02-13-2021, 11:05 PM RE: Serious Puzzles for The Research Folks Q
(This post was last modified: 02-13-2021, 11:57 PM by Max.)

Solving Section III

The following is 100% stolen from this site.

The key is 0362514 (KRYPTOS).

And the encryption process is Route Transposition followed by a Keyed Columnar Transposition.

Step 1, Route Transposition:

First we pad the message fitting it into a 86xN box.

Why padding it? To make the text in all the columns line up leaving columns of only two different lengths for the person decrypting to deal with, who is expected to know exactly how many of them there are and which ones they are. We are just being considerate of the guy with the key on the other end.

How many letters to add? The message length is 336 and we are fitting it into a box of width 86. 86 mod 7 = 2.

It means that every line except for the last one will have 2 extra columns. 336 mod 86 = 78.

The last line will be 78 letters long and 78 mod 7 = 1.

And since the number of the last line's "extra" columns has to be the same as the first lines to make the columns line up, we only need one extra Q to make it 2 for all the lines. Clear enough?

Now to the transposition itself:

In by Rows backwards into 86x4, Out by Columns in groups of 7 which is the length of the key:

SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGEDEBRISTHATENCUM

BEREDTHELOWERPARTOFTHEDOORWAYWASREMOVEDWITHTREMBLINGHAND

SIMADEATINYBREACHINTHEUPPERLEFTHANDCORNERANDTHENWIDENING

THEHOLEALITTLEIINSERTEDTHECANDLEANDPEEREDINTHEHOTAIRESCA

PINGFROMTHECHAMBERCAUSEDTHEFLAMETOFLICKERBUTPRESENTLYDET

AILSOFTHEROOMWITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ?

->

?QGNIHTYNAEESUOYNACXTSIMEHTMORFDEGREMENIHTIWMOOREHTFOSLIATEDYLTNESERPTUBREKCILFOTEMALF

EHTDESUACREBMAHCEHTMORFGNIPACSERIATOHEHTNIDEREEPDNAELDNACEHTDETRESNIIELTTILAELOHEHTGNI

NEDIWNEHTDNARENROCDNAHTFELREPPUEHTNIHCAERBYNITAEDAMISDNAHGNILBMERTHTIWDEVOMERSAWYAWROO

DEHTFOTRAPREWOLEHTDEREBMUCNETAHTSIRBEDEGASSAPFOSNIAMEREHTYLWOLSYLTARAPSEDYLWOLS

What makes me think it was written backwards? Because the extra space is not at the end of the message but before the first letter. Who would bother calculating the position of the first letter and start writing the message beginning with the 7th column just to make it fill up the rectangle perfectly at the end?

I think it is easier to simply fill out the rectangle backwards if you are doing it with pen and paper.

Either way, it is 86x4 with 7 spaces in front of the first letter.

?QGNIHT YNAEESU OYNACXT SIMEHTM ORFDEGR EMENIHT IWMOORE HTFOSLI ATEDYLT NESERPT UBREKCI LFOTEMA LF

EHTDESU ACREBMA HCEHTMO RFGNIPA CSERIAT OHEHTNI DEREEPD NAELDNA CEHTDET RESNIIE LTTILAE LOHEHTG NI

NEDIWNE HTDNARE NROCDNA HTFELRE PPUEHTN IHCAERB YNITAED AMISDNA HGNILBM ERTHTIW DEVOMER SAWYAWR OO

DEHTFOT RAPREWO LEHTDER EBMUCNE TAHTSIR BEDEGAS SAPFOSN IAMEREH TYLWOLS YLTARAP SEDYLWO LS

Whichever way the text was written initially, after we restack it into 7 columns, it will result in:

?QGNIHT

EHTDESU

NEDIWNE

DEHTFOT

YNAEESU

ACREBMA

HTDNARE

RAPREWO

OYNACXT

HCEHTMO

NROCDNA

LEHTDER

SIMEHTM

RFGNIPA

HTFELRE

EBMUCNE

ORFDEGR

CSERIAT

PPUEHTN

TAHTSIR

EMENIHT

OHEHTNI

IHCAERB

BEDEGAS

IWMOORE

DEREEPD

YNITAED

SAPFOSN

HTFOSLI

NAELDNA

AMISDNA

IAMEREH

ATEDYLT

CEHTDET

HGNILBM

TYLWOLS

NESERPT

RESNIIE

ERTHTIW

YLTARAP

UBREKCI

LTTILAE

DEVOMER

SEDYLWO

LFOTEMA

LOHEHTG

SAWYAWR

LS

LF

NI

OO

Now write the key on top and proceed with...

Step 2, The Keyed Columnar Transposition:

KRYPTOS KOPRSTY

0362514 -> 0123456

?QGNIHT ?HNQTIG

EHTDESU ESDHUET

NEDIWNE NNIEEWD

DEHTFOT DOTETFH

YNAEESU YSENUEA

ACREBMA AMECABR

HTDNARE HRNTEAD

RAPREWO RWRAOEP

OYNACXT OXAYTCN

HCEHTMO HMHCOTE

NROCDNA NNCRADO

LEHTDER LETERDH

SIMEHTM STEIMHM

RFGNIPA RPNFAIG

HTFELRE HRETELF

EBMUCNE ENUBECM

ORFDEGR OGDRREF

CSERIAT CARSTIE

PPUEHTN PTEPNHU

TAHTSIR TITARSH

EMENIHT EHNMTIE

OHEHTNI ONHHITE

IHCAERB IRAHBEC

BEDEGAS BAEESGD

IWMOORE IROWEOM

DEREEPD DPEEDER

YNITAED YETNDAI

SAPFOSN SSFANOP

HTFOSLI HLOTISF

NAELDNA NNLAADE

AMISDNA ANSMADI

IAMEREH IEEAHRM

ATEDYLT ALDTTYE

CEHTDET CETETDH

HGNILBM HBIGMLN

TYLWOLS TLWYSOL

NESERPT NPEETRS

RESNIIE RINEEIS

ERTHTIW EIHRWTT

YLTARAP YAALPRT

UBREKCI UCEBIKR

LTTILAE LAITELT

DEVOMER DEOERMV

SEDYLWO SWYEOLD

LFOTEMA LMTFAEO

LOHEHTG LTEOGHH

SAWYAWR SWYARAW

LS L S

LF L F

NI N I

OO O O

Now to the last...

Step 3, Out by columns downwards, left to right resulting in:

?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

Reminds you of anything? ;)

The decryption process requires knowing the key and the rectangle size for the route transposition:

KRYPTOS and 86.

First we determine the line lengths to split the message:

86 mod 7 = 2. It means that two of the columns are going to be longer.

Which two and by how much?

The first two in our system (they are 0 and 3 for the person decrypting the message), with lengths 51 and 47.

The difference between those lengths will be the same (4) for 86 mod 7 regardless of the message length.

You may want to find out why as an excercise.

So we...

Step 1, Split the input as follows:

0 ?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

1 HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

2 NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

3 QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

4 TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

5 IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

6 GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

Step 2, Write it in columns... (i omitted it to make Step 3 clearer, hence the following text is on its side)

Step 3, Reorder the columns according to the key:

0 ?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

3 QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

6 GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

2 NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

5 IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

1 HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

4 TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

Step 4, Chop them into 86-letter long lines

(in groups of 4 of course, since 337/86 is > 3 but is <= 4)

0 ?END YAHR OHNL SRHE OCPT EOIB IDYS HNAI ACHT NREY ULDS LLSL LNO

3 QHEE NCTA YCRE IFTB RSPA MHHE WENA TAMA TEGY EERL BTEE FOAS FIO

6 GTDH ARDP NEOH MGFM FEUH EECD MRIP FEIM EHNL SSTT RTVD OHW

2 NDIT EENR AHCT ENEU DRET NHAE OETF OLSE DTIW ENHA EIOY TEY

5 IEWF EBAE CTDD HILC EIHS ITEG OEAO SDDR YDLO RITR KLML EHA

1 HSNO SMRW XMNE TPRN GATI HNRA RPES LNNE LEBL PIIA CAEW MTW

4 TUET UAEO TOAR MAEE RTNR TIBS EDDN IAAH TTMS TEWP IERO AGR

Step 5, Read the resulting 4 lines of the message backwards (reverse of the Step 1 of encryption). Done.

If the same key KRYPTOS=0362514 was used to encrypt the 4th part, the decryption process would be as follows:

Let's say the number of columns for the route transposition was 49 or 21...

Step 1:

?OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

->

?OBKRUOXOGHULB

SOLIFBBWFLRVQQ

PRNGKSSOTWTQSJ

QSSEKZZWATJKLU

DIAWINFBNYPVTT

MZFPKWGDKZXTJC

DIGKUHUAUEKCAR

Step 2:

?SPQDMD

OORSIZI

BLNSAFG

KIGEWPK

RFKKIKU

UBSZNWH

OBSZFGU

XWOWBDA

OFTANKU

GLWTYZE

HRTJPXK

UVQKVTC

LQSLTJA

BQJUTCR

Step 3:

KOPRSTY KRYPTOS

0123456 -> 0362514

?SPQDMD ?QDPMSD

OORSIZI OSIRZOI

BLNSAFG BSGNFLA

KIGEWPK KEKGPIW

RFKKIKU RKUKKFI

UBSZNWH UZHSWBN

OBSZFGU OZUSGBF

XWOWBDA XWAODWB

OFTANKU OAUTKFN

GLWTYZE GTEWZLY

HRTJPXK HJKTXRP

UVQKVTC UKCQTVV

LQSLTJA LLASJQT

BQJUTCR BURJCQT

Step 4:

For 49 columns:

?QDPMSD BSGNFLA RKUKKFI OZUSGBF OAUTKFN HJKTXRP LLASJQT

OSIRZOI KEKGPIW UZHSWBN XWAODWB GTEWZLY UKCQTVV BURJCQT

For 21 columns:

?QDPMSD UZHSWBN HJKTXRP

OSIRZOI OZUSGBF UKCQTVV

BSGNFLA XWAODWB LLASJQT

KEKGPIW OAUTKFN BURJCQT

RKUKKFI GTEWZLY

Step 5:

For 49:

TQCJRUBVVTQCKUYLZWETGBWDOAWXNBWSHZUWIPGKEKIOZRISOTQJSALLPRXTKJHNFKTUAOFBGSUZOIFKKUKRALFNGSBDSMPDQ?

For 21:

YLZWETGIFKKUKRTQCJRUBNFKTUAOWIPGKEKTQJSALLBWDOAWXALFNGSBVVTQCKUFBGSUZOIOZRISOPRXTKJHNBWSHZUDSMPDQ?

Step 6: Breaking the cipher (most probably the same double-key Vigenere) and reading the message.

The final decrypted message

SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGEDEBRIST

HATENCUMBEREDTHELOWERPARTOFTHEDOORWAYWASREMOVEDW

ITHTREMBLINGHANDSIMADEATINYBREACHINTHEUPPERLEFTH

ANDCORNERANDTHENWIDENINGTHEHOLEALITTLEIINSERTEDT

HECANDLEANDPEEREDINTHEHOTAIRESCAPINGFROMTHECHAMB

ERCAUSEDTHEFLAMETOFLICKERBUTPRESENTLYDETAILSOFTH

EROOMWITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ

Add in a sprinkling of punctuation, and it comes out to be

Slowly, desparatly slowly, the remains of passage debris that encumbered the lower

Part of the doorway was removed. With trembling hands i made a tiny breach in the upper

Lefthand corner and then widening the hole a little i inserted the candle and peered

in. The hot air escaping from the chamber caused the flame to flicker but presently

Details of the room within emerged from the mist x can you see anything q?

The above solution's came from the University of California, San Diego Math Dept.

https://www.math.ucsd.edu/

The following is 100% stolen from this site.

The key is 0362514 (KRYPTOS).

And the encryption process is Route Transposition followed by a Keyed Columnar Transposition.

Step 1, Route Transposition:

First we pad the message fitting it into a 86xN box.

Why padding it? To make the text in all the columns line up leaving columns of only two different lengths for the person decrypting to deal with, who is expected to know exactly how many of them there are and which ones they are. We are just being considerate of the guy with the key on the other end.

How many letters to add? The message length is 336 and we are fitting it into a box of width 86. 86 mod 7 = 2.

It means that every line except for the last one will have 2 extra columns. 336 mod 86 = 78.

The last line will be 78 letters long and 78 mod 7 = 1.

And since the number of the last line's "extra" columns has to be the same as the first lines to make the columns line up, we only need one extra Q to make it 2 for all the lines. Clear enough?

Now to the transposition itself:

In by Rows backwards into 86x4, Out by Columns in groups of 7 which is the length of the key:

SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGEDEBRISTHATENCUM

BEREDTHELOWERPARTOFTHEDOORWAYWASREMOVEDWITHTREMBLINGHAND

SIMADEATINYBREACHINTHEUPPERLEFTHANDCORNERANDTHENWIDENING

THEHOLEALITTLEIINSERTEDTHECANDLEANDPEEREDINTHEHOTAIRESCA

PINGFROMTHECHAMBERCAUSEDTHEFLAMETOFLICKERBUTPRESENTLYDET

AILSOFTHEROOMWITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ?

->

?QGNIHTYNAEESUOYNACXTSIMEHTMORFDEGREMENIHTIWMOOREHTFOSLIATEDYLTNESERPTUBREKCILFOTEMALF

EHTDESUACREBMAHCEHTMORFGNIPACSERIATOHEHTNIDEREEPDNAELDNACEHTDETRESNIIELTTILAELOHEHTGNI

NEDIWNEHTDNARENROCDNAHTFELREPPUEHTNIHCAERBYNITAEDAMISDNAHGNILBMERTHTIWDEVOMERSAWYAWROO

DEHTFOTRAPREWOLEHTDEREBMUCNETAHTSIRBEDEGASSAPFOSNIAMEREHTYLWOLSYLTARAPSEDYLWOLS

What makes me think it was written backwards? Because the extra space is not at the end of the message but before the first letter. Who would bother calculating the position of the first letter and start writing the message beginning with the 7th column just to make it fill up the rectangle perfectly at the end?

I think it is easier to simply fill out the rectangle backwards if you are doing it with pen and paper.

Either way, it is 86x4 with 7 spaces in front of the first letter.

?QGNIHT YNAEESU OYNACXT SIMEHTM ORFDEGR EMENIHT IWMOORE HTFOSLI ATEDYLT NESERPT UBREKCI LFOTEMA LF

EHTDESU ACREBMA HCEHTMO RFGNIPA CSERIAT OHEHTNI DEREEPD NAELDNA CEHTDET RESNIIE LTTILAE LOHEHTG NI

NEDIWNE HTDNARE NROCDNA HTFELRE PPUEHTN IHCAERB YNITAED AMISDNA HGNILBM ERTHTIW DEVOMER SAWYAWR OO

DEHTFOT RAPREWO LEHTDER EBMUCNE TAHTSIR BEDEGAS SAPFOSN IAMEREH TYLWOLS YLTARAP SEDYLWO LS

Whichever way the text was written initially, after we restack it into 7 columns, it will result in:

?QGNIHT

EHTDESU

NEDIWNE

DEHTFOT

YNAEESU

ACREBMA

HTDNARE

RAPREWO

OYNACXT

HCEHTMO

NROCDNA

LEHTDER

SIMEHTM

RFGNIPA

HTFELRE

EBMUCNE

ORFDEGR

CSERIAT

PPUEHTN

TAHTSIR

EMENIHT

OHEHTNI

IHCAERB

BEDEGAS

IWMOORE

DEREEPD

YNITAED

SAPFOSN

HTFOSLI

NAELDNA

AMISDNA

IAMEREH

ATEDYLT

CEHTDET

HGNILBM

TYLWOLS

NESERPT

RESNIIE

ERTHTIW

YLTARAP

UBREKCI

LTTILAE

DEVOMER

SEDYLWO

LFOTEMA

LOHEHTG

SAWYAWR

LS

LF

NI

OO

Now write the key on top and proceed with...

Step 2, The Keyed Columnar Transposition:

KRYPTOS KOPRSTY

0362514 -> 0123456

?QGNIHT ?HNQTIG

EHTDESU ESDHUET

NEDIWNE NNIEEWD

DEHTFOT DOTETFH

YNAEESU YSENUEA

ACREBMA AMECABR

HTDNARE HRNTEAD

RAPREWO RWRAOEP

OYNACXT OXAYTCN

HCEHTMO HMHCOTE

NROCDNA NNCRADO

LEHTDER LETERDH

SIMEHTM STEIMHM

RFGNIPA RPNFAIG

HTFELRE HRETELF

EBMUCNE ENUBECM

ORFDEGR OGDRREF

CSERIAT CARSTIE

PPUEHTN PTEPNHU

TAHTSIR TITARSH

EMENIHT EHNMTIE

OHEHTNI ONHHITE

IHCAERB IRAHBEC

BEDEGAS BAEESGD

IWMOORE IROWEOM

DEREEPD DPEEDER

YNITAED YETNDAI

SAPFOSN SSFANOP

HTFOSLI HLOTISF

NAELDNA NNLAADE

AMISDNA ANSMADI

IAMEREH IEEAHRM

ATEDYLT ALDTTYE

CEHTDET CETETDH

HGNILBM HBIGMLN

TYLWOLS TLWYSOL

NESERPT NPEETRS

RESNIIE RINEEIS

ERTHTIW EIHRWTT

YLTARAP YAALPRT

UBREKCI UCEBIKR

LTTILAE LAITELT

DEVOMER DEOERMV

SEDYLWO SWYEOLD

LFOTEMA LMTFAEO

LOHEHTG LTEOGHH

SAWYAWR SWYARAW

LS L S

LF L F

NI N I

OO O O

Now to the last...

Step 3, Out by columns downwards, left to right resulting in:

?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

Reminds you of anything? ;)

The decryption process requires knowing the key and the rectangle size for the route transposition:

KRYPTOS and 86.

First we determine the line lengths to split the message:

86 mod 7 = 2. It means that two of the columns are going to be longer.

Which two and by how much?

The first two in our system (they are 0 and 3 for the person decrypting the message), with lengths 51 and 47.

The difference between those lengths will be the same (4) for 86 mod 7 regardless of the message length.

You may want to find out why as an excercise.

So we...

Step 1, Split the input as follows:

0 ?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

1 HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

2 NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

3 QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

4 TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

5 IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

6 GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

Step 2, Write it in columns... (i omitted it to make Step 3 clearer, hence the following text is on its side)

Step 3, Reorder the columns according to the key:

0 ?ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIACHTNREYULDSLLSLLNO

3 QHEENCTAYCREIFTBRSPAMHHEWENATAMATEGYEERLBTEEFOASFIO

6 GTDHARDPNEOHMGFMFEUHEECDMRIPFEIMEHNLSSTTRTVDOHW

2 NDITEENRAHCTENEUDRETNHAEOETFOLSEDTIWENHAEIOYTEY

5 IEWFEBAECTDDHILCEIHSITEGOEAOSDDRYDLORITRKLMLEHA

1 HSNOSMRWXMNETPRNGATIHNRARPESLNNELEBLPIIACAEWMTW

4 TUETUAEOTOARMAEERTNRTIBSEDDNIAAHTTMSTEWPIEROAGR

Step 4, Chop them into 86-letter long lines

(in groups of 4 of course, since 337/86 is > 3 but is <= 4)

0 ?END YAHR OHNL SRHE OCPT EOIB IDYS HNAI ACHT NREY ULDS LLSL LNO

3 QHEE NCTA YCRE IFTB RSPA MHHE WENA TAMA TEGY EERL BTEE FOAS FIO

6 GTDH ARDP NEOH MGFM FEUH EECD MRIP FEIM EHNL SSTT RTVD OHW

2 NDIT EENR AHCT ENEU DRET NHAE OETF OLSE DTIW ENHA EIOY TEY

5 IEWF EBAE CTDD HILC EIHS ITEG OEAO SDDR YDLO RITR KLML EHA

1 HSNO SMRW XMNE TPRN GATI HNRA RPES LNNE LEBL PIIA CAEW MTW

4 TUET UAEO TOAR MAEE RTNR TIBS EDDN IAAH TTMS TEWP IERO AGR

Step 5, Read the resulting 4 lines of the message backwards (reverse of the Step 1 of encryption). Done.

If the same key KRYPTOS=0362514 was used to encrypt the 4th part, the decryption process would be as follows:

Let's say the number of columns for the route transposition was 49 or 21...

Step 1:

?OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

->

?OBKRUOXOGHULB

SOLIFBBWFLRVQQ

PRNGKSSOTWTQSJ

QSSEKZZWATJKLU

DIAWINFBNYPVTT

MZFPKWGDKZXTJC

DIGKUHUAUEKCAR

Step 2:

?SPQDMD

OORSIZI

BLNSAFG

KIGEWPK

RFKKIKU

UBSZNWH

OBSZFGU

XWOWBDA

OFTANKU

GLWTYZE

HRTJPXK

UVQKVTC

LQSLTJA

BQJUTCR

Step 3:

KOPRSTY KRYPTOS

0123456 -> 0362514

?SPQDMD ?QDPMSD

OORSIZI OSIRZOI

BLNSAFG BSGNFLA

KIGEWPK KEKGPIW

RFKKIKU RKUKKFI

UBSZNWH UZHSWBN

OBSZFGU OZUSGBF

XWOWBDA XWAODWB

OFTANKU OAUTKFN

GLWTYZE GTEWZLY

HRTJPXK HJKTXRP

UVQKVTC UKCQTVV

LQSLTJA LLASJQT

BQJUTCR BURJCQT

Step 4:

For 49 columns:

?QDPMSD BSGNFLA RKUKKFI OZUSGBF OAUTKFN HJKTXRP LLASJQT

OSIRZOI KEKGPIW UZHSWBN XWAODWB GTEWZLY UKCQTVV BURJCQT

For 21 columns:

?QDPMSD UZHSWBN HJKTXRP

OSIRZOI OZUSGBF UKCQTVV

BSGNFLA XWAODWB LLASJQT

KEKGPIW OAUTKFN BURJCQT

RKUKKFI GTEWZLY

Step 5:

For 49:

TQCJRUBVVTQCKUYLZWETGBWDOAWXNBWSHZUWIPGKEKIOZRISOTQJSALLPRXTKJHNFKTUAOFBGSUZOIFKKUKRALFNGSBDSMPDQ?

For 21:

YLZWETGIFKKUKRTQCJRUBNFKTUAOWIPGKEKTQJSALLBWDOAWXALFNGSBVVTQCKUFBGSUZOIOZRISOPRXTKJHNBWSHZUDSMPDQ?

Step 6: Breaking the cipher (most probably the same double-key Vigenere) and reading the message.

The final decrypted message

SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGEDEBRIST

HATENCUMBEREDTHELOWERPARTOFTHEDOORWAYWASREMOVEDW

ITHTREMBLINGHANDSIMADEATINYBREACHINTHEUPPERLEFTH

ANDCORNERANDTHENWIDENINGTHEHOLEALITTLEIINSERTEDT

HECANDLEANDPEEREDINTHEHOTAIRESCAPINGFROMTHECHAMB

ERCAUSEDTHEFLAMETOFLICKERBUTPRESENTLYDETAILSOFTH

EROOMWITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ

Add in a sprinkling of punctuation, and it comes out to be

Slowly, desparatly slowly, the remains of passage debris that encumbered the lower

Part of the doorway was removed. With trembling hands i made a tiny breach in the upper

Lefthand corner and then widening the hole a little i inserted the candle and peered

in. The hot air escaping from the chamber caused the flame to flicker but presently

Details of the room within emerged from the mist x can you see anything q?

The above solution's came from the University of California, San Diego Math Dept.

https://www.math.ucsd.edu/

Half the population at Base Camp was clinically delusional.

-- John Krakauer 1996 Everest Expedition.

Apophenia is : “the tendency to perceive a connection or meaningful pattern between unrelated or random things (such as objects or ideas)”