4
$\begingroup$

This question is follow up to Something is wrong with the equation

This time our numbers are given as below included each numbers on a keyboard order and some basic math operators next to each other once.

enter image description here

By just swapping two squares at a time, find the equality with the least amount of swapping.

Note that when you put your result to Wolfram Alpha, it should say "True" as a result. so PEMDAS is necessary.

$\endgroup$
4
$\begingroup$

I can do it in

5 exchanges

The swaps are:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
1 2 3 4 5 6 7 8 ^ 0 + - * / ! 9 = (Swapped 9 and ^)
1 2 3 4 5 6 7 8 ^ 0 + - * / ! = 9 (Swapped 9 and =)
6 2 3 4 5 1 7 8 ^ 0 + - * / ! = 9 (Swapped 6 and 1)
6 / 3 4 5 1 7 8 ^ 0 + - * 2 ! = 9 (Swapped 2 and /)
6 / 3 - 5 1 7 8 ^ 0 + 4 * 2 ! = 9 (Swapped 4 and -)
To get 2 - 1 + 8 = 9

Alternate answer in the same number of swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
1 7 3 4 5 6 2 8 9 0 + - * / ! ^ = (Swapped 2 and 7)
1 7 - 4 5 6 2 8 9 0 + 3 * / ! ^ = (Swapped 3 and -)
1 7 - 4 5 + 2 8 9 0 6 3 * / ! ^ = (Swapped 6 and +)
1 7 - 4 5 + 2 8 = 0 6 3 * / ! ^ 9 (Swapped 9 and =)
1 7 - 4 5 + 2 8 = 0 / 3 * 6 ! ^ 9 (Swapped 6 and /)
To get 17 - 45 + 28 = 0

Third answer using the same number of swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
1 9 3 4 5 6 7 8 2 0 + - * / ! ^ = (Swapped 2 and 9)
1 9 3 4 5 6 7 8 * 0 + - 2 / ! ^ = (Swapped 2 and *)
1 9 3 4 5 6 7 8 * 0 + - ! / 2 ^ = (Swapped 2 and !)
1 9 = 4 5 6 7 8 * 0 + - ! / 2 ^ 3 (Swapped 3 and =)
1 9 = 4 - 6 7 8 * 0 + 5 ! / 2 ^ 3 (Swapped 5 and -)
to get 19 = 4 - 0 + 120 / 8

Fourth answer using the same number of swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
4 2 3 1 5 6 7 8 9 0 + - * / ! ^ = (Swapped 1 and 4)
4 2 3 / 5 6 7 8 9 0 + - * 1 ! ^ = (Swapped 1 and /)
4 2 0 / 5 6 7 8 9 3 + - * 1 ! ^ = (Swapped 0 and 3)
4 2 0 / 5 = 7 8 9 3 + - * 1 ! ^ 6 (Swapped 0 and 3)
4 2 0 / 5 = 7 8 - 3 + 9 * 1 ! ^ 6 (Swapped - and 9)
To get 84 = 84

Fifth answer using the same number of swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
8 2 3 4 5 6 7 1 9 0 + - * / ! ^ = (Swapped 1 and 8)
8 2 3 4 5 6 7 + 9 0 1 - * / ! ^ = (Swapped 1 and +)
8 2 3 4 5 6 7 + 9 0 / - * 1 ! ^ = (Swapped 1 and /)
8 2 - 4 5 6 7 + 9 0 / 3 * 1 ! ^ = (Swapped 3 and -)
8 2 - 4 5 = 7 + 9 0 / 3 * 1 ! ^ 6 (Swapped 6 and =)
to get 37 = 37.

A quite amazing sixth answer using the same number of swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
1 6 3 4 5 2 7 8 9 0 + - * / ! ^ = (Swapped 6 and 2)
1 6 ^ 4 5 2 7 8 9 0 + - * / ! 3 = (Swapped 3 and ^)
1 6 ^ 4 / 2 7 8 9 0 + - * 5 ! 3 = (Swapped 5 and /)
1 6 ^ 4 / 2 - 8 9 0 + 7 * 5 ! 3 = (Swapped 7 and -)
1 6 ^ 4 / 2 - 8 = 0 + 7 * 5 ! 3 9 (Swapped 9 and =)
to arrive at 32760 = 32760. (Entering "16^4/2-8=0+7*5!39" gives "True" on Wolfram Alpha)

————————————————————————————————————————————

Edit: So I realized that the question statement says that

Note that when you put your result to Wolfram Alpha, it should say "True" as a result.

Hence we can cheese the question and use just:

3 exchanges

Via the following swaps:

1 2 3 4 5 6 7 8 9 0 + - * / ! ^ = (Original)
1 2 3 ! 5 6 7 8 9 0 + - * / 4 ^ = (Swapped 4 and !)
1 2 3 ! = 6 7 8 9 0 + - * / 4 ^ 5 (Swapped 5 and =)
1 2 3 ! = 6 7 * 9 0 + - 8 / 4 ^ 5 (Swapped 8 and *)
if you enter "123!=67*90+-8/4^5" on Wolfram Alpha it will return "True" as 123 != 6029.9921875 (Though it is noted that the question does mention "equality", hence this funny answer isn't valid)

$\endgroup$
  • $\begingroup$ :) pretty good answers $\endgroup$ – Oray Sep 7 '18 at 16:06
  • $\begingroup$ @Oray is it optimal though? $\endgroup$ – lovemathboy Sep 7 '18 at 16:06
  • $\begingroup$ i found it 5 swaps as well, not sure 4 is possible, i will wait a day, then accept your answer $\endgroup$ – Oray Sep 7 '18 at 19:34
2
$\begingroup$

A possible solution using 13 swaps:

1 x 3 ! + 9 - 2 ^ 0 = 6 5 8 / 4 7

Which boils down to

14 = 14

Swaps:

1 2 3 4 5 6 7 8 9 0 + - x / ! ^ =
1 x 3 4 5 6 7 8 9 0 + - 2 / ! ^ =
1 x 3 ! 5 6 7 8 9 0 + - 2 / 4 ^ =
1 x 3 ! + 6 7 8 9 0 5 - 2 / 4 ^ =
1 x 3 ! + 9 7 8 6 0 5 - 2 / 4 ^ =
1 x 3 ! + 9 - 8 6 0 5 7 2 / 4 ^ =
1 x 3 ! + 9 - 2 6 0 5 7 8 / 4 ^ =
1 x 3 ! + 9 - 2 ^ 0 5 7 8 / 4 6 =
1 x 3 ! + 9 - 2 ^ 0 = 7 8 / 4 6 5
1 x 3 ! + 9 - 2 ^ 0 = 6 8 / 4 7 5
1 x 3 ! + 9 - 2 ^ 0 = 6 5 / 4 7 8
1 x 3 ! + 9 - 2 ^ 0 = 6 5 8 4 7 /
1 x 3 ! + 9 - 2 ^ 0 = 6 5 8 / 7 4
1 x 3 ! + 9 - 2 ^ 0 = 6 5 8 / 4 7

I'm sure there's a solution using less swaps, so consider this the benchmark to break.

$\endgroup$
1
$\begingroup$

I think I can do it in

$9$ exchanges

Swaps

1 2 3 4 5 6 7 8 9 0 + - x / ! ^ =
1 ^ 3 4 5 6 7 8 9 0 + - x / ! 2 =
1 ^ 3 ! 5 6 7 8 9 0 + - x / 4 2 =
1 ^ 3 ! - 6 7 8 9 0 + 5 x / 4 2 =
1 ^ 3 ! - 6 5 8 9 0 + 7 x / 4 2 =
1 ^ 3 ! - 6 5 8 x 0 + 7 9 / 4 2 =
1 ^ 3 ! - 6 5 8 x 0 + 2 9 / 4 7 =
1 ^ 3 ! - 6 5 8 x 0 + 2 7 / 4 9 =
1 ^ 3 ! - 6 5 8 x 0 + 2 7 / 9 4 =
1 ^ 3 ! - 6 5 8 x 0 + 2 7 / 9 = 4

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.