I have been asked how I solve cryptograms. I thought the best way to illustrate how I do it was to present some some sample cryptograms and write down my thoughts as I solve them. Here are 25 cryptograms, presented in roughly decreasing length.
Solving cryptograms depends on one's ability to spot patterns. You'll often spot the digram TH if the cryptogram contains words like THE, THAT, THERE, THOSE, etc. You will often see common suffixes such as -TION, -SION, -ING, -INE, -NESS, and -LESS.
Solving can sometimes be easier if you can determine which letters represent vowels and which represent consonant. In Cryptograms: A Study of Ciphers and their Solution, Helen Fouché Gaines gives some guidelines for spotting vowels. These are general observations, not hard and fast rules.
Here are the 25 cryptograms I have chosen.
1. ZPL TZOCT SVAA TZVAA CLWOVD ZPLD ZPL TPOMUST UG UXC RCLTLDYL ODM UXC MLLMT POIL IODVTPLM GCUW ZPL LOCZP. ZPLCL VT DU WOD SPU MULT DUZ JDUS ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA KXAFOJUI)
The words ZPL, ZPLD, ZPLCL, and ZPOZ strongly suggest that ZP is the digram TH. In particular, ZPL is the word THE, and ZPOZ is the word THAT. This gives Z=T, P=H, L=E, and O=A. Filling those in, we get
THE TA T E A THE THE
ZPL TZOCT SVAA TZVAA CLWOVD ZPLD ZPL
HA E E E A
TPOMUST UG UXC RCLTLDYL ODM UXC
EE HA E A HE THE EA TH.
MLLMT POIL IODVTPLM GCUW ZPL LOCZP.
THE E A H E T
ZPLCL VT DU WOD SPU MULT DUZ JDUS
THAT. H THE E T T
ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC
E E T A THE TA H HA
LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA
A
KXAFOJUI)
The word LOCZP, EA*TH, is only missing one letter. It's easy to see that it's EARTH, and C=R.
THE TAR T RE A THE THE
ZPL TZOCT SVAA TZVAA CLWOVD ZPLD ZPL
HA R RE E E A R
TPOMUST UG UXC RCLTLDYL ODM UXC
EE HA E A HE R THE EARTH.
MLLMT POIL IODVTPLM GCUW ZPL LOCZP.
THERE A H E T
ZPLCL VT DU WOD SPU MULT DUZ JDUS
THAT. H THE E T T R R
ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC
E E T AR THE TAR H HA
LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA
A
KXAFOJUI)
The word POIL (HA*E) is HAVE. So I=V.
THE TAR T RE A THE THE
ZPL TZOCT SVAA TZVAA CLWOVD ZPLD ZPL
HA R RE E E A R
TPOMUST UG UXC RCLTLDYL ODM UXC
EE HAVE VA HE R THE EARTH.
MLLMT POIL IODVTPLM GCUW ZPL LOCZP.
THERE A H E T
ZPLCL VT DU WOD SPU MULT DUZ JDUS
THAT. H THE E T T R R
ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC
E E T AR THE TAR H HA
LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA
A V)
KXAFOJUI)
The words ZPLD (THE*) and must be one of THEM, THEN, or THEY. When compared with the word DUZ (**T), we see that D=N fits best. This means that DUZ is NOT. So D=N and U=O.
THE TAR T RE A N WHEN THE
ZPL TZOCT SVAA TZVAA CLWOVD SPLD ZPL
HA O O O R RE EN E AN O R
TPOMUST UG UXC RCLTLDYL ODM UXC
EE HAVE VAN HE RO THE EARTH.
MLLMT POIL IODVTPLM GCUW ZPL LOCZP.
THERE NO AN HO OE NOT NO
ZPLCL VT DU WOD SPU MULT DUZ JDUS
THAT. H THEN E NOT T RN O R
ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC
E E TO AR THE TAR H HA
LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA
A OV)
KXAFOJUI)
We have several words from which only one letter is missing. These are UXC (O*R), ODM (AN*), WOD (*AN), SPU (*HO), SL (*E), and ZXCD (T*RN). These words our OUR, AND, MAN, WHO, and TURN, giving X=U, M=D, W=M, and S=W.
THE TAR W T REMA N WHEN THE ZPL TZOCT SVAA TZVAA CLWOVD SPLD ZPL HADOW O OUR RE EN E AND OUR TPOMUST UG UXC RCLTLDYL ODM UXC DEED HAVE VAN HED ROM THE EARTH. MLLMT POIL IODVTPLM GCUW ZPL LOCZP. THERE NO MAN WHO DOE NOT NOW ZPLCL VT DU WOD SPU MULT DUZ JDUS THAT. WH THEN W WE NOT TURN OUR ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC E E TOWARD THE TAR WH (M HA LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA U A OV) KXAFOJUI)
It's now easy to see that TZOCT is STARS, SVAA is WILL, CLWOVD is REMAIN, OG is OF, GCUW is FROM, JDUS is KNOW, and SPE is WHY. This means T=S, V=I, A=L, G=F, J=K, and E=Y.
THE STARS WILL STILL REMAIN WHEN THE ZPL TZOCT SVAA TZVAA CLWOVD SPLD ZPL SHADOWS OF OUR RESEN E AND OUR TPOMUST UG UXC RCLTLDYL ODM UXC DEEDS HAVE VANISHED FROM THE EARTH. MLLMT POIL IODVTPLM GCUW ZPL LOCZP. THERE IS NO MAN WHO DOES NOT KNOW ZPLCL VT DU WOD SPU MULT DUZ JDUS THAT. WHY, THEN, WILL WE NOT TURN OUR ZPOZ. SPE, ZPLD, SVAA SL DUZ ZXCD UXC EYES TOWARDS THE STARS? WHY? (MIKHAIL LELT ZUSOCMT ZPL TZOCT? SPE? (WVJPOVA UL AKOV) KXAFOJUI)
It's easy to see that the word RCLTLDYL (*RESEN*E) is PRESENCE, giving R=P and Y=C. We have the letters K and F left to deal with. The only letters that we have not used in our plaintext are B, G, J, Q, X, and Z. To me, who doesn't speak Russian, GULBAKOV and BULGAKOV seem equally likely. I had to look it up. It's BULGAKOV. The solution is:
THE STARS WILL REMAIN WHEN THE SHADOWS OF OUR PRESENCE ANDOUR DEEDS HAVE VANISHED FROM THE EARTH. THERE IS NO MAN WHO DOES NOT KNOW THAT. WHY, THEN, WILL WE NOT TURN OUR EYES TOWARDS THE STARS? WHY? (MIKHAIL BULGAKOV).
2. Y NZC RUKO RQ EFX CXZ ZOH Y FZJX OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ. WQ DYKCE YHXZ UD WUJXWXOE, UD EFX HZOLX, LXKEZYOGQ LZWX DKUW EFX KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
A single letter word must be one of A, I, or more rarely, O. Sometimes we can decide in favor of A when the single letter occurs in a place where the subject pronoun I is impossible. If we can't decide right away, we at least know that the single letter must be a vowel. As ever in the plaintext of longer cryptograms, the digram TH is likely to occur frequently. The words EFX and EFZE, along with the frequency of the digram EF makes it likely that EF is the digram TH. This means that E=T, F=H, X=E, and Z=A. Now, since we can't also have Y=A, then Y=I.
I A THE EA A I HA E
Y NZC RUKO RQ EFX CXZ ZOH Y FZJX
TI E THAT A THE EAT E E T
OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD
I E HA E TA E A E THE EA.
WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ.
I T I EA E E T, THE
WQ DYKCE YHXZ UD WUJXWXOE, UD EFX
A E, E TAI A E THE
HZOLX, LXKEZYOGQ LZWX DKUW EFX
H TH THE A E (I A A A
KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
The words CXZ (*EA), FZJX (HA*E), ZGG (A**), nd YHXZ (I*EA) appear to be SEA, HAVE, ALL, and IDEA. This gives C=S, J=V, G=L, and H=D.
I AS THE SEA A D I HAVE
Y NZC RUKO RQ EFX CXZ ZOH Y FZJX
TI ED THAT ALL THE EAT EVE TS
OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD
LI E HAVE TA E LA E THE SEA.
WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ.
I ST IDEA VE E T, THE
WQ DYKCE YHXZ UD WUJXWXOE, UD EFX
DA E, E TAI L A E THE
HZOLX, LXKEZYOGQ LZWX DKUW EFX
H TH THE AVES. (ISAD A D A
KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
NZC (*AS) and ZOH (A*D) appear to be the words WAS and AND. This gives N=W and O=N.
I WAS N THE SEA AND I HAVE
Y NZC RUKO RQ EFX CXZ ZOH Y FZJX
N TI ED THAT ALL THE EAT EVENTS
OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD
LI E HAVE TA EN LA E THE SEA.
WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ.
I ST IDEA VE ENT, THE
WQ DYKCE YHXZ UD WUJXWXOE, UD EFX
DAN E, E TAINL A E THE
HZOLX, LXKEZYOGQ LZWX DKUW EFX
H TH THE WAVES. (ISAD A D N AN)
KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
The word EZPXO (TA*EN) appears to be TAKEN, giving P=K. GYDX (LI*E) has several possibilities, but when compared with the word UD, LIFE is most likely. This means D=F, and U must be O (I is already used).
I WAS O N THE SEA AND I HAVE
Y NZC RUKO RQ EFX CXZ ZOH Y FZJX
NOTI ED THAT ALL THE EAT EVENTS OF
OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD
LIFE HAVE TAKEN LA E THE SEA.
WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ.
I ST IDEA OF OVE ENT, OF THE
WQ DYKCE YHXZ UD WUJXWXOE, UD EFX
DAN E, E TAINL A E F O THE
HZOLX, LXKEZYOGQ LZWX DKUW EFX
H TH OF THE WAVES. (ISADO A D N AN)
KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
The word DKUW (F*O*) fits the pattern FROM, making K=R and W=M. KFQEFW (RH*THM) is RHYTHM, making Q=M.
I WAS ORN Y THE SEA AND I HAVE Y NZC RUKO RQ EFX CXZ ZOH Y FZJX NOTI ED THAT ALL THE REAT EVENTS OF OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD MY LIFE HAVE TAKEN LA E Y THE SEA. WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ. MY FIRST IDEA OF MOVEMENT, OF THE WQ DYKCE YHXZ UD WUJXWXOE, UD EFX DAN E, ERTAINLY AME FROM THE HZOLX, LXKEZYOGQ LZWX DKUW EFX RHYTHM OF THE WAVES. (ISADORA D N AN) KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
The words RUKO (*ORN), RQ (*Y), OUEYLXH (NOTI*ED), and TKXZE (*REAT) appear to be BORN, BY, NOTICED, and GREAT. This gives R=B, L=C, H=D, and T=G.
I WAS BORN BY THE SEA AND I HAVE Y NZC RUKO RQ EFX CXZ ZOH Y FZJX NOTICED THAT ALL THE GREAT EVENTS OF OUEYLXH EFZE ZGG EFX TKXZE XJXOEC UD MY LIFE HAVE TAKEN LACE BY THE SEA. WQ GYDX FZJX EZPXO AGZLX RQ EFX CXZ. MY FIRST IDEA OF MOVEMENT, OF THE WQ DYKCE YHXZ UD WUJXWXOE, UD EFX DANCE, CERTAINLY CAME FROM THE HZOLX, LXKEZYOGQ LZWX DKUW EFX RHYTHM OF THE WAVES. (ISADORA D NCAN) KFQEFW UD EFX NZJXC. (YCZHUKZ HSOLZO)
We can see now that the remaining letters are A=P and S=U. The solution is:
I WAS BORN BY THE SEA AND I HAVE NOTICED THAT ALL THE GREAT EVENTS OF MY LIFE HAVE TAKEN PLACE BY THE SEA. MY FIRST IDEA OF MOVEMENT, OF THE DANCE, CERTAINLY CAME FROM THE RHYTHM OF THE WAVES. (ISADORA DUNCAN).
3. KW ANX LF WJNW WJF ARVW KZWFPFVWKZI NAFPKDNZ VWPOIIUF KV WJF VWPOIIUF WR VFW RZFVFUC CPFF CPRA WJF UKAKWV RZF KV LRPZ WR, NZE WJFZ WR UFNPZ VRAFWJKZI RC WJF GNUOF RC WJRVF UKAKWV. (IPFKU ANPDOV)
The words WJNW and WJF, and the frequency of the digram WJ, indicate thst WJ stands for TH. So WJNW is THAT and WJF is THE.
T A E THAT THE T TE E T
KW ANX LF WJNW WJF ARVW KZWFPFVWKZI
A E A T E THE T E T
NAFPKDNZ VWPOIIUF KV WJF VWPOIIUF WR
ET E E EE THE T E
VFW RZFVFUC CPFF CPRA WJF UKAKWV RZF
T A THE T EA
KV LRPZ WR, NZE WJFZ WR UFNPZ
ETH THE A E TH E
VRAFWJKZI RC WJF GNUOF RC WJRVF
T E A
UKAKWV. (IPFKU ANPDOV)
The words KW (*T) and WR (T*) indicate that K=I and R=O.
IT A E THAT THE O T I TE E TI KW ANX LF WJNW WJF ARVW KZWFPFVWKZI A E I A T E I THE T E TO NAFPKDNZ VWPOIIUF KV WJF VWPOIIUF WR ET O E E EE O THE I IT O E VFW RZFVFUC CPFF CPRA WJF UKAKWV RZF I O TO, A THE TO EA KV LRPZ WR, NZE WJFZ WR UFNPZ O ETHI O THE A E O THO E VRAFWJKZI RC WJF GNUOF RC WJRVF I IT EI A UKAKWV. (IPFKU ANPDOV)
The word RZF (O*E) indicates that Z=N. The word WJRVF (THO*E) indicates that V=S. Note the word CPFF (**EE). One common word ending in EE is FREE. This would make CPRA (**O*) FROM. C=F, P=R, A=M.
IT MA E THAT THE MOST INTERESTIN KW ANX LF WJNW WJF ARVW KZWFPFVWKZI AMERI AN STR E IS THE STR E TO NAFPKDNZ VWPOIIUF KV WJF VWPOIIUF WR SET ONESE F FREE FROM THE IMITS ONE VFW RZFVFUC CPFF CPRA WJF UKAKWV RZF IS ORN TO, AN THEN TO EARN KV LRPZ WR, NZE WJFZ WR UFNPZ SOMETHIN OF THE A E OF THOSE VRAFWJKZI RC WJF GNUOF RC WJRVF IMITS. REI MAR S) UKAKWV. (IPFKU ANPDOV)
We can see ANX is MAY, LF is BE, KZWFPFVWKZI is INTERESTING, NAFPKDNZ is AMERICAN, RZFVFUC is ONESELF. So X=Y, L=B, I=G, D=C, and U=L.
IT MAY BE THAT THE MOST INTERESTING KW ANX LF WJNW WJF ARVW KZWFPFVWKZI AMERICAN STR GGLE IS THE STR GGLE TO NAFPKDNZ VWPOIIUF KV WJF VWPOIIUF WR SET ONESELF FREE FROM THE LIMITS ONE VFW RZFVFUC CPFF CPRA WJF UKAKWV RZF IS BORN TO, AN THEN TO LEARN KV LRPZ WR, NZE WJFZ WR UFNPZ SOMETHING OF THE AL E OF THOSE VRAFWJKZI RC WJF GNUOF RC WJRVF LIMITS. (GREIL MARC S) UKAKWV. (IPFKU ANPDOV)
We can see that NZE is AND, so E=D. We can see that VWPOIIUF is STRUGGLE, so O=U. This makes the word GNUOF the word VALUE, so G=V, and the puzzle is solved.
IT MAY BE THAT THE MOST INTERESTING AMERICAN STRUGGLE IS THE STRUGGLE TO SET ONESELF FREE FROM THE LIMITS ONE IS BORN TO, AND THEN TO LEARN SOMETHING OF THE VALUE OF THOSE LIMITS. (GREIL MARCUS).
4. VUMH-XUVQUEG ICMM WUUQ N ANB HXSA DUCBY NDOUEG IKUB KU CV CB GKU QSIUX SH UBUACUV, NBT ICMM UBNDMU KCA GS HUUM GKNG KU ANF DU CB GKU XCYKG IKUB GKU ISXMT CV NYNCBVG KCA. (DUXGXNBT XPVVUMM)
The doubling of U in the word WUUQ indicates that U is either E or O. N is most likely A. Note that NBT is a conjunction, most likely the word AND. The word ICMM and the words CV CB in sequence indicate that C=I, and ICMM is WILL. Let's start with N=A, B=N, T=D, I=W, C=I, and M=L.
L WILL A AN
VUMH-XUVQUEG ICMM WUUQ N ANB HXSA
IN A W N I IN W
DUCBY NDOUEG IKUB KU CV CB GKU QSIUX
N I AND WILL NA L I
SH UBUACUV, NBT ICMM UBNDMU KCA GS
L A A IN I
HUUM GKNG KU ANF DU CB GKU XCYKG
W N W LD I A AIN I
IKUB GKU ISXMT CV NYNCBVG KCA.
AND LL)
(DUXGXNBT XPVVUMM)
GKNG appears to be the word THAT. If K=H, then IKUB is the word WHEN. So G=T, K=H, U=E.
EL E E T WILL EE A AN VUMH-XUVQUEG ICMM WUUQ N ANB HXSA EIN A E T WHEN HE I IN THE WE DUCBY NDOUEG IKUB KU CV CB GKU QSIUX ENE IE AND WILL ENA LE HI T SH UBUACUV, NBT ICMM UBNDMU KCA GS EEL THAT HE A E IN THE I HT HUUM GKNG KU ANF DU CB GKU XCYKG WHEN THE W LD I A AIN T HI IKUB GKU ISXMT CV NYNCBVG KCA. E T AND ELL) (DUXGXNBT XPVVUMM)
GS (T*) shows that S=O. ANB (*AN) shows that A=M. CV (I*) must be IS, because N and T already used. So V=S. UBNDMU is the word ENABLE, so D=B. HUUM is FEEL, so H=F. Notice the word XCYKG (*I*HT). When H comes just before T like this, it's a good indication that the letter before H is G. So Y=G.
SELF- ES E T WILL EE A MAN F OM VUMH-XUVQUEG ICMM WUUQ N ANB HXSA BEING AB E T WHEN HE IS IN THE OWE DUCBY NDOUEG IKUB KU CV CB GKU QSIUX OF ENEMIES, AND WILL ENABLE HIM TO SH UBUACUV, NBT ICMM UBNDMU KCA GS FEEL THAT HE MA BE IN THE IGHT HUUM GKNG KU ANF DU CB GKU XCYKG WHEN THE WO LD IS AGAINST HIM. IKUB GKU ISXMT CV NYNCBVG KCA. (BE T AND SSELL) (DUXGXNBT XPVVUMM)
HXSA (F*OM) indicates that X=R. This means that VUMH-XUVQUEG is SELF-RESPECT (Q=P AND E=C), and the author is BERTRAND RUSSEL (P=U). If E=C, then NDOUEG (AB*ECT) is ABJECT. So O=J. WUUQ is KEEP; W=K. ANF (MA*) is the word MAY; F=Y. The solution is:
SELF-RESPECT WILL KEEP A MAN FROM BEING ABJECT WHEN HE IS IN THE POWER OF ENEMIES, AND WILL ENABLE HIM TO FEEL THAT HE MAY BE IN THE RIGHT WHEN THE WORLD IS AGAINST HIM. (BERTRAND RUSSELL)
5. O LSGOX ATGO AR DRDOFFV YSALAIOF XY NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR)
O is either A or I. Note the word KYSFT'R. The apostrophe indicates that R is either S or T. In either case, the word AR indicates that A=I. Hence O=A.
A A I A I A I I A
O LSGOX ATGO AR DRDOFFV YSALAIOF XY
A I A I A
NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR
JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV
I A A
RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT
A I I (A I
ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR)
Note the word DRDOFFV. Since O is A, -OFFV is probably the common suffix -ALLY. So F=L and V=Y. The word ATGO (I**A) is most likely IDEA, and ATGOR would then be IDEAS. So T=D, G=E, and R=S.
A EA IDEA IS S ALLY I I AL O LSGOX ATGO AR DRDOFFV YSALAIOF XY E A E DIS E E EA IDEAS NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR E E E LD EEDS E EY JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV S D E LD'S I A E A D RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT ESS AD ISSI (A S I EL S) ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR)
DRDOFFV (*S*ALLY) is USUALLY, so D=U. OIT (A?D) is AND, so I=N.
A EA IDEA IS USUALLY I INAL O LSGOX ATGO AR DRDOFFV YSALAIOF XY E AN NE DIS E E EA IDEAS NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR E EN E LD NEEDS E EY JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV SU UND E LD'S I N AN E AND RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT ESS AD ISSI N (AUS IN EL S) ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR) The author's first name is AUSTIN, so X=T. XPG is then T*E, so P=H. The word YIG (*NE) is ONE, so Y=O. A EAT IDEA IS USUALLY O I INAL TO O LSGOX ATGO AR DRDOFFV YSALAIOF XY O E THAN ONE DIS O E E EAT IDEAS NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR O E HEN THE O LD NEEDS THE THEY JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV SU OUND THE O LD'S I NO AN E AND RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT ESS O AD ISSION (AUSTIN HEL S) ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR)
The author's last name is PHELPS, so Z=P. RDSSYDIT (SU**OUND) is SURROUND, so S=R. KPGI (*HEN) is WHEN, so K=W.
A REAT IDEA IS USUALLY ORI INAL TO O LSGOX ATGO AR DRDOFFV YSALAIOF XY ORE THAN ONE DIS O ERER. REAT IDEAS NYSG XPOI YIG TARJYUGSGS. LSGOX ATGOR O E WHEN THE WORLD NEEDS THE THEY JYNG KPGI XPG KYSFT IGGTR XPGN. XPGV SURROUND THE WORLD'S I NORAN E AND RDSSYDIT XPG KYSFT'R ALIYSOIJG OIT PRESS OR AD ISSION (AUSTIN PHELPS) ZSGRR CYS OTNARRAYI. (ODRXAI ZPGFZR)
It's easy to see now that L=G and N=M. JYNG (*OME) is COME, so J=C. TARJYUGSGS (DISCO*ERER) is then DISCOVERER, so J=V. CYS (*OR) is FOR, so C=F. This solution is:
A GREAT IDEA IS USUALLY ORIGINAL TO MORE THAN ONE DISCOVERER. GREAT IDEAS COME WHEN THE WORLD NEEDS THEM. THEY SURROUND THE WORLD'S IGNORANCE AND PRESS FOR ADMISSION. (AUSTIN PHELPS).
6. EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH ZHN GXN DB AXHZN GJ WDBWHS RBP EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ MDZ LGXSBHV. (ROOH SHDBRC)
Normally, when the inner letters of a four letter word double, they indicate vowels. But, when that word is a name, it is just as likely that they indicate consonants. ANNE is a common example. The author's first name is ROOH, which fits the pattern of ANNE. So let's try R=A, O=N, and H=E.
E E A E A E E
EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH
E E E A
ZHN GXN DB AXHZN GJ WDBWHS RBP
A A A A E
EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC
ENA E E NE E
PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ
E (ANNE E A
MDZ LGXSBHV. (ROOH SHDBRC)
The word RSH (A*E) is ARE. S=R.
ERE A ER A E E
EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH
E E ER A
ZHN GXN DB AXHZN GJ WDBWHS RBP
A A R A ARE
EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC
ENA E ER NE E
PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ
R E (ANNE RE A
MDZ LGXSBHV. (ROOH SHDBRC)
The word RQHSDER (A*ER**A) appears to be AMERICA. Q=M, D=I, E=C.
C M I C ERE AMERICA E E EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH E I E I ER A ZHN GXN DB AXHZN GJ WDBWHS RBP I AM A I RIA ARE I EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC ENA I E ERM NE E I PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ I R E (ANNE REI A MDZ LGXSBHV. (ROOH SHDBRC)
The words NMH and MH fit THE and HE. N=T and M=H.
C M I C ERE AMERICA HE HE EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH ET T I E T I ER A ZHN GXN DB AXHZN GJ WDBWHS RBP CI AM A HI T RIA ARE TI EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC ENATI THE TERM NE E IT PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ HI R E (ANNE REI A MDZ LGXSBHV. (ROOH SHDBRC)
CGBW-NHSQ (****-TERM) fits the pattern LONG-TERM. C=L, G=O, B=N, and W=G. Note that this means that my assumption that O=N is wrong.
COL M I CO ERE AMERICA HEN HE EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH ET O T IN E T O GINGER AN ZHN GXN DB AXHZN GJ WDBWHS RBP CINNAMON, AN HI TORIAN ARE TILL EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC E ATING THE LONG-TERM ENE IT O PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ HI O RNE (A E REINAL) MDZ LGXSBHV. (ROOH SHDBRC)
KMHB (*HEN) shows that K=W. GXN (O*T) shows that X=U. RBP (AN*) shows that P=D. ZNDCC (*TILL) shows that Z=S. GJ is OF, since R is already used. So J=F.
COLUM US DISCO ERED AMERICA WHEN HE EGCXQOXZ PDZEGIHSHP RQHSDER KMHB MH SET OUT IN UEST OF GINGER AND ZHN GXN DB AXHZN GJ WDBWHS RBP CINNAMON, AND HISTORIANS ARE STILL EDBBRQGB, RBP MDZNGSDRBZ RSH ZNDCC DE ATING THE LONG-TERM ENEFITS OF PHORNDBW NMH CGBW-NHSQ OHBHJDNZ GJ HIS OURNE (A E REINAL) MDZ LGXSBHV. (ROOH SHDBRC)
Inspection of what has been filled in so far shows that O=B, I=V, A=Q, L=J, and V=Y. The solution is:
COLUMBUS DISCOVERED AMERICA WHEN HE SET OUT IN QUEST OF GINGER AND CINNAMON, AND HISTORIANS ARE STILL DEBATING THE BENEFITS OF HIS JOURNEY. (ABBE REINAL)
7. OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM, HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO RXM QXHWTOZ RQHWZ QE XWNMXB, YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE CHHNB. (RYQXHBM QGMXDM)
The word SOGDO follows the pattern of WHICH. There are other words which follow this pattern, such as ORDER, but the letter O doesn't occur very frequently and is probably not R. The letter S occurs only once and is probably not a vowel. We will start with S=W, O=H, G=I, D=C.
HI I CC
OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM,
I WHICH
HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO
H
RXM QXHWTOZ RQHWZ QE XWNMXB,
I
YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE
I C
CHHNB. (RYQXHBM QGMXDM)
RDDHWJZ probably begins with ACC. RJ then would be AN. So R=A and J=N. RJK is AND, so K=D.
HI I AN ACC N A
OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM,
N NI AN WHICH
HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO
A H A
RXM QXHWTOZ RQHWZ QE XWNMXB,
NA AND DI
YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE
(A I C
CHHNB. (RYQXHBM QGMXDM)
GB is more likely to be IS than IT, and the first word appears to be HISTORY. B=S, Z=T, H=O, X=R, and E=Y.
HISTORY IS AN ACCO NT, OST Y A S OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM, O NTS, OST Y NI ORTANT, WHICH HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO AR RO HT A O T Y R RS, RXM QXHWTOZ RQHWZ QE XWNMXB, OST Y NA S AND SO DI RS, OST Y YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE OO S. (A ROS I RC CHHNB. (RYQXHBM QGMXDM)
QE (*Y) must be either BY or MY. If Q is M, then QXHWTOZ is MRO**HT, which doesn't make any sense. So Q=B, and QXHWTOZ is BROUGHT. W=U and T=G.
HISTORY IS AN ACCOUNT, OST Y A S OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM, O NTS, OST Y UNI ORTANT, WHICH HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO AR BROUGHT ABOUT BY RU RS, RXM QXHWTOZ RQHWZ QE XWNMXB, OST Y NA S AND SO DI RS, OST Y YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE OO S. (A BROS BI RC CHHNB. (RYQXHBM QGMXDM)
Since R has already been used, HC must be OF. C=H. CHHNB (FOO*S) is FOOLS. N=L. YHBZNE (*OSTLY) is MOSTLY. Y=M.
HISTORY IS AN ACCOUNT, MOSTLY FALS OGBZHXE GB RJ RDDHWJZ, YHBZNE CRNBM, OF NTS, MOSTLY UNIM ORTANT, WHICH HC MPMJZB, YHBZNE WJGYIHXZRJZ, SOGDO AR BROUGHT ABOUT BY RUL RS, RXM QXHWTOZ RQHWZ QE XWNMXB, MOSTLY NA S AND SOLDI RS, MOSTLY YHBZNE AJRPMB, RJK BHNKGMXB, YHBZNE FOOLS. (AMBROS BI RC CHHNB. (RYQXHBM QGMXDM)
By now, your eye should be practiced enough to fill the rest in without further hints. The solution:
HISTORY IS AN ACCOUNT, MOSTLY FALSE, OF EVENTS, MOSTLY UNIMPORTANT, WHICH ARE BROUGHT ABOUT BY RULERS. MOSTLY KNAVES, AND SOLDIERS, MOSTLY FOOLS. (AMBROSE BIERCE)
8. GC WRT NBPS SR OPRN WRTZ SZTK RMGPGRP RC VRHKRPK, NBSQE SEK KCCKQS MZRATQKA GP WRT DW SEK CGZVS VGXES RC B UKSSKZ CZRH EGH. (BZSETZ VQERMKPEBTKZ)
Note the words SR and RC. Letters that can appear as either the first or the second letter in a two letter word are N, O, S, and T. If R is one of N, S, or T, then C must be O. But note the word CZRH. This word would have to fit O*N*, O*S*, or O*T*. The only thing that really fits is OATH, but H can't stand for itself. R=O is the best fit. Note the words GC and RC. IF and OF are the best fit. G=I, C=F.
IF O O O O GC WRT NBPS SR OPRN WRTZ SZTK O I IO OF O O RMGPGRP RC VRHKRPK, NBSQE SEK FF O I O FI KCCKQS MZRATQKA GP WRT DW SEK CGZVS I OF F O I VGXES RC B UKSSKZ CZRH EGH. (BZSETZ O VQERMKPEBTKZ)
CZRH is the word FROM. Z=R and H=M. EGH is *IM, so E=H. B=A, since I is already used.
IF O A O O O R R GC WRT NBPS SR OPRN WRTZ SZTK O I IO OF OM O A H H RMGPGRP RC VRHKRPK, NBSQE SEK FF RO I O H FIR KCCKQS MZRATQKA GP WRT DW SEK CGZVS I H OF A R FROM HIM. (AR H R VGXES RC B UKSSKZ CZRH EGH. (BZSETZ HO HA R) VQERMKPEBTKZ)
Learn to spot words like WRT (*O*) and WRTZ (*O*R) when you've used O and R. These are the words YOU and YOUR. W=Y and T=U. This makes the author's first name ARTHUR. S=T. SEK is THE. S=T and K=E. CGZVS is FIRST. V=S. Let's fill these in before continuing.
IF YOU A T TO O YOUR TRUE GC WRT NBPS SR OPRN WRTZ SZTK O I IO OF SOMEO E, AT H THE RMGPGRP RC VRHKRPK, NBSQE SEK EFFE T RO U E I YOU Y THE FIRST KCCKQS MZRATQKA GP WRT DW SEK CGZVS SI HT OF A ETTER FROM HIM. (ARTHUR VGXES RC B UKSSKZ CZRH EGH. (BZSETZ S HO E HAUER) VQERMKPEBTKZ)
VRHKRPK is SOMEONE. P=N. RMGPGRP, then, is OPINION. M=P. KCCKQS is EFFECT. Q=C. DW is BY. D=B VGXES is SIGHT. X=G.
IF YOU ANT TO NO YOUR TRUE GC WRT NBPS SR OPRN WRTZ SZTK OPINION OF SOMEONE, ATCH THE RMGPGRP RC VRHKRPK, NBSQE SEK EFFECT PRO UCE IN YOU BY THE FIRST KCCKQS MZRATQKA GP WRT DW SEK CGZVS SIGHT OF A ETTER FROM HIM. (ARTHUR VGXES RC B UKSSKZ CZRH EGH. (BZSETZ SCHOPENHAUER) VQERMKPEBTKZ)
The word OPRN (*NO*) follows the pattern of KNOW. With that, it's easy to fill the rest in. The solution is:
IF YOU WANT TO KNOW YOUR TRUE OPINION OF SOMEONE, WATCH THE EFFECT PRODUCED IN YOU BY THE FIRST SIGHT OF A LETTER FROM HIM. (ARTHUR SCHOPENHAUER)
9. S WPAN SP XQVA VLMBACQYX NLJ S ZLG EQYDZAG EZA ZAV QU EZA YPWPQNP. LPG KASPH VA, S NLPEAG EQ CSUE EZLE ZAVCSPA L CSEECA KSE VQMA. (VLA NAXE)
S and L are both single letter words. One is A and one is I. Note the words EZA and EZLE. If L is A, these appear to be THE and THAT. So let S=I, L=A, E=T, Z=H, and A=E.
I E I E A E A I HA S WPAN SP XQVA VLMBACQYX NLJ S ZLG T HE THE HE THE A EQYDZAG EZA ZAV QU EZA YPWPQNP. LPG EI E, I A TE T I T THAT KASPH VA, S NLPEAG EQ CSUE EZLE HE I E A ITT E IT E. AE E T) ZAVCSPA L CSEECA KSE VQMA. (VLA NAXE)
EQ is TO. Q=O. CSEECA (*ITT*E) is LITTLE. C=L. What first name fits *AE? If you're old enough you've heard of MAE WEST. V=M, N=W, X=S.
I EW I SOME MA ELO S WA I HA S WPAN SP XQVA VLMBACQYX NLJ S ZLG TO HE THE HEM O THE OW A EQYDZAG EZA ZAV QU EZA YPWPQNP. LPG EI ME, I WA TE TO LI T THAT KASPH VA, S NLPEAG EQ CSUE EZLE HEMLI E A LITTLE IT MO E. (MAE WEST) ZAVCSPA L CSEECA KSE VQMA. (VLA NAXE)
The phrase L CSEECA KSE VQMA is A LITTLE BIT MORE. K=B and M=R. ZAVCSPA is HEMLINE. P=N. WPAN is KNEW. W=K. ZLG is HAD because S is already used. G=D.
I KNEW IN SOME MAR ELO S WA I HAD S WPAN SP XQVA VLMBACQYX NLJ S ZLG TO HED THE HEM O THE NKNOWN. AND EQYDZAG EZA ZAV QU EZA YPWPQNP. LPG BEIN ME, I WANTED TO LI T THAT KASPH VA, S NLPEAG EQ CSUE EZLE HEMLINE A LITTLE BIT MORE. (MAE WEST) ZAVCSPA L CSEECA KSE VQMA. (VLA NAXE)
YPWPQNP is the word UNKNOWN. Taking Y=U, the rest is easy to fill in. The solution is:
I KNEW IN SOME MARVELOUS WAY I HAD TOUCHED THE HEM OF THE UNKNOWN. AND BEING ME, I WANTED TO LIFT THAT HEMLINE A LITTLE BIT MORE. (MAE WEST)
10. UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY EO YDA UTXYHBAX T UHFA NDTKA KAHZSTSP YE XAA YDTSPX OZEU YDA GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
Since H and T are single letter words, one is A and one is I. GKHSY'X indicates that X is either S or T. Consider the words HX H. If H is I, then we have either IS I or IT I, neither of which make sense. so H=A and T=I.
A A A
UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY
I A I A I
EO YDA UTXYHBAX T UHFA NDTKA
A I I
KAHZSTSP YE XAA YDTSPX OZEU YDA
A I I A
GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
Take a look at the words KAHZSTSP and YDTSPX. Both contain the trigram TSP, which stands for I**. This suggests an -ING ending. Let S=N and P=G.
G N A N A A
UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY
I A I A I
EO YDA UTXYHBAX T UHFA NDTKA
A NING ING
KAHZSTSP YE XAA YDTSPX OZEU YDA
AN IN I A
GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
PZASS (G***N) fits the pattern of GREEN. Let Z=R and A = E.
GREEN A E N A A RE
UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY
E I A E I A E I E
EO YDA UTXYHBAX T UHFA NDTKA
EARNING EE ING R E
KAHZSTSP YE XAA YDTSPX OZEU YDA
AN IN IE RE A E)
GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
XAA (*EE) shows that X=S. KAHZSTSP is either LEARNING or YEARNING. Trying K=Y doesn't work (consider NDTKA). So K=L. YDTSPX is **INGS, which fits THINGS. Y=T and D=H.
GREEN TH A E NL AS A RES LT UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY THE ISTA ES I A E HILE EO YDA UTXYHBAX T UHFA NDTKA LEARNING T SEE THINGS R THE KAHZSTSP YE XAA YDTSPX OZEU YDA LANT'S INT IE (H. RE ALE) GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
ZAXCKY is RESULT. C=U. YE is TO. E=O. NDTKA is WHILE. N=W.
GREEN THU A E ONL AS A RESULT UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY O THE ISTA ES I A E WHILE EO YDA UTXYHBAX T UHFA NDTKA LEARNING TO SEE THINGS RO THE KAHZSTSP YE XAA YDTSPX OZEU YDA LANT'S OINT O IEW. (H. RE ALE) GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
EO is OF, since R has already been used. O=F. ESKQ is ONLY. Q=Y. ITAN is VIEW. I=V. OZEU is FROM. U=M.
MY GREEN THUM AME ONLY AS A RESULT UQ PZAAS YDCUJ WHUA ESKQ HX H ZAXCKY OF THE MISTA ES I MA E WHILE EO YDA UTXYHBAX T UHFA NDTKA LEARNING TO SEE THINGS FROM THE KAHZSTSP YE XAA YDTSPX OZEU YDA LANT'S OINT OF VIEW. (H. FRE ALE) GKHSY'X GETSY EO ITAN. (D. OZAF HKA)
The solution is:
MY GREEN THUMB CAME ONLY AS A RESULT OF THE MISTAKES I MADE WHILE LEARNING TO SEE THINGS FROM THE PLANT'S POINT OF VIEW. (H. FRED ALE)
11. BS'G DHGGBOAC SH HLF SHH KXQV. P KPF LBSV HFC LPSQV MFHLG LVPS SBKC BS BG; P KPF LBSV SLH LPSQVCG BG FCRCT NXBSC GXTC. (ACC GCEPAA)
BS'G is most likely IT'S. B=I, S=T, G=S. Since P stands by itself, P=A.
IT'S SSI T T A A BS'G DHGGBOAC SH HLF SHH KXQV. P KPF IT AT S AT TI IT IS; LBSV HFC LPSQV MFHLG LVPS SBKC BS BG; A A IT T AT S IS P KPF LBSV SLH LPSQVCG BG FCRCT IT S S A NXBSC GXTC. (ACC GCEPAA)
H=O. Since SLH is T*O, L=W. LBSV is WITH, so V=H.
IT'S OSSI TO OW TOO H. A A BS'G DHGGBOAC SH HLF SHH KXQV. P KPF WITH O WAT H OWS WHAT TI IT IS; LBSV HFC LPSQV MFHLG LVPS SBKC BS BG; A A WITH TWO WAT H S IS P KPF LBSV SLH LPSQVCG BG FCRCT IT S S A NXBSC GXTC. (ACC GCEPAA)
HLF is OWN; F=N. MFLHG is KNOWS; M=K. LPSQV is WATCH; Q=C. LPSQVCG is WATCHES; C=E.
IT'S POSSIBLE TO OWN TOO MUCH. A MAN BS'G DHGGBOAC SH HLF SHH KXQV. P KPF WITH ONE WATCH KNOWS WHAT TIME IT IS; LBSV HFC LPSQV MFHLG LVPS SBKC BS BG; A MAN WITH TWO WATCHES IS NE E P KPF LBSV SLH LPSQVCG BG FCRCT UITE SU E. (LEE SE ALL) NXBSC GXTC. (ACC GCEPAA)
DHGGBOAC is POSSIBLE; D=P, O=B, and A=L. KXQV is MUCH; K=M and X=U. The solution is:
IT'S POSSIBLE TO OWN TOO MUCH. A MAN WITH ONE WATCH KNOWS WHAT TIME IT IS; A MAN WITH TWO WATCHES IS NEVER QUITE SURE. (LEE SEGALL)
12. SG JDT KOLH FHDFBH CRSWL CRHJ'YH CRSWLSWE, CRHJ'BB BDQH JDT; UTC SG JDT YHOBBJ KOLH CRHK CRSWL, CRHJ'BB ROCH JDT. (IDW KOYXTSZ)
CRHJ'BB is THEY'LL. C=T, R=H, H=E, J=Y, and B=L. CRHJ'YH can be either THEY'RE or THEY'VE. Y=R fits best, because then YHOBBJ (RE*LLY) is REALLY. O=A.
Y A E E LE TH THEY'RE SG JDT KOLH FHDFBH CRSWL CRHJ'YH TH THEY'LL L E Y T CRSWLSWE, CRHJ'BB BDQH JDT; UTC SG Y REALLY A E THE TH THEY'LL JDT YHOBBJ KOLH CRHK CRSWL, CRHJ'BB HATE Y AR ROCH JDT. (IDW KOYXTSZ)
JDT is YOU; D=O and T=U. BDQH is LOVE; Q=V. UTC is BUT; U=B. S=I; all other vowels have been used.
I YOU A E EO LE THI THEY'RE SG JDT KOLH FHDFBH CRSWL CRHJ'YH THI I THEY'LL LOVE YOU; BUT I CRSWLSWE, CRHJ'BB BDQH JDT; UTC SG YOU REALLY A E THE THI THEY'LL JDT YHOBBJ KOLH CRHK CRSWL, CRHJ'BB HATE YOU O AR UI ROCH JDT. (IDW KOYXTSZ)
FHDFBH is PEOPLE; F=P. CRHK must be THEM, because Y is already used; K=M. KOLH is MAKE; L=K. CRSWL is THINK; W=N. SG is IF; G=F.
IF YOU MAKE PEOPLE THINK THEY'RE SG JDT KOLH FHDFBH CRSWL CRHJ'YH THINKIN THEY'LL LOVE YOU; BUT IF CRSWLSWE, CRHJ'BB BDQH JDT; UTC SG YOU REALLY MAKE THEM THINK, THEY'LL JDT YHOBBJ KOLH CRHK CRSWL, CRHJ'BB HATE YOU. ON MAR UI ROCH JDT. (IDW KOYXTSZ)
We need to choose two letters for the author's name. Let's try placing a Q before the U. MARQUI* fits MARQUIS. So X=Q and Z=S. Here is the solution:
IF YOU MAKE PEOPLE THINK THEY'RE THINKING, THEY'LL LOVE YOU; BUT IF YOU REALLY MAKE THEM THINK, THEY'LL HATE YOU. (DON MARQUIS)
13. FJA OACLQV GJQ LFCRIL RGRI HCQN FJA LQTCXA DL TVCQQFAM RVM DL WDZA MTLF EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA RLRVFA)
FJA is frequent enough to be the word THE. F=T, J=H, and A=E.
THE E H T THE
FJA OACLQV GJQ LFCRIL RGRI HCQN FJA
E TE E T
LQTCXA DL TVCQQFAM RVM DL WDZA MTLF
T THE E ETE
EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA
TE)
RLRVFA)
GJQ is WHO; G=W and Q=O.
THE E O WHO T W O THE
FJA OACLQV GJQ LFCRIL RGRI HCQN FJA
O E OOTE E T
LQTCXA DL TVCQQFAM RVM DL WDZA MTLF
OW O T THE W O E ETE
EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA
TE)
RLRVFA)
RGRI (*W**) is AWAY. R=A and I=Y.
THE E O WHO T AY AWAY O THE FJA OACLQV GJQ LFCRIL RGRI HCQN FJA O E OOTE A E T LQTCXA DL TVCQQFAM RVM DL WDZA MTLF OW A O T Y THE W O E ETE EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA A A TE) RLRVFA)
LFCRIL is STRAYS; L=S and C=R. HCQN is FROM; H=F and N=M. REQTF is ABOUT; E=B and T=U.
THE ERSO WHO STRAYS AWAY FROM THE FJA OACLQV GJQ LFCRIL RGRI HCQN FJA SOUR E S U ROOTE A S E UST LQTCXA DL TVCQQFAM RVM DL WDZA MTLF OW ABOUT BY THE W (MO EF ETE EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA ASA TE) RLRVFA)
OACLQV is PERSON; O=P and V=N. DL is IS; D=I. RVM is AND; M=D. LQTCXA is SOURCE; X=C.
THE PERSON WHO STRAYS AWAY FROM THE FJA OACLQV GJQ LFCRIL RGRI HCQN FJA SOURCE IS UNROOTED AND IS I E DUST LQTCXA DL TVCQQFAM RVM DL WDZA MTLF OWN ABOUT BY THE WIND. (MO EFI ETE EWQGV REQTF EI FJA GDVM. (NQWAHD ZAFA ASANTE) RLRVFA)
WDZA is LIKE; W=L and Z=K. EWQGV is BLOWN; E=B. The solution is:
THE PERSON WHO STRAYS AWAY FROM THE SOURCE IS UNROOTED AND IS LIKE DUST BLOWN ABOUT BY THE WIND. (MOLEFI KETE ASANTE)
14. L MCKX PIRXD GLVMXT L MCT RLUX RP HFARLKCRX UPTXVRQ, EFR L CU RPP EFVQ RMLDBLDO CEPFR UQVXAI. (XTLRM VLRGXAA)
RP and RPP suggest TO and TOO. If R=T, then EFR probably BUT. Let's start with R=T, P=O, E=B, and F=U. Note that L is either A or I.
O T T TO
L MCKX PIRXD GLVMXT L MCT RLUX RP
U T T O T BUT TOO BU
HFARLKCRX UPTXVRQ, EFR L CU RPP EFVQ
T BOUT T T
RMLDBLDO CEPFR UQVXAI. (XTLRM VLRGXAA)
CEPFR is ABOUT, which means C=A and L=I. CU is AM, so U=M. EFVQ is BUSY; V=S and Q=Y.
I A O T IS I A TIM TO L MCKX PIRXD GLVMXT L MCT RLUX RP U TI AT MO STY, BUT I AM TOO BUSY HFARLKCRX UPTXVRQ, EFR L CU RPP EFVQ T I I ABOUT MYS IT SIT RMLDBLDO CEPFR UQVXAI. (XTLRM VLRGXAA)
RLUX is TIME; X=E. UPTXVRQ is MODESTY; T=D. UQVXAI is MYSELF; A=L and I=F.
I A E OFTE IS ED I AD TIME TO L MCKX PIRXD GLVMXT L MCT RLUX RP ULTI ATE MODESTY, BUT I AM TOO BUSY HFARLKCRX UPTXVRQ, EFR L CU RPP EFVQ T I I ABOUT MYSELF. (EDIT SIT ELL) RMLDBLDO CEPFR UQVXAI. (XTLRM VLRGXAA)
The author's name is EDITH SITWELL; M=H and G=W. MCKX is HAVE; K=V. PIRXD is OFTEN; D=N. HFARLKCRX (*ULTIVATE) is CULTIVATE; H=C. RMLDBLDO (THIN*IN*) is THINKING; B=K and O=G. The solution is:
I HAVE OFTEN WISHED I HAD TIME TO CULTIVATE MODESTY, BUT I AM TOO BUSY THINKING ABOUT MYSELF. (EDITH SITWELL)
15. HF HK SHKZ FT VZQZQCZV FBOF LTW OVZ TDZ TE FBTKZ SBT XOD CZ ETTGZJ KTQZ TE FBZ FHQZ. (GOWVZDXZ U. RZFZV)
When ever something with the pattern HF HK appears at the beginning of a sentence, it is very probable that is is the phrase "IT IS". If F=T, then FBOF is THAT. Also, FT is TO. Let's start with H=I, F=T, K=S, B=H, O=A, and T=O.
IT IS IS TO THAT O A HF HK SHKZ FT VZQZQCZV FBOF LTW OVZ O O THOS HO A OO SO O TDZ TE FBTKZ SBT XOD CZ ETTGZJ KTQZ TE TH TI A T FBZ FHQZ. (GOWVZDXZ U. RZFZV)
FBZ is THE; Z=E. SBT is WHO; S=W. OVZ (A*E) is ARE; V=R. CZ (*E) is BE; C=B.
IT IS WISE TO RE E BER THAT O ARE HF HK SHKZ FT VZQZQCZV FBOF LTW OVZ O E O THOSE WHO A BE OO E SO E O TDZ TE FBTKZ SBT XOD CZ ETTGZJ KTQZ TE THE TI E. A RE E ETER) FBZ FHQZ. (GOWVZDXZ U. RZFZV)
VZQZQCZV is REMEMBER; Q=M. TDZ is ONE; D=N. XOD (*AN) is CAN; X=C. LTW fits YOU; L=Y and W=U.
IT IS WISE TO REMEMBER THAT YOU ARE HF HK SHKZ FT VZQZQCZV FBOF LTW OVZ ONE O THOSE WHO CAN BE OO E SOME O TDZ TE FBTKZ SBT XOD CZ ETTGZJ KTQZ TE THE TIME. AURENCE ETER) FBZ FHQZ. (GOWVZDXZ U. RZFZV)
After you've been solving cryptograms for a while, you will come to recognize frequently quoted authors. This author is LAURENCE J. PETER. The solution is:
IT IS WISE TO REMEMBER THAT YOU ARE ONE OF THOSE WHO CAN BE FOOLED SOME OF THE TIME. (LAURENCE J. PETER)
16. FMT RBCGFPE'W GNTP SJ MTAA GW P RAPIT OMTCT TQTCZHSNZ MPW FS DGEN MGW SOE HBWGETWW. (OTENTAA RMGAAGRW)
Whenever an apostrophe followed by a letter occurs at the end of a long word, it's a good bet that the letter is an S. Since we have RBCGFPE'W, W is probably S. Note the last word, HBWGETWW. -LESS and -NESS are common suffixes. We'll start with W=S and T=E. E is either L or N. P is either A or I.
E 'S E E S E
FMT RBCGFPE'W GNTP SJ MTAA GW P RAPIT
E E E E S S
OMTCT TQTCZHSNZ MPW FS DGEN MGW
S ESS. E E S)
SOE HBWGETWW. (OTENTAA RMGAAGRW)
GW P is IS A; G=I and P=A. HBWGETWW is BUSINESS; H=B, B=U, E=N. FMT is THE; F=T and M=H. OMTCT is WHERE; O=W and C=R.
THE URITAN'S I EA HE IS A A E FMT RBCGFPE'W GNTP SJ MTAA GW P RAPIT WHERE E ER B HAS T IN HIS OMTCT TQTCZHSNZ MPW FS DGEN MGW WN BUSINESS. (WEN E HI I S) SOE HBWGETWW. (OTENTAA RMGAAGRW)
MTAA is HELL and RAPIT is PLACE. The solution is:
THE PURITAN'S IDEA OF HELL IS A PLACE WHERE EVERYBODY HAS TO MIND HIS OWN BUSINESS. (WENDELL PHILLIPS)
17. IW ESOW G CNPO. PGED HZ BKW QNLTGPW, INB GEFGAQ RGCCESZV ESOW BKW CSPOWZQ NZCWLZWGBK. (DSPKGWE PGSZW)
G=A. G can't be I because it doesn't fit as a subject pronoun. Besides, GEFGAQ fits the pattern of ALWAYS. We'll start with G=A, E=L, F=W, A=Y, and Q=S.
L A AL S A
IW ESOW G CNPO. PGED HZ BKW QNLTGPW,
ALWAYS A L L
INB GEFGAQ RGCCESZV ESOW BKW
S A A L A
CSPOWZQ NZCWLZWGBK. (DSPKGWE PGSZW)
The digram BK is common, and BKW seems to be THE. B=T, K=H, and W=E. IW appears to be WE at first glance, but then INB (W*T) doesn't fit. IW is BE and INB is BUT. I=B and N=U.
BE L E A U AL THE SU A E,
IW ESOW G CNPO. PGED HZ BKW QNLTGPW,
BUT ALWAYS A L L E THE
INB GEFGAQ RGCCESZV ESOW BKW
E S U E EATH. HAEL A E)
CSPOWZQ NZCWLZWGBK. (DSPKGWE PGSZW)
The author's first name appears to be MICHAEL. D=M, S=I, P=C. ESOW is LIKE; O=K.
BE LIKE A UCK. CALM THE SU ACE, IW ESOW G CNPO. PGED HZ BKW QNLTGPW, BUT ALWAYS A LI LIKE THE INB GEFGAQ RGCCESZV ESOW BKW ICKE S U E EATH. (MICHAEL CAI E) CSPOWZQ NZCWLZWGBK. (DSPKGWE PGSZW)
The author's last name is CAINE; Z=N. CNPO is DUCK; C=D. QNLTGPW is SURFACE; L=R and T=F.
BE LIKE A DUCK. CALM N THE SURFACE, IW ESOW G CNPO. PGED HZ BKW QNLTGPW, BUT ALWAYS ADDLIN LIKE THE INB GEFGAQ RGCCESZV ESOW BKW DICKENS UNDE NEATH. (MICHAEL CAINE) CSPOWZQ NZCWLZWGBK. (DSPKGWE PGSZW)
The solution is:
BE LIKE A DUCK. CALM ON THE SURFACE, BUT ALWAYS PADDLING LIKE THE DICKENS UNDERNEATH. (MICHAEL CAINE)
18. F ZFDD XHCUFL BG UOB LG BOCCGZ OBV VHPCOVH UM RGQD YM UOWFBP UH IOLH IFU. (YGGWHC L. ZORIFBPLGB)
F ZFDD looks like I WILL. F=I, Z=W, and D=L.
I WILL I W
F ZFDD XHCUFL BG UOB LG BOCCGZ OBV
L I
VHPCOVH UM RGQD YM UOWFBP UH IOLH
I W I
IFU. (YGGWHC L. ZORIFBPLGB)
The fact that BOCCGZ ends in plaintext W, and that G doubles in the author's first name indicates that G is E or O. I think that -OW fits better than -EW here, so I'll go with G=O.
I WILL I O O OW
F ZFDD XHCUFL BG UOB LG BOCCGZ OBV
O L I
VHPCOVH UM RGQD YM UOWFBP UH IOLH
I OO W I O
IFU. (YGGWHC L. ZORIFBPLGB)
-TON is a common suffix for a last name, and this makes the words BG and LG the words NO and TO. B=N and L=T.
I WILL IT NO N TO N OW N
F ZFDD XHCUFL BG UOB LG BOCCGZ OBV
O L IN T
VHPCOVH UM RGQD YM UOWFBP UH IOLH
I OO T. W IN TON)
IFU. (YGGWHC L. ZORIFBPLGB)
The word BOCCGZ is NARROW. O=A and C=R. OBV is AND; V=D. UOB is MAN; U=M.
I WILL RMIT NO MAN TO NARROW AND F ZFDD XHCUFL BG UOB LG BOCCGZ OBV D RAD M O L MA IN M AT VHPCOVH UM RGQD YM UOWFBP UH IOLH IM. OO R T. WA IN TON) IFU. (YGGWHC L. ZORIFBPLGB)
XHCUFL is PERMIT; X=P and H=E. VHPCOVH is DEGRADE; P=G. UM is MY; M=Y. UOWFBP is MAKING; W=K. IOLH is HATE; I=H.
I WILL PERMIT NO MAN TO NARROW AND F ZFDD XHCUFL BG UOB LG BOCCGZ OBV DEGRADE MY O L Y MAKING ME HATE VHPCOVH UM RGQD YM UOWFBP UH IOLH HIM. OOKER T. WA HINGTON) IFU. (YGGWHC L. ZORIFBPLGB)
The solution is:
I WILL PERMIT NO MAN TO NARROW AND DEGRADE MY SOUL BY MAKING ME HATE HIM. (BOOKER T. WASHINGTON)
19. KRG PYC DJ PB PIESCGBK DI LYXTSXXYDB XRDSVL BDK MG AYTKDIU, MSK QIDEIGXX. (ZDXGQR ZDSMGIK)
Note the suffix -GXX in the last word. This is probably -ESS. Let G=E and X=S. With G=E, the first word is probably THE. Let K=T, R=H, and G=E.
THE E T S SS KRG PYC DJ PB PIESCGBK DI LYXTSXXYDB SH T E T T ESS. XRDSVL BDK MG AYTKDIU, MSK QIDEIGXX. SE H E T) (ZDXGQR ZDSMGIK)
The word MSK is probably BUT. NOT is also a conjunction that would fit, but that would make MG to be NE, which is not a word. Let M=B and S=U. Making S=U makes XRDSVL look like SHOULD. Let D=O, V=L, and L=D.
THE O U E T O D S USS O KRG PYC DJ PB PIESCGBK DI LYXTSXXYDB SHOULD OT BE TO BUT O ESS. XRDSVL BDK MG AYTKDIU, MSK QIDEIGXX. OSE H OUBE T) (ZDXGQR ZDSMGIK)
The word BDK is NOT; B=N. LYXTSXXYDB is DISCUSSION; Y=I and T=C. PN is AN; P=A. The author's first name is JOSEPH; Z=J and Q=P. QIDEIGXX is PROGRESS; I=R and E=G.
THE AI O AN ARGU ENT OR DISCUSSION KRG PYC DJ PB PIESCGBK DI LYXTSXXYDB SHOULD NOT BE ICTOR BUT PROGRESS. XRDSVL BDK MG AYTKDIU, MSK QIDEIGXX. (JOSEPH JOUBERT) (ZDXGQR ZDSMGIK)
The solution is:
THE AIM OF AN ARGUMENT OR DISCUSSION SHOULD NOT BE VICTORY, BUT PROGRESS. (JOSEPH JOUBERT)
20. QOII, XM X BSIIOK WRO QDHVL VNCAOD, QRU KXK UHN SVJQOD WRO TRHVO? (GSCOJ WRNDAOD)
X is either A or I. KXK is probably DID. Let K=D and X=I.
I I D
QOII, XM X BSIIOK WRO QDHVL VNCAOD,
DID
QRU KXK UHN SVJQOD WRO TRHVO?
(GSCOJ WRNDAOD)
This cryptogram is a question. XM X is probably IF I, and QRU XKX is probably WHO DID. If QRU is WHO, then WRO is probably THE. This would make QOII the word WELL. Let M=F, Q=W, R=H, U=O, W=T, O=E, and I=L.
WELL, IF I LLED THE W E
QOII, XM X BSIIOK WRO QDHVL VNCAOD,
WHO DID O WE THE H E?
QRU KXK UHN SVJQOD WRO TRHVO?
E TH E
(GSCOJ WRNDAOD)
BSIIOK looks like CALLED, and S=A makes SVJQOD look like ANSWER. Let B=C, S=A, V=N, J=S, and D=R.
WELL, IF I CALLED THE WR N N ER, QOII, XM X BSIIOK WRO QDHVL VNCAOD, WHO DID O ANSWER THE H NE? QRU KXK UHN SVJQOD WRO TRHVO? A ES TH R ER) (GSCOJ WRNDAOD)
TRHVO looks like PHONE. This shows that I made a wrong choice. What I thought was WHO DID is actually WHY DID. But the identification of WH was enough to help us proceed. In any case, we're pretty close to the solution. The solution is:
WELL, IF I CALLED THE WRONG NUMBER, WHY DID YOU ANSWER THE PHONE? (JAMES THURBER)
21. PU ORXAS FMISSI RMH QSSC HSWXVSH VX DSSYJCE YSXYAS ZIXP DCXOJCE PS. (AXC FRMCSU)
Note QSSC. S is probably E or O. E fits better in YSXYAS than O does. Let S=E. YSXYAS looks like PEOPLE. Let Y=P, X=O, and A=L.
OLE EE EE E O E O
PU ORXAS FMISSI RMH QSSC HSWXVSH VX
EEP PEOPLE O O E. (LO
DSSYJCE YSXYAS ZIXP DCXOJCE PS. (AXC
E
FRMCSU)
The suffix JCE occurs twice and is probably ING. Let J=I, C=N and E=G.
OLE EE EEN E O E O
PU ORXAS FMISSI RMH QSSC HSWXVSH VX
EEPING PEOPLE O NO ING E. (LON
DSSYJCE YSXYAS ZIXP DCXOJCE PS. (AXC
NE
FRMCSU)
The last three words seem to be FROM KNOWING ME. Let Z=F, I=R, P=M, D=K, and O=W. The word PU is MY; U=Y.
MY W OLE REER EEN E O E O PU ORXAS FMISSI RMH QSSC HSWXVSH VX KEEPING PEOPLE FROM KNOWING ME. (LON DSSYJCE YSXYAS ZIXP DCXOJCE PS. (AXC FRMCSU) NEY)
ORXAS is WHOLE; R=H. FMISSI is CAREER; F=C and M=A.
MY WHOLE CAREER HA EEN E O E O PU ORXAS FMISSI RMH QSSC HSWXVSH VX KEEPING PEOPLE FROM KNOWING ME. (LON DSSYJCE YSXYAS ZIXP DCXOJCE PS. (AXC CHANEY) FRMCSU)
The solution is:
MY WHOLE CAREER HAD BEEN DEVOTED TO KEEPING PEOPLE FROM KNOWING ME. (LON CHANEY)
22. CGIP LNWW CH CGIP, OTZ PG LNWW O WGU GB QNZZWH-OVHZ QHT. (MNT KDCCOSZ)
O must be the letter A. I wouldn't fit the conjunction OTZ. OTZ is probably AND. Let O=A, T=N, and Z=D.
AND A
CGIP LNWW CH CGIP, OTZ PG LNWW O
DD A D N. N A D)
WGU GB QNZZWH-OVHZ QHT. (MNT KDCCOSZ)
Note the word LNWW. W is probably L. This would make WGU the word LET or LOT. Note the word GB. This makes it more likely that G=O. Let W=L, G=O, and U=T.
O LL O AND O LL A CGIP LNWW CH CGIP, OTZ PG LNWW O LOT O DDL A D N. N A D) WGU GB QNZZWH-OVHZ QHT. (MNT KDCCOSZ)
WGU GB appears to be LOT OF, and OTZ PG LNWW appears to be AND SO WILL. B=F, P=S, L=W, and N=I.
O S WILL O S, AND SO WILL A CGIP LNWW CH CGIP, OTZ PG LNWW O LOT OF IDDL A D N. IN A D) WGU GB QNZZWH-OVHZ QHT. (MNT KDCCOSZ)
QNZZWH-OVHZ appears to be MIDDLE-AGED. Let Q=M, H=E, and V=G.
O S WILL E O S, AND SO WILL A CGIP LNWW CH CGIP, OTZ PG LNWW O LOT OF MIDDLE-AGED MEN. IN A D) WGU GB QNZZWH-OVHZ QHT. (MNT KDCCOSZ)
The author is KIN HUBBARD. The solution is:
BOYS WILL BE BOYS, AND SO WILL A LOT OF MIDDLE-AGED MEN. (KIN HUBBARD)
23. OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
MY D looks like IS A. Let M=I, Y=S, and D=A.
I A IS A I I IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A SSI S D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
JUBUYYMVE seems to end in -SSION. Let V=O and E=N.
IO AO IS A N; I OI IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A SSION. SO O D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
U seems to be a vowel, and A, I, and O are already used. U is excluded, of course, because a letter can't stand for itself. E is the only vowel left. U=E.
IOE AO E IS A N; I OI IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A E ESSION. ESOE O D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
Given the occurrence of OE in the first word and the author's last name, my choice of V=O seems wrong. I could at this point erase things and start over, thus hiding the mistake. It's a human tendency to try to hide a mistake. The better course is to acknowledge it, admit it, and correct it. So let's backtrack and try again.
I A IS A I I IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A SSI S D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
We know that O is a consonant, otherwise OZAZHE would be preceded by AN. Z is probably a vowel. Z contacts H twice, so H is probably a consonant. H also contacts D, which we know is a vowel. The pattern of U in JUBUYYMVE and the fact that H contacts U suggests that U is a vowel. U can't be A, I, or U. O doesn't seem to fit. Let's try U=E.
I E A E IS A I I IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A E ESSI ES E D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
V contacts D, Z, M and U, all vowels. It's obviously a consonant. Let's try V=T.
ITE AT E IS A I TI IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A E ESSIT ESTE T D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
JUBUYYMVE is NECESSITY; J=N, B=C, and E=Y.
ITE AT E IS A Y; ICTI N IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A NECESSITY. C ESTE T N) D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
G=O. The only choice left for Z is Z=U.
ITE ATU E IS A U U Y; ICTION IS OMVUHDVZHU MY D OZAZHE; NMBVMGJ MY A NECESSITY. C ESTE TON) D JUBUYYMVE. (R. Q. BTUYVUHVGJ)
The first word is LITERATURE; O=L and H=R. OZAZHE is LUXURY. The author is G. K. CHESTERTON. The solution is:
LITERATURE IS A LUXURY; FICTION IS A NECESSITY. (G. K. CHESTERTON)
Note: As you practice solving, you'll become familiar with authors who are often quoted.
24. PYN MSG AJ HOGMJEJ SGV HDOCC AJ HDNLOV. (MFSECJH Z. UJDDJEOGK)
The pattern that J follows seems to indicate that it's a vowel. Note the author's last name. JDDJ seems to indicate that J is a vowel and D is a consonant. The word AJ seems to indicate that A is a consonant. I have enough practice to read YOU CAN BE ... AND STILL BE ... So let's make P=Y, Y=O, N=U, M=C, S=A, G=N, A=B, J=E, V=D, H=S, D=T, O=I and C=L.
YOU CAN BE SINCE E AND STILL BE PYN MSG AJ HOGMJEJ SGV HDOCC AJ STU ID. (C A LES ETTE IN HDNLOV. (MFSECJH Z. UJDDJEOGK)
E=R and L=P. The author's first name is CHARLES; F=H. The author's last name seems to be KETTERING; U=K and K=G. I confess that I don't know the author's middle initial, and had to look it up. The solution is:
YOU CAN BE SINCERE AND STILL BE STUPID. (CHARLES F. KETTERING)
25. BM GWM DVQ ATOVSWZQMC ZD VXW CJMMO. (BZJJZGF AGRJZQQ)
The pattern of Z throughout the cryptogram suggests that it's a vowel. J is a consonant. So is D. With this in mind, I read BM GWM DVQ as WE ARE NOT. Let B=W, M=E, G=A, W=R, D=N, V=O, and Q=T.
WE ARE NOT O R TE N O R BM GWM DVQ ATOVSWZQMC ZD VXW EE (W A A TT) CJMMO. (BZJJZGF AGRJZQQ)
VXU is OUR; X=U. The author's first name is WILLIAM; Z=I, J=L, and F=M.
WE ARE NOT O RITE IN OUR BM GWM DVQ ATOVSWZQMC ZD VXW LEE (WILLIAM A LITT) CJMMO. (BZJJZGF AGRJZQQ)
CJMMO appears to be SLEEP. Let C=S and O=P. The author's last name is HAZLITT (again, familiarity with authors helps when solving short cryptograms.) A=H and R=Z.
WE ARE NOT H PO RITES IN OUR BM GWM DVQ ATOVSWZQMC ZD VXW SLEEP. (WILLIAM HAZLITT) CJMMO. (BZJJZGF AGRJZQQ)
The solution is:
WE ARE NOT HYPOCRITES IN OUR SLEEP. (WILLIAM HAZLITT)