## No Card Game I Know Of!

Posted by Karl Sharman on July 29, 2011 – 10:45 am

You have a hand of cards that depict each of the letters of the alphabet. You fan the cards and see they are in the following order:

F, W, X, Y, Z, K, L, A, T, U, J, N, O, S, D, E, P, V, C, B, G, H, I, Q, R, M

Your job is to put the cards in alphabetical order in the minimum number of moves. You may remove any number of consecutive cards and insert them anywhere else in the fan; this counts as one move.

Of course, you must preserve the order of the cards you are moving, and keep all cards facing the same way.

Describe how you rearrange the cards in this (example) format:

1. Move D-K to between S and T.

July 29th, 2011 at 12:42 pm

I was able to do it in 7 moves.

July 30th, 2011 at 6:01 am

That’s a lot of moves cazayoux – it can be done in less than that!

August 1st, 2011 at 11:33 am

I was able to do it in 6 moves various ways… however, still searching for an optimal way.

I have in my head how to map the possible moves in a manner that reveals the quickest route… just need to get it on the whiteboard.

fun one!

August 2nd, 2011 at 7:24 am

Some of the letters are already in a run, such as WXYZ.

I would consider this one segment.

In total, we begin with 16 segments

F,WXYZ,KL,A,TU,J,NO,S,DE,P,V,C,B,GHI,QR,M

The best case scenarios is if we move a series of segments and, after combining congruous segments, we can reduce the number of segments by three. Of the series of segments moving, each end would reduce the segment count. As well, the gap left behind by the series of segments closes would reduce the segment count as well.

For example, when we move J-P between IQ, then we have

F,WXYZ,KL,A,TU,V,C,B,GHI,J,NO,S,DE,P,QR,M

we can combine at UV, IJ, and PQ

F,WXYZ,KL,A,TUV,C,B,GHIJ,NO,S,DE,PQR,M

We have reduced the segment count by 3 to now have 13 segments.

If we reduce by 3 each time, then it will take four more moves to get to 1 (goal!).

16 – 3 – 3 – 3 – 3 – 3 = 1 in five moves.

So I’m looking for a way to reduce the segment count by three each time.

However, I’m having trouble doing this.

If any move results in a segment reduction of only two, then the best I can do is accomplish the goal in 6 moves.

Now that I think about it, to get the “A” on the left or the “Z” on the right, a 3-segment-reduction is not possible.

Does this mean 6 moves is the minimum possible to accomplish the task?

………F,WXYZ,KL,A,TU,V,C,B,GHI,J,NO,S,DE,P,QR,M

1. Move J-P to between I and Q. (same as V-I to U..J) [-3]

………F,WXYZ,KL,A,TUV,C,B,GHIJ,NO,S,DE,PQR,M

2. Move T-C to between S and D. (same as B-S to A..T) [-3]

………F,WXYZ,KL,AB,GHIJ,NO,STUV,CDE,PQR,M

3. Move W-B to between V and C. (same as G-V to F..W) [-3]

………FGHIJ,NO,STUVWXYZ,KL,ABCDE,PQR,M

4. Move F-O to between E and P (same as S-E to _..F) [-2]

………STUVWXYZ,KL,ABCDEFGHIJ,NOPQR,M

5. Move S-L to between R and M (same as A-R to _..S) [-2]

………ABCDEFGHIJ,NOPQRSTUVWXYZ,KLM

6. Move K-M to between J and N (same as N-Z to M.._) [-2]

………ABCDEFGHIJKLMNOPQRSTUVWXYZ

There are many scenarios for the last three steps.

(For the moment) I’m convinced it cannot be done in less than 6 moves.

August 2nd, 2011 at 9:25 am

Cazayoux – as you seem to be the only one trying this puzzle, I will tell you that you are correct!

J-P between I and Q

B-S between A and T

G-V between F and W

N-Z after M

P-R between O and S

F-L between E and M

I like the logical approach you took!

August 2nd, 2011 at 12:59 pm

Thanks.

It may not have been the most popular, but I enjoyed it.

August 3rd, 2011 at 6:57 am

I was attempting it, but my methods were quite crude and could only achieve 10 moves…

Nice work Cazayoux.