Vexa & Repin
Vexa
I was just comparing the spectral data of old oil pigments to the RGB values in early 8-bit games. Ever wonder how a 17th‑century Venetian would have mapped their palette into a 256‑color cartridge? I bet you could spot the exact hue shift if you overlay a palette from a 1700s canvas onto a NES sprite sheet. Want to try it? Also, quick puzzle: how many ways can you arrange eight distinct browns in a 3x3 grid so no row or column repeats a shade?
Repin
Ah, the folly of mixing a 17th‑century canvas with an 8‑bit cartridge. In Venice, the palette was built from real pigments—ochre, umber, lead‑white—none of which can be reduced to a single RGB triplet without loss of depth. If you insist on overlaying a 1700s color scheme onto a NES sprite sheet, just remember that every hue is a spectrum, not a single channel value.
As for the puzzle: you have eight distinct browns to place in nine cells, with no repeats in any row or column. Since each shade appears at most once, the only choice is which square remains empty. Pick one of the nine squares for the void, then arrange the eight colors in the remaining eight positions. That gives 9 × 8! = 362 880 arrangements. Simple, but a reminder that in the old masters’ world, the number of options is limited by the available pigments, not by a 256‑color palette.
Vexa
You’re right about the loss of nuance, but I’m not looking for an art history lecture. Focus: how many RGB values can map to a single Venetian ochre? Pick a target hue and list the RGB triples that fall within a 2‑point tolerance. Also, here’s a new challenge—can you find a 4×4 Latin square with the digits 1‑4 where no row or column contains the same number twice, but each 2×2 subgrid has a different set of numbers? Give me the solution or tell me you’re stuck.
Repin
Venetian ochre is usually around RGB (194,154,98). If you allow a tolerance of ±2 in each channel, every channel can be 192,193,194,195,or 196, and the same for the other two channels. That gives 5 × 5 × 5 = 125 possible RGB triples. A few examples that fall inside that band are: (192,152,96), (194,156,100), (196,154,98), (193,153,99), (195,155,97).
As for the 4 × 4 Latin square with the 2 × 2 sub‑blocks all different, I’m still trying to piece one together. I haven’t found a satisfactory example yet, so I’m stuck for now.
Vexa
125 triples, fine. That’s just a brute‑force list, no art insight. For the 4×4 Latin square with distinct 2×2 blocks, here’s one:
1 2 3 4
3 4 1 2
2 3 4 1
4 1 2 3
Check the 2×2 blocks: (1,2,3,4), (3,4,1,2), (2,3,4,1), (4,1,2,3)—all different. Try it out. If you can’t see why it works, I’ll point out the pattern.
Repin
The grid you gave is a perfectly valid 4×4 Latin square. Each row and column has the digits 1‑4 once. The 2×2 sub‑blocks are:
Top left: 1 2 / 3 4
Top right: 3 4 / 1 2
Bottom left: 2 3 / 4 1
Bottom right: 4 1 / 2 3
All four blocks contain each digit once, so they are distinct as sets. The pattern is just a cyclic shift by two positions each row, which guarantees the block condition. So you’re right; it works.
Vexa
Nice, you nailed it. Next: if you want to tweak it, shift the first row by one instead of two and see what breaks. That will force a conflict in the sub‑blocks—good exercise.
Repin
If you shift the first row only one step, the neat 2×2 pattern falls apart. The top‑left block will now contain 1,2,4,3 – you get two 3s in that block, and the block no longer has each digit exactly once. The same happens with the other blocks; they all end up repeating a number, so the distinct‑set requirement is broken.
Vexa
Right, a single shift breaks the sub‑block uniqueness. If you want a harder test, try constructing a 6×6 Latin square where every 3×3 block has a different multiset. It’ll expose the same symmetry issue.