am »
Can we do 88888?

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #377 on: September 09, 2021, 11:44:24 am »
It's not strictly the highest but it also goes along with the earlier 4s strategy

Anthony Adams

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Re: Camp Ivanhoe
« Reply #378 on: September 09, 2021, 11:44:32 am »
4+2+3+1+1 is 11

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #379 on: September 09, 2021, 11:45:22 am »
I have not wrapped my head around the whole clicks on wheel counting method huh

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #380 on: September 09, 2021, 11:45:43 am »
In that case I endorse 99976

Oracle

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Re: Camp Ivanhoe
« Reply #381 on: September 09, 2021, 11:47:16 am »
The next highest is 99985, then 99983, 99976, 99972


99976, might be promising for disguising?

I still think we should avoid giving away the pair of 9s if at all possible.

26599

89098
(4)(3)(5)(0)(1) = 13 clicks.
79097
(5)(3)(5)(0)(2) = 15 clicks

Though that is no where near the highest. a palindrome would be tricky to undo though.

Anthony Adams

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Re: Camp Ivanhoe
« Reply #382 on: September 09, 2021, 11:48:45 am »
80000 gets 15 with 4+4+5+1+1?

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #383 on: September 09, 2021, 11:49:28 am »
The numbers we put in our code aren't going to be what leads them to the solution though, they'll do that through the logic puzzle once they start entering guesses and getting data on X correct numbers with Y in the correct position right?  So it shouldn't be giving anything away for us to have a series of 9s in our shuffle.

Oracle

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Re: Camp Ivanhoe
« Reply #384 on: September 09, 2021, 11:49:36 am »
80000 gets 15 with 4+4+5+1+1?
The sum of 80000 is 8 though which is a terrible tiebreaker.

I also suggested 99866 earlier which could help hide the pair of 9s at the end potentially, but also if they figure it out could give it away.

Anthony Adams

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Re: Camp Ivanhoe
« Reply #385 on: September 09, 2021, 11:50:19 am »
87654 as well with 4+1+1+4+5?

Herman Blume

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Re: Camp Ivanhoe
« Reply #386 on: September 09, 2021, 11:50:26 am »
I've only seen this challenge in the last 5 minutes so I am trying to binge the last few pages. In general I like the idea of duplicating at least one number in our submission.

Anthony Adams

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Re: Camp Ivanhoe
« Reply #387 on: September 09, 2021, 11:50:56 am »
Oh, I thought we were doing it as 1 5 digit number for the tie-breaker

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #388 on: September 09, 2021, 11:51:04 am »
Neither team is going to get info for correct numbers or correct positions out of the shuffle the other tribe creates unless they enter that shuffle as a guess, which wouldn't be an efficient use of a guess since that is absolutely not a possible solution based on the 15 click shuffle requirement

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #389 on: September 09, 2021, 11:51:46 am »
Yeah tie breaker is sum of digits.  9 + 9 + 9 + 7 + 6 would be 40 which is pretty solid.

Anthony Adams

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Re: Camp Ivanhoe
« Reply #390 on: September 09, 2021, 11:52:14 am »
Neither team is going to get info for correct numbers or correct positions out of the shuffle the other tribe creates unless they enter that shuffle as a guess, which wouldn't be an efficient use of a guess since that is absolutely not a possible solution based on the 15 click shuffle requirement

Why not, it's not like we're gonna get it on our first guess? :I

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #391 on: September 09, 2021, 11:53:36 am »
First guess should be something that is valid 15 clicks away from whatever scramble they give us, no?

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #392 on: September 09, 2021, 11:54:04 am »
Anyway we're getting ahead of ourselves talking about solving strategies when we still need a scramble

Anthony Adams

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Re: Camp Ivanhoe
« Reply #393 on: September 09, 2021, 11:55:30 am »
First guess should be something that is valid 15 clicks away from whatever scramble they give us, no?

I think it should be more information based than guess based since we have nothing previous to go on?

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #394 on: September 09, 2021, 11:56:09 am »
Well it'll be a combination of both.  It'll be Mastermind strategy, but with a side of remembering this curious requirement about the clicks in the decoy we're given.

Young Writer

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Re: Camp Ivanhoe
« Reply #395 on: September 09, 2021, 11:56:19 am »
First guess should be something that is valid 15 clicks away from whatever scramble they give us, no?

Technically yes, but we'd need a second guess with all the numbers not in that first guess also 15 away which is...difficult.

Oracle

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Re: Camp Ivanhoe
« Reply #396 on: September 09, 2021, 11:57:40 am »
Okay so. What I've seen so far in suggestions
Number & Tiebreaker amount
99949 or 99994 = 40
99976 = 40
99866 = 38
79097 = 32
44444 = 20

Have there been any other suggestions or ideas I've missed?

Young Writer

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Re: Camp Ivanhoe
« Reply #397 on: September 09, 2021, 11:58:52 am »
What if we motivate ourselves to do better by going for the lowest number we can possibly make within 15 clicks.

Richie Tenenbaum

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Re: Camp Ivanhoe
« Reply #398 on: September 09, 2021, 11:59:08 am »
First guess should be something that is valid 15 clicks away from whatever scramble they give us, no?

Technically yes, but we'd need a second guess with all the numbers not in that first guess also 15 away which is...difficult.
If we get >1 correct number in correct position off of our first guess it's better to iterate and lock in those correct numbers as fast as possible so we have fewer permutations to solve in the other positions

See now the Mastermind stuff is coming back to me, ah nostalgia

Sam Shakusky

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Re: Camp Ivanhoe
« Reply #399 on: September 09, 2021, 12:00:02 pm »
My problem with 99994 is that they may correctly deduce that that was the highest sum of digits we could reach, whereas 44444 doesn't really give any deductions and it's highly unlikely for the tiebreaker to come into play.

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For our guessing, the strategy I've come up with is:

First guess; +1 to all digits
(For example, if we receive 37966 guess 48077)
Then -1 to all digits
+2 to all digits
-2 to a digits
+0 to all digits
+3 to all digits
+4 to all digits
+5 to all digits

Because we know that the differences must sum to 15 we may be able to skip one or more of those. What this does is gets us to at most 120 combinations left after at most 8 guesses, because we will know the amount slots need to change by, we just need to arrange these changes in the correct order.

Suppose our changes our A, B, C, D, and E; Our next guess would apply them in order ABCDE

If we have 5 match we're done (9 total).
4 match is impossible.
If we have 3 match it takes at most an additional 5 guesses (14 total)
If we get 2 match it takes at most an additional 6 or 8 (idk which) guesses (15 or 17 total)
If we get 1 match it at most an additional 8 guesses. (17 total)
If we get 0 matches we than submit BCDEA and go to that many matches (18 total)


Which means unless I messed up the numbers or misunderstood something it will take at most 18 guesses.