In this illustration I arrange the signs in groups that depict: humans, animals, crops, tools, and various artifacts. All signs suit fine into those catagories. What is more spectacular is that they, in accordance with the true inscription, belong together two by two, in what I call "stem-elements". which furthermore are in compliance with three stem-groups. The stems turn up to be a very efficient instrument for the understanding of the disc. -Before you'll find no reliable indicators, in whatever enumerations you try, whereas the inscription literally 'explodes' in categories of divisibility by eleven, after the acceptance of my 22 stem-forms, 'never known of before'!. A classic break-through, what riddles concern.
I define a stem as: A sequence of two 'and only two signs', which are repeated in at least one more signgroup. The stem is not allowed to overlap any other stem, which is already confirmed by a higher frequency.
NB! If you try this definition with some paragraphs based on language, you typically shall get clusters of unbreakable signs, and only rather few non-inverted stems. In great contrast to the tendency of an order of signs inside my 22 stem-forms; only two out of the thirthy stem-signs occur in both positions inside the various stems, they are "sign F and Q".
It is naive, if not suspect, to argue that the unravelling of the 70 stems is just a two way choice 'either calendar or language'. They were a million way choice.There are no comparison between the common process of getting familiar with an inscription, for instance by recognizing that B22 and B29 are somehow related groups, or to observe that B21 is a repetiton of B26 etc., contra the process of realising the absolute amount of 70 stems, which is a countable and sophisticated product.
Primary stands my essential discovery of the turning point "the 22 stem-forms" and the immediate shocking results they bring along. - Secondary we got all other of my perfectionist illustrations and the aphoristic remarks to match, those should be taken as 'in spirit sketches' to be called "Ariadne's thread"-Making us assured, that we are on the right track.



In this way I've arranged all the 61 dissimilar elements into four groups. Essential three stem-groups. As seen, the inscription consist of 60+1 dissimilar elements, the same number as the total amount of signgroups; and a sixth part of 366 days.
If you accept my stems, you have to accept the whole package, because all my observations are just consequences of the stems. So if I have had a mild fortune, my humble request complied with from the start for a rapid publication of my coded alphabetic notation of the stems, those exact same consequences of the stems would have been confirmed from every quarter of the globe by people in a matter of months.
Everyone who was profoundly occupied in the past done with this very same topic must have had the same alert over and over again, like I did, that they were dealing with some mathematical construction. then overhearing this signal, they turned back to their established grammatical studies over the topic. Those are my witnesses.
After a span of one-hundred years still no 'mutual consistency' among the many linguistic attempts to cut the knot, did show. A suspicion thrust forward, that 'the linguistic perception' was out of course, that the very disc might depict a calculation of some kind instead. On that notion a new generation af noblemen groped their ways with highly theoretic calendar-proposals, zodiacs, and astrolabia (only fantasy fix a limit). No-one ought to disagree, when I claim, that only by my 'breaking discovery' of abbreviated stems, a mathematical foundation appears, and at the same time filling in the gap from an amputated model to a 365-days calendar. -This vessel left for others to polish to perfection.

"Probably the diversified look of the glyphs has no other purpose, than to differentiate them from one another into 46 dissimilar species for counting reasons"


A first-rate demonstration of a non-grammatical content in a structure is this tabulation of the signgroups, which marks 30 pair of signgroups; Each pair, by components, tied together by an identical sign (marked blue). The two components of the pair have moreover each an offshoot (red letters ) to the 'next-floor' pairs, which possess exactly the same qualities 'one sign in common, each component with an offshoot to the next pair'. In this way all 61 signgroups are linked together into one single string. The most frequent elements are 'BA' with thirteen occurrences, and 'the thorns' with seventeen. Those two are isolated from the above countings. This chaining together make use of as much as 30 out of a total of 46 dissimilar signs (2/3) as links. Out of 61 signgroups as many as 53 are unrepeated.
If all the words in a piece of poetry or a prosaic text, chosen by random, could be linked together in one unbroken chain by double stitch! same as could those clusters of signs; Would'nt we then expect, that unknown disciplines inside philology had developed during time? For instance metaphorical translations of texts into unambiguous patterns, like the harmony of the spheres, or like Mandelbrot music? This is not the case in texts, otherwise opponents of my discovery would long ago have dished up with competitive examples from Literature, but they stay mute, whereas in systems alias protokols, ledgers, book-keeping by double entrance, various tables, such symmetri like 'a hollow pyramide' or 'a gnomonical arrangement' is given by itself. Thereby this classical break-through is in reality settled. Follow-up-questions like, what is the origin of the subject? How old? An interwoven text simultaneously? Or is it a fake? Etc. - you name it - are of course secondary questions. "Discovery" is without comparative degree!
Shame to the highly educated, but immature persons, excuses inadequate, who tempt such backward things to happen, whatever reasons they bring forward. Kind of childish excommunication out of envy; as if my advancements into the heart of this topic was on black-list. The truth is rather that my discovery is covetuosly pursued. As you all know: I did announce my break-through to this famous riddle, at the age of 34 years old, in 1985.


The 50 cartouches in the top, on the grey-coloured floor of this drawing, are the 22 pair of stem-signgroups, and the six unpaired stem-signgroups. together containing 290 units, inclusive absent units (in blue) and thorns. The succession of those 50 cartouches is modest re-arranged 'from the original succession on the disc', by the purpose to keep the 22 pairs close connected, and also to gain five double months of 58 unit-days on line. The fifth double month uses the 16 thorns of the stem-signgroups to obtain the count up to its 58 units.
Now the lower deck on the drawing with a pink floor holds the eleven non-stem-signgroups 'curiously located on uneven numbers', out of which eigth constitute a last sixth periode of 58 days. Is this sixth double month accumulating a reservoir for the superior calculation going on in the 50 stem-signgroups? Finally I try out the three non-stem signgroups in excess :B03, A13, and A18 as representing 17 orange epigomenal days in this 29 x12 moon-calendar. Curiously giving 29 signgroups on side A, and 29 signgroups on side B. Conclusion: Several interlaced systems framed by a calendar. This is as close as we get by now: A verifiable approximation of the truth to the question "What is the Phaistos disc about?". On this you can rely . Think about the gnomonical arrangement , think about the hierachical order of the elements, and do not forget how all the signgroups will be stitched together into one string 'no criss-cross'. Consider the splitting of time in shifts between 12 and 17. Remind it all!



If those seventy stems really reflect the secret about this old enigma, if they so were the key and the first sure entry to this Minoan secret, you would expect some kind of correspondence between the stem-signs inside and outside of the stems; and yes, this was exatly, what was found.
Comparing 'the unfolded arrangement' with a picture-lottery seems significant, in which the pair of stems, subsidiary all stems, are playing-squares (point 1, 2), and all other signs are pieces (skin colour). It is a familiar situation, that one or more pieces are missing. Let us imagine conversely, that we have had a mingling together with some outside pieces. We therefore charge ourselves with the task to sort out the pieces, that are in excess. The remaindersigns (point 5) have no correspondence with the playing squares, and are so to be sorted out without further ado. Putting down the pieces (point 3), where they fit into the squares (sky blue), will finaly leave us with nine pieces (point 4), which were represented in the squares, but they are now in excess, because the playing squares are already occupied. The dolphin (sign K) for instance is present in two stems, but it occurs four times independently, there are so to say two uncovering dolphins in excess. Those nine stemsigns are to be sorted out too, like the remaindersigns. - So are the thorns (point 6). Of cause, this is not an argue in favour of the inscription as a picture-lottery, I just want to indicate, that a principle of a similar kind can be applied with success; that some complicated, but symmetrical proportions, between the signs on the basis of their functions, are unveiled.
By observing the numbers 18, 22 and 26, it struck me, that square-sides on respectively 10, 12 and 14 hold circumferences, which are related to these numbers. This compels an almost unambiguous and most expressive way, in which 'the unfolded situation' is to be arranged, as the signs of the 33 pair of stems are to be set up as circumferences in a quadratic framework 'the folded arrangement'. The upper half of the framework is uncovered components of stems, and the lowest part is those signs in stems, which have cover from the reduced stemsigns (point 3). As it does show, the covering signs do not only make up a gemination of the lowest part of the framework, they keep themselves within the areas in their half, which are marked out by the three stemgroups. I have divided the frame into six zones, each containing 22 signs. The six zones are symbolized by Aa, Ab, Ba, Bb and Ca, Cb. - Bb for instance is those 22 signs in second position, which are gathered in the bottom left-hand corner of the frame. Together with Ba, Bb is able to establish 22 covering stems, of which 11 are dissimilar.* the covering stemsigns (Bb, Cb, Ba) consist of 33 signs in first position and 33 signs in second position. The upper half (Aa, Ca, Ab) then get the same bisection of the positions, obviously.The zones Ca and Cb hold each 11 signs in both first and second position. Together they compose 22 stems crosswise of the median line. It is seen, that multiples of eleven are reflected in a lot of new facets, although there are some limited ways to castle the signs, within those by positions and stemgroups restricted areas. This gnomonical arrangement, I believe, is the most ideal way to illustrate those symmetrical proportions, which are unquestionably available in the inscription. Especially the arrangement substantiate the legitimacy of the three stemgroups. There are however irregularities to be mentioned: Stemgroup II holds 18 covering and 18 covered, but 22 uncovered signs; While stemgroup III has 22 and 22, but 18 signs. If you consider the inscription as a numerical system, it would probably had made a more convincing impression, if the equal conditions had been respectively: 18, 18 and 18 plus 22, 22 and 22, together with the 26, 26 and 26 signs of signgroup I; on the other hand the very irregularities may be promising for a more complex application, than a mere ornamental play with some imprints and their quantities, on the part of the designer of the Phaistos disc.
*As regards positions, the right sign in a stem holds first position and vice versa.


You'll find 70 stems in the inscription as a whole. As much as 66 of these stems are entering into stem-pairs, leaving 4 unpaired stems. Those 66 stems fit into a frame, which I call the gnomonical arrangement (the figure above). - Stemgroup I is aqua, stemgroup II is yellow, stemgroup III is pink.
In the same way. You will find 75 stem signs in total, which are disconnected from stems. 66 of those isolated stem signs suit into pairs with the signs inside of the 70 stems, giving 9 signs in excess. In other words: They will be covering exactly one half part of the components in the 66 stem pairs, as a duplicate of the lower part of the arrangement. -All in all comparable to a picture-lottery. Harmony everywhere! manifesting itself as multiples of eleven. - Good-bye language!


In this diagram, you have, in the upper left the frame of the seventy stems, called "the gnomonical arrangement". The components separated in black and red letters. All the parted stem-components in the six bricks of the frame are paired into three bricks in the middle section of the diagram. The black letters are the concrete components of the reduced stems in the lower left corner White letters are the calculated missing components. Again assempled in the middle section.
To the right is seen the intended different-, or related arrangement. The six colored zones are identical two by two. You may argue against this idea, that the vessel is fragmentary in its left corner, but still it is recognizable as what it is. So is Venus de Milo.
If you turn the wrong side outwards on a piece of knitting fabrics, a topological discipline, you will better understand its construction process. Likewise 'the related arrangement' add to the clarification of 29 in the inscrription. Though the upper left corner of this arrangement has seven stems, which seem to be misplaced, but they are interchangable with seven of the nine reduced stems in excess bringing perfection to the frame. Amazing!



Comprehensive view : Ergo, the 70 stem-elements are arranged in the upper half of the above "related gnomonical arrangement". Now the inscription shows to be mischievous, this is probably the reason why its symmetries never have been acknowledged before because the sign-materiale within the stem-elements also exists outside the stems, as isolated and singular symbols.
My great idea is then, that the isolated signs are abbreviations from stem-elements. I've placed those "reduced stems" in the lower part of the arrangement. All visible signs in the reduced stems are hold in black, as were the stems, and the abbreviated signs are painted light greyish.
Each of the six colored corners (gnomons) in the above, by functions determined arrangement consists of total 22 stems and 11 different forms. You're looking at a minutely duplicate-system, nothing less than a revelation concerning this topic !
Now the credibility of these 70 elements lies on the structures that they reveal, for instance the perfect symmetrical interrelations between the stems and the reduced stems, as seen in these gnomonical arrangements, which undoubtly can not be explained from language or linear metrics. Under the device : Nothing new under the sun, a reader would soon have brought to light a similar splendid linguistic analogy, had it been language. They stay mute ...
On the other hand, calculations and counter-checks are fundamental principles in building up ledgers. Or if the calendar is the primary factor of the inscription, then perhaps the hieroglyphs could be describing the movements of the sun versus the moon through the year, or a similar system of congruency.


By replacing the seven red-coloured stems by the seven reduced stems in excess the related gnomonical arrangement became a sublime symmetrical construction. For instance its six gnomons are identical two by two.
On the other hand if you remove the seven red-coloured stems without any replacements, and count the amount of pair of stem-signs instead of the pair of stems, then the upper part has 29 such pairs, consisting of total 58 signs, so has the lower part 'the reduced stems', and simultaneously the 58 absent stem units that they hide.
Outside the frame are now superfluous 29 stem-elements of two units (58). Finally 29 non-stem signs to be redoubled to 58 signs, rounding the total numbers of units up to 348, not to forget the 17 thorns. - The calendar of the Minoans! With this I have reverted to the calendar possibility. Each of the four halves, the fourth half being the absent units in the reduced stems, is systems of 29 well-organized pair of signs (58 components) and 8 immediately unpaired characters. Although, there is one condition for this complete symmetrical arrangement: That the obliterated reduced element in A08 unambiguously is set to be the sign "P". Thus the Gnomonic arrangement compels the identity of the missing final character in A08.
To end up the calculation. The inscription moreover contains 29 'non-stem elements', which is my designation to the left over signs different from the 215 stem signs. (29x7, +12) Giving: 132+8 +66+9 +29 =244 signs in total (29x8, +12). If you add the 17 thorns, then the total is 261 units. (29x9)
When you include the absent units, the calculation looks this way: 132+8 +132+18 +58, or 70 +70 +75 +75 +29 +29 +17 thorns =365, .
How come the torso "Venus de Milo" be an ideal of beauty? Is'nt it because our sense of proportions approximately tell us, what she looked like? Everyones sense of proportions should tell them too, that the only mystery left in the case of the Phaistos disc is the lack of response to my structural analysis. 17 +12 +17 +12 +17, 12 +17 +12 +17 +12, 17 +12 +17 +12 +17, 12 +17 + 12 +17 +12, 17 +12 + 17 +12 +17 =365. (29x12, +17)
Doctor of Philology Professor, Emeritus Ernst Doblhofer, Die Entzifferung vershollener Schriften und Sprache (1961). About the Phaistos disk:
"Possibly a professional investigator will sooner or later win the laurels promised to the one who solves the riddle of this clay plaque, which can be seen today in the Heraclion Museum. Or perhaps a brilliant amateur will solve the mystery of the spiral images and, like a modern Theseus, find the way out of this new labyrinth of the island of Minos"


I better comment this important illustration with both some words and some numbers. If I call, what is left of the actual inscription, the eight piles of two colours in the upper left, for the tableau, then the figure to the right is referred to as the foundation. First the tableau is made free of all 29 non-stem elements, these are the two waste piles in the bottom. Secondly I have also removed 12 stems and 17 reduced stem-elements, together containing 12 +17 +12 =41 signs ( the red and violet waste piles in the lower left). Now the tableau is composed of 58 stems and 58 expanded reduced stems with the total of 232 units (29 x8)* As seen, the waste is marked on the foundation too, except for the 4 unpaired stems and the 9 reduced stems in excess, that can be found in the lower right figure. Another detail of interest: Having sorted out the waste from the inscription results in that one signgroup on side B, and two on side A are emptied of signs, giving 29 sign-groups on both sides. -B19, -A07. and -A19
*If the reduced stems were not expanded, this shortform tableau would be of 174 visible signs (29 x6). This is 70 units from the actual Phaistos disc inscriptions 244 signs. (29x8, +12). Or you could say that if the 70 genuine stems were reduced each with one sign, the result would be the same number, 244 -70 =174. (29x6) -If 174 is redoubled, you end up with 348 (29 x12)
"The science of counting things is basically the discipline of turning problems into enumeration statements to find out if they can be solved this way".


Simply define 22 stem forms of two signs, and find all 70 stems hidden in the sign material of 244 signs. Consequently take a census of the isolated 75 stem signs, which do not possess the second units. The remaining materiale is of 29 non-stem signs, and 17 thorns. The Minoan calendar is reborn -: Those four numbers are all expressible through a shift between 12 and 17 -, (5,7,5,7,5)
70 stems =12 +17 +12 +17 +12 -, 75 reduced stems =17 +12 +17 +12 +17 -, 29 non-stem elements =17 +12 -, 17 thorns =17. Perhaps the diversified looks of the glyphs have no other purpose, than to differentiate them from one another into 46 dissimilar species ((17 +12 +17) for counting reasons.
- Now consider this inscription solved.



This above linear transcription, which constitute an eight month calendar, is the first step towards my demonstration of a 364-day calendar in the Phaistos disc. I ought to settle, that without adding the two initial "pearl strings" to the inscription (sign g) as genuine units, such possibility, as to proceed, by the way of trial and error, to a similar regular eight months calendar, do not exist.


(1) A08 only contains 4 signs, and not 5.
(2) The northern or southern area contains an additional 18th thorn.
Signs and units are days, signgroups are equal to weeks. 7 or 8 signgroups with alternately 29 or 31 signs are months.

Move the 61th signgroup "A31" into the bull's eye, and unfold the two spirals to a single grand circle, that A30 is followed by B01 and B30 by A01. Such circle has a periphery of 240 signs.
Now displace the starting point from A01 one step back to B30, and count foreward 8 signgroups, accordingly {B30,A01,A02,A03,A04,A05,A06,A07}, and continue with the next 7 signgroups. Make this shift eight times in total for the 61-1 signgroups. A count over will ensure you that each area of 8 and 7 signgroups contains 29 and 31 visible signs by turn! Finally A31.
This was all about the 243 signs which are visible on the Disc in this productive variation of my 244-day prototype calendar.

If you are aware of my interpretation of this hieroglyphic inscription, you'll know that I plead that all signs form part of elements of always two signs. My method thereby determines 70 stem-elements together with 104 incomplete elements, because these only contain a single sign. In other words the inscription is in lack of 104 units. The same figure as the number of reduced elements.
Prototype: 70 x2 stem-coponenets and 104 incomplete elements consequently 104 absent units, plus 17 strokes. -365-.
Variation : 70 x2 stem-components and 103 reduced elements consequently 103 invisible units, plus 18 thorns. -364-.

Imagine a compass card being placed in the middle of the grand-circle, appointing {B30,A01,A02,A03,A04,A05,A06,A07}as the western area, and {A30,B01,B02,B03,B04,B05,B06,B07} as the eastern area. The signgroups in those two corners contain together 31 reduced stems or 31 absent units (if you can possess something absent.) The same pattern repeats itself for the 8 northern and 8 southern signgroups, while it is somehow different with the 7 signgroups sited northwest and the 7 in southwest with 29 units together, and finally for the northeast and southeast signgroups with 29 unit-days, thus forming a fleur-de-lis. As to symmetry, not unlike the pointer figure of the various Maya and Actez calendar stones.
This was my introduction to the missing season of the Phaistos Disc Calendar.


In the above figure "the packed circle" the two spirals on the disc are unfolded to a single "grand circle", which contains the 243-3 visible signs at the rim apart from the three signs of A31
At the same time the absent parts of the 103-1 reduced elements emerge as, previously covered, feathers into a median circle.
Finally an inner circle shows 18 thorns, which are concrete strokes that are made before the clay disc did set.
Every eight part of the circle is made up by alternating 7 and 8 signgroups.
It is important for me to state that this observation of a calendar has never been done before by anyone else, by that simple fact that no-one has thought out earlier that the two dotted dividers in front of A01 and B01 have values as normal signs. This discovery takes you to the conclussion that the signs are not syllables.
Now asking the question: what is then the situation? make you arrive to my discovery of the 22 stem-forms, verified by the gnomonical arrangement.
Definitely the very most plausible interpretation of this famous disc from near Hagia Triada in southern Crete; though overheard!



The above arrangement is comparable to Dr. Christoph Henke's experiment:The hierarchy of the characters on the diskus of Phaistos (file.pdf).
In his paper, as I read it, he actual prove, that my results cannot be reached, if the hieroglyphs represent grammar. Very pleasant to know.The solution to the whole problem was to gather the signs two by two, instead of regarding each sign alone. The signs inside the 22 stems-forms possess an order of precedence in themselves too. Perhaps his analysis could be an example of a result indirectly gained by my decipherment -a back-door. That would be an example of why it is so dissatisfactory (cf.  Marianna Ridderstad. Anistoriton vol.12. 2010), that my decipherment was never received in the expected way back in 1985. - Never referred to. By chance I found this website from 1999 ,using two highly sophisticated figures of mine that announce the 22 stemforms. - I do not see my name mentioned. Neither on this blogspot.
N.B.: It would be nice to receive some royalties for my popular mantelpiece in LEGO of an hollow calendar step-pyramid. :´ )

The divisibility by eleven, through this formular [Commensurable to a 5.5 * n table with an allowed inacurracy on plus/minus 0.5, when n is an unequal number ]:
{5,6,11,16,17,22,27,28,33,38,39,44,49,50,55,60,61}, is continued (far beyond its certification through the gnomonical arrangement with its already omnipresency of eleven) in this " The alternative calendar ". Here I choose to combine 44 consequtive reduced elements into 22 so-called "combies", leaving only 60 reduced elements instead of the otherwise 104. Following these metrical footsteps : _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ .

Carrying the revealed system to a logical conclusion, could be, to consider the 70 stems together with the 22 combined reduced elements, as the one type of elements: contra the 60 reduced elements and the 22 medial absent elements as the second type. Thus forming a periodical alternation between long and shortform elements. The open spaces between the sign-groups would take part in this interaction. The 22 "combies" borrow 44 reduced elements, to become like stems, leaving over 60 of the before 104 reduced elements, but this is compensated for by the 22 absent medial elements. This variant is called the "Alternative calendar".




Som skaberen af dette projekt har jeg en naturlig uvilje imod at dekorere mine fyldestgørende illustrationer med overflødige ord, fordi:
1) Min undersøgelse er baseret på tegninger og geometri (topologi).
2) Mine beviser og ideer er "født" som billeder, derfor finder jeg det æstetisk imperfekt overdrevent at anvende det skrevne ord, på noget der netop intet har med sprog at gøre.