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Wednesday, September 2, 2026

Vibrant Houston Flower

Vibrant Houston Flower by Jean Claude Constantin
Vibrant Houston Flower was created by Jean Claude Constantin and used by Wil Strijbos as his exchange puzzle at the 41st International Puzzle Party (IPP).  It consists of a tray and 24 metallic beads.  The beads are divided into 11 pairs with each pair having a different color.  The tray consists of an outer ring, a middle ring, and a color key in the center.  The outer ring has spaces to hold all 12 pairs of beads while the middle ring can accommodate 9.  The color key in the center has a fixed set of 12 smaller beads matching the colors of the pairs along with how many spaces are required to exist between the pair of beads of that color when they are placed within a ring.  The number of spaces run incrementally from 0 to 11.

There are 2 challenges.  The first challenge is to place the first 9 pairs with required spacing from 0 to 8 within the middle ring.  The second challenge is to place all 12 pairs in the outer ring so that they meet the spacing requirements.

Vibrant Houston Flower Beads
Lacking a real strategy, I started with the first challenge by placing the pairs from the shortest to longest distance between them.  Eventually, you get to the point where it’s not possible to add the final beads in the spaces that remain.  However, instead of backing up, it is possible to start shifting pairs around into the available spaces to arrange the vacant spaces for the remaining beads.

Having solved the first challenge, I proceeded with the second using the same approach.  After shifting a lot of pairs, I finally got it down to 1 pair left to place.  And after a lot more shifting, I still had not placed that last pair.  So I put it on the shelf.  Several days later, I took it down for another session and noticed after a few minutes that shifting 2 pairs at the same time opened up the space that I needed and I was done.

I wouldn’t be surprised if there was a nice algorithm to arrange any number of pairs.  However, the 9-pair and 12-pair challenges seemed to be good choice for solving it without such an algorithm.

Only 2 more pairs to go!