Jotto Logic

In math today, we started a "year long tournament" for something called Jotto. The rules are simple:

1. Each team chooses a secret, 5 letter word. No repeating characters, no proper nouns, and it must be an English word.
   
2. Each team takes turns guessing a word. The other team tells them how many letters it shares with their secret word. ie: My word is blond. They guess other. I tell them 1, because both words have an o. Order does not matter.
   
3. Using logic and / or cheating methods, you need to guess the other team's word first.
   

I Googled a bit to find some strategy on how to play it, but was let down. I thought up a simple system of guessing a word then modifying it slightly to rule out letters. Vowels are the easiest, not sure where to go after that. Unfortunately, this was after I got home and we had gotten halfway through the game. I stored my thoughts in a text file for organization and make a branch of a word, it's mutations, and what each possible outcome of each mutation means for the letters.

"I" or "Impossible" means there is no way these letters can be in their word.
"K" or "Known" means I know these letters are in the word.
Everything else goes in limbo.

When two lines follow one number and a colon, it means one of those lines will happen but there is no way to know which. It's much cleaner than an "OR".

"AND" means both letters or both statements.
"OR" means one of the group, unless otherwise stated. It never means all statements can be true. 'This or that'.

 Spoiler

Code:

Impossible:
    s; t; i; c; k;
Known:
    
~~~~~~~~~~~~~~~~~
    
bland (2):
    blond
        1:    o = I AND a = K
        2:    o AND a = I
            o AND a = K
        3:    a = I AND o = K
    blind
        1:    i = I AND a = K
        2:    i AND a = I
    blend
        1:    e = I AND a = K
        2:
            e AND a = I
            e AND a = K
        3:    a = I AND e = K
    brand
        1:    r = I AND l = K
        2:     r AND l = I
            r AND l = K
        3:    l = I AND r = K
    blunt
        0:    (u AND t = I) AND (a AND d = K)
        1:    a AND d AND u = K
            ( u = I) AND (a OR d = K)
        2:    (a OR d = K) AND (b or l or n = K) AND (u = K)
            (a AND d AND u = I) AND (b or l or n = K)(2/3)
        3:    (a AND d = I) AND (u = K) AND (b OR n OR l = K)(2/3 must be K)

List of words and their number from class.

Code:

shirt    1
flirt    1
stick    0
bland    2
nodes    1
dance    2
dorky    1
barns    2
bored    3

I'm looking for some help. Is my logic off? Did I miss an absolute when looking over the words, or did I mistype something in my mutations tree? Is there a better way to go about this?

I don't want to spend an hour working out what words to do in what order if I have a giant error right here.

Code:

shirt    3
flirt    1
stick    0
bland    2
nodes    1
dance    2
dorky    1
barns    2
bored    3

Can SHIRT = 3 if STICK = 0? Does that mean H or R appear more than once?
Because then the result is a word containing HHRBE. wut?

That was a typo copying it down from my paper. It's suppose to be Shirt = 1. Nice catch.

reminds me of this game on Newgrounds. much a similar ruleset actually:
http://www.newgrounds.com/portal/view/427859

nvm

Getting a zero word is a good move.

If you can come up with 5 words of 5 letters each, all unique, you greatly narrow the possibilities. Since there are only 5 vowels though this is kinda tough.

Code:

hr    1
flr    1
bland    2
node    1
dane    2
dorky    1
barn    2
bored    3

Since the letters 'stick' don't appear, I've taken the liberty of removing them from the remaining words.

It's not a massive jump from here to assume that R is in the correct solution. flh are not as common letters, and well, if you're wrong, you just re-examine this assumption.
So taking R out, and subtracting 1 from each word containing it, you get:
s, t, i, c, k, f, l, h - eliminated
r - in the word

Code:

band    2
node    1
dane    2
doky    0
ban    1
boed    2

Then doky are obviously not in it either
eliminated: s, t, i, c, k, f, l, h, d, o, k, y
in the solution: r
(unknown: a, b, e, g, j, m, n, p, q, u, v, w, x, z)

Running another pass to remove doky,

Code:

ban    2
ne    1
ane    2
ban    1
be    2

There's a contradiction here: ban 2 - ban 1
Conclusion: R is not in the solution.

Dropping back a bit,
eliminated: s t i c k r

Code:

h    1
fl    1
bland    2
node    1
dane    2
doky    1
ban    2
boed    3

h is obviously in the solution, as is either f or l.
Another less immediate conclusion - ban has 2, bland has 2. So l and d are not in the solution. Going back...
eliminated: s t i c k r l d
in the solution: h, f

Code:

ban    2
noe    1
ane    2
oky    1
ban    2
boe    3

in the solution: h f b o e - again, not consistent.


I think you must still have something wrong with the original numbering, cause as it stands there is no solution, or I slipped up on my logic somewhere.

Is it possible that there isn't enough information yet to decode it?

My words and their numbers in the post match the paper, so either they screwed up or I wrote something down wrong. I wouldn't doubt either one to be honest. I'll check with my team to make sure the numbers are right Friday (next time the class meets), but until then I'd still like to work out an effective way to work through this so I don't have to make a new thread or post for every match up. It can't take 5 minutes of work or take 20 guesses.

5 words of 5 letters with no repeats given a 26 letter alphabet is going to be tricky, but I'll give it a shot.

You can rule out several letters by asking about "slick". This will also make your existing solutions much more useful.

With limited information like this, the best way to tackle it is to make an assumption, then test it.

EG.
Hypothesis: R is in the solution
Conclusion: bland (2) and barns (2) contradict one another

Hypothesis: R is not in the solution
Conclusion: Again, a contradiction is reached.



For an easy demonstration these clues are impossible:

Consider nodes (1) vs. bored (3) - once you eliminate the 's', due to 'stick', you're left with node (1) vs. bored (3).
In order to be true, B and R must both be in the solution. One of "ode" is in the solution, and 'n' is not in the solution.
Anyway,
S T I C K N - not in solution
B R - in solution

Code:

h    0
fl    0
land    1
node    1
dane    2
doky    0
an    0
oed    1

a, n, d, o, k, y, f, l, h - not in the solution
land - contains 1 letter
Contradiction!


Hypothetical error: nodes incorrect?

I used a python script to brute force this (21^5 is really easy for a computer to accomplish) and it came up with no possible solutions.

Well, I mean, I could have checked for 21 * 20 * 19 * 18 * 17, but that would have increased the amount of coding needed. So I just did 21 ^ 5 and wiped out all the impossible answers.

I didn't read the other posts in the thread.

Just based on the OP, it sounds like you need to determine which letters are in the word before you can do any meaningful guessing. Ideally, you get this done in at most 14 guesses, with each guess containing entirely different letters from each other at first, then cross-referencing from there on. Words like "stick" in your on-going session are extremely valuable, in that they tell you there's no relationship between your guess and the other team's secret word.

After that, is the hard/easy part. It can be hard if the other team chose a word with several possible permutations, but it can be easy if they did something stupid like "Zebra". At this point, the only information you get from each guess is that the word you chose isn't the other team's word.

This line of logic is meant for efficient yet as sure-fire as possible. It's most likely not the quickest method, as there's obviously better secret words to be used when considering letter combinations.

It's just like digit place. Or mastermind. Except not all combinations of 5 letters are valid. And no repeats (like digit place, unlike mastermind). Fascinating.

It's just like digit place. Or mastermind. Except not all combinations of 5 letters are valid. And no repeats (like digit place, unlike mastermind). Fascinating.

It also doesn't tell you if the letters are in the correct place. That's Mastermind factorial or something.

Depends how you play Mastermind, really. There are a few rulesets of different difficulty:

black and white pegs, black tells you the peg is correct, white tells you it's in the wrong spot.

Easy: pegs in the same position as they refer to.
Medium: pegs are in any position
Hard: only 1 type of peg (black/white) - that's what this game is.

Plus IIRC Mastermind only has 8-10 colours, this game has 26. Reducing that by limiting it to 1 of a colour just means 10^5 vs. 26!/21!, which is still much much larger.