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November 29

Escape room balance puzzle

The four colours represent integer weights, not necessarily distinct, from 0 to 9 inclusive. What's a simple way to deduce the weight of each colour without trial-and-error or backtracking? Thanks, cmɢʟee⎆τaʟκ 15:23, 29 November 2021 (UTC)[reply]

If I'm interpreting the diagram correctly, we get the following relations: b < g, y < g, 2r < y, b+g = y+2r, y+g = 3b. Looking at the second equation, since y and g are distinct (by the second inequality), their sum is at most 17. So b is at most 5.

Combining the first equation with the second inequality gives b < 2r.

Subtracting the two equations and simplifying gives y+r=2b. Combining with the third inequality gives 3r < 2b.

So our possibilities for (b, 2b) are (0, 0), (1, 2), (2, 4), (3, 6), (4, 8) or (5, 10). Only (5, 10) contains both a multiple of 2 and a multiple of 3, so b = 5, and r = 3. Plugging this into the earlier equations gives y = 7 and g = 8.--108.36.85.111 (talk) 16:41, 29 November 2021 (UTC)[reply]

You can approach such problems more systematically using integer programming techniques. Combining the two equations gives g = b + r. Also, because the values are integers, b + 1 ≤ g, y + 1 ≤ g, 2r + 1 ≤ y. This gives

0=−b+g−r

0=−3b+g+y

1=−b+g−s₁

1=g−y-s₂

1=y−2r-s₃

where all variables are ≥0. Solving for b, g, y, r, s₁ in terms of s₂ and s₃ gives

b=5+3s₂+2s₃

g=8+5s₂+3s₃

y=7+4s₂+3s₃

r=3+2s₂+s₃

s₁=2+2s₂+s₃

This gives all non-negative integers solutions as s₂ and s₃ range over the non-negative integers. (The magic is deciding which variables to solve for. There are 7 choose 2 = 21 possibilities but the simplex algorithm can be used to avoid trial and error.) From the second equation it's clear that the only solution with g<10 has s₂ = s₃ = 0, making the only solution b=5, g=8, y=7, r=3. --RDBury (talk) 22:23, 29 November 2021 (UTC)[reply]

Thank you very much, RDBury and user at 108.36.85.111. Good insights, in particular that blue <= 5. Cheers, cmɢʟee⎆τaʟκ 00:27, 30 November 2021 (UTC)[reply]