# Hard 4x8 chocolate bar Riddle v2

This is inspired from this puzzle:

You have $$4$$x$$8$$ chocolate, you can cut only straight with the knife.

What is the least amount of cutting required to have $$32$$ pieces of 1x1 chocolates?

and

What if putting chocolates onto each other was not allowed?

• Just to clarify, when you say least amount of cutting, does that mean the total length of all cuts? – hexomino Sep 26 '18 at 8:57
• @hexomino total amount of cutting... not total length of cutting :) – Oray Sep 26 '18 at 8:59
• My answer doesn't involve putting them on top of each other, but rearranging the cut pieces next to each other to achieve the same effect of cutting through multiple separate pieces in a single cutting action. Does this count? – AHKieran Sep 26 '18 at 9:21
• Is some part of the puzzle missing? Now the first question has the same number of cuts as "cut any convex shape into 32 parts of any shape", and the second part doesn't change the puzzle at all. – Bass Sep 26 '18 at 10:20
• @AHKieran - If you think about it, the result is the same. If you have 2 * 1x4 bars, in both cases - placing them on top of each other, and placing them next to each other, one cut in the middle has the same results (separating both). – Battle Sep 26 '18 at 12:46

Total Cuts:

$$5$$

Method:

Cut in half vertically, creating 2, $$2\times8$$ pieces, then place these end to end to get a $$2\times16$$ piece. Cut again vertically through both pieces, to get 4, $$1\times8$$ pieces. Place these side by side to form a $$4\times8$$ piece again, this time, cut in half horizontally, and move the pieces to form an $$8\times4$$ shape, cut and move into $$16\times2$$, then cut once more and you have 32 individual pieces. This totals 5 cuts.

Animated:

Explanation:

This puzzle works because with each cut, we half the size of every piece. to work this out quickly, we could do $$\log_2(32) = 5$$.

• animation is so cool :) – Oray Sep 29 '18 at 12:34

It can be done in

5 cuts.

because

Simply imagine repeatedly folding it in half like a piece of paper until it is $$1\times1$$. Instead of folding, you cut and stack the pieces on top of each other. If you are not allowed to stack on top of each other for the cut, then you can put them next to each other instead.
$$4\times 8 = 2^2 \times 2^3$$ so it takes 5 cuts to reduce to $$2^0 \times 2^0$$.

It may be done in

5 cuts if you are allowed to stack the pieces on top of each other after each cut.

Counting on this

First cut → two 4x4 pieces
Second cut → four 2x4 pieces
Third cut → eight 2x2 pieces
Fourth cut → sixteen 1x2 pieces
Fifth cut → thirty-two 1x1 squares

• This answer can't be valid because the question stated What if putting chocolates onto each other was not allowed? – Draco18s no longer trusts SE Sep 26 '18 at 22:42
• @Draco18s, the question was updated to add that after this answer was posted. – Tom Sep 27 '18 at 8:31
• Ah, didn't notice that. – Draco18s no longer trusts SE Sep 27 '18 at 16:50

31 times

Because

you can either cut rows of 8 (3 cuts, you now have 4 rows of 8), then separate each row with 7 cuts each. $$3 + (4*7) = 31$$.
or cut vertically first seven times to create 8 columns of 4, then cut each 3 times. $$7 + (8*3) = 31$$.
changing between cutting horizontally and vertically each time will not help getting a lower amount of cuts. it always results in 31.

Unless..

we are allowed to cut 4 rows, put them together as if the bar were still whole, and cut 7 times vertically. In that case we get 11 cuts. I don't think this is allowed in this puzzle

After Q-Edit:

In this puzzle, we don't have to separate the pieces and cut each piece on its own. As others now have said, it is possible with 5 cuts.
What if putting chocolates onto each other was not allowed? - You don't even have to stack the halves on top of each other, just arrange them next to each other so you can cut each piece the same way with one cut.

• There are other possibilities. e.g. cut into two 2x8, then those into 2x2s, etc. – Jaap Scherphuis Sep 26 '18 at 9:02
• @JaapScherphuis see edit – kscherrer Sep 26 '18 at 9:03
• @Cashbee there is no restriction in the question. – Oray Sep 26 '18 at 9:05
• @Oray well yes there is. from the inspiration-puzzle: and every time we cut the chocolate we separate the pieces and cut each piece on its own. you did not say that this is not valid anymore for your puzzle. Anyway, I covered both possibilities – kscherrer Sep 26 '18 at 9:08
• To be fair, you had that exact sentence in your own question when I read it. I see it is now removed. Well played – kscherrer Sep 26 '18 at 9:13

5 cuts (or 10 if rearranging is not allowed as well)

because

1. Cut into 2 2x8 pieces.
2. Rearrange into 2x16 and cut into 2 1x16 (actually 4 1x8) pieces.
3. Rearrange into 4x8 and cut into 2 4x4 (actually 8 1x4) pieces.
4. Rearrange into 8x4 and cut into 2 8x2 (actually 16 1x2) pieces.
5. Rearrange into 16x2 and cut into 2 16x1 (actually 32 1x1) pieces.

and

If rearranging is not allowed, we can cut only 1 line at once, and you need to cut 10 lines (3 in the "north-south" and 7 in the "east-west" direction).

0

because

Chocolate is usually already divided into 1 x 1 blocks. You can easily crush it with your hand and you don't need any help of a knife.

I think that it is already divided into 32 blocks. Why would you provide dimensions 4 x 8 otherwise? Not 2 x 4 nor 1 x 2?

• The title says that the bar is hard, so you probably cannot crush it with hands :) and seriously, the puzzle is not about it, it asks how many times you should divide the 4x8 bar (cutting, crushing - it doesn't matter) to get 32 "unit" blocks. – trolley813 Sep 26 '18 at 12:07

There is another way that hasn't been mentioned.

The puzzle says that you have to cut straight with the knife, not that the cut has to be straight. So if you move the chocolate while cutting you can cut it into 32 pieces in one cut.

• nice idea, but is would have to be a very long and possibly curved knife – tom Sep 27 '18 at 9:32