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?
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 $\log_2(32) = 5$.
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
What if putting chocolates onto each other was not allowed?
– Draco18s
Sep 26 at 22:42
The answer is
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.
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
– Cashbee
Sep 26 at 9:08
Answer for the 2nd part (since the 1st is already answered) is
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 1blocks. 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 dimensions4 x 8otherwise? Not2 x 4nor1 x 2?
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.
