You have the following 4x6 chocolate bar. The question is what is the least amount of cuts you have to do in order to create 24 pieces of 1x1 chocolates. the cuts can have any shape they want as long as they start from a point in the perimeter and end on another point of the perimeter. The line cant hit itself and every time we cut the chocolate we separate the pieces and cut each piece on its own.
3$\begingroup$ Does the statement "the line cant hit itself" imply that the line can't cross a point twice? Also, does "perimeter" also mean the edges of the shapes after they have been separated from the larger piece or just the 4x6 perimeter in the original shape? $\endgroup$– GraySep 26, 2018 at 19:32
The answer is that
you need 23 cuts, no matter how you do it.
Each cut adds one piece to the total number of pieces you have. You need to go from 1 piece to 24 pieces, so you need 23 cuts total.
If cuts are allowed to meet the perimeter between the start and end, as in Peregrine Rook's answer, the best you can do is
3. To do this, first cut into two spiral pieces as shown below, then deal with each piece with a single wiggly cut.
You can't do it with fewer.
The first cut can't completely cut off any square that's not on an edge, because it would have to hit itself to separate that piece. If the first cut leaves at least two larger pieces, you need to cut each of those at least once. So the only option is that you separate some single squares and one larger piece with the first cut. Since the first cut doesn't meet itself, you can't separate two adjacent single squares with it, and so the large piece must have all the interior plus some squares on the boundary. Now you have the same problem that some squares have adjacent squares on all four sides, and these can't be separated with a single, non-self-intersecting cut.
Of course, if cuts aren't allowed to hit the perimeter in between (and I suspect that is the intended interpretation), Deusovi's answer is correct.
1$\begingroup$ Nice answer, was just about to post a similar one, but with different cuts $\endgroup$– KrugaSep 26, 2018 at 7:57
$\begingroup$ @Kruga Thanks! Yours is a nice solution too - I wouldn't have expected it to be possible without the first cut leaving two large pieces. $\endgroup$ Sep 26, 2018 at 8:27
$\begingroup$ The way I understand it is that as soon as a piece is separated in two smaller pieces the cut ends. Otherwise you could do it with a single cut. $\endgroup$– Daniel PSep 26, 2018 at 12:12
1$\begingroup$ @DanielP no, you couldn't do it with one cut, since OP said the cut can't meet itself. As I said, Deusovi has given the solution for your interpretation of the rules. My answer solves Peregrine Rook's interpretation. $\endgroup$ Sep 26, 2018 at 12:35
1$\begingroup$ @DanielP if it doesn't meet itself then there will still be a thin section connecting the two "pieces". And see the comments under Peregrine's answer for discussion on interpreting the rules. As I said in my answer, I suspect yours is the intended interpretation. However, Peregrine's interpretation makes an interesting question, so I answered it. $\endgroup$ Sep 26, 2018 at 12:50
$\begingroup$ I think count is not correct. In Cut#2, as soon as you reach to the perimeter again, the cut is complete. So, picture Cut#2 is not just one but 4 cuts. $\endgroup$– PM.Sep 27, 2018 at 4:08
$\begingroup$ @PM. The problem statement is at best unclear: "the cuts can have any shape they want as long as they start from a point in the perimeter and end on another point of the perimeter." Nowhere does the spec say that the perimeter can't be visited en route to the final destination. $\endgroup$ Sep 28, 2018 at 19:36
$\begingroup$ By reading it two-three times again, I agree. You are right, the question does not say that what I had inferred. $\endgroup$– PM.Oct 3, 2018 at 23:35
$\begingroup$ This looks like a good method for the case when cuts can visit the perimeter but not run along it. $\endgroup$ Oct 10, 2018 at 7:41
I submit that I can do it in
Start from the 6×4 bar,
Now make these $2$ cuts:
You now have twelve 1×1 pieces and a 6×2 chunk (from the center of the 6×4 bar). That can trivially be cut into 1×1 pieces with $11$ cuts, so the entire job is done with $6+2+11=19$ cuts.
I’m sure that this can be improved further.
3$\begingroup$ I cant approve this as an answer because the second cut hits the perimeter 6 times instead of 2 as the rule said $\endgroup$ Sep 26, 2018 at 6:23
1$\begingroup$ The problem with this method is that you cant prove that the answer you have is the one with the least cuts possible, you can make guesses but not explain why your guess has the least cuts. I hope someone can prove me wrong on this comment $\endgroup$ Sep 26, 2018 at 6:50
2$\begingroup$ " the cuts can have any shape they want as long as they start from a point in the perimeter and end on another point of the perimeter." Although not very well written, that sentence can be interpreted to mean that any movement of the knife that starts on the perimeter and ends on another point on the perimeter is a cut. In that case, his method involves 23 cuts, not 19. $\endgroup$ Sep 26, 2018 at 6:58
1$\begingroup$ @stelioball Carrying the formalism to it's absurd xtreme, I can separate the whole bar into 24 pieces with a single cut. $\endgroup$ Sep 26, 2018 at 7:15
1$\begingroup$ Please note that I explicitly said that this wasn’t optimal; it was offered as a proof-of-concept that Deusovi’s answer was not optimal. $\endgroup$ Sep 26, 2018 at 15:54
My formalist solution, in which I take no pleasure.
Five cuts. I hope my color guide is clear.
1$\begingroup$ "the line cant hit itself" $\endgroup$ Sep 26, 2018 at 7:38
$\begingroup$ Oops, Back to the drawing board. $\endgroup$ Sep 26, 2018 at 7:58