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What is the minimum number of steps required to cut this chocolate bar which is a single piece into at least 30 single pieces?

1)Image for chocolate bar

Chocolate bar

2)Image for single piece

enter image description here

Edit:
Your knife is only as long as the width of a single piece and stacking (laying rows or columns on top of each other when you've cut those lose) is not allowed.

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  • $\begingroup$ I have edited it i am looking for at least 30 single pieces $\endgroup$
    – user2408578
    Apr 29, 2015 at 13:50
  • $\begingroup$ if we cut all the columns so you have 5 separate columns of 8, can we lay them next to each other to cut the rows, so cutting a row takes 1 cut instead of 5 seperate ones. $\endgroup$
    – Vincent
    Apr 29, 2015 at 13:55
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    $\begingroup$ OOooooooh I get it , lol I just noticed theres 40 pieces, and the question is how many cuts for 30. I just assumed right off the start there was 30 and never checked. lol. The world makes sense again, all is good $\endgroup$
    – Spacemonkey
    Apr 29, 2015 at 14:25
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    $\begingroup$ You should change the wording to not redefine "single piece". The sentence "cut this chocolate bar which is a single piece into at least 30 single pieces" is confusing. $\endgroup$
    – mbeckish
    Apr 29, 2015 at 16:47
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    $\begingroup$ Wait... are we supposed to ignore the fact that the width is longer than the height? $\endgroup$
    – kaine
    Apr 29, 2015 at 16:54

6 Answers 6

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My answer is

54, because almost every piece needs to be cut at least 2 times starting from second top piece and go around bar, 3 corner pieces will be ignored as they will be cut because of other pieces.

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The answer is:

54. my previous answer still hold but you have to make 5 cuts instead of 1 when trying to seperate a row. so 6 * 5 + 4 * 6 = 54.

I would say (after edit of the question this answer is no longer viable):

24 cuts first 6 cuts to get 6 seperate rows (you will keep 1 piece with 2 rows attached) then you have to cut each column which requires 4 cuts, so 4 * 6 = 24.

edited calculation error.

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  • $\begingroup$ i have edited it....but you answered it so fast...give it another try $\endgroup$
    – user2408578
    Apr 29, 2015 at 14:00
  • $\begingroup$ @user2408578 I've edited it, is this correct? $\endgroup$
    – Vincent
    Apr 29, 2015 at 14:06
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    $\begingroup$ If anything, wouldn't it be 54? And also, how is this better than the most basic way of cutting from the top? $\endgroup$
    – mdc32
    Apr 29, 2015 at 14:09
  • $\begingroup$ @mdc32 6 * 5 = 30 and 4 * 5 = 20, how's 30 + 20 = 54? Also you could do it cutting from the left top to the bottom (giving you 8 pieces after 15 cuts) this would be 57 cuts for 30 pieces) $\endgroup$
    – Vincent
    Apr 29, 2015 at 14:22
  • $\begingroup$ @mdc32 you were right, i've fixed my mistake, well spotted $\endgroup$
    – Vincent
    Apr 29, 2015 at 14:27
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All this thinking made me hungry so I figured I'd eat the chocolate pieces we didn't need. (sorry)

I ended up with

40
enter image description here

Disclaimer: This is not meant as a serious answer, it's meant to make you smile.

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  • $\begingroup$ It made me smile so +1, to be fair it did not say you could or could not eat some of the pieces ;) $\endgroup$
    – Vincent
    Apr 29, 2015 at 16:11
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"What is the minimum number of steps required to cut this chocolate bar which is a single piece into at least 30 single pieces?"

I hope this question isn't about word play, but until it's stated I'll keep this as an option.

My answer is:

3. There are only three steps required. 1. Cut a chocolate bar edge. 2. Check/Stop if we have 30 pieces. 3. Repeat previous steps.

Alternatively if you are some sort of savage, you can use just 1 step: Cut all the chocolate bar edges.

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  • $\begingroup$ This is the most interesting answer, I think. $\endgroup$
    – JLee
    Apr 29, 2015 at 16:44
  • $\begingroup$ Why not just: 1) Cut all chocolate bar edges. $\endgroup$
    – mbeckish
    Apr 30, 2015 at 14:39
  • $\begingroup$ @mbeckish, That's just being a savage! $\endgroup$
    – Mark N
    Apr 30, 2015 at 14:40
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The simplist solution is 54:

enter image description here

Note that there are 6 rows of 5 horizontal cuts and 4 rows of 6 vertical cuts.

We want to maximize the remaining number of uncut lengths. For an $X*Y$ rectangle, the number of uncut lengths is equal to $(4*X*Y-2*(Y+X))/2$. For $X*Y=10$ this becomes $20-10/X-X$. The maximum integer value for this is $X=3$ but Y must be an integer so this doesn't work. This best solution, therefore, is for $X=2$ and $Y=5$.

If we assume that the vertical cuts are $5/6$ the length of horizontal cuts, we can improve this even further. The image above would only use 50 cuts as the long vertical lengths need one less cut to complete them.

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Same answer as everyone else:

54

You can come to this conclusion without trying all of the different cutting patterns.

The total number of edges that connect one piece to another is 67. Minimizing the number of cuts is equivalent to maximizing the number of edges that remain uncut.

So the problem reduces to choosing 10 or less pieces that are part of your "remainder" such that their total number of uncut edges is maximized.

The most uncut edges I have been able to find amongst 10 or less pieces is:

13

which gives me the same result as everyone else.

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